Question 1

A poll of 1700 randomly selected students in grades 6 through 8 was conducted and found that 37% enjoy playing sports. Would confidence in the results increase if the sample size were 3800 instead of 1700? Why or why not?

Accepted Answer

The answer of A poll of 1700 randomly selected students...

Question 2

A history teacher assigns letter grades on a test according to the following scheme: A: Top 10% B: Scores below the top 10% and above the bottom 60% C: Scores below the top 40% and above the bottom 20% D: Scores below the top 80% and above the bottom 10% F: Bottom 10% Scores on the test are normally distributed with a mean of 69 and a standard deviation of 13.4. Find the numerical limits for each letter grade.

Accepted Answer

To find the numerical limits for each letter grade, we use the Z-score formula $Z = \frac{X - \mu}{\sigma}$, where $X$ is the score, $\mu$ is the mean, and $\sigma$ is the standard deviation. For a normally distributed set of scores, the top 10% corresponds to a Z-score of approximately 1.28, the top 40% to a Z-score of approximately 0.25, the bottom 60% (or top 40%) to a Z-score of approximately -0.25, the top 80% to a Z-score of approximately -0.84, and the bottom 10% to a Z-score of approximately -1.28. Using these Z-scores and the given mean (69) and standard deviation (13.4), we calculate the score limits for each grade category. For example, for A (top 10%), the score limit is $69 + 1.28(13.4) \approx 86$. Similar calculations for other grades lead to the limits provided in option A.

Question 3

The systolic blood pressures of the patients at a hospital are normally distributed with a mean of 136 mm Hg and a standard deviation of 13.8 mm Hg. Find the two blood pressures having these properties: the mean is midway between them and 90% of all blood pressures are between them.

Accepted Answer

Given that 90% of the values lie between these two blood pressures and the distribution is normal, we use the concept of standard deviations in a normal distribution. For 90% of the data to be within these values, we are looking at the blood pressures that lie within ±1.645 standard deviations from the mean (since the z-score for 90% confidence interval is approximately ±1.645). Calculating the values: Mean ± 1.645(Standard Deviation) = 136 ± 1.645(13.8) = 136 ± 22.7 = 113.3 mm Hg, 158.7 mm Hg.

Question 4

Identify the probability of each sample, and describe the sampling distribution of the sample means.
-Personal phone calls received in the last three days by a new employee were 2, 6, and 8. Assume that samples of size 2 are randomly selected with replacement from this population of three values.

Accepted Answer

Since samples of size 2 are selected with replacement from a population of three values (2, 6, 8), there are 3x3=9 possible samples. Each sample has an equal chance of being selected, making the probability of each sample $1 / 9$.

Question 5

Assume that the weights of quarters are normally distributed with a mean of 5.67 g and a standard deviation 0.070 g. A vending machine will only accept coins weighing between 5.48 g and 5.82 g. What percentage of legal quarters will be rejected?

Accepted Answer

The answer of Assume that the weights of quarters are...

Question 6

List the different possible samples, and find the mean of each of them
-Personal phone calls received in the last three days by a new employee were 3, 1, and 8. Assume that samples of size 2 are randomly selected with replacement from this population of three values.

Accepted Answer

The answer of List the different possible samples, and find...

Question 7

The diameters of pencils produced by a certain machine are normally distributed with a mean of 0.30 inches and a standard deviation of 0.01 inches. What is the probability that the diameter of a randomly selected pencil will be less than 0.285 inches?

Accepted Answer

To find the probability that the diameter of a randomly selected pencil will be less than 0.285 inches, we use the Z-score formula: Z = (X - μ) / σ, where X is the value in question, μ is the mean, and σ is the standard deviation. Plugging in the values: Z = (0.285 - 0.30) / 0.01 = -1.5. Looking up the Z-score of -1.5 in a standard normal distribution table gives a probability of approximately 0.0668.

Question 8

A bank's loan officer rates applicants for credit. The ratings are normally distributed with a mean of 200 and a standard deviation of 50. If an applicant is randomly selected, find the probability of a rating that is between 170 and 220.

Accepted Answer

To find the probability of a rating between 170 and 220, we use the Z-score formula: Z = (X - μ) / σ, where X is the score, μ is the mean, and σ is the standard deviation. For X=170, Z1 = (170 - 200) / 50 = -0.6. For X=220, Z2 = (220 - 200) / 50 = 0.4. Looking these Z-scores up in a standard normal distribution table or using a calculator, the probability between Z1 and Z2 is approximately 0.3811.

Question 9

A poll of 1200 randomly selected students in grades 6 through 8 was conducted and found that 39% enjoy playing sports. What is the sampling distribution suggested by the given data?

Accepted Answer

The answer of A poll of 1200 randomly selected students...

Question 10

Scores on a test have a mean of 73 and $Q _ { 3 }$ is 80. The scores have a distribution that is approximately normal.
Find P90. (You will need to first find the standard deviation.)

Accepted Answer

To find P90, we first need to determine the standard deviation. In a normal distribution, Q3 (the 75th percentile) is approximately 0.675 standard deviations above the mean. Given Q3 = 80 and mean = 73, the standard deviation is approximately (80 - 73) / 0.675 = 10.37. For P90, which is 1.28 standard deviations above the mean, we calculate 73 + 1.28 * 10.37 ≈ 86.4.

Question 11

The volumes of soda in quart soda bottles are normally distributed with a mean of 32.3 oz and a standard deviation of 1.2 oz. What is the probability that the volume of soda in a randomly selected bottle will be less than 32 oz?

Accepted Answer

The answer of The volumes of soda in quart soda...

Question 12

Scores on a test have a mean of 62 and $\mathrm { Q } _ { 3 }$ is 84. The scores have a distribution that is approximately normal. Find the standard deviation. Round your answer to the nearest tenth.

Accepted Answer

The answer of Scores on a test have a mean...

Question 13

A recent survey based on a random sample of n = 500 voters, predicted that the Independent candidate for the mayoral election will get 24% of the vote, but he actually gets 28%. Can it be concluded that the survey was done incorrectly?

Accepted Answer

The answer of A recent survey based on a random...

Question 14

A math teacher gives two different tests to measure students' aptitude for math. Scores on the first test are normally distributed with a mean of 23 and a standard deviation of 5.1. Scores on the second test are normally distributed with a mean of 68 and a standard deviation of 10.4. Assume that the two tests use different scales to measure the same aptitude. If a student scores 28 on the first test, what would be his equivalent score on the second test? (That is, find the score that would put him in the same percentile.)

Accepted Answer

To find the equivalent score, first calculate the z-score for the first test: z = (28 - 23) / 5.1 = 0.98. Then, use this z-score to find the equivalent score on the second test: score = 68 + (0.98 * 10.4) = 78.192, which rounds to 78.

Question 15

A poll of 1300 randomly selected students in grades 6 through 8 was conducted and found that 44% enjoy playing sports. Is the 44% result a statistic or a parameter? Explain.

Accepted Answer

The answer of A poll of 1300 randomly selected students...

Question 16

A bank's loan officer rates applicants for credit. The ratings are normally distributed with a mean of 200 and a standard deviation of 50. If an applicant is randomly selected, find the probability of a rating that is between 200 and 275.

Accepted Answer

To find the probability of a rating between 200 and 275, we use the Z-score formula: Z = (X - μ) / σ, where X is the value, μ is the mean, and σ is the standard deviation. For X=275, Z = (275-200)/50 = 1.5. Looking up 1.5 in the Z-table gives us the probability of 0.9332. However, since we're looking for the probability between the mean (200) and 275, we subtract the probability of being below the mean (0.5) from 0.9332, resulting in 0.4332.

Question 17

The weekly salaries of teachers in one state are normally distributed with a mean of $490 and a standard deviation of $45. What is the probability that a randomly selected teacher earns more than $525 a week?

Accepted Answer

The probability that a randomly selected teacher earns more than $525 a week is the area under the normal distribution curve to the right of $525. This area is equal to 1 - P(X < 525), where X is the random variable representing the weekly salary of a teacher. Using the standard normal distribution (z-distribution), we have:z = (525 - 490) / 45 = 0.78P(X < 525) = P(Z < 0.78) = 0.7823Therefore, the probability that a randomly selected teacher earns more than $525 a week is 1 - 0.7823 = 0.2177.

Question 18

In one region, the September energy consumption levels for single-family homes are found to be normally distributed with a mean of 1050 kWh and a standard deviation of 218 kWh. For a randomly selected home, find the probability that the September energy consumption level is between 1100 kWh and 1225 kWh.

Accepted Answer

To find the probability, calculate the z-scores for 1100 kWh and 1225 kWh, then find the probability between these z-scores using the standard normal distribution table. The z-score for 1100 kWh is (1100-1050)/218 ≈ 0.2294, and for 1225 kWh is (1225-1050)/218 ≈ 0.8028. The probability between these z-scores is approximately 0.1971.

Question 19

The lengths of human pregnancies are normally distributed with a mean of 268 days and a standard deviation of 15 days. What is the probability that a pregnancy lasts at least 300 days?

Accepted Answer

To find the probability that a pregnancy lasts at least 300 days, we use the Z-score formula: Z = (X - μ) / σ, where X is the value (300 days), μ is the mean (268 days), and σ is the standard deviation (15 days). Plugging in the values, we get Z = (300 - 268) / 15 = 32 / 15 ≈ 2.13. Looking up 2.13 in a standard normal distribution table or using a calculator, we find the area to the right (since we're interested in "at least") is approximately 0.0166.

Question 20

List the different possible samples, and find the mean of each of them
-The number of books sold over the course of the four-day book fair were 139, 199, 241, and 61. Assume that samples of size 2 are randomly selected with replacement from this population of four values.

Accepted Answer

The answer of List the different possible samples, and find...

Quiz 6: Normal Probability Distributions

A poll of 1700 randomly selected students in grades 6 through 8 was conducted and found that 37% enjoy playing sports. Would confidence in the results increase if the sample size were 3800 instead of 1700? Why or why not?

The systolic blood pressures of the patients at a hospital are normally distributed with a mean of 136 mm Hg and a standard deviation of 13.8 mm Hg. Find the two blood pressures having these properties: the mean is midway between them and 90% of all blood pressures are between them.

Identify the probability of each sample, and describe the sampling distribution of the sample means. -Personal phone calls received in the last three days by a new employee were 2, 6, and 8. Assume that samples of size 2 are randomly selected with replacement from this population of three values.

Assume that the weights of quarters are normally distributed with a mean of 5.67 g and a standard deviation 0.070 g. A vending machine will only accept coins weighing between 5.48 g and 5.82 g. What percentage of legal quarters will be rejected?

List the different possible samples, and find the mean of each of them -Personal phone calls received in the last three days by a new employee were 3, 1, and 8. Assume that samples of size 2 are randomly selected with replacement from this population of three values.