2Major 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
2.1 To find the probability that the mean salary of the 100 players exceeded R3.5 million, we need to calculate the z-score first:
z = (3.5 - 3.26) / (1.2 / sqrt(100)) = 2.08
Now, we can use a standard normal distribution table to find the probability that z > 2.08, which is approximately 0.0197. Therefore, the probability that the mean salary of the 100 players exceeded R3.5 million is 0.0197.
2.2 To find the probability that the mean salary of the 100 players exceeded R4.0 million, we follow the same steps:
z = (4.0 - 3.26) / (1.2 / sqrt(100)) = 5.67
Using the standard normal distribution table, we find that the probability that z > 5.67 is almost 0. Therefore, the probability that the mean salary of the 100 players exceeded R4.0 million is nearly 0.
2.3 To find the probability that the mean salary of the 100 players was no more than R3.0 million:
z = (3.0 - 3.26) / (1.2 / sqrt(100)) = -2.08
Using the standard normal distribution table, we find that the probability that z < -2.08 is approximately 0.0197. Therefore, the probability that the mean salary of the 100 players was no more than R3.0 million is 0.0197.
2.4 To find the probability that the mean salary of the 100 players was less than R2.5 million:
z = (2.5 - 3.26) / (1.2 / sqrt(100)) = -6.67
Using the standard normal distribution table, we find that the probability that z < -6.67 is nearly 0. Therefore, the probability that the mean salary of the 100 players was less than R2.5 million is almost 0.
2.5 To find the value of A such that p(X < A) = 0.9616, we look for the z-score that corresponds to a cumulative probability of 0.9616 in the standard normal distribution table. That z-score is approximately 1.75.
Then we find the corresponding salary value:
A = 1.75 * (1.2 / sqrt(100)) + 3.26 = 3.563
Therefore, p(X < 3.563) = 0.9616.