Aldo rented a truck for one day. There was a base fee of $20.99, and there was an additional charge of 86 cents for each mile driven. Aldo had to pay $155.15
when he returned the truck. For how many miles did he drive the truck?
​

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.

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.

To learn more about Confidence interval

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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.

The quantity of oil in the first 180 minutes was 62.78 million gallons.

It 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.

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.

More about the integration link is given below.

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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.

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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Answer:

The probability is [tex]\frac{5}{9}[/tex]

Step-by-step explanation:

The total number of students are 200.

number of full timers is 110 and number of part timers is 90.

number of male students is 90 and number of female students is 110.

Let the probability of part timers be P(B).

P(B) = [tex]\frac{90}{200}[/tex] = [tex]\frac{9}{20}[/tex]

Let the probability of female part timers be P(A)

P(A) = [tex]\frac{50}{200} = \frac{5}{20}[/tex]

now, the final probability is

= [tex]\frac{P(A)}{P(B)}[/tex]

=[tex]\frac{5/20}{9/20} = \frac{5}{9}[/tex]

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

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.

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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Answer:

y=3 or y-3=0

Step-by-step explanation:

-2y=-6

-2y÷(-2)=-6÷(-2)

y=-6÷(-2)

y=-6÷2

y=3

or

-2y=-6

-2y+6=0

-y+3=0

y-3=0

See picture for answer and solution steps.

1km___100000cm

0.7km___X=7000cm

(0.7*100000)/1=7000

25cm___1 step

7000cm___X=280 steps

(7000*1)/25= 280

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].

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.

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.

Answer:  C

Step-by-step explanation:

Rejecting the null hypothesis means we've found a significant difference in the means.  That means the probability that we'd see means so far apart by chance is less than our threshold 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

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%.

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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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]

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.

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.

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.

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.

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.

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:

This is the t-value you use to test your hypothesis on a t-distribution curve.

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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

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

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.

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.        

The confidence interval for population mean (when population standard deviation is unknown) is given by :-

[tex]\overline{x}-t^*\dfrac{s}{\sqrt{n}}< \mu<\overline{x}+z^*\dfrac{s}{\sqrt{n}}[/tex]

, where n= sample size

[tex]\overline{x}[/tex] = Sample mean

s= sample size

t* = Critical value.

Given : n= 25

Degree of freedom : [tex]df=n-1=24[/tex]

[tex]\overline{x}= \$93.36[/tex]

[tex]s=\ $19.95[/tex]

Significance level for 98% confidence interval : [tex]\alpha=1-0.98=0.02[/tex]

Using t-distribution table ,

Two-tailed critical value for 98% confidence interval :

[tex]t^*=t_{\alpha/2,\ df}=t_{0.01,\ 24}=2.4922[/tex]

⇒ The critical value that should be used in constructing the confidence interval = 2.4922

Then, the 95% confidence interval would be :-

[tex]93.36-(2.4922)\dfrac{19.95}{\sqrt{25}}< \mu<93.36+(2.4922)\dfrac{19.95}{\sqrt{25}}[/tex]

[tex]=93.36-9.943878< \mu<93.36+9.943878[/tex]

[tex]=93.36-9.943878< \mu<93.36+9.943878[/tex]

[tex]=83.416122< \mu<103.303878\approx83.4161<\mu<103.3039[/tex]

Hence, the 98% confidence interval for the mean repair cost for the dryers. = [tex]83.4161<\mu<103.3039[/tex]

Answer:

Since the p–value is less than the significance level, the null hypothesis is rejected. The true proportion of engineering students planning graduate studies significantly different from 0.5 at  α=0.10

Step-by-step explanation:

Please see attachment

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.

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.

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:

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.

To calculate the test statistic for a hypothesis test about a population proportion, several pieces of information are required:

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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Answer:

249.6×432.6=107,976.96

Answer:

Step-by-step explanation:

2.4*104=249.6

4.2*103=432.6

(249.6)(432.6)=107976.96

All a matter of simple multiplication. ;)

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.

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:

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.

Mrs. Maxwell needs to buy 18.66 packs of pencil containing 336 pencils for 84 students.

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.