Question 1

A population has a proportion equal to 0.30. Calculate the following probabilities with n = 100. Find P( ≥ 0.27).

Accepted Answer

The probability of getting a sample proportion of 0.27 or more is the same as the probability of getting a sample proportion of less than 0.27. The probability of getting a sample proportion of less than 0.27 is 1 - P(X ≥ 0.27) = 1 - 0.7422 = 0.2578.

Question 2

Suppose the life of a particular brand of calculator battery is approximately normally distributed with a mean of 75 hours and a standard deviation of 10 hours. What is the probability that a single battery randomly selected from the population will have a life between 70 and 80 hours?

Accepted Answer

To find the probability that a single battery will have a life between 70 and 80 hours, we use the properties of the normal distribution. First, we calculate the z-scores for 70 and 80 hours:For 70 hours: $Z = \frac{70 - 75}{10} = -0.5$For 80 hours: $Z = \frac{80 - 75}{10} = 0.5$Using standard normal distribution tables or a calculator, we find the probabilities corresponding to these z-scores:The probability of a z-score being less than 0.5 is approximately 0.6915, and the probability of a z-score being less than -0.5 is approximately 0.3085. The probability of a battery lasting between 70 and 80 hours is the difference between these two probabilities:$0.6915 - 0.3085 = 0.3830$Therefore, the correct answer is 0.3830.

Question 3

The proportion of items in a population that possess a specific attribute is known to be 0.70. If a simple random sample of size n = 100 is selected and the proportion of items in the sample that contain the attribute of interest is 0.65, what is the sampling error?

Accepted Answer

Sampling error is calculated as the difference between the sample statistic and the population parameter. Here, it is 0.65 (sample proportion) - 0.70 (population proportion) = -0.05.

Question 4

If a random sample of 200 items is taken from a population in which the proportion of items having a desired attribute is p = 0.30, what is the probability that the proportion of successes in the sample will be less than or equal to 0.27?

Accepted Answer

The answer of If a random sample of 200 items...

Question 5

A normally distributed population has a mean of 500 and a standard deviation of 60. Determine the probability that a random sample of size 25 selected from the population will have a sample mean greater than or equal to 515.

Accepted Answer

The correct answer is A because the probability that a random sample of size 25 selected from the population will have a sample mean greater than or equal to 515 is 0.1056.

Question 6

A population has a proportion equal to 0.30. Calculate the following probabilities with n = 100. Find

Accepted Answer

The answer of A population has a proportion equal to...

Question 7

A population has a proportion equal to 0.30. Calculate the following probabilities with n = 100. Find

Accepted Answer

The answer of A population has a proportion equal to...

Question 8

A random sample of 100 items is selected from a population of size 350. What is the probability that the sample mean will exceed 200 if the population mean is 195 and the population standard deviation equals 20? (Hint: Use the finite correction factor since the sample size is more than 5% of the population size.)

Accepted Answer

The answer of A random sample of 100 items is...

Question 9

A normally distributed population has a mean of 500 and a standard deviation of 60. Determine the probability that a random sample of size 16 selected from this population will have a sample mean less than 475.

Accepted Answer

The probability that a sample mean is less than a certain value can be found using the Z-score formula for a sample mean: Z = (X̄ - μ) / (σ/√n), where X̄ is the sample mean, μ is the population mean, σ is the population standard deviation, and n is the sample size. Here, Z = (475 - 500) / (60/√16) = -25 / (60/4) = -25 / 15 = -1.67. Looking up -1.67 on the Z-table gives a probability of approximately 0.0475, indicating the probability that a random sample of size 16 will have a sample mean less than 475.

Question 10

Suppose the life of a particular brand of calculator battery is approximately normally distributed with a mean of 75 hours and a standard deviation of 10 hours. What is the probability that 16 randomly sampled batteries from the population will have a sample mean life of between 70 and 80 hours?

Accepted Answer

To find the probability that the sample mean life of 16 batteries is between 70 and 80 hours, we use the formula for the standard error of the mean: $SE = \frac{\sigma}{\sqrt{n}}$, where $\sigma$ is the population standard deviation and $n$ is the sample size. Here, $\sigma = 10$ and $n = 16$, so $SE = \frac{10}{\sqrt{16}} = \frac{10}{4} = 2.5$. Next, we calculate the Z-scores for 70 and 80 hours: $Z_{70} = \frac{70 - 75}{2.5} = -2$ and $Z_{80} = \frac{80 - 75}{2.5} = 2$. Looking up these Z-scores in a standard normal distribution table or using a calculator, we find the probabilities corresponding to Z = -2 and Z = 2, and the area between these Z-scores represents the probability we're interested in, which is approximately 0.9544. However, due to rounding or a typo in the options, the closest provided option is A) 0.9444.

Question 11

The branch manager for United Savings and Loan in Seaside, Virginia, has worked with her employees in an effort to reduce the waiting time for customers at the bank. Recently, she and the team concluded that average waiting time is now down to 3.5 minutes with a standard deviation equal to 1.0 minute. However, before making a statement at a managers' meeting, this branch manager wanted to double-check that the process was working as thought. To make this check, she randomly sampled 25 customers and recorded the time they had to wait. She discovered that mean wait time for this sample of customers was 4.2 minutes. Based on the team's claims about waiting time, what is the probability that a sample mean for n = 25 people would be as large or larger than 4.2 minutes?

Accepted Answer

The answer of The branch manager for United Savings and...

Question 12

A population has a proportion equal to 0.30. Calculate the following probabilities with n = 100. Find

Accepted Answer

The answer of A population has a proportion equal to...

Question 13

In an article entitled "Fuel Economy Calculations to Be Altered," James R. Healey indicated that the government planned to change how it calculates fuel economy for new cars and trucks. This is the first modification since 1985. It is expected to lower average mileage for city driving in conventional cars from 10% to 20%. AAA has forecast that the 2008 Ford F-150 would achieve 15.7 mile per gallon (mpg). The 2008 Ford F-150 was tested by AAA members driving the vehicle themselves and was found to have an average of 14.3 mpg. Assume that the mean obtained by AAA members is the true mean for the population of 2008 Ford F-150 trucks and that the population standard deviation is 5 mpg. The current method of calculating the mpg forecasts that the 2008 F-150 will average 16.8 mpg. Determine the probability that these same 100 AAA members would average more than 16.8 mpg while testing the 2008 F-150.

Accepted Answer

The probability is 0 because the sample mean (14.3 mpg) is significantly lower than 16.8 mpg, and given the population standard deviation (5 mpg) and the sample size (100), it's statistically improbable for the sample mean to exceed the forecasted mean of 16.8 mpg under normal distribution assumptions.

Question 14

Suppose that a population is known to be normally distributed with mean = 2,000 and standard deviation = 230. If a random sample of size n = 8 is selected, calculate the probability that the sample mean will exceed 2,100.

Accepted Answer

The answer of Suppose that a population is known to...

Question 15

SeeClear Windows makes windows for use in homes and commercial buildings. The standards for glass thickness call for the glass to average 0.375 inches with a standard deviation equal to 0.050 inch. Suppose a random sample of n = 50 windows yields a sample mean of 0.392 inches. What is the probability if the windows meet the standards?

Accepted Answer

To find the probability, we calculate the z-score using the formula: z = (X̄ - μ) / (σ/√n), where X̄ = 0.392, μ = 0.375, σ = 0.050, and n = 50. The z-score is approximately 2.40. Using a standard normal distribution table, the probability of a z-score greater than 2.40 is approximately 0.0082.

Question 16

Given a population where the proportion of items with a desired attribute is p = 0.25, if a sample of 400 is taken, what is the standard deviation of the sampling distribution of ?

Accepted Answer

The standard deviation of the sampling distribution of the sample proportion is calculated using the formula: sqrt(p(1-p)/n), where p = 0.25 and n = 400. This gives sqrt(0.25*0.75/400) = 0.0217.

Question 17

Suppose the life of a particular brand of calculator battery is approximately normally distributed with a mean of 75 hours and a standard deviation of 10 hours. If the manufacturer of the battery is able to reduce the standard deviation of battery life from 10 to 9 hours, what would be the probability that 16 batteries randomly sampled from the population will have a sample mean life of between 70 and 80 hours?

Accepted Answer

To find the probability that the sample mean life of 16 batteries is between 70 and 80 hours, we use the formula for the standard error of the mean, which is the standard deviation divided by the square root of the sample size (n). With the new standard deviation of 9 hours, the standard error is 9 / sqrt(16) = 9 / 4 = 2.25 hours. We then use the Z-score formula for each boundary: Z = (X - μ) / σ, where X is the boundary value, μ is the mean, and σ is the standard error. For X = 70, Z = (70 - 75) / 2.25 = -2.22. For X = 80, Z = (80 - 75) / 2.25 = 2.22. Looking up these Z-scores in the standard normal distribution table or using a calculator, we find the probabilities corresponding to these Z-scores and subtract the lower from the upper to find the probability of the sample mean being between 70 and 80 hours. The result is approximately 0.9736, indicating a high probability.

Question 18

Suppose nine items are randomly sampled from a normally distributed population with a mean of 100 and a standard deviation of 20. The nine randomly sampled values are:  Calculate the probability of getting a sample mean that is smaller than the sample mean for these nine sampled values.

Accepted Answer

The answer of Suppose nine items are randomly sampled from...

Question 19

Given a population where the proportion of items with a desired attribute is p = 0.25, if a sample of 400 is taken, what is the probability the proportion of successes in the sample will be greater than 0.22?

Accepted Answer

This problem can be solved using the normal approximation to the binomial distribution. The mean of the sampling distribution of the proportion is $p = 0.25$, and the standard deviation is $\sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.25(0.75)}{400}} = 0.02165$. We want to find the probability that the sample proportion is greater than 0.22, so we calculate the z-score for 0.22: $z = \frac{0.22 - 0.25}{0.02165} = -1.387$. Looking up this z-score in a standard normal distribution table or using a calculator gives us the probability of being less than 0.22. Since we want the probability of being greater, we subtract this value from 1. The exact probability might vary slightly depending on the method of calculation, but the closest answer provided is 0.9162, indicating a high likelihood of the sample proportion being greater than 0.22.

Question 20

In an article entitled "Fuel Economy Calculations to Be Altered," James R. Healey indicated that the government planned to change how it calculates fuel economy for new cars and trucks. This is the first modification since 1985. It is expected to lower average mileage for city driving in conventional cars from 10% to 20%. AAA has forecast that the 2008 Ford F-150 would achieve 15.7 mile per gallon (mpg). The 2008 Ford F-150 was tested by AAA members driving the vehicle themselves and was found to have an average of 14.3 mpg. Assume that the mean obtained by AAA members is the true mean for the population of 2008 Ford F-150 trucks and that the population standard deviation is 5 mpg. Suppose 100 AAA members were to test the 2008 F-150. Determine the probability that the average mpg would be at least 15.7.

Accepted Answer

The probability can be calculated using the standard normal distribution. The standard error is 5/sqrt(100) = 0.5. The z-score for 15.7 is (15.7 - 14.3) / 0.5 = 2.8. The probability of a z-score being at least 2.8 is approximately 0.0026.

Quiz 7: Introduction to Sampling Distributions

A population has a proportion equal to 0.30. Calculate the following probabilities with n = 100. Find P( ≥ 0.27) .

Suppose the life of a particular brand of calculator battery is approximately normally distributed with a mean of 75 hours and a standard deviation of 10 hours. What is the probability that a single battery randomly selected from the population will have a life between 70 and 80 hours?

The proportion of items in a population that possess a specific attribute is known to be 0.70. If a simple random sample of size n = 100 is selected and the proportion of items in the sample that contain the attribute of interest is 0.65, what is the sampling error?

If a random sample of 200 items is taken from a population in which the proportion of items having a desired attribute is p = 0.30, what is the probability that the proportion of successes in the sample will be less than or equal to 0.27?

A normally distributed population has a mean of 500 and a standard deviation of 60. Determine the probability that a random sample of size 25 selected from the population will have a sample mean greater than or equal to 515.

A population has a proportion equal to 0.30. Calculate the following probabilities with n = 100. Find

A population has a proportion equal to 0.30. Calculate the following probabilities with n = 100. Find

A normally distributed population has a mean of 500 and a standard deviation of 60. Determine the probability that a random sample of size 16 selected from this population will have a sample mean less than 475.

Suppose the life of a particular brand of calculator battery is approximately normally distributed with a mean of 75 hours and a standard deviation of 10 hours. What is the probability that 16 randomly sampled batteries from the population will have a sample mean life of between 70 and 80 hours?

A population has a proportion equal to 0.30. Calculate the following probabilities with n = 100. Find

Suppose that a population is known to be normally distributed with mean = 2,000 and standard deviation = 230. If a random sample of size n = 8 is selected, calculate the probability that the sample mean will exceed 2,100.

Given a population where the proportion of items with a desired attribute is p = 0.25, if a sample of 400 is taken, what is the standard deviation of the sampling distribution of ?

Suppose nine items are randomly sampled from a normally distributed population with a mean of 100 and a standard deviation of 20. The nine randomly sampled values are: Calculate the probability of getting a sample mean that is smaller than the sample mean for these nine sampled values.

Given a population where the proportion of items with a desired attribute is p = 0.25, if a sample of 400 is taken, what is the probability the proportion of successes in the sample will be greater than 0.22?