A roller coaster is being designed that will accommodate 60 riders. The maximum weight the coaster can hold safely is 12000 pounds. According to the National Health Statistics Reports, the weights of adult U.S. men have mean 190 pounds and standard deviation 64 pounds, and the weights of adult U.S. women have mean 166 pounds and standard deviation 73 pounds.

Question

(b) If a random sample of 60 adult men ride the coaster, what is the probability that the maximum safe weight will be exceeded? Round the answer to at least four decimal places.

If a random sample 60 adult men ride the coaster, the probability that the maximum safe weight will be exceeded is [blank]

Answer 1

To find the probability that the maximum safe weight of the roller coaster (12,000 pounds) will be exceeded when 60 adult men ride it, we can use the Central Limit Theorem. The theorem tells us that the sampling distribution of the sample mean will be normally distributed if the sample size is sufficiently large (which it is in this case with \( n = 60 \)).

Given:

• Mean weight of adult men: \( \mu = 190 \) pounds
• Standard deviation of adult men: \( \sigma = 64 \) pounds
• Sample size: \( n = 60 \)
• Maximum safe weight: \( W = 12,000 \) pounds

Step 1: Calculate the mean and standard deviation of the sample mean.

The mean of the sample means (expected mean weight) is the same as the population mean: \[ \mu_{\bar{x}} = \mu = 190 \text{ pounds} \]

The standard deviation of the sample mean (standard error) is calculated as follows: \[ \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{64}{\sqrt{60}} \approx \frac{64}{7.746} \approx 8.25 \text{ pounds} \]

Step 2: Convert the maximum weight limit to the mean weight of the sample.

We need to find the sample mean weight that corresponds to the maximum weight limit of 12,000 pounds. Since there are 60 riders, we divide the total weight limit by the number of riders: \[ \bar{x}_{\text{max}} = \frac{12000}{60} = 200 \text{ pounds} \]

Step 3: Calculate the z-score.

Using the z-score formula: \[ z = \frac{\bar{x} - \mu_{\bar{x}}}{\sigma_{\bar{x}}} \] where \( \bar{x} = 200 \) pounds, \( \mu_{\bar{x}} = 190 \) pounds, and \( \sigma_{\bar{x}} \approx 8.25 \) pounds.

Calculating the z-score: \[ z = \frac{200 - 190}{8.25} \approx \frac{10}{8.25} \approx 1.2121 \]

Step 4: Find the probability.

We need to find the probability that the sample mean exceeds 200 pounds, which is \( P(Z > 1.2121) \).

Using the standard normal distribution table, we find: \[ P(Z < 1.2121) \approx 0.8871 \]

Thus, the probability that \( Z \) is greater than 1.2121 is: \[ P(Z > 1.2121) = 1 - P(Z < 1.2121) \approx 1 - 0.8871 \approx 0.1129 \]

Final Answer:

The probability that the maximum safe weight will be exceeded when 60 adult men ride the coaster is approximately: \[ \boxed{0.1129} \]

Answer 2

what about a random sample of 60 women

Answer 3

To find the probability that the maximum safe weight of the roller coaster (12,000 pounds) will be exceeded when 60 adult women ride it, we'll follow a similar process as with the men.

Given:

• Mean weight of adult women: \( \mu = 166 \) pounds
• Standard deviation of adult women: \( \sigma = 73 \) pounds
• Sample size: \( n = 60 \)
• Maximum safe weight: \( W = 12,000 \) pounds

Step 1: Calculate the mean and standard deviation of the sample mean.

The mean of the sample means (expected mean weight) remains the same as the population mean: \[ \mu_{\bar{x}} = \mu = 166 \text{ pounds} \]

The standard deviation of the sample mean (standard error) is calculated as follows: \[ \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{73}{\sqrt{60}} \approx \frac{73}{7.746} \approx 9.42 \text{ pounds} \]

Step 2: Convert the maximum weight limit to the mean weight of the sample.

As before, we need to find the sample mean weight that corresponds to the maximum weight limit of 12,000 pounds, which remains the same: \[ \bar{x}_{\text{max}} = \frac{12000}{60} = 200 \text{ pounds} \]

Step 3: Calculate the z-score.

Using the z-score formula: \[ z = \frac{\bar{x} - \mu_{\bar{x}}}{\sigma_{\bar{x}}} \] where \( \bar{x} = 200 \) pounds, \( \mu_{\bar{x}} = 166 \) pounds, and \( \sigma_{\bar{x}} \approx 9.42 \) pounds.

Calculating the z-score: \[ z = \frac{200 - 166}{9.42} \approx \frac{34}{9.42} \approx 3.605 \]

Step 4: Find the probability.

We need to find the probability that the sample mean exceeds 200 pounds, which is \( P(Z > 3.605) \).

Using the standard normal distribution table (or a calculator), we determine: \[ P(Z < 3.605) \approx 1.0000 \] Thus, the probability that \( Z \) is greater than 3.605 is: \[ P(Z > 3.605) = 1 - P(Z < 3.605) \approx 1 - 1.0000 = 0.0000 \]

Final Answer:

The probability that the maximum safe weight will be exceeded when 60 adult women ride the coaster is approximately: \[ \boxed{0.0000} \]

This indicates that it is extremely unlikely for the total weight of 60 adult women to exceed the weight limit of 12,000 pounds.

Answer 4

(c) If a random sample of 60 adult women ride the coaster, what is the probability that the maximum safe weight will be exceeded? Round the answer to at least four decimal places.

If a random sample 60 adult women ride the coaster, the probability that the maximum safe weight will be exceeded is [blank]

Answer 5

We need to calculate the probability that the total weight of a random sample of 60 adult women exceeds the maximum safe weight of the roller coaster, which is 12,000 pounds.

Given Information:

• Mean weight of adult women: \( \mu = 166 \) pounds
• Standard deviation of adult women: \( \sigma = 73 \) pounds
• Sample size: \( n = 60 \)
• Maximum safe weight: \( W = 12,000 \) pounds

Step 1: Calculate the mean and standard deviation of the sample mean.

The mean of the sample means (\( \mu_{\bar{x}} \)): \[ \mu_{\bar{x}} = \mu = 166 \text{ pounds} \]

The standard deviation of the sample mean (standard error, \( \sigma_{\bar{x}} \)): \[ \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{73}{\sqrt{60}} \approx \frac{73}{7.746} \approx 9.42 \text{ pounds} \]

Step 2: Find the sample mean that corresponds to the maximum weight limit.

We convert the maximum weight limit to the mean weight per individual in the sample: \[ \bar{x}_{\text{max}} = \frac{12000}{60} = 200 \text{ pounds} \]

Step 3: Calculate the z-score.

We calculate the z-score using the formula: \[ z = \frac{\bar{x} - \mu_{\bar{x}}}{\sigma_{\bar{x}}} \] where \( \bar{x} = 200 \) pounds, \( \mu_{\bar{x}} = 166 \) pounds, and \( \sigma_{\bar{x}} \approx 9.42 \) pounds.

Calculating the z-score: \[ z = \frac{200 - 166}{9.42} \approx \frac{34}{9.42} \approx 3.605 \]

Step 4: Find the probability.

Now we need to find the probability that the sample mean exceeds 200 pounds (i.e., \( P(Z > 3.605) \)).

Consulting the z-table for \( z = 3.605 \): \[ P(Z < 3.605) \approx 1.0000 \] Thus, \[ P(Z > 3.605) = 1 - P(Z < 3.605) \approx 1 - 1.0000 = 0.0000 \]

Final Answer:

The probability that the maximum safe weight will be exceeded when 60 adult women ride the coaster is approximately: \[ \boxed{0.0000} \]

This indicates that it is extremely unlikely for the total weight of 60 adult women to exceed the weight limit of 12,000 pounds.