Question
The average score of all pro golfers for a particular course has a mean of 70 and a standard deviation of 3.0. Suppose 36 pro golfers played the course today. Find the probability that the average score of the 36 pro golfers exceeded 71.
2. Major league baseball salaries averaged R3.26 million with a standard deviation of R 1.2 million in a recent year. Suppose a sample of 100 major league players was taken.
2.1 Find the probability that the mean salary of the 100 players exceeded R 3.5 million.
2.2 Find the probability that the mean salary of the 100 players exceeded R 4.0 million.
2.3 Find the probability that the mean salary of the 100 players was no more than R 3.0 million.
2.4 Find the probability that the mean salary of the 100 players was less than R 2.5 million.
2.5 Find P(X ̅<A)=0.9616 find A.
2. Major league baseball salaries averaged R3.26 million with a standard deviation of R 1.2 million in a recent year. Suppose a sample of 100 major league players was taken.
2.1 Find the probability that the mean salary of the 100 players exceeded R 3.5 million.
2.2 Find the probability that the mean salary of the 100 players exceeded R 4.0 million.
2.3 Find the probability that the mean salary of the 100 players was no more than R 3.0 million.
2.4 Find the probability that the mean salary of the 100 players was less than R 2.5 million.
2.5 Find P(X ̅<A)=0.9616 find A.
Answers
GPT-4o mini
To solve these problems, we will use the properties of the sampling distribution of the sample mean. For a sample size \( n \), if the population mean is \( \mu \) and the population standard deviation is \( \sigma \), then the standard deviation of the sample mean (also known as the standard error) is given by:
\[
\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}
\]
For the first question regarding pro golfers' average scores:
1. **Given**: \(\mu = 70\), \(\sigma = 3.0\), \(n = 36\).
- Calculate the standard error:
\[
\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{3.0}{\sqrt{36}} = \frac{3.0}{6} = 0.5
\]
- We want to find the probability that the average score (\(\bar{x}\)) exceeds 71:
- First, calculate the z-score:
\[
z = \frac{\bar{x} - \mu}{\sigma_{\bar{x}}} = \frac{71 - 70}{0.5} = \frac{1}{0.5} = 2
\]
- Now, we look up the z-score of 2 in the standard normal distribution table or use a calculator. The area to the left of \( z = 2 \) is approximately 0.9772. Thus, the area to the right (which is what we want) is:
\[
P(\bar{x} > 71) = 1 - 0.9772 = 0.0228
\]
Therefore, the probability that the average score of the 36 pro golfers exceeded 71 is **0.0228**.
---
Now, for the second question regarding major league baseball salaries:
1. **Given**: \(\mu = 3.26\) million, \(\sigma = 1.2\) million, \(n = 100\).
- Calculate the standard error:
\[
\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{1.2}{\sqrt{100}} = \frac{1.2}{10} = 0.12
\]
### 2.1 Probability that the mean salary exceeds R 3.5 million:
- Calculate the z-score:
\[
z = \frac{3.5 - 3.26}{0.12} = \frac{0.24}{0.12} = 2.0
\]
- The area to the left (from z-table) is approximately 0.9772, so:
\[
P(\bar{x} > 3.5) = 1 - 0.9772 = 0.0228
\]
### 2.2 Probability that the mean salary exceeds R 4.0 million:
- Calculate the z-score:
\[
z = \frac{4.0 - 3.26}{0.12} = \frac{0.74}{0.12} \approx 6.167
\]
- The area to the left is practically 1 (very close to it), thus:
\[
P(\bar{x} > 4.0) \approx 0
\]
### 2.3 Probability that the mean salary was no more than R 3.0 million:
- Calculate the z-score:
\[
z = \frac{3.0 - 3.26}{0.12} = \frac{-0.26}{0.12} \approx -2.167
\]
- The area to the left (from z-table) is approximately 0.0150, so:
\[
P(\bar{x} \leq 3.0) \approx 0.0150
\]
### 2.4 Probability that the mean salary was less than R 2.5 million:
- Calculate the z-score:
\[
z = \frac{2.5 - 3.26}{0.12} = \frac{-0.76}{0.12} \approx -6.333
\]
- The area to the left is practically 0, thus:
\[
P(\bar{x} < 2.5) \approx 0
\]
### 2.5 Find \( P(\bar{X} < A) = 0.9616 \):
- The z-score associated with 0.9616 is approximately 1.75 (from z-tables).
- Now, solve for \( A \):
\[
1.75 = \frac{A - 3.26}{0.12}
\]
Rearranging gives:
\[
A = 1.75 \cdot 0.12 + 3.26 \approx 0.21 + 3.26 = 3.47 \text{ million}
\]
Thus, the value of \( A \) is approximately **R 3.47 million**.
\[
\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}
\]
For the first question regarding pro golfers' average scores:
1. **Given**: \(\mu = 70\), \(\sigma = 3.0\), \(n = 36\).
- Calculate the standard error:
\[
\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{3.0}{\sqrt{36}} = \frac{3.0}{6} = 0.5
\]
- We want to find the probability that the average score (\(\bar{x}\)) exceeds 71:
- First, calculate the z-score:
\[
z = \frac{\bar{x} - \mu}{\sigma_{\bar{x}}} = \frac{71 - 70}{0.5} = \frac{1}{0.5} = 2
\]
- Now, we look up the z-score of 2 in the standard normal distribution table or use a calculator. The area to the left of \( z = 2 \) is approximately 0.9772. Thus, the area to the right (which is what we want) is:
\[
P(\bar{x} > 71) = 1 - 0.9772 = 0.0228
\]
Therefore, the probability that the average score of the 36 pro golfers exceeded 71 is **0.0228**.
---
Now, for the second question regarding major league baseball salaries:
1. **Given**: \(\mu = 3.26\) million, \(\sigma = 1.2\) million, \(n = 100\).
- Calculate the standard error:
\[
\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{1.2}{\sqrt{100}} = \frac{1.2}{10} = 0.12
\]
### 2.1 Probability that the mean salary exceeds R 3.5 million:
- Calculate the z-score:
\[
z = \frac{3.5 - 3.26}{0.12} = \frac{0.24}{0.12} = 2.0
\]
- The area to the left (from z-table) is approximately 0.9772, so:
\[
P(\bar{x} > 3.5) = 1 - 0.9772 = 0.0228
\]
### 2.2 Probability that the mean salary exceeds R 4.0 million:
- Calculate the z-score:
\[
z = \frac{4.0 - 3.26}{0.12} = \frac{0.74}{0.12} \approx 6.167
\]
- The area to the left is practically 1 (very close to it), thus:
\[
P(\bar{x} > 4.0) \approx 0
\]
### 2.3 Probability that the mean salary was no more than R 3.0 million:
- Calculate the z-score:
\[
z = \frac{3.0 - 3.26}{0.12} = \frac{-0.26}{0.12} \approx -2.167
\]
- The area to the left (from z-table) is approximately 0.0150, so:
\[
P(\bar{x} \leq 3.0) \approx 0.0150
\]
### 2.4 Probability that the mean salary was less than R 2.5 million:
- Calculate the z-score:
\[
z = \frac{2.5 - 3.26}{0.12} = \frac{-0.76}{0.12} \approx -6.333
\]
- The area to the left is practically 0, thus:
\[
P(\bar{x} < 2.5) \approx 0
\]
### 2.5 Find \( P(\bar{X} < A) = 0.9616 \):
- The z-score associated with 0.9616 is approximately 1.75 (from z-tables).
- Now, solve for \( A \):
\[
1.75 = \frac{A - 3.26}{0.12}
\]
Rearranging gives:
\[
A = 1.75 \cdot 0.12 + 3.26 \approx 0.21 + 3.26 = 3.47 \text{ million}
\]
Thus, the value of \( A \) is approximately **R 3.47 million**.