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.
Find the probability that the mean salary of the 100 players was no more than R 3.0 million.
To find the probability that the mean salary of the 100 players was no more than R3.0 million, we need to calculate the z-score and then find the corresponding probability using the standard normal distribution.
First, calculate the z-score:
\[ z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}} \]
where:
- X = R3.0 million
- μ = R3.26 million
- σ = R1.2 million
- n = 100
\[ z = \frac{3.0 - 3.26}{\frac{1.2}{\sqrt{100}}} \]
\[ z = \frac{-0.26}{0.12} \]
\[ z = -2.1667 \]
Next, find the probability using a standard normal distribution table or calculator. The probability that the mean salary of the 100 players was no more than R3.0 million is the same as finding the probability that a z-score is less than -2.1667.
Looking up the z-score of -2.1667 on a standard normal distribution table, we find that the probability is approximately 0.0152 or 1.52%.
Therefore, the probability that the mean salary of the 100 players was no more than R3.0 million is 1.52%.