(a) If 60 people are riding the coaster, and their total weight is 12,000 pounds, what is their average weight?
9 answers
To find the average weight of the 60 riders if their total weight is 12,000 pounds, you would use the formula for average weight, which is:
\[ \text{Average weight} = \frac{\text{Total weight}}{\text{Number of riders}} \]
Substituting the given values:
\[ \text{Average weight} = \frac{12,000 \text{ pounds}}{60 \text{ riders}} = 200 \text{ pounds} \]
So, the average weight of the riders would be 200 pounds.
To find the probability that a random sample of 60 adult men exceeds the maximum safe weight of 12,000 pounds, we can use the Central Limit Theorem (CLT). According to CLT, the distribution of the sample mean will be approximately normally distributed if the sample size is large, which it is in this case (60 riders).
From the problem statement, we know the following for adult men:
The standard deviation of the sample mean (\( \sigma_{\bar{x}} \)) for a sample of size \( n \) is given by:
\[ \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{68}{\sqrt{60}} \]
Calculating this:
\[ \sigma_{\bar{x}} \approx \frac{68}{7.75} \approx 8.74 \]
Next, we want to find the probability that the average weight of a sample of 60 adult men exceeds 200 pounds. First, we need to compute the Z-score for 200 pounds:
\[ Z = \frac{\bar{x} - \mu}{\sigma_{\bar{x}}} = \frac{200 - 194}{8.74} \approx \frac{6}{8.74} \approx 0.686 \]
Now we will find the probability that the sample mean exceeds 200 pounds, which corresponds to finding \( P(Z > 0.686) \).
Using a Z-table or standard normal distribution calculator:
\[ P(Z > 0.686) = 1 - P(Z < 0.686) \approx 1 - 0.7549 \approx 0.2451 \]
Thus, the probability that the maximum safe weight will be exceeded is approximately:
\[ \boxed{0.2451} \]
To determine the probability that a random sample of 60 adult women will exceed the maximum safe weight of 12,000 pounds, we'll again use the Central Limit Theorem.
According to the problem statement, the parameters for adult women are:
Using these values, we can calculate the standard deviation of the sample mean (\( \sigma_{\bar{x}} \)) for a sample of size \( n = 60 \):
\[ \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{77}{\sqrt{60}} \]
Calculating this:
\[ \sigma_{\bar{x}} \approx \frac{77}{7.75} \approx 9.87 \]
Next, we want to find the probability that the average weight of a sample of 60 adult women exceeds 200 pounds. We calculate the Z-score for 200 pounds:
\[ Z = \frac{\bar{x} - \mu}{\sigma_{\bar{x}}} = \frac{200 - 164}{9.87} \approx \frac{36}{9.87} \approx 3.644 \]
Now we'll find the probability that the sample mean exceeds 200 pounds, which corresponds to finding \( P(Z > 3.644) \).
Using a Z-table or standard normal distribution calculator:
Using a standard normal distribution calculator, we find:
\[ P(Z < 3.644) \approx 0.9999 \]
Thus,
\[ P(Z > 3.644) = 1 - P(Z < 3.644) \approx 1 - 0.9999 = 0.0001 \]
Therefore, the probability that the maximum safe weight will be exceeded when a sample of 60 adult women rides the coaster is approximately:
\[ \boxed{0.0001} \]
To find \( P(Z > 0.684) \), we can use the standard normal distribution table or a calculator to determine \( P(Z < 0.684) \) first and then calculate \( P(Z > 0.684) \).
Find \( P(Z < 0.684) \): Looking up \( Z = 0.684 \) in the Z-table or using a standard normal calculator, we find:
\[ P(Z < 0.684) \approx 0.7534 \]
Calculate \( P(Z > 0.684) \): Since the total area under the normal curve is 1, we can find \( P(Z > 0.684) \):
\[ P(Z > 0.684) = 1 - P(Z < 0.684) = 1 - 0.7534 \approx 0.2466 \]
Thus, the probability \( P(Z > 0.684) \) is approximately:
\[ \boxed{0.2466} \]
To find \( P(Z > 3.62) \), we first need to determine the area to the left of \( Z = 3.62 \), which is represented as \( P(Z < 3.62) \).
Find \( P(Z < 3.62) \): The Z-score of 3.62 is quite high, and the area associated with such a value is very close to 1. Using a standard normal distribution table or calculator, we can find:
\[ P(Z < 3.62) \approx 0.99984 \]
Calculate \( P(Z > 3.62) \): To find \( P(Z > 3.62) \), we subtract the value from 1:
\[ P(Z > 3.62) = 1 - P(Z < 3.62) = 1 - 0.99984 \approx 0.00016 \]
Thus, the probability \( P(Z > 3.62) \) is approximately:
\[ \boxed{0.00016} \]
