Answer:
78.52% probability that the sample mean is greater than 320 minutes
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal probability distribution
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
In this problem, we have that:
[tex]\mu = 330, \sigma = 80, n = 40, s = \frac{80}{\sqrt{40}} = 12.65[/tex]
What is the likelihood the sample mean is greater than 320 minutes?
This is 1 subtracted by the pvalue of Z when X = 320. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{320 - 330}{12.65}[/tex]
[tex]Z = -0.79[/tex]
[tex]Z = -0.79[/tex] has a pvalue of 0.2148
1 - 0.2148 = 0.7852
78.52% probability that the sample mean is greater than 320 minutes
The Harrisons drove 304.2 miles in 6.25 hours. What was their average speed, to the nearest tenth?
Answer:
The average speed was 48.7 miles per hour.
Step-by-step explanation:
The average speed v is given by the following formula:
[tex]v = \frac{d}{t}[/tex]
In which d is the distance, and t is the time.
The Harrisons drove 304.2 miles in 6.25 hours
This means that [tex]d = 304.2, t = 6.25[/tex]
We have the distance is miles and the time in hours, so the distance is in miles per hour.
So
[tex]v = \frac{304.2}{6.25} = 48.7[/tex]
The average speed was 48.7 miles per hour.
A small post office has two open windows. Customers arrive according to a Poisson distribution at the rate of 1 every 3 minutes. However, only 80% of them seek service at the windows. The service time per customer is exponential, with a mean of 5 minutes. All arriving customers form one line and access available windows on a FIFO basis.
(a) What is the probability that an arriving customer will wait in line?
(b) What is the probability that both windows are idle?
(c) What is the average length of the waiting line?
(d) Would it be possible to offer reasonable service with only one window? Explain.
Answer:
A) probability that an arriving customer will wait in line is 67%
B)the probability that both windows are idle is 0.33
C) the average length of the waiting line is 1.33
D)it would not be possible to offer a reasonable service with only one window
Step-by-step explanation:
arrival rate: δ = 20 x 0.80 = 16 customers per hour
service rate: μ = 2 × (60/5) = 24 customers/hour
Utilization factor is given as;
Φ = δ/μ
So, Φ = 16/24 ≈ 0.67
A) the probability that an arriving customer will wait in line is;
16/24 x 100% ≈ 67%
B) probability that both windows are idle is;
P(x=0) = 1 - 0.67 = 0.33
C) The average number of customers in the post office will be;
L_s = Φ/(1 - Φ)
L_s = 0.67/(1 - 0.67)
L_s = 0.67/0.33
L_s ≈ 2 customers
Thus, the average length of the waiting line is;
L_w = L_s - Φ
L_w = 2 - 0.67
L_w = 1.33
D) this part demands that we find the utilization factor with only one window.
Thus;
arrival rate: δ = 20 x 0.80 = 16 customers per hour
And
service rate: μ = 1 × (60/5) = 12 customers/hour
Thus, Utilization factor = 16/12 = 1.33
Thus, it would not be possible to offer a reasonable service with only one window
A guitar had been marked down by 34% and sold for $825.
What is the original price of the guitar?
The original price of the guitar was $1250, calculated by taking the sale price of $825 and dividing it by 0.66, which represents the remaining percentage of the price after a 34% discount.
Explanation:To find the original price of the guitar that has been marked down by 34% and sold for $825, we first need to consider that after the markdown, the guitar's price is equivalent to 100% - markdown percentage of the original price. In this case, it's 100% - 34% = 66% of the original price.
Let's denote the original price by P. Since 66% of P equates to $825, we can set up the following equation:
0.66 imes P = $825
Solving for P:
P = $825 / 0.66
P = $1250
Therefore, the original price of the guitar was $1250.
PLEASE HELP ASAP!!!!!!!!!!!!!!!!!!!! 28 POINTS!
Cassie received a 5%-off coupon and a $5.00-off coupon from a department store. She visits the department store during a tax-free sale and plans to spend no more than $33.95. She also plans to use both of the coupons she received on her purchase. If this situation is modeled by the inequality below, what must be the original purchase total, x, before the discounts are applied?
0.95x - $5.00 < $33.95
A.
x < $38.95
B.
x < $30.74
C.
x < $30.47
D.
x < $41.00
Answer:
D.
Step-by-step explanation:
Rearrange
[tex]0.95x - 5.00 < 33.95[/tex]
[tex]0.95x<38.95[/tex]
[tex]x<\frac{38.95}{0.95}[/tex]
[tex]x<41[/tex]
Suppose that the quality control manager for a cereal manufacturer wants to ensure that bags of cereal are being filled correctly. The equipment is calibrated to fill bags with a mean of 17 oz of cereal with a standard deviation of 0.2 oz. The quality control inspector selects a random sample of 52 boxes and finds that the mean amount of cereal for these boxes is 17.04 oz. He uses this data to conduct a one-sample z ‑test with a null hypothesis of H 0 : μ = 17 against the alternative hypothesis H 1 : μ ≠ 17 , where μ is the mean amount of cereal in each box. He calculates a z ‑score of 1.44 and a P -value of 0.1499 .
Are these results statistically significant at a significance level of 0.05?
No, these results are not statistically significant because p>0.05,
No, these results are not statistically significant because p < 0.05.
Yes, these results are statistically significant because p < 0.05.
Yes, these results are statistically significant because p > 0.05.
Answer: No, these results are not statistically significant because
p > 0.05
Step-by-step explanation:
The null hypothesis is
H0 : μ = 17
The alternative hypothesis is
H 1 : μ ≠ 17
where μ is the mean amount of cereal in each box.
The p value that he got is 0.1499. This is greater than alpha = 0.05 which is the given level of significance.
If the level of significance is lesser than the p value, we would accept the null hypothesis.
Therefore, the correct option is
No, these results are not statistically significant because p>0.05
this uses pythagorean theorem
Step-by-step explanation:
x²=a²+b²
x=√6²+12²
x=√180
x=3√2v
y²=16²+12²
y=√400
y=20
Answer: the answer for rafter 1 is 13.4 and the answer for rafter 2 is 20
Step-by-step explanation: I just know
What is the y intercept of this line and how did you find it?
(1,4) (6,-1)
Answer:
The slope is -1
Step-by-step explanation:
Let's find the slope between your two points.
(1,4);(6,−1)
(x1,y1)=(1,4)
(x2,y2)=(6,−1)
Use the slope formula:
m= y2−y1/x2−x1 = −1−4/6−1
= −5/5
= −1
Hope this is a good explanation :)
At a large university, the mean amount spent by students for cell phone service is $58.90 per month with a standard deviation of $3.64 per month. Consider a group of 44 randomly chosen university students. What is the probability that the mean amount of their monthly cell phone bills is more than $60?
Answer:
2.28% probability that the mean amount of their monthly cell phone bills is more than $60
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal probability distribution
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
In this problem, we have that:
[tex]\mu = 58.90, \sigma = 3.64, n = 44, s = \frac{3.64}{\sqrt{44}} = 0.54875[/tex]
What is the probability that the mean amount of their monthly cell phone bills is more than $60?
This is 1 subtracted by the pvalue of Z when X = 60. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{60 - 58.90}{0.54875}[/tex]
[tex]Z = 2[/tex]
[tex]Z = 2[/tex] has a pvalue of 0.9772
1 - 0.9772 = 0.0228
2.28% probability that the mean amount of their monthly cell phone bills is more than $60
The average rainfall in Phoenix is 8.29 inches per year. The table shows recent data on the difference in annual rainfall from the average. Phoenix Annual Total Rainfall Year Rainfall compared to average yearly rainfall 2008 +6.57 inches 2009 –2.68 inches 2010 +12.26 inches 2011 –4.38 inches 2012 –4.46 inches Which list represents the years from driest to wettest?
Answer:
c
Step-by-step explanation:
cuanto es dos cuartos mas un octavo
Answer:
0.625
Step-by-step explanation:
Write out the following sums, one term for each value of k. Simplify each term as much as possible, but do not enter decimals. For example, enter 1+4+9 instead of 12+22+32 or 14, or enter 1/2+1/2 instead of 0.5+0.5 or 1. The purpose of this problem is for you to show that you know how to interpret summation notation and write all of the terms in a sum, which is why you are being told not to reduce your answers very much.
The correct question is:
Write out the following sums, one term for each value of k. Simplify each term as much as possible, but do not enter decimals. For example, enter 1 + 4 + 9 instead of 1² + 2² + 3² or 14, or enter 1/2 + 1/2 instead of 0.5 + 0.5 or 1.
The purpose of this problem is for you to show that you know how to interpret summation notation and write all of the terms in a sum, which is why you are being told not to reduce your answers very much.
[tex](a) \sum_{k=0}^5 2^k \\ \\(b) \sum_{k=2}^7 \frac{1}{k} \\ \\(c) \sum_{k=1}^5 k^2 \\ \\(d) \sum_{k=1}^6 \frac{1}{6} \\ \\(e) \sum_{k=1}^6 2k[/tex]
Answer:
[tex](a) \sum_{k=0}^5 2^k = $1 + 2 + 4 + 8 + 16 + 32$ \\ \\(b) \sum_{k=2}^7 \frac{1}{k} = \frac{1}{2} + \frac{1}{3} + \frac{1}{4}+ \frac{1}{5}+ \frac{1}{6}+ \frac{1}{7} \\ \\(c) \sum_{k=1}^5 k^2 = 1 + 4 + 9 + 16 + 25 \\ \\(d) \sum_{k=1}^6 \frac{1}{6} = \frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{1}{6} \\ \\(e) \sum_{k=1}^6 2k = 2 +4 +6 +8 +10 +12[/tex]
Step-by-step explanation:
[tex](a) \sum_{k=0}^5 2^k\\For k = 0: 2^k = 2^0 = 1\\For k = 1: 2^1 = 2\\For k = 2: 2^2 = 4\\For k = 3: 2^3 = 8\\For k = 4: 2^4 = 16\\For k = 5: 2^5 = 32\\\sum_{k=0}^5 2^k = 1 + 2 + 4 + 8 + 16 + 32[/tex]
[tex](b) \sum_{k=2}^7 \frac{1}{k}\\For k = 2: 1/2\\For k = 3: 1/3\\For k = 4: 1/4\\For k = 5: 1/5\\For k = 6: 1/6\\For k = 7: 1/7\\ \sum_{k=2}^7 \frac{1}{k} = 1/2 + 1/3 + 1/4 + 1/5 + 1/6+ 1/7[/tex]
[tex](c) \sum_{k=1}^5 k^2\\For k = 1: 1^2 = 1\\For k = 2: 2^2 = 4\\For k = 3: 3^2 = 9\\For k = 4: 4^2= 16\\For k = 5: 5^2 = 25\\\sum_{k=1}^5 k^2 = 1 + 4 + 9 + 16 + 25[/tex]
[tex](d) \sum_{k=1}^6 \frac{1}{6}\\For k = 1: 1/6\\For k = 2: 1/6\\For k = 3: 1/6\\For k = 4: 1/6\\For k = 5: 1/6\\For k = 6: 1/6\\ \sum_{k=1}^6 \frac{1}{6} = 1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6[/tex]
[tex](e) \sum_{k=1}^6 2k\\For k = 1: 2\times1 = 2\\For k = 2: 2\times2 = 4\\For k = 3: 2\times3 = 6\\For k = 4: 2\times4 = 8\\For k = 5: 2\times5 = 10\\For k = 6: 2\times6 = 12\\\sum_{k=1}^6 2k = 2 +4 +6 +8 +10 +12[/tex]
A biologist is trying to determine the average age of a local forest. She cuts down 18 randomly selected trees and counts the number of tree rings, which can be used to estimate the age of the tree. What critical value should she use to construct a 99% confidence interval
Answer:
The critical value of t for 99% confidence interval is 2.898.
Step-by-step explanation:
The complete question is:
A biologist is trying to determine the average age of a local forest. She cuts down 18 randomly selected trees and counts the tree rings. They find the average number of tree rings to be 83 with a variance of 320. What is the critical value for the 99% confidence interval?
The population variance is not known and the sample size is too small. So a t-confidence interval will be used to estimate the population mean age of a local forest.
The (1 - α)% confidence interval for population mean is:
[tex]CI=\bar x\pm t_{\alpha/2, (n-1)}\times \frac{s}{\sqrt{n}}[/tex]
The information provided is:
n = 18
(1 - α)% = 99%
The degrees of freedom of the critical value of t is:
n - 1 = 18 - 1 = 17
Compute the critical value of t as follows:
[tex]t_{\alpha/2, (n-1)}=t_{0.01/2, (18-1)}=t_{0.005, 17}=2.898[/tex]
*Use a t-table.
Thus, the critical value of t for 99% confidence interval is 2.898.
If 22x = 23, what is the value of x?
Answer:
1.045
Step-by-step explanation:
Answer:
3/2
Step-by-step explanation:
It is estimated that the city of Houston, Texas, will double in size every 35 years. If the population is currently 2.4 million people, how many
people will there be in 20 years?
OA.
4.8 million
OB. 8.36 million
OC. 3.77 million
D.
3.57 million
Final answer:
To find out how many people will be in Houston, Texas in 20 years, we can use the doubling time of 35 years. Currently, the population is 2.4 million. Using the formula New Population = Current Population x 2^(Number of Doubling Cycles), the new population will be approximately 4.8 million. Option A .
Explanation:
To find out how many people will be in Houston, Texas in 20 years, we can use the doubling time of 35 years. Currently, the population is 2.4 million. Since the city doubles in size every 35 years, in 20 years it will go through 20/35 of a doubling cycle.
To calculate how many people there will be, we use the formula:
New Population = Current Population x 2^(Number of Doubling Cycles)
Plugging in the values, we have:
New Population = 2.4 million x 2^(20/35)
Using a calculator, we find that the new population will be approximately 4.8 million. Therefore, the answer is 4.8 million.
Use DeMoivre's Theorem to find the indicated power of the complex number. Write answers in rectangular form. [one half (cosine StartFraction pi Over 16 EndFraction plus i sine StartFraction pi Over 16 EndFraction )]Superscript 8
Answer:
[tex](\, \cos(\frac{\pi}{16}) + i\sin(\frac{\pi}{16}) \,)^{1/2} = \cos(\frac{\pi}{32}) + i\sin(\frac{\pi}{32}) = 0.99 + i0.09[/tex]
Step-by-step explanation:
The complex number given is
[tex]z = (\, \cos(\frac{\pi}{16}) + i\sin(\frac{\pi}{16}) \,)^{1/2}[/tex]
Now, remember that the DeMoivre's theorem states that
[tex]( \cos(x) + i\sin(x) )^n = \cos(nx) + i\sin(nx)[/tex]
Then for this case we have that
[tex](\, \cos(\frac{\pi}{16}) + i\sin(\frac{\pi}{16}) \,)^{1/2} = \cos(\frac{\pi}{32}) + i\sin(\frac{\pi}{32}) = 0.99 + i0.09[/tex]
A total of 58 trophies will be given out at the swim meet each box holds 6 trophies how many boxes are needed to carry the trophies
Answer:
You will need 10 boxes
Step-by-step explanation:
Nine boxes will only hold 54 trophies so you need one more than nine.
Answer:
You will need 348 boxes for the trophies
Step-by-step explanation:
just multiply 58 and 6
A box is 4 inches wide, 5 inches long and 3 inches tall. What’s the equation that would be used to find the surface area of the box?
Equation used to determine the surface area of the box : 2( lb + bh + lh)
Surface area of the box : 94in²
Given, A box is 4 inches wide, 5 inches long and 3 inches tall.
Formula of surface area of cuboid : 2( lb + bh + lh)
Here,
l = 5in
b = 4in
h = 3in
Substitute the values,
Surface area = 2(5×4 + 4×3 + 5×3)
Surface area = 94in²
Know more about surface area,
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Solve the right triangle. Round to two decimal places
A=20 , b=6.00
Answer:14
Step-by-step explanation:
(1 point) Write the given second order equation as its equivalent system of first order equations. u′′+6.5u′−1.5u=8sin(3t),u(1)=−3,u′(1)=−1.5 Use v to represent the "velocity function", i.e. v=u′(t). Use u and v for the two functions, rather than u(t) and v(t). (The latter confuses webwork. Functions like sin(t) are OK.)
Answer:
The equivalent system of first-order differential equations is:
[tex]\[ \begin{cases}u' = v \\v' = 8\sin(3t) - 6.5v + 1.5u\end{cases} \][/tex]
with the initial conditions:
[tex]\[ \begin{cases}u(1) = -3 \\v(1) = -1.5\end{cases} \][/tex]
To convert the given second-order differential equation into an equivalent system of first-order equations, we follow these steps:
1. Identifying the original second-order differential equation:
u'' + 6.5u' - 1.5u = 8sin(3t)
2. Then we introduce a new variable v to represent the first derivative of u:
v = u'
3. Now ,we write the second-order equation as a system of first-order equations:
The original equation is:
u'' + 6.5u' - 1.5u = 8sin(3t)
Using the new variable v = u' , the second derivative u'' can be written as v'. Therefore, we have:
v' + 6.5v - 1.5u = 8sin(3t)
4. Formulating the system of first-order equations:
The first equation comes directly from the definition of v:
u' = v
The second equation comes from the rewritten second-order equation:
v' = 8sin(3t) - 6.5v + 1.5u
5. Let us include the initial conditions:
The initial conditions provided are:
u(1) = -3
u'(1) = -1.5
Since v = u', this translates to:
v(1) = -1.5
6. Writing the system with initial conditions:
The system of first-order equations is:
[tex]\[ \begin{cases} u' = v \\ v' = 8\sin(3t) - 6.5v + 1.5u \end{cases} \][/tex]
With the initial conditions:
[tex]\[ \begin{cases} u(1) = -3 \\ v(1) = -1.5 \end{cases} \][/tex]
Find the area of the figure.
5.5 cm
20 cm
The area of the figure is
Answer:
25.5 cm²
Step-by-step explanation:
5.5 cm × 20 cm = 25.5 cm²
Answer:i think the answer is 110 but it would be more helpful if i knew what kind of shape it is
Step-by-step explanation:
to find the area of a square or rectangle(assuming this is a square or rectangle) you multiply the base by the height
When someone is on trial for suspicion of committing a crime, the hypotheses are: H0 : innocent HA : guilty Which of the following is correct? Question 40 options: Type I error is convicting an innocent person. Type I error is acquitting a guilty person. Type II error is acquitting an innocent person. Type II error is convicting an innocent person.
Answer:
(A) Type I error is convicting an innocent person.
Step-by-step explanation:
A type I error is the rejection of a true null hypothesis.A type II error is the non-rejection of a false negative/null hypothesis.When someone is on trial for suspicion of committing a crime, the hypotheses are:
[tex]H_0[/tex] : innocent
[tex]H_A[/tex] : guilty
From the given options, A Type I error is convicting an innocent person. If the null hypothesis holds (i.e. a person is innocent) nut we still go ahead to convict the person, we have rejected a true null hypothesis.
The Colorado Mining and Mineral Company has 1000 employees engaged in its mining operations. It has been estimated that the probability of a worker meeting with an accident during a 1-yr period is 0.08. What is the probability that more than 70 workers will meet with an accident during the 1-yr period
Answer:
86.65% probability that more than 70 workers will meet with an accident during the 1-yr period
Step-by-step explanation:
Binomial probability distribution
Probability of exactly x sucesses on n repeated trials, with p probability.
Can be approximated to a normal distribution, using the expected value and the standard deviation.
The expected value of the binomial distribution is:
[tex]E(X) = np[/tex]
The standard deviation of the binomial distribution is:
[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]
Normal probability distribution
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
When we are approximating a binomial distribution to a normal one, we have that [tex]\mu = E(X)[/tex], [tex]\sigma = \sqrt{V(X)}[/tex].
In this problem, we have that:
[tex]p = 0.08, n = 1000[/tex]
So
[tex]\mu = E(X) = np = 1000*0.08 = 80[/tex]
[tex]\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{1000*0.08*0.92} = 8.58[/tex]
What is the probability that more than 70 workers will meet with an accident during the 1-yr period
Using continuity correction, this is [tex]P(X \geq 70 + 0.5) = P(X \geq 70.5)[/tex], which is 1 subtracted by the pvalue of Z when X = 70.5. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{70.5 - 80}{8.58}[/tex]
[tex]Z = -1.11[/tex]
[tex]Z = -1.11[/tex] has a pvalue of 0.1335
1 - 0.1335 = 0.8665
86.65% probability that more than 70 workers will meet with an accident during the 1-yr period
The probability that more than 70 workers will be involved in an accident is 0.8665
The given parameters are:
[tex]\mathbf{n = 1000}[/tex] --- population
[tex]\mathbf{p = 0.08}[/tex] --- the probability that a worker meets an accident
[tex]\mathbf{x = 70}[/tex] -- the number of workers
Start by calculating the mean and the standard deviation
[tex]\mathbf{\mu = np}[/tex] --- mean
So, we have:
[tex]\mathbf{\mu = 1000 \times 0.08}[/tex]
[tex]\mathbf{\mu = 80}[/tex]
[tex]\mathbf{\sigma = \sqrt{\mu(1 - p)}}[/tex]
So, we have:
[tex]\mathbf{\sigma = \sqrt{80 \times (1 - 0.08)}}[/tex]
[tex]\mathbf{\sigma = \sqrt{73.6}}[/tex]
[tex]\mathbf{\sigma = 8.58}[/tex]
The probability is then represented as
[tex]\mathbf{P(x > 70) = P(x > 70.5)}[/tex] ---- By continuity correction
Calculate the z-score for x = 70.5
[tex]\mathbf{z = \frac{x - \mu}{\sigma}}[/tex]
So, we have:
[tex]\mathbf{z = \frac{70.5 - 80}{8.58}}[/tex]
[tex]\mathbf{z = -1.11}[/tex]
So, we have:
[tex]\mathbf{P(x > 70) = P(z > -1.11)}[/tex]
Using z-scores of probabilities, we have:
[tex]\mathbf{P(x > 70) = 0.8665}[/tex]
Hence, the probability that more than 70 workers will be involved in an accident is 0.8665
Read more about probabilities at:
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A statistics lecturer poses the following question to her students as homework: 'Suppose I collected a sample and calculated the sample proportion. If I construct a 90% confidence interval for the population proportion and a 95% confidence interval for the population proportion, which of these intervals will be wider?' Three students provide their answers as follows: Tim: 'The 90% confidence interval will be wider.' Trevor: 'The 95% confidence interval will be wider.' Tracy: 'There is not enough information to tell. Either interval could be wider.'
Answer: The 95% confidence interval will be wider.
Step-by-step explanation:
Confidence interval for population proportion is written as
Sample proportion ± margin of error
margin of error = z score × √pq/n
The z score is determined by the confidence level. The z score for a confidence level of 95% is higher than the z score for a confidence level of 90%
This means that with all other things being equal, a 95% confidence level will give a higher margin of error compared to a 90% confidence level.
The higher the margin of error, the wider the confidence interval. Therefore,
The 95% confidence interval will be wider.
The correct statement is provided by Trevor: 'The 95% confidence interval will be wider.'
Given that,
If construct a 90% confidence interval for the population proportion,
And a 95% confidence interval for the population proportion,
Tim: 'The 90% confidence interval will be wider.'
Trevor: 'The 95% confidence interval will be wider.
Tracy: 'There is not enough information to tell.
Either interval could be wider.'
We have to determine,
Which of these intervals will be wider.
According to the question,
Three students provide their answers as follows:
Tim: 'The 90% confidence interval will be wider.
'Trevor: 'The 95% confidence interval will be wider.
'Tracy: 'There is not enough information to tell. Either interval could be wider.'
Therefore, Confidence interval for population proportion is written as
Sample proportion ± margin of error
Margin of error [tex]= z score \times \frac{ \sqrt{pq}}{n}[/tex]
The z score for a confidence level of 95% is higher than the z score for a confidence level of 90%.
Other things being equal, a 95% confidence level will give a higher margin of error compared to a 90% confidence level.
The higher the margin of error, the wider the confidence interval.
Therefore, The 95% confidence interval will be wider.
Hence, The correct statement is provided by Trevor: 'The 95% confidence interval will be wider.'
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Write two different word problems that can be represented by the following equation 5.25x+7.50=75.75
Answer:
12.75
Step-by-step explanation:
A comparative study of organic and conventionally grown produce was checked for the presence of E. coli. Results are summarized below. The Prevalence of E. Coli in Organic and Conventional Produce Sample Size E. Coli Prevalence Organic 200 5 Conventional 500 25 Is there a significant difference in the proportion of E. Coli in organic vs. conventionally grown produce? Test at α = 0.10. Be sure to report your hypotheses, show all work, and explain the meaning of your answer.
Answer:
The calculated z- value = 1.479 < 1.645 at 0.10 or 90% level of significance.
The null hypothesis is accepted at 90% level of significance.
There is no significant difference in the proportion of E. Coli in organic vs. conventionally grown produce.
Step-by-step explanation:
Step:-(i)
Given first sample size n₁ = 200
The first sample proportion [tex]p_{1} = \frac{5}{200} = 0.025[/tex]
Given first sample size n₂= 500
The second sample proportion [tex]p_{2} = \frac{25}{500} = 0.05[/tex]
Step:-(ii)
Null hypothesis :H₀:There is no significant difference in the proportion of E. Coli in organic vs. conventionally grown produce
Alternative hypothesis:-H₁
There is significant difference in the proportion of E. Coli in organic vs. conventionally grown produce
level of significance ∝=0.10
Step:-(iii)
The test statistic
[tex]Z =\frac{p_{1} - p_{2} }{\sqrt{pq(\frac{1}{n_{1} }+\frac{1}{n_{2} } } }[/tex]
where p = [tex]\frac{n_{1} p_{1} + n_{2}p_{2} }{n_{1}+n_{2} }= \frac{200X0.025+500X0.05 }{500+200}[/tex]
p = 0.0428
q = 1-p =1-0.0428 = 0.9572
[tex]Z =\frac{0.025- 0.05}{\sqrt{0.0428X0.9571(\frac{1}{200 }+\frac{1}{500 } } }[/tex]
Z = -1.479
|z| = |-1.479|
z = 1.479
The tabulated value z= 1.645 at 0.10 or 90% level of significance.
The calculated z- value = 1.479 < 1.645 at 0.10 or 90% level of significance.
The null hypothesis is accepted at 90% level of significance.
Conclusion:-
There is no significant difference in the proportion of E. Coli in organic vs. conventionally grown produce
A random sample of ten 2011 sports cars is taken and their city mileage is recorded. The results are as follows: 20 21 25 21 21 23 31 32 28 26 Assuming the population distribution is normal, compute E, the margin of error for the t interval, for a 90% confidence interval for m, the population mean of the city mpg for 2011 sports cars.
Answer:
[tex] ME = 1.833 * \frac{4.367}{\sqrt{10}}= 2.531[/tex]
Step-by-step explanation:
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[/tex] represent the sample mean for the sample
[tex]\mu[/tex] population mean (variable of interest)
s represent the sample standard deviation
n represent the sample size
Solution to the problem
The confidence interval for the mean is given by the following formula:
[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)
In order to calculate the mean and the sample deviation we can use the following formulas:
[tex]\bar X= \sum_{i=1}^n \frac{x_i}{n}[/tex] (2)
[tex]s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}[/tex] (3)
The mean calculated for this case is [tex]\bar X=24.8[/tex]
The sample deviation calculated [tex]s=4.367[/tex]
In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:
[tex]df=n-1=10-1=9[/tex]
Since the Confidence is 0.90 or 90%, the value of [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.005[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,9)".And we see that [tex]t_{\alpha/2}=1.833[/tex]
And the margin of error would be given by:
[tex] ME = 1.833 * \frac{4.367}{\sqrt{10}}= 2.531[/tex]
A box contains 6 black socks and 4 red socks. What is the probability of picking two black socks from the bag without replacement?
Answer:
43/91 or 47%
Step-by-step explanation:
In 2014, the Community College Survey of Student Engagement reported that 32% of the students surveyed rarely or never use academic advising services. Suppose that in reality, 42% of community college students rarely or never use academic advising services at their college. In a simulation we select random samples from this population. For each sample we calculate the proportion who rarely or never use academic advising services. If we randomly sample 200 students from this population repeatedly, the standard error is approximately 3.5%. Is it unusual to see 32% who rarely or never use academic advising services in one of these samples
Answer:
[tex]Z = -2.865[/tex] means that it would be unusual to see 32% who rarely or never use academic advising services in one of these samples
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
Z scores below -2 are considered unusually low.
Z scores above 2 are considered unusually high.
For a sample proportion p in a sample of size n, we have that [tex]\mu = p, \sigma = \sqrt{\frac{p(1-p)}{n}}[/tex]
In this problem, we have that:
[tex]\mu = 0.42, \sigma = \sqrt{\frac{0.42*0.58}{200}} = 0.0349[/tex]
Is it unusual to see 32% who rarely or never use academic advising services in one of these samples
What is the z-score for X = 0.32?
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{0.32 - 0.42}{0.0349}[/tex]
[tex]Z = -2.865[/tex]
[tex]Z = -2.865[/tex] means that it would be unusual to see 32% who rarely or never use academic advising services in one of these samples
No, it is not unusual to see 32% of students rarely or never using academic advising services in one of these samples.
Explanation:To determine if it is unusual to see 32% of students rarely or never using academic advising services in one of these samples, we can compare it to the range of values that would be considered usual. In this case, we can use the 95% confidence interval provided, which states that the true proportion of community college students who rarely or never use academic advising services is between 0.113 and 0.439. If the observed proportion falls within this interval, it would be considered usual; otherwise, it would be considered unusual.
Since 32% falls within the range of 0.113 and 0.439, it is considered a usual value. Therefore, it is not unusual to see 32% of students rarely or never use academic advising services in one of these samples.
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A florist charges $12.00 for delivery plus an additional $1.50 per mile from the flower shop. The florist pays the delivery driver $0.75 per mile and $4.50 for gas per delivery. If x is the number of miles a delivery location is from the flower shop, what expression models the amount of money the florist earns for each delivery?
The amount the florist earns for each delivery, with 'x' being the miles away from the flower shop, can be modeled with the equation: Earnings = (7.5 + 0.75x). This represents the fixed net income of $7.5 and $0.75 per mile after paying the driver.
Explanation:The florist charges $12.00 for delivery and an additional $1.50 per mile from the flower shop. However, the florist also has costs to cover, namely $0.75 per mile to pay the driver, and $4.50 for gas per delivery. The net earning per delivery, with 'x' representing the number of miles a delivery location is from the flower shop, can be modeled by the following algebraic expression: Earnings = (12 + 1.5x) - (0.75x + 4.5).
This actually simplifies to: Earnings = (7.5 + 0.75x). The 7.5 is the fixed net income for each delivery (gross earnings minus the gasoline cost) and 0.75x is the per-mile net income after the driver is paid.
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Researchers are investigating the effectiveness of leg-strength training on cycling performance. A sample of 7 men will be selected to participate in a training program that lasts for one month. Peak power during cycling will be recorded for each man both before training and after training. The mean difference in times will be used to construct a 95 percent confidence interval for the mean difference in the population.
Answer:
A. The interval will be narrower if 15 men are used in the sample.
Step-by-step explanation:
Hello!
When all other things remain the same, which of the following statements about the width of the interval is correct?
A. The interval will be narrower if 15 men are used in the sample.
B. The interval will be wider if 15 men are used in the sample.
C. The interval will be narrower if 5 men are used in the sample.
D. The interval will be narrower if the level is increased to 99% confidence.
E. The interval will be wider if the level is decreased to 90% confidence.
Consider that the variable of interest "Xd: Difference between the peak power of a cyclist before training and after training" has a normal distribution. To construct the confidence interval for the population mean of the difference you have to use a pooled t-test.
The general structure for the CI is "point estimate"±" margin fo error"
Any modification to the sample size, sample variance and/or the confidence level affect the length of the interval (amplitude) and the margin of error (semiamplitude)
The margin of error of the interval is:
d= [tex]t_{n-1;1-\alpha /2}[/tex] * (Sd/n)
1) The sample size changes, all other terms of the interval stay the same.
As you can see the margin of error and the sample size (n) have an indirect relationship. This means, that when the sample size increases, the semiamplitude decreases, and when the sample size decreases, the semiamplitude increases.
↓d= [tex]t_{n-1;1-\alpha /2}[/tex] * (Sd/↑n)
↑d= [tex]t_{n-1;1-\alpha /2}[/tex] * (Sd/↓n)
Correct option: A. The interval will be narrower if 15 men are used in the sample.
2) The confidence level has a direct relationship with the semiamplitude of the interval, this means that when the confidence level increases, so do the semiamplitude, and if the level decreases, so do the semiamplitude:
↓d= ↓[tex]t_{n-1;1-\alpha /2}[/tex] * (Sd/n)
↑d= ↑[tex]t_{n-1;1-\alpha /2}[/tex] * (Sd/n)
I hope it helps!
Answer:
The interval will be narrower if 15 men are used in the sample.
Step-by-step explanation: