Question 1

Suppose a population has a mean of 400 and a standard deviation of 24.If a random sample of size 144 is drawn from the population, the probability of drawing a sample with a mean less than 402 is closest to _______.

Accepted Answer

To find the probability of drawing a sample with a mean less than 402, we use the Central Limit Theorem, which states that the sampling distribution of the sample mean will be normally distributed if the sample size is large enough (n ≥ 30 is a common rule of thumb), regardless of the shape of the population distribution. Given a population mean (μ) of 400, a standard deviation (σ) of 24, and a sample size (n) of 144, we first calculate the standard error of the mean (σM) as σ/√n = 24/√144 = 24/12 = 2. We then calculate the Z-score for a sample mean of 402, which is (X̄ - μ) / σM = (402 - 400) / 2 = 1. A Z-score of 1 corresponds to the area to the left of it in the standard normal distribution, which is approximately 0.8413. This means there's about an 84.13% chance of drawing a sample with a mean less than 402.

Question 2

Albert Abbasi, VP of Operations at Ingleside International Bank, is evaluating the service level provided to walk-in customers.Accordingly, he plans a sample of waiting times for walk-in customers.If the population of waiting times has a mean of 15 minutes and a standard deviation of 4 minutes, the probability that Albert's sample of 64 will have a mean less than 15 minutes is ________.

Accepted Answer

Given the symmetry of the normal distribution, the probability that the sample mean is less than the population mean (15 minutes) is 0.5000, as the distribution of sample means is centered around the population mean.

Question 3

A carload of steel rods has arrived at Cybermatic Construction Company.The car contains 50,000 rods.Claude Ong, Manager of Quality Assurance, directs his crew to measure the lengths of 100 randomly selected rods.If the population of rods has a mean length of 120 inches and a standard deviation of 0.05 inch, the probability that Claude's sample has a mean greater than 120.0125 inches is _____________.

Accepted Answer

To find the probability that the sample mean is greater than 120.0125 inches, we use the Central Limit Theorem. The standard error (SE) of the sample mean is the population standard deviation divided by the square root of the sample size, which is $0.05 / \sqrt{100} = 0.005$. We then calculate the Z-score for 120.0125 inches, which is $(120.0125 - 120) / 0.005 = 2.5$. Looking up 2.5 on the Z-table gives us the probability of 0.9938 that a Z-score is less than 2.5, so the probability of being greater than 2.5 is $1 - 0.9938 = 0.0062$.

Question 4

Suppose 30% of the U.S.population has green eyes.If a random sample of size 1200 U.S.citizens is drawn, then the probability that less than 348 of those U.S.citizens have green eyes is closest to _______.

Accepted Answer

The expected number of people with green eyes in the sample is 0.3 * 1200 = 360. The standard deviation is sqrt(1200 * 0.3 * 0.7) ≈ 15.87. Using the normal approximation, the z-score for 348 is (348 - 360) / 15.87 ≈ -0.756. The probability of a z-score less than -0.756 is approximately 0.2236.

Question 5

A random sample of size 100 is drawn from a population with a standard deviation of 10.If only 5% of the time a sample mean greater than 20 is obtained, the mean of the population is closest to ______

Accepted Answer

The sample mean follows a normal distribution with mean equal to the population mean and standard deviation equal to the population standard deviation divided by the square root of the sample size. Given that 5% of the time the sample mean is greater than 20, we can use the z-score for 5% in the upper tail (z = 1.645) to find the population mean. Solving the equation 20 = μ + 1.645*(10/√100) gives μ ≈ 18.35.

Question 6

If the population proportion is 0.90 and a sample of size 64 is taken, which is closest to the probability that the sample proportion is less than 0.88?

Accepted Answer

The answer of If the population proportion is 0.90 and...

Question 7

Albert Abbasi, VP of Operations at Ingleside International Bank, is evaluating the service level provided to walk-in customers.Accordingly, he plans a sample of waiting times for walk-in customers.If the population of waiting times has a mean of 15 minutes and a standard deviation of 4 minutes, the probability that Albert's sample of 64 will have a mean between 13.5 and 16.5 minutes is closest to ________.

Accepted Answer

Using the Central Limit Theorem, we can calculate the probability that the sample mean falls within a certain range. The formula for the z-score is $Z = \frac{\bar{X} - \mu}{\sigma/\sqrt{n}}$, where $\bar{X}$ is the sample mean, $\mu$ is the population mean, $\sigma$ is the population standard deviation, and $n$ is the sample size. For the lower limit (13.5 minutes), $Z = \frac{13.5 - 15}{4/\sqrt{64}} = \frac{-1.5}{0.5} = -3$. For the upper limit (16.5 minutes), $Z = \frac{16.5 - 15}{4/\sqrt{64}} = \frac{1.5}{0.5} = 3$. Looking up these Z-scores in the standard normal distribution table, the area between Z = -3 and Z = 3 is approximately 0.9974, indicating the probability.

Question 8

If the population proportion is 0.90 and a sample of size 64 is taken, which is closest to the probability that the sample proportion is more than 0.89?

Accepted Answer

The sample proportion follows a normal distribution with mean 0.90 and standard deviation sqrt((0.90*0.10)/64). Using the standard normal distribution, the probability that the sample proportion is more than 0.89 is approximately 0.6064.

Question 9

Albert Abbasi, VP of Operations at Ingleside International Bank, is evaluating the service level provided to walk-in customers.Accordingly, he plans a sample of waiting times for walk-in customers.If the population of waiting times has a mean of 15 minutes and a standard deviation of 4 minutes, the probability that Albert's sample of 64 will have a mean less than 14 minutes is closest to________.

Accepted Answer

The correct answer is found using the Central Limit Theorem, which states that the sampling distribution of the sample mean will be normally distributed if the sample size is large enough (n ≥ 30 in this case, so the condition is met). The standard error of the mean (SEM) is calculated as the population standard deviation divided by the square root of the sample size, which is 4 / sqrt(64) = 4 / 8 = 0.5. To find the probability that the sample mean is less than 14 minutes, we calculate the Z-score: (14 - 15) / 0.5 = -2. The Z-score of -2 corresponds to a probability of 0.0228, indicating the likelihood that a sample mean is less than 14 minutes given the population parameters.

Question 10

Albert Abbasi, VP of Operations at Ingleside International Bank, is evaluating the service level provided to walk-in customers.Accordingly, he plans a sample of waiting times for walk-in customers.If the population of waiting times has a mean of 15 minutes and a standard deviation of 4 minutes, the probability that Albert's sample of 64 will have a mean less than 16 minutes is closest to ________.

Accepted Answer

The standard error of the mean is 4/√64 = 0.5. The z-score for a sample mean of 16 is (16-15)/0.5 = 2. The probability of a z-score less than 2 is approximately 0.9772.

Question 11

A carload of steel rods has arrived at Cybermatic Construction Company.The car contains 50,000 rods.Claude Ong, manager of Quality Assurance, directs his crew to measure the lengths of 100 randomly selected rods.If the population of rods has a mean length of 120 inches and a standard deviation of 0.05 inch, the probability that Claude's sample has a mean between 119.985 and 120.0125 inches is closest to ____________.

Accepted Answer

By the Central Limit Theorem, the sample mean will be approximately normally distributed with a mean of 120 and a standard deviation of 0.05/sqrt(100) = 0.005. So, the probability that the sample mean is between 119.985 and 120.0125 is approximately P(119.985 < X < 120.0125) = P((119.985 - 120)/0.005 < Z < (120.0125 - 120)/0.005) = P(-3 < Z < 5) = 0.9999 - 0.0001 = 0.9998.

Question 12

Suppose 40% of the population of pre-teens have a TV in their bedroom.If a random sample of 500 pre-teens is drawn from the population, then the probability that 44% or more of the pre-teens have a TV in their bedroom is closest to _______.

Accepted Answer

Using the normal approximation to the binomial distribution, we first calculate the mean (μ) and standard deviation (σ) for the number of pre-teens with a TV in their bedroom. The mean is given by μ = np = 500 * 0.40 = 200. The standard deviation is σ = sqrt(np(1-p)) = sqrt(500 * 0.40 * 0.60) ≈ 10.95. To find the probability of having 44% or more with a TV, we convert 44% of 500 to a number, which is 220. We then use the z-score formula: z = (X - μ) / σ = (220 - 200) / 10.95 ≈ 1.83. Looking up the z-score of 1.83 in the z-table gives us the area to the left, which is approximately 0.9664. Since we want the area to the right, we subtract this from 1, giving us 1 - 0.9664 = 0.0336.

Question 13

Suppose a population has a mean of 450 and a variance of 900.If a random sample of size 100 is drawn from the population, the probability that the sample mean is between 448 and 453 is closest to _______.

Accepted Answer

The answer of Suppose a population has a mean of...

Question 14

Suppose 40% of the population of pre-teens have a TV in their bedroom.If a random sample of 500 pre-teens is drawn from the population, then the probability that 44% or fewer of the pre-teens have a TV in their bedroom is closest to _______.

Accepted Answer

Using the normal approximation to the binomial distribution, we first find the mean (μ) and standard deviation (σ) of the distribution. The mean is given by μ = np, where n is the sample size (500) and p is the probability of success (0.40). Thus, μ = 500 * 0.40 = 200. The standard deviation is given by σ = sqrt(np(1-p)), so σ = sqrt(500 * 0.40 * 0.60) ≈ 10.95.We are interested in the probability that the proportion is 44% or fewer, which means 220 or fewer pre-teens out of 500 have a TV in their bedroom (since 44% of 500 is 220). To find this probability, we convert the number of pre-teens to a z-score. The z-score is given by z = (X - μ) / σ, where X is the value of interest (220), μ is the mean (200), and σ is the standard deviation (10.95). Thus, z = (220 - 200) / 10.95 ≈ 1.83.Looking up the z-score of 1.83 in a standard normal distribution table or using a calculator, we find that the area to the left of z = 1.83 is approximately 0.9664. This area represents the probability that 44% or fewer of the pre-teens in a sample of 500 have a TV in their bedroom, which is closest to choice A.

Question 15

Suppose 40% of the population of pre-teens have a TV in their bedroom.If a random sample of 500 pre-teens is drawn from the population, then the probability that between 36% and 44% of the pre-teens have a TV in their bedroom is closest to _______.

Accepted Answer

The sample proportion follows a normal distribution with mean 0.4 and standard deviation sqrt(0.4*0.6/500). Using the normal approximation, the probability that the sample proportion is between 0.36 and 0.44 is approximately 0.9328.

Question 16

Suppose 40% of all college students have a computer at home and a sample of 64 is taken.Which is closest to the probability that more than 30 of those in the sample have a computer at home?

Accepted Answer

The probability that more than 30 students have a computer at home can be found using the normal approximation to the binomial distribution. The mean (μ) of the distribution is n*p = 64*0.40 = 25.6, and the standard deviation (σ) is sqrt(n*p*(1-p)) = sqrt(64*0.40*0.60) ≈ 3.92. To find the probability of more than 30 students, we calculate the z-score for 30.5 (for continuity correction) as (30.5-25.6)/3.92 ≈ 1.25. Looking up this z-score in a standard normal distribution table or using a calculator gives a right-tail probability of approximately 0.1314, indicating that the probability more than 30 students have a computer at home is closest to 0.1314.

Question 17

Suppose 40% of all college students have a computer at home and a sample of 100 is taken.Which is closest to the probability that more than 50 of those in the sample have a computer at home?

Accepted Answer

The probability of more than 50 students having a computer at home is very low since only 40% are expected to have one, making the expected number 40 out of 100. The correct choice is the one that represents a small probability, which is 0.0207.

Question 18

Suppose a population has a mean of 870 and a variance of 8,100.If a random sample of size 36 is drawn from the population, the probability that the sample mean is between 840 and 900 is closest to _______.

Accepted Answer

The standard deviation of the population is the square root of the variance, which is sqrt(8100) = 90. The standard error of the mean (SEM) for a sample size of 36 is the standard deviation divided by the square root of the sample size, which is 90/sqrt(36) = 15. The Z-scores for 840 and 900, given a mean of 870, are (840-870)/15 = -2 and (900-870)/15 = 2, respectively. Looking up these Z-scores in a standard normal distribution table or using a calculator, the probability of being between -2 and 2 standard deviations from the mean is approximately 0.9544.

Question 19

A carload of steel rods has arrived at Cybermatic Construction Company.The car contains 50,000 rods.Claude Ong, manager of Quality Assurance, directs his crew to measure the lengths of 100 randomly selected rods.If the population of rods have a mean length of 120 inches and a standard deviation of 0.05 inch, the probability that Claude's sample has a mean less than 119.985 inches is closest to _____________.

Accepted Answer

To find the probability that the sample mean is less than 119.985 inches, we use the formula for the z-score of a sample mean: $z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}}$, where $\bar{x}$ is the sample mean, $\mu$ is the population mean, $\sigma$ is the population standard deviation, and $n$ is the sample size. Plugging in the given values: $z = \frac{119.985 - 120}{0.05/\sqrt{100}} = \frac{-0.015}{0.005} = -3$. Looking up a z-score of -3 in the standard normal distribution table gives a probability of approximately 0.0013.

Question 20

Suppose a population has a mean of 870 and a variance of 1,600.If a random sample of size 64 is drawn from the population, the probability that the sample mean is between 860 and 875 is closest to _______.

Accepted Answer