Answer:
90% confidence interval: (2186.53;2881.47)
95% confidence interval: (2118.73;2949.27)
99% confidence interval: (1987.37;3080.63)
For the last part is not the best way say : "This interval describes the price of 95% of the rents of all the unfurnished one-bedroom apartments in the Boston area."
The best interpretation is this one: "We are 95% confident that the actual mean for the rents of unfurnished one-bedroom apartments in the Boston area is between (2118.73;2949.27)"
Step-by-step explanation:
1) Previous concepts
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
[tex]\bar X=2534[/tex] represent the sample mean
[tex]\mu[/tex] population mean (variable of interest)
[tex]\sigma=670[/tex] represent the population standard deviation
n=10 represent the sample size
90% confidence interval
The confidence interval for the mean is given by the following formula:
[tex]\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex] (1)
Since the Confidence is 0.90 or 90%, the value of [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.05[/tex], and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-NORM.INV(0.05,0,1)".And we see that [tex]z_{\alpha/2}=1.64[/tex]
Now we have everything in order to replace into formula (1):
[tex]2534-1.64\frac{670}{\sqrt{10}}=2186.53[/tex]
[tex]2534+1.64\frac{670}{\sqrt{10}}=2881.47[/tex]
So on this case the 90% confidence interval would be given by (2186.53;2881.47)
95% confidence interval
The confidence interval for the mean is given by the following formula:
[tex]\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex] (1)
Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that [tex]z_{\alpha/2}=1.96[/tex]
Now we have everything in order to replace into formula (1):
[tex]2534-1.96\frac{670}{\sqrt{10}}=2118.73[/tex]
[tex]2534+1.96\frac{670}{\sqrt{10}}=2949.27[/tex]
So on this case the 95% confidence interval would be given by (2118.73;2949.27)
99% confidence interval
The confidence interval for the mean is given by the following formula:
[tex]\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex] (1)
Since the Confidence is 0.99 or 99%, the value of [tex]\alpha=0.01[/tex] and [tex]\alpha/2 =0.005[/tex], and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-NORM.INV(0.005,0,1)".And we see that [tex]z_{\alpha/2}=2.58[/tex]
Now we have everything in order to replace into formula (1):
[tex]2534-2.58\frac{670}{\sqrt{10}}=1987.37[/tex]
[tex]2534+2.58\frac{670}{\sqrt{10}}=3080.63[/tex]
So on this case the 99% confidence interval would be given by (1987.37;3080.63)
For the last part is not the best way say : "This interval describes the price of 95% of the rents of all the unfurnished one-bedroom apartments in the Boston area."
The best interpretation is this one: "We are 95% confident that the actual mean for the rents of unfurnished one-bedroom apartments in the Boston area is between (2118.73;2949.27)"
A random sample of 50 cars in the drive-thru of a popular fast food restaurant revealed an average bill of $18.21 per car. The population standard deviation is $5.92. Round your answers to two decimal places.
(a) State the point estimate for the population mean cost of fast food bills at this restaurant $
(b) Calculate the 95% margin of error. $
(c) State the 95% confidence interval for the population mean cost of fast food bills at this restaurant. $ ≤ µ ≤ $
(d) What sample size is needed if the error must not exceed $1.00? n =
Answer:
a) [tex]\bar{x} = 18.21[/tex]
b) 1.64
c) (16.57,19.85)
d) The sample size must be 135 or greater if the error must not exceed $1.00.
Step-by-step explanation:
We are given the following information in the question:
Sample size, n = 50
Sample mean = $18.21
population standard deviation = $5.92
a) Point estimate for the population mean cost
[tex]\bar{x} = 18.21[/tex]
b) Margin of error =
[tex]z_{critical}\displaystyle\frac{\sigma}{\sqrt{n}}[/tex]
[tex]z_{critical}\text{ at}~\alpha_{0.05} = 1.96[/tex]
Margin of error = [tex]1.96\displaystyle\frac{5.92}{\sqrt{50}} = 1.64[/tex]
c) 95% Confidence interval
[tex]\mu \pm z_{critical}\displaystyle\frac{\sigma}{\sqrt{n}}[/tex]
Putting the values, we get,
[tex]18.21 \pm 1.64) = (16.57,19.85)[/tex]
d) Marginal error less than $1.00
[tex]1.96\displaystyle\frac{\sigma}{\sqrt{n}} \leq 1\\\\\sqrt{n} \geq 1.96\times 5.92\\n \geq (11.6)^2\\n \geq 134.56 \approx 135[/tex]
Thus, the sample size must be 135 or greater if the error must not exceed $1.00.
(A) Point estimate for the population mean cost: $18.21. (b) Margin of error: $1.64. (c) 95% Confidence interval: $16.57 ≤ μ ≤ $19.85. (d) The sample size must be 135 or greater if the error must not exceed $1.00.
(a) Point estimate for the population mean cost
The point estimate for the population mean (μ) is the sample mean ([tex]\bar{x}[/tex]). In this case:
Point estimate ([tex]\bar{x}[/tex]) = $18.21
(b) Margin of error
The formula for the margin of error (E) in a confidence interval is given by:
E = Z * (σ / √n)
where:
Z is the Z-score corresponding to the desired confidence level,
σ is the population standard deviation,
n is the sample size.
For a 95% confidence interval, Z is approximately 1.96. Let's plug in the values:
E ≈ 1.96 * (5.92 / √50)
E ≈ 1.96 * (5.92 / 7.07)
E ≈ 1.64
So, the margin of error is approximately $1.64.
(c) 95% Confidence interval
The confidence interval is given by:
Confidence interval = [tex]\bar{x}[/tex] - E ≤ μ ≤ [tex]\bar{x}[/tex] + E
Plugging in the values:
$18.21 - 1.64 ≤ μ ≤ $18.21 + 1.64
$16.57 ≤ μ ≤ $19.85
So, the 95% confidence interval for the population mean cost is $16.57 ≤ μ ≤ $19.85.
(d) Sample size needed if the error must not exceed $1.00
The formula for the margin of error is:
E = Z * (σ / √n)
We want the error (E) to be less than $1.00, so:
$1.00 = 1.96 * (5.92 / √n)
Solving for n:
√n = 1.96 * (5.92 / 1.00)
n = [(1.96 * 5.92) / 1.00]^2
For a 95% confidence interval, Z is approximately 1.96:
n ≈ [(1.96 * 5.92) / 1.00]^2
n ≈ (11.5872 / 1.00)^2
n ≈ (11.5872)^2
n ≈ 134.52
Since the sample size must be a whole number, we round up to the nearest whole number. Therefore, n = 135.
So, the correct answer for (d) is:
The sample size must be 135 or greater if the error must not exceed $1.00.
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Lindsay has saved $134.75 to buy a tablet. How much more money does she need to save before she can buy the tablet? Write an equation and solve it by using the information in the table.
Answer:
The answer is $64.25 or you could say its B.
Hope this helps :)
The average Math SAT score of all incoming freshman at a large university is 490. A simple random sample of 200 of these incoming freshmen has an average Math SAT score of 495 with a standard deviation of 60. Set up an approximate 95% confidence interval for the average Math SAT score of all incoming freshmen at the university.Select only one of the boxes below.A. It is appropriate to compute a confidence interval for this problem using the Normal curve.B. Making the requested confidence interval does not make any sense.C. The Normal curve cannot be used to make the requested confidence interval.
Answer:
Which is the output of the formula =AND(12>6;6>3;3>9)?
A.
TRUE
B.
FALSE
C.
12
D.
9
Step-by-step explanation:
Our environment is very sensitive to the amount of ozone in the upper atmosphere. The level of ozone normally found is 6.5 parts/million (ppm). A researcher believes that the current ozone level is not at a normal level. The mean of 27 samples is 6.9 ppm with a variance of 1.21. Assume the population is normally distributed. A level of significance of 0.05 will be used. Find the value of the test statistic. Round your answer to three decimal places.
The value of the test statistic is 1.583.
Explanation:To find the value of the test statistic, we need to calculate the z-score using the formula:
z = (x - μ) / (σ / √n)
where x is the sample mean, μ is the population mean, σ is the population standard deviation, and n is the sample size.
In this case, the sample mean is 6.9 ppm, the population mean is 6.5 ppm, the population variance is 1.21, and the sample size is 27. Calculate the test statistic:
z = (6.9 - 6.5) / (√(1.21 / 27)) = 0.4 / 0.252951 = 1.583
Rounding to three decimal places, the value of the test statistic is 1.583.
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An oil tanker breaks apart and starts leaking. As time goes on, the rate at which the oil is leaking out will diminish. Suppose that "t" hours after the tanker breaks apart, the oil is leaking out at a rate of R(t)=(0.7)/(1+t^2) million gallons per minute. Then ___ million gallons of oil will leak out in the first 3 hours after the shipwreck.
The quantity of oil in the first 180 minutes was 62.78 million gallons.
IntegrationIt is the reverse of differentiation.
Given
An oil tanker breaks apart and starts leaking. As time goes on, the rate at which the oil is leaking out will diminish.
The tanker breaks apart, the oil is leaking out at a rate of R(t).
[tex]\rm R(t) = \dfrac{0.7}{1 + t^2}[/tex], where t is time in minutes.
How much a million gallons of oil will leak out in the first 3 hours after the shipwreck?Convert the hours into minutes.
1 hours = 60 minutes
3 hours = 180 minutes
Then the quantity of oil in the first 180 minutes will be
V(t) = R(t)
Integrate the function.
[tex]\rm V(t) = \int_0^{180} R(t) dt\\\\V(t) = \int_0^{180} \dfrac{0.7}{1+t^2} dt\\\\V(t) = 0.7 \int_0^{180} \dfrac{1}{1 + t^2}dt\\\\V(t) = 0.7[tan^{-1}t]_0^{180}\\\\V(t) = 0.7 [tan^{-1}180 -tan^{-1}0]\\\\V(t) = 0.7 * 89.682\\\\V(t) = 62.78[/tex]
Thus, the quantity of oil in the first 180 minutes was 62.78 million gallons.
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The number of million gallons of oil that will leak out in the first 3 hours after the shipwreck can be found by evaluating the integral of the rate function over the given interval.
Explanation:To find the number of million gallons of oil that will leak out in the first 3 hours after the shipwreck, we need to find the integral of the rate function R(t) over the interval t = 0 to t = 3.
The integral of R(t) = (0.7)/(1+t^2) with respect to t is:
∫R(t) dt = 0.7 ∫(1+t^2)^-1 dt
Integrating this expression gives:
∫R(t) dt = 0.7 ln|1+t^2| + C
Evaluating this expression from t = 0 to t = 3, we get:
∫03R(t) dt = 0.7 ln|1+3^2| - 0.7 ln|1+0^2|
Simplifying further:
∫03R(t) dt = 0.7 ln(10) - 0.7 ln(1)
Since ln(1) = 0, we have:
∫03R(t) dt = 0.7 ln(10) - 0 = 0.7 ln(10)
Finally, converting the result to million gallons:
∫03R(t) dt = 0.7 ln(10) million gallons
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18. Young millennials, adults aged 18 to 34, are viewed as the future of the restaurant industry. During 2011, this group consumed a mean of 192 restaurant meals per person (NPD Group website, November 7, 2012). Conduct a hypothesis test to determine if the poor economy caused a change in the frequency of consuming restaurant meals by young millennials in 2012. Formulate hypotheses that can be used to determine whether the annual mean number of restaurant meals per person has changed for young millennials in 2012.
To determine if the poor economy caused a change in the frequency of consuming restaurant meals by young millennials in 2012, a hypothesis test can be conducted by comparing the mean number of meals consumed in 2012 to the mean from 2011.
Explanation:To determine if the poor economy caused a change in the frequency of consuming restaurant meals by young millennials in 2012, we can conduct a hypothesis test.
The null hypothesis, denoted as H0, states that there is no change in the mean number of restaurant meals per person for young millennials in 2012.
The alternative hypothesis, denoted as Ha, states that there is a change in the mean number of restaurant meals per person for young millennials in 2012.
We would use statistical data from 2012 to compare the mean number of restaurant meals consumed by young millennials to the mean from 2011 (192 meals per person) to determine if there is a significant change.
Final answer:
To determine if there has been a change in the mean number of restaurant meals consumed by young millennials in 2012 from the 192 meals reported in 2011, a two-tailed hypothesis test is required. The null hypothesis states no change (H0: μ = 192), while the alternative hypothesis suggests a change (H1: μ ≠ 192). A statistical test like a t-test or z-test, based on sample data, can be employed to assess this hypothesis with a commonly used significance level of 0.05.
Explanation:
Hypothesis Testing for Changes in Restaurant Meal Consumption
To investigate whether the economic downturn resulted in a change in the frequency of consuming restaurant meals by young millennials in 2012, compared to the mean of 192 meals reported in 2011, we should conduct a two-tailed hypothesis test. The null hypothesis (H0) states that there is no change in the annual mean number of restaurant meals consumed per person among young millennials. In formulaic terms, this is H0: μ = 192. The alternative hypothesis (H1) suggests that there has been a change, which can be phrased as H1: μ ≠ 192.
To conduct the test, we would need sample data from young millennials in 2012 regarding their restaurant meal consumption. Statistical software or methods such as a t-test or z-test would be used depending on the sample size and variance. The test will compare the sample mean to the known mean of 192, with a chosen level of significance, usually 0.05. If the p-value obtained from the test is less than the significance level, we reject the null hypothesis, indicating a significant change in the mean number of meals consumed.
To analyze trends related to meal consumption, factors such as technology use, employment, and economic factors (such as NEET rates) may also provide contextual insights on the behaviors of young millennials. While some young adults are increasingly engaged with online avenues for food purchase, a change in eating-out behavior may be attributable to these broader lifestyle and economic shifts.
Jill’s bowling scores are approximately normally distributed with mean 170 and standard deviation 20, while Jack’s scores are approximately normally distributed with mean 160 and standard deviation 15. If Jack and Jill each bowl one game, then assuming that their scores are independent random variables, approximate the probability that
(a) Jack’s score is higher;
(b) the total of their scores is above 350.
Answer:
0.3446,0.2119
Step-by-step explanation:
Given that Jill’s bowling scores are approximately normally distributed with mean 170 and standard deviation 20, while Jack’s scores are approximately normally distributed with mean 160 and standard deviation 15.
If X represents Jill scores and Y Jack scores we have
X is N(170,20) and Y is N(160,15)
since x and y are independent we have the difference
X-Y is [tex]N(170-16,\sqrt{20^2+15^2} )\\=N(10, 25)[/tex]
a) Prob that Jack’s score is higher
= P(-x+y>0)
=[tex]P(Z<\frac{10}{25} )\\= P(Z<0.4)\\\\= =0.3446[/tex]
b) X+Y is Normal with (330, 25)
[tex]P(X+Y>350) = P(Z>\frac{350-330}{25} )\\=P(Z>0.8)\\\\= =0.2119[/tex]
To find the probability that Jack's score is higher than Jill's, calculate the z-scores and compare them. For the probability that the total of their scores is above 350, find the z-score for the sum and use a standard normal distribution table. The probability of Jack's score being higher is 50% and the probability of the sum being above 350 is 78.81%.
Explanation:To approximate the probability that Jack's score is higher than Jill's, we can use the concept of the z-score. The z-score measures how many standard deviations a value is from the mean. For Jack's score, we calculate his z-score as (his score - his mean) / his standard deviation, which is (160 - 160) / 15 = 0. For Jill's score, the z-score is (170 - 170) / 20 = 0.
Since both z-scores are 0, we can conclude that the probability of Jack's score being higher than Jill's is 0.5, or 50%.
To find the probability that the total of their scores is above 350, we need to find the z-score for the sum. The sum of their scores is 160 + 170 = 330. The mean of the sum is 160 + 170 = 330, and the standard deviation of the sum is sqrt((15^2) + (20^2)) = sqrt(625) = 25.
Therefore, the z-score for the sum is (350 - 330) / 25 = 0.8. Using a standard normal distribution table or calculator, we can find that the probability of the sum being above 350 is approximately 0.7881, or 78.81%.
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A company's accounts payable department is trying to reduce the time payment of invoices and has recentily completed a flowchart. Which of the fo be the best for them to use next?
(A) Fishbone diagram
(B) Scatter diagram
(C) Box and whisker plot
(D) Histogram
Answer:
A"fishbone" diagram
Step-by-step explanation:
A"fishbone" diagram, may help to determine potential causes of an issue and organise concepts into valuable groups of brain storming. A depiction of the fishbone is a graphical way of looking at causes and effects.
It's a more organised solution than other brain storming tools that cause problems. The issue or effect is shown on the fish's head or mouth.
A hypothesis test is to be performed in order to test the proportion of people in a population that have some characteristic of interest. Select all of the pieces of information that are needed in order to calculate the test statistic for the hypothesis test:
A. the size of the sample selected
B. the sample proportion calculated
C. the level of significance used
D. the proposed population proportion
E. the characteristic of interest
F. the actual population proportion
To calculate the test statistic for a hypothesis test on population proportion, you need the size of the sample, sample proportion, level of significance, and proposed population proportion.
Explanation:In order to calculate the test statistic for a hypothesis test on the proportion of people with a characteristic of interest, the following pieces of information are needed:
The size of the sample selected (A) The sample proportion calculated (B) The level of significance used (C) The proposed population proportion (D)
These pieces of information are necessary to calculate the test statistic, which is used to determine whether the sample data provides enough evidence to support or reject the hypothesis about the population proportion.
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The information needed to calculate the test statistic for a hypothesis test of a population proportion includes the size of the sample selected, the sample proportion calculated, and the proposed population proportion.
Explanation:To calculate the test statistic for a hypothesis test about a population proportion, several pieces of information are required:
A. the size of the sample selected: You need to know how many observations you have collected.B. the sample proportion calculated: This is the proportion of the sample that exhibits the characteristic of interest, calculated from the data.D. the proposed population proportion: This is the proportion that the null hypothesis is proposing to test against.Note that while C. the level of significance used and F. the actual population proportion are important for different hypothesis testing steps such as defining the null and alternative hypotheses, calculating the p-value, or making the final decision for the hypothesis test, they do not directly factor into the calculation of the test statistic. Also, E. the characteristic of interest is not required to calculate the test statistic itself, but it is necessary to determine which is the appropriate test to apply and to structure the hypotheses correctly.
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University degree requirements typically are different for Bachelor of Science degrees and Bachelor of Arts degrees. Some students get a Bachelor of Arts and Science degree, which requires meeting graduation criteria for both degrees. A student advisor needs to know the probability a newly admitted student is interested in such a program, so that the student can be properly advised. A study of previous years finds that the probability a student gets a Bachelor of Science degree is P(Science)=0.3 and the probability a student gets a Bachelor of Arts degree is P(Arts)=0.6 . The study also shows that the probability a student gets no degree is P(no)=0.2 .
A - The probability a student gets a Bachelor of Arts and Science degree is:
a) 0.3.
b) 0.6.
c) 0.1.
d) 0.2.
B - The probability a student gets a Bachelor of Arts degree or a Bachelor of Science degree is:
a)0.9.
b) 0.3.
c) 0.6.
d) 0.8.
C - The probability a student gets only a Bachelor of Arts degree is:
a) 0.1.
b) 0.5.
c) 0.3.
d) 0.6.
The probability a student gets a Bachelor of Arts and Science degree is 0.18, the probability a student gets a Bachelor of Arts degree or a Bachelor of Science degree is 0.9, and the probability a student gets only a Bachelor of Arts degree is 0.5.
Explanation:The probability a student gets a Bachelor of Arts and Science degree can be found by multiplying the probabilities of getting a Bachelor of Arts degree and a Bachelor of Science degree.
So, P(Arts and Science) = P(Arts) * P(Science) = 0.6 * 0.3 = 0.18. Therefore, the probability a student gets a Bachelor of Arts and Science degree is 0.18, which is not listed as an option in the given choices.
The probability a student gets a Bachelor of Arts degree or a Bachelor of Science degree can be found by adding the probabilities of getting each degree. So, P(Arts or Science) = P(Arts) + P(Science) = 0.6 + 0.3 = 0.9.
Therefore, the probability a student gets a Bachelor of Arts degree or a Bachelor of Science degree is 0.9, which is listed as option a) 0.9.
The probability a student gets only a Bachelor of Arts degree can be found by subtracting the probability of getting a Bachelor of Science degree and the probability of getting no degree from 1.
So, P(Only Arts) = 1 - P(Science) - P(no) = 1 - 0.3 - 0.2 = 0.5. Therefore, the probability a student gets only a Bachelor of Arts degree is 0.5, which is listed as option b) 0.5.
A certain type of automobile battery is known to last an average of 1140 days with a standard deviation of 80 days. If 400 of these batteries are selected, find the following probabilities for the average length of life of the selected batteries. (Round your answers to four decimal places.)
(a) The average is between 1128 and 1140.
(b) The average is greater than 1152.
(c) The average is less than 940.
Answer:
Mean = [tex]\mu = 1140[/tex]
Standard deviation = [tex]\sigma = 80[/tex]
Find the probabilities for the average length of life of the selected batteries.
A)The average is between 1128 and 1140.
We are supposed to fidn P(1128<x<1140)
Formula : [tex]Z=\frac{x-\mu}{\sigma}[/tex]
At x = 1128
[tex]Z=\frac{1128-1140}{80}[/tex]
[tex]Z=-0.15[/tex]
Refer the z table for p value
P(x<1128)=0.4404
At x = 1140
[tex]Z=\frac{1140-1140}{80}[/tex]
[tex]Z=0[/tex]
Refer the z table for p value
P(x<1140)=0.5
So,P(1128<x<1140)=P(x<1140)-P(x<1128)=0.5-0.4404=0.0596
Hence the probabilities for the average length of life of the selected batteries is between 1128 and 1140 is 0.0596
B)The average is greater than 1152.
P(x>1152)
At x = 1128
[tex]Z=\frac{1152-1140}{80}[/tex]
[tex]Z=0.15[/tex]
Refer the z table for p value
P(x<1152)=0.5596
So,P(x>1152)=1-P(x<1152)=1-0.5596=0.4404
Hence the probabilities for the average length of life of the selected batteries is greater than 1152 is 0.4404
C) The average is less than 940.
P(x<940)
At x = 940
[tex]Z=\frac{940-1140}{80}[/tex]
[tex]Z=-2.5[/tex]
Refer the z table for p value
P(x<940)=0.5596
Hence the probabilities for the average length of life of the selected batteries is less than 940 is 0.5596
The question involves calculating the probability of the lifespan of car batteries based on given statistical data using the standard normal distribution.
The student is asking about calculating probabilities related to the life expectancy of car batteries using the principles of statistics. They provided information about the average lifespan of the batteries (1140 days), the standard deviation (80 days), and the sample size (400).
For part (a):
To find the probability that the average lifespan of the selected batteries is between 1128 and 1140 days, you would use the sampling distribution of the sample mean. The mean of the sampling distribution is the same as the population mean, 1140, and the standard deviation (often termed the standard error) of this distribution is the population standard deviation divided by the square root of the sample size: σ/√n or 80/√400 = 4 days. You would then use a standard normal distribution to find the probability that a normally distributed random variable falls between the z-scores corresponding to 1128 and 1140 days.
For part (b):
To determine the probability that the average is greater than 1152, again find the z-score for 1152 and use the standard normal distribution.
For part (c):
Asking for the probability of the average being less than 940 is a theoretical scenario far from the mean. Given the standard deviation, this would likely yield a probability close to 0.
Suppose that Stephen is the quality control supervisor for a food distribution company. A shipment containing many thousands of apples has just arrived. Unknown to Stephen, 15% of the apples are damaged due to bruising, worms, or other defects. If Stephen samples 10 apples from the shipment, use the binomial distribution to estimate the probability that his sample will contain at least one damaged apple. Give your answer as a decimal precise to at least four decimal places.
Answer: 0.8031
Step-by-step explanation:
Binomial distribution:
For a binomial variable x,
The probability of getting success in x trials = [tex]P(X=x)=^nC_xp^x(1-p)^{n-x}[/tex]
, where n = Total trials .
p= Probability of getting success in each trial.
For the given situation , we take x as the number of damaged apple .
Given : The proportion of damaged apples : p=0.15
n= 10
Then, the probability that his sample will contain at least one damaged apple. :-
[tex]P(x\geq1)=1-P(x<1)\\\\=1-P(x=0)\\\\=1-^{10}C_{0}(0.15)^0(1-0.15)^{10}\\\\=1-(1)(0.85)^{10}\ \ [\because\ ^nC_0=1]\\=1-0.196874404341=0.803125595659\approx0.8031[/tex]
Hence, the required probability = 0.8031
Final answer:
To find the probability of at least one damaged apple in a sample of 10 from a shipment with a 15% damage rate, we calculate the complement (no damaged apples) and subtract it from 1, resulting in a probability of 0.8031.
Explanation:
The question asks us to use the binomial distribution to estimate the probability that a sample of 10 apples will contain at least one damaged apple, given that 15% of the apples are known to be damaged. To find the probability of at least one damaged apple, it's easiest to calculate the probability of the opposite event (no damaged apples) and subtract that from 1.
Using the binomial distribution formula, the probability of getting exactly k successes (damaged apples) in n trials (sampled apples) is given by P(X=k) = (nCk)*(p^k)*((1-p)^(n-k)), where p is the probability of success (0.15 in this case), and nCk is the binomial coefficient.
For k=0 (no damaged apples), the calculation becomes P(X=0) = (10C0)*(0.15^0)*((1-0.15)^(10-0)) = 1*(1)*(0.85^10) ≈ 0.1969. Therefore, the probability of finding at least one damaged apple is 1 - P(X=0) = 1 - 0.1969 = 0.8031.
Mean birthweight is studied because low birthweight is an indicator of infant mortality. A study of babies in Norway published in the International Journal of Epidemiology shows that birthweight of full-term babies (37 weeks or more of gestation) are very close to normally distributed with a mean of 3600 g and a standard deviation of 600 g. Suppose that Melanie is a researcher who wishes to estimate the mean birthweight of full-term babies in her hospital. What is the minimum number of babies she should sample if she wishes to be at least 95% confident that the mean birthweight of the sample is within 100 grams of the the mean birthweight of all babies? Assume that the distribution of birthweights at her hospital is normal with a standard deviation of 600 g. n =
Melanie, as a researcher, needs to sample at least 139 full-term newborn babies at her hospital to be 95% confident that the mean birthweight of the sample is within 100 grams of the mean of all babies.
Explanation:To estimate the mean birthweight of full-term babies in her hospital with an error of at most 100 grams and a 95% confidence level, Melanie can use the formula for sample size in a normal population: n = (Z^2 * σ^2) / E^2 where Z is the Z-value from the Z-table for the desired level of confidence (for 95%, Z = 1.96), σ is the standard deviation of the population (600 grams), and E is the maximum allowable error (100 grams).
Plugging in these values, we get n = (1.96^2 * 600^2) / 100^2 = 138.2976, which we round up to 139 since we can't have a fractional number of babies.
So, Melanie should sample at least 139 babies to be at least 95% confident that the mean birthweight of the sample is within 100 grams of the mean birthweight of all babies.
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Suppose 42% of politicians are lawyers. If a random sample of size 628 is selected, what is the probability that the proportion of politicians who are lawyers will differ from the total politicians proportion by less than 5%? Round your answer to four decimal places.
The probability that the proportion of politicians who are lawyers differs from the total politician proportion by less than 5% is approximately 0.9793 (rounded to four decimal places).
1. **Calculate the standard deviation [tex](\( \sigma \))[/tex]:**
[tex]\[ \sigma = \sqrt{\frac{p(1-p)}{n}} \][/tex]
[tex]\[ \sigma = \sqrt{\frac{0.42(1-0.42)}{628}} \][/tex]
[tex]\[ \sigma \approx \sqrt{\frac{0.42 \times 0.58}{628}} \][/tex]
[tex]\[ \sigma \approx \sqrt{\frac{0.2436}{628}} \][/tex]
[tex]\[ \sigma \approx \sqrt{0.0003882} \][/tex]
[tex]\[ \sigma \approx 0.019705 \][/tex]
2. **Find the z-scores for [tex]\( p + 0.05 \)[/tex] and [tex]\( p - 0.05 \)[/tex]:**
[tex]\[ Z_{\text{upper}} = \frac{0.42 + 0.05 - 0.42}{0.019705} \][/tex]
[tex]\[ Z_{\text{upper}} = \frac{0.05}{0.019705} \][/tex]
[tex]\[ Z_{\text{upper}} \approx 2.53 \][/tex]
[tex]\[ Z_{\text{lower}} = \frac{0.42 - 0.05 - 0.42}{0.019705} \][/tex]
[tex]\[ Z_{\text{lower}} = \frac{-0.05}{0.019705} \][/tex]
[tex]\[ Z_{\text{lower}} \approx -2.53 \][/tex]
3. **Find the probability using the standard normal distribution table:**
[tex]\[ \text{Probability} = P(-2.53 < Z < 2.53) \][/tex]
By looking up the values in the standard normal distribution table or using a calculator, the probability is approximately 0.9793.
Minor surgery on horses under field conditions requires a reliable short-term anesthetic producing good muscle relaxation, minimal cardiovascular and respiratory changes, and a quick, smooth recovery with minimal aftereffects so that horses can be left unattended. An article reports that for a sample of n = 69 horses to which ketamine was administered under certain conditions, the sample average lateral recumbency (lying-down) time was 18.94 min and the standard deviation was 8.3 min. Does this data suggest that true average lateral recumbency time under these conditions is less than 20 min? Test the appropriate hypotheses at level of significance 0.10.State the appropriate null and alternative hypotheses.
Answer:
Null hypothesis:[tex]\mu \geq 20[/tex]
Alternative hypothesis:[tex]\mu < 20[/tex]
[tex]P(t_{68}<-1.06)=0.146[/tex]
If we compare the p value and the significance level given [tex]\alpha=0.1[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we FAIL reject the null hypothesis, and the the actual sample average lateral recumbency is not significantly lower than 20.
Step-by-step explanation:
1) Data given and notation
[tex]\bar X=18.94[/tex] represent the mean for the sample
[tex]s=8.3[/tex] represent the standard deviation for the sample
[tex]n=69[/tex] sample size
[tex]\mu_o =20[/tex] represent the value that we want to test
[tex]\alpha[/tex] represent the significance level for the hypothesis test.
t would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value for the test (variable of interest)
2) State the null and alternative hypotheses.
We need to conduct a hypothesis in order to determine if the true average lateral recumbency time under these conditions is less than 20 min:
Null hypothesis:[tex]\mu \geq 20[/tex]
Alternative hypothesis:[tex]\mu < 20[/tex]
We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:
[tex]t=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}[/tex] (1)
t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".
3) Calculate the statistic
We can replace in formula (1) the info given like this:
[tex]t=\frac{18.94-20}{\frac{8.3}{\sqrt{69}}}=-1.06[/tex]
4) Calculate the P-value
First we need to calculate the degrees of freedom
[tex]df=n-1=69-1=68[/tex]
The critical value for this case would be :
[tex]P(t_{68}<-1.06)=0.146[/tex]
5) Conclusion
If we compare the p value and the significance level given [tex]\alpha=0.1[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we FAIL reject the null hypothesis, and the the actual sample average lateral recumbency is not significantly lower than 20.
A recent study was conducted to determine the curing efficiency (time to harden) of dental composites (resins for the restoration of damaged teeth) using two different types of lights. Independent random samples of lights were obtained and a certain composite was cured for 40 seconds. The depth of each cure (in mm) was measured using a penetrometer. The summary statistics for the Halogen light were n_1 = 10, x_1 = 5.35, and s_1 = 0.7. The summary statistics for the LuxOMax light were n_2 = 10, x_2 = 3.90, and s_2 = 0.8. Assume the underlying populations are normal, with equal variances. a. The maker of the Halogen light claims that they produce a larger cure depth after 40 seconds than LuxOMax lights. Is there any evidence to support this claim? Use alpha = 0.01. b. Construct a 99% confidence interval for the difference in population mean cure depths.
Answer:
(1.4328, 1.622)
Claim is supported by evidence.
Step-by-step explanation:
Given that a recent study was conducted to determine the curing efficiency (time to harden) of dental composites (resins for the restoration of damaged teeth) using two different types of lights.
Let X be the halogen and y the Luxomax light
[tex]H_0: \bar x =\bar y\\H_a: \bar x > \bar Y[/tex]
(Two tailed test)
we are given data as:
Group Group One Group Two
Mean 5.3500 3.9000
SD 0.7000 0.8000
SEM 0.2214 0.2530
N 10 10
The mean of Group One minus Group Two equals 1.4500
Std error for difference = 0.336
Test statistic t=4.3135
df = 18
p value = 0.0004
Since p <0.01 at 1% level, reject H0
There is significant difference and hence the claim is valid.
There is evidence to support this claim at 1% significance level
Margin of error =1.172
99% confidence interval = [tex](1.45-1.172, 1.45+1.172)\\\\=(1.4328, 1.622)[/tex]
To determine if the Halogen light produces a larger cure depth, a two-sample t-test with equal variances is used. The test statistic is calculated and compared with the critical value at α = 0.01 to test the claim. Additionally, a 99% confidence interval for the difference in mean cure depths is constructed.
Explanation:To assess whether the Halogen light produces a larger cure depth after 40 seconds than LuxOMax lights, we perform a two-sample t-test assuming equal variances. The hypotheses are:
H0: μ1 - μ2 = 0 (no difference in cure depths)Ha: μ1 - μ2 > 0 (Halogen light produces a larger cure depth)Using the given data, we calculate the test statistic:
μ1 = 5.35, μ2 = 3.90, n1 = n2 = 10, s1 = 0.7, s2 = 0.8.
The pooled standard deviation (Sp) is computed as: Sp = √[((n1-1)s1² + (n2-1)s2²) / (n1+n2-2)] = √[(9×0.7² + 9×0.8²) / 18].
The test statistic t is calculated by: t = (x1 - x2) / (Sp×√(1/n1 + 1/n2)).
Using α = 0.01, we compare the calculated t value to the critical t value from a t-distribution table. If t > t_critical, we reject H0, providing evidence to support the Halogen light's claim.
Next, we construct a 99% confidence interval (CI) for the difference in population mean cure depths:
CI = (x1 - x2) ± (t_critical×Sp×√(1/n1 + 1/n2)).
This interval estimates the range within which the true difference in mean cure depths between the two lights lies with 99% confidence.
Let x be a random variable that represents the pH of arterial plasma (i.e., acidity of the blood). For healthy adults, the mean of the x distribution is μ = 7.4.† A new drug for arthritis has been developed. However, it is thought that this drug may change blood pH. A random sample of 31 patients with arthritis took the drug for 3 months. Blood tests showed that x = 8.6 with sample standard deviation s = 2.9. Use a 5% level of significance to test the claim that the drug has changed (either way) the mean pH level of the blood
(b) What sampling distribution will you use? Explain the rationale for your choice of sampling distribution.
What is the value of the sample test statistic? (Round your answer to three decimal places.)
We are using a t-test for hypothesis testing given a population mean, sample mean, sample size, and sample standard deviation. After calculations, the test-statistic was found to be approximately 2.88.
Explanation:This is a question that involves statistical hypothesis testing. We're given a population mean (μ = 7.4), a sample mean (x = 8.6), the sample size (n = 31), and the sample standard deviation (s = 2.9). We are asked to test the claim that the new drug changes the blood's pH level using a 5% significance level.
This is a case of a two-tailed test, because we're interested in whether the drug changes, which means it could be either increase or decrease the blood's pH level.
To answer part (b) of the question, you would use a t-distribution as your sampling distribution. In many cases, especially in health sciences, Student's t-distribution is used when the sample size is less than 30 and the population standard deviation is not known. It is also used when the sample follow a normal distribution or when the sample size is large.
For the test statistic using the t-distribution, we would use the formula: t = (x - μ) / (s / sqrt(n)). Plugging in our numbers, we get: t = (8.6 - 7.4) / (2.9 / sqrt(31)), which yields a test-statistic of approximately 2.88 when rounded to three decimal places.
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This hypothesis test utilises t-distribution, given the small sample size. First, find the standard error (SE) via dividing the standard deviation by the root of the sample size. Then, calculate the t-statistic by dividing the means difference by the SE.
Explanation:The question is related to the field of statistics. Specifically, it is about Hypothesis Testing. The hypothesis we are testing is whether the mean pH in blood has been changed by a new drug from its accepted norm of 7.4.
Because we have a small sample size (n<30), we will use the t-distribution for our hypothesis test. This is due to the Central Limit Theorem, which states that for samples of size 30 or more, the sampling distribution will approximate a normal distribution. But for less than 30, we should use the t-distribution.
To calculate the test statistic:
First, find the standard error (SE) by dividing the standard deviation (s = 2.9) by the square root of the sample size: SE = 2.9 / √31.Next, calculate the t-statistic by taking the difference in means (sample mean - population mean = 8.6 - 7.4) and dividing by the SE.This is the t-value you use to test your hypothesis on a t-distribution curve.
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A random sample of 65 bags of white cheddar popcorn weighed, on average, 5.74 ounces with a standard deviation of 0.26 ounce. Test the hypothesis that muequals5.8 ounces against the alternative hypothesis, muless than5.8 ounces, at the 0.10 level of significance.
Answer:
We failed to accept null hypothesis
Step-by-step explanation:
x= 5.74
Standard deviation = [tex]0.26[/tex]
[tex]H_0:\mu = 5.8\\H_a:\mu < 5.8[/tex]
n = 65
We will use z test
Formula : [tex]z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
Substitute the values :
[tex]z=\frac{5.74-5.8}{\frac{0.26}{\sqrt{65}}}[/tex]
[tex]z=-1.86[/tex]
Refer the z table for p value
p value = 0.0314
α = 0.10
p value < α
So, we failed to accept null hypothesis
The mean score of adult men on a psychological test that measures "masculine stereotypes" is 4.88. A researcher studying hotel managers suspects that successful managers score higher than adult men in general. A random sample of 48 managers of large hotels has mean x-bar = 5.91. Assume the population standard deviation is sigma = 3.2.
Using the null and alternative hypotheses that you set up in problem 5, the value of the test statistic for this hypothesis test is ______
Answer:
We use z-test for this hypothesis.
[tex]z_{stat} = 2.23[/tex]
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = 4.88
Sample mean, [tex]\bar{x}[/tex] = 5.91
Sample size, n = 48
Alpha, α = 0.05
Population standard deviation, σ = 3.2
First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu = 4.88\\H_A: \mu > 4.88[/tex]
The null hypothesis states that the mean score of successful managers on a psychological test is 4.88 and the alternate hypothesis says that the mean score of successful managers on a psychological test is greater than 4.88.
We use One-tailed z test to perform this hypothesis.
Formula:
[tex]z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }[/tex]
Putting all the values, we have
[tex]z_{stat} = \displaystyle\frac{5.91 - 4.88}{\frac{3.2}{\sqrt{48}} } = 2.23[/tex]
Last semester you took 3 final exams and you want to figure out which exam score really is your best score. Here are the scores from your classes. Which of your three exam scores is the worst when compared to the other students in your classes? Math: Your score = 78, Class mean = 82, Standard deviation = 5 History: Your score = 85, Class mean = 87, Standard deviation = 4 English: Your score = 80, Class mean = 76, Standard deviation = 2.5.
Answer:
76
Step-by-step explanation:
Final answer:
The worst exam score when compared to the other students is the one with the lowest z-score. In this case, with a z-score of -0.8, the 78 in Math is the worst score.
Explanation:
To determine which exam score was the worst compared to the other students, we need to calculate the number of standard deviations the student's score falls from the class mean. This calculation is known as a z-score. A z-score allows us to understand the position of a score within the context of a distribution.
For the Math exam:
Z = (Your score - Class mean) / Standard deviation
= (78 - 82) / 5
= -0.8.
For the History exam:
Z = (85 - 87) / 4
= -0.5.
For the English exam:
Z = (80 - 76) / 2.5
= 1.6.
The score with the lowest z-score is the worst because it is the farthest below the mean. Comparing the z-scores, the Math exam has the lowest z-score at -0.8, so the 78 in Math is the worst score when compared to the other students in your classes.
The proportion of students in a psychology experiment who could remember an eight-digit number correctly for t minutes was :
0.9 − 0.3 ln(t) (for t > 1).
Find the proportion that remembered the number for 5 minutes. (Round your answer to one decimal place.)
Answer:
Step-by-step explanation:
The proportion of students in a psychology experiment who could remember an eight-digit number correctly for t minutes was :
0.9 − 0.3 ln(t)
the proportion that remembered the number for 5 minutes, we would substitute t = 5 into expression, 0.9 − 0.3 ln(t). It becomes
0.9 − 0.3 ln5
= 0.9 - 0.3 × 1.60943791243
= 0.9 - 0.4828
= 0.4172
Approximately 0.4 to 1 decimal place
An elementary school art class teacher plans to display artwork next to the door of each of the classrooms in the school. Each classroom door will only have one piece of artwork displayed, and the school has 22 such doors. If the teacher has 12 sketches and 16 oil paintings, what is the probability that 10 sketches and 12 oil paintings are chosen to be displayed?
Answer: Our required probability is 0.32.
Step-by-step explanation:
Since we have given that
Number of doors to be selected in a manner = 22
Number of sketches = 12
Number of oil paintings = 16
Total number of doors we have =12+16 =28
We need to find the probability that 10 sketches and 12 oil paintings are chosen to be displayed.
So, probability would be
[tex]\dfrac{^{12}C_{10}\times ^{16}C_{12}}{^{28}C_{22}}\\\\=\dfrac{66\times 1820}{376740}\\\\=\dfrac{120120}{376740}\\\\=0.3188\\\\\approx 0.32[/tex]
Hence, our required probability is 0.32.
A random sample of 25 employees of a local company has been taken. A 95% confidence interval estimate for the mean systolic blood pressure for all employees of the company is 123 to 139. Which of the following statements is valid?
(A) 95% of the sample of employees has a systolic blood pressure between 123 and 139.
(B) If the sampling procedure were repeated many times, 95% of the resulting confidence intervals would contain the population mean systolic blood pressure.
(C) If the sampling procedure were repeated many times, 95% of the sample means would be between 123 and 139.
(D) 95% of the population of employees has a systolic blood pressure between 123 and 139.
Answer:
(B) If the sampling procedure were repeated many times, 95% of the resulting confidence intervals would contain the population mean systolic blood pressure.
Step-by-step explanation:
A confidence interval of 95% means that there is a 95% certainty that for a given sample, the population mean will be within the confidence interval estimated.
This is the same as saying that if he sampling procedure were repeated many times, 95% of the time the population mean would be contained in the resulting confidence interval.
Therefore, the answer is B)
The correct statement is (B) If the sampling procedure were repeated many times, 95% of the resulting confidence intervals would contain the population mean systolic blood pressure.
A 95% confidence interval is a range of values that is likely to contain the population mean. This does not mean that there is a 95% probability that the population mean falls within the interval calculated from a single sample. Instead, it means that if we were to take many samples and calculate a confidence interval from each sample, approximately 95% of those intervals would contain the true population mean.
Let's evaluate the other options:
(A) 95% of the sample of employees has a systolic blood pressure between 123 and 139. - This statement is incorrect because the confidence interval refers to the population mean, not the distribution of individual sample values. It does not imply that 95% of the sample has values within the interval.
(C) If the sampling procedure were repeated many times, 95% of the sample means would be between 123 and 139. - This statement is incorrect because it misinterprets the confidence interval. The confidence interval gives us a range where we expect the population mean to lie, not the sample means. While the sample means will vary, the confidence interval tells us about the variability of the estimate of the population mean, not the variability of the sample means themselves.
(D) 95% of the population of employees has a systolic blood pressure between 123 and 139. - This statement is incorrect because it generalizes the confidence interval to the entire population. The confidence interval does not tell us what proportion of the population falls within the interval; it only provides a range that is likely to contain the population mean.
Therefore, the correct interpretation of the 95% confidence interval is that if we were to take many samples from the population and calculate a 95% confidence interval for each, approximately 95% of those intervals would contain the true population mean systolic blood pressure. This is why option (B) is the valid statement.
48000 fair dice are rolled independently. Let X count the number of sixes that appear. (a) What type of random variable is X? (b) Write the expression for the probability that between 7500 and 8500 sixes show. That is Pp7500 ď X ď 8500q. (c) The sum you wrote in part b) is ridiculous to evaluate. Instead, approximate the value by a normal distribution and evaluate in terms of the distribution Φpxq " PpNp0, 1q ď xq of a standard normal random variable. (d) Why do you think a normal distribution is a good choice for approximation
Answer:
1 because almost certain event
Step-by-step explanation:
whenever a fair die is rolled the number of getting a 6 is having probability 1/6. Each throw is independent of the other
Hence no of sixes would be binomial with p=1/6
when 48000 dice are rolled, using binomial would be a hectic task.
Hence we approximate to normal distribution
X - no of sixes in 48000 throws would be normal
with mean = np = 8000
Var =npq = 6666.667
Std dev = 81.650
Now it is easier to find out
[tex]P(7500<x<8500)\\=P(7499.5<x<8499.5)[/tex](using continuity correctin)
=[tex]P(|z|<6.1295)\\=1[/tex]
we get this probability almost equal to 1
d) Normal distribution is a good choice because when no of trials increase using binomial and combination formulae would not be easy
John and James live 0.7 km apart. If John takes 25cm steps, then how many steps would it take him to walk from his house to James house?
1km___100000cm
0.7km___X=7000cm
(0.7*100000)/1=7000
25cm___1 step
7000cm___X=280 steps
(7000*1)/25= 280
There is a contest being held at Company A. Company A employs forty people. The first 3 employees to correctly complete a riddle each receives an extra vacation day that year. What is the probability that those first 3 employees are in the order of Jack, Bill, and John?
Answer:
The probability that those first 3 employees are in the order of Jack, Bill, and John is [tex]\frac{1}{59280}[/tex]
Step-by-step explanation:
Consider the provided information.
Company A employs forty people. We need to find the probability that those first 3 employees are in the order of Jack, Bill, and John?
First find the total number of ways;
We have 40 employees out of which we need to select only 3.
This can be written as: [tex]40\times 39 \times 38=59280[/tex]
The number of ways of selecting jack, bill and john in the same order = 1
[tex]Probability=\frac{\text{Favorable outcomes}}{\text{Total number of outcomes}}[/tex]
[tex]Probability=\frac{1}{59280}[/tex]
Therefore, the probability that those first 3 employees are in the order of Jack, Bill, and John is [tex]\frac{1}{59280}[/tex].
An article presents voltage measurements for a sample of 66 industrial networks in Estonia. Assume the rated voltage for these networks is 232 V. The sample mean voltage was 231.7 V with a standard deviation of 2.19 V. Let μμ represent the population mean voltage for these networks.
Answer:
a sample of 29 network will have a voltage of 232V
Step-by-step explanation:
N = 66
variate = 232 V.
mean voltage = 231.7 V
standard deviation = 2.19 V.
The z-value given by = (variate -mean)/ standard deviation
= (232-231.7)/2.19 =
z = 0.3/2.19 = 0.137
From of normal distribution table, the z of 0.137 found between the area of z=0 and z=0.5 is given as 0.0557
Thus the area to the right of the z=0.0557 ordinate is 0.5000−0.0557=0.4443 = 44.43%
thus, this is the probability of Network voltage = 44.43%.
for sample of 66 network, it is likely that 66×0.4443 = 29.32, i.e.
a sample of 29 network will have a voltage of 232V
:
Mrs. Maxwell is buying pencils for her students. She has 3 classes of 28 students each. For Valentine’s Day, she wants to give each of her students 4 pencils. She can purchase packs of 18 pencils for $2.52. How many packs of pencils will Mrs. Maxwell need to purchase? Justify your answer.
Mrs. Maxwell needs to buy 18.66 packs of pencil containing 336 pencils for 84 students.
Solution:Given that
Mrs. Maxwell is buying pencils for her students.
She has 3 classes of 28 students each.
She wants to give each of her students 4 pencils.
She can purchase packs of 18 pencils for $2.52
Need to determine number of packs of pencils Mrs. Maxwell need to purchase.
Let’s first determine total number of students
As there are 3 classes of 28 students each means
Number of students in 1 class = 28
=> Number of student in 3 classes = 28 x 3 = 84
Mrs. Maxwell wants to give each of her students 4 pencils.
So number of pencils required for 1 student = 4
=> number of pencils required for 84 student = 4 x 84 = 336
Given that number of pencils in one pack = 18
So number of pack containing 336 pencils [tex]=\frac{336}{18}=18.66 \text { packs }[/tex]
So Mrs. Maxwell needs to buy 18.66 packs of pencil containing 336 pencils for 84 students.
Final answer:
Mrs. Maxwell needs to purchase 19 packs of pencils to provide 4 pencils to each of her 84 students, as pencil packs come in sets of 18 and she needs a total of 336 pencils.
Explanation:
The question requires us to calculate the total number of pencil packs Mrs. Maxwell needs to purchase for her students for Valentine’s Day. Mrs. Maxwell has 3 classes with 28 students each, which totals to 3 * 28 = 84 students. Since she wants to give each student 4 pencils, she will need 84 * 4 = 336 pencils in total. Pencil packs come in sets of 18, so to find out how many packs she will need, we divide the total number of pencils by the number in each pack: 336 pencils ÷ 18 pencils per pack = 18.67 packs. Since she cannot buy a fraction of a pack, she needs to round up to the nearest whole number, which is 19 packs of pencils.
From where the shoes spilled (48°N, 161°W) to where they were found on May 22nd 1996 (54°N, 133°W), how many kilometers did they travel?
How many days did they take to travel that distance (use April 30 as the date found)?
What was their rate of travel in kilometers per hour?
Answer:
a) Around 2,052.583 Km.
b) Around 22 days.
c) Around 3.887 Km/h
Step-by-step explanation:
a)
In order to find and approximation of the distance traveled, we have to make some assumptions:
They traveled directly from the starting point to the end point without detours. There where no high hills or deep depressions between the points.If these assumptions hold, then the distance d in Km can be calculated by using the haversine formula:
[tex]\large d=2R*arcsin\left(\sqrt{sin^2((\varphi_2-\varphi_1)/2)+cos(\varphi_1)cos(\varphi_2)sin^2((\lambda_2-\lambda_1)/2)}\right)[/tex]
where
[tex]\large (\varphi_1, \lambda_1)[/tex] are the latitude and longitude of the starting point in radians.
[tex]\large (\varphi_2, \lambda_2)[/tex] are the latitude and longitude of the end point in radians.
R = radius of Earth in kilometers.
Be careful to convert the angles into radians before computing the trigonometric functions. This can be done by cross-multiplication knowing that 180° ≅ 3.141592654 radians.
In the problem we have
[tex]\large \varphi_1[/tex] = 48° = 0.837758041 radians
[tex]\large \lambda_1[/tex] = 161° = 2.809980096 radians
[tex]\large \varphi_2[/tex] = 54° = 0.942477796 radians
[tex]\large \lambda_2[/tex] = 133° = 2.321287905 radians
so
[tex]\large (\varphi_2-\varphi_1)/2[/tex] = 3° = 0.052359878 radians
[tex]\large (\lambda_2-\lambda_1)/2[/tex] = -14° = -0.2443461 radians
Replacing in the formula for the distance:
[tex]\large d=2R*arcsin\left(\sqrt{sin^2(0.052359878)+cos(0.837758041)cos(0.942477796)sin^2(-0.2443461)}\right)=\\\\=0.32237835*R[/tex]
According to NASA, the radius of Earth at the poles is around 6,356 Km and at the equator is 6,378 Km.
Since they traveled around the middle point between the equator and the North pole, a better estimate of the radius in this case would be the average (6,378+6,356)/2 = 6,367 Km
We have that an approximation to the distance traveled would be
0.32237835*6,367 = 2,052.583 Km
b)
Assuming that the shoes where left and found the same year, there are 22 days from April 30th to May 22nd , so they traveled for around 22 days (this may vary slightly depending on the exact time the shoes were left and found)
c)
They traveled 2,052.583 Km in 22 days.
22 days equals 22*24 = 528 hours.
By cross-multiplication
528 h _______ 2,052.583 Km
1 h __________ x Km
x = 2,052.583/528 ≅ 3.887 Km
So they traveled at a rate of 3.887 Km/h
Listed below are the number of years it took for a random sample of college students to earn bachelor's degrees (based on data from the National Center for Education Statistics). 4, 4, 4, 4, 4, 4, 4.5, 4.5, 4.5, 4.5, 4.5, 4.5, 6, 6, 8, 9, 9, 13, 13, 15
(a) Calculate the sample mean and standard deviation.
(b) Calculate the standard error, SE.
(c) What is the point estimate for the mean time required for all college students to earn bachelor's degrees?
(d) Construct the 90% confidence interval estimate of the mean time required for all college students to earn bachelor's degrees.
(e) Does the confidence interval contain the value of 4 years? Is there anything about the data that would suggest that the confidence interval might not be a good result?
Answer:
a) Mean = 6.5, sample standard deviation = 3.50
b) Standard error = 0.7826
c) Point estimate = 6.5
d) Confidence interval: (5.1469 ,7.8531)
Step-by-step explanation:
We are given the following data set for students to earn bachelor's degrees.
4, 4, 4, 4, 4, 4, 4.5, 4.5, 4.5, 4.5, 4.5, 4.5, 6, 6, 8, 9, 9, 13, 13, 15
a) Formula:
[tex]\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}[/tex]
where [tex]x_i[/tex] are data points, [tex]\bar{x}[/tex] is the mean and n is the number of observations.
[tex]Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}[/tex]
[tex]Mean =\displaystyle\frac{130}{20} = 6.5[/tex]
Sum of squares of differences = 6.25 + 6.25 + 6.25 + 6.25 + 6.25 + 6.25 + 4 + 4 + 4 + 4 + 4 + 4 + 0.25 + 0.25 + 2.25 + 6.25 + 6.25 + 42.25 + 42.25 + 72.25 = 233.5
[tex]S.D = \sqrt{\frac{233.5}{19}} = 3.50[/tex]
b) Standard Error
[tex]= \displaystyle\frac{s}{\sqrt{n}} = \frac{3.50}{\sqrt{20}} = 0.7826[/tex]
c) Point estimate for the mean time required for all college is given by the sample mean.
[tex]\bar{x} = 6.5[/tex]
d) 90% Confidence interval:
[tex]\bar{x} \pm t_{critical}\displaystyle\frac{s}{\sqrt{n}}[/tex]
Putting the values, we get,
[tex]t_{critical}\text{ at degree of freedom 19 and}~\alpha_{0.10} = \pm 1.729[/tex]
[tex]6.5 \pm 1.729(\frac{3.50}{\sqrt{20}} ) = 6.5 \pm 1.3531 = (5.1469 ,7.8531)[/tex]
e) No, the confidence interval does not contain the value of 4 years. Thus, confidence interval is not a good estimator as most of the value in the sample is of 4 years. Most of the sample does not lie in the given confidence interval.
Final answer:
The question requires calculating the sample mean, standard deviation, standard error, point estimate for the population mean, and constructing a confidence interval for the mean time to earn bachelor's degrees. The value of 4 years will need to be checked against the calculated confidence interval, and the reliability of these results may be scrutinized based on the distribution of the data.
Explanation:
The question involves calculating various statistical measures for a dataset representing the number of years college students took to earn bachelor's degrees. To address these parts:
To calculate the sample mean, you add up all the numbers and divide by the total count of numbers. The sample standard deviation measures the amount of variation or dispersion in a set of values. You use the formula for standard deviation for a sample.
The standard error (SE) is calculated by dividing the sample standard deviation by the square root of the sample size.
The point estimate for the mean time is the sample mean, as it provides the best estimate of the population mean based on the sample data.
To construct the confidence interval, you would typically use the sample mean ± (critical value from the t-distribution * standard error). For 90% confidence, you can find the critical t-value for 19 degrees of freedom (since sample size minus one equals degrees of freedom).
To answer whether the confidence interval contains the value of 4 years and discuss the reliability of this interval, you'll examine the calculated interval and consider factors like the presence of outliers and the shape of the distribution.