Question

The average score for all basketball players in a particular season has a mean of 75 and a standard deviation of 3.5. Suppose 49 played basketball today. Find the probability that the average score of the 49 players exceeded 76

Answer 1

Answer:

0.0228

Step-by-step explanation:

Given that:

mean (μ) = 75, Standard deviation (σ) = 3.5

The z score is a score used in statistics to measure by how many standard deviations the raw score is above or below the mean. If the z score is positive, it means that the raw sore is above the mean and if the z score is negative then the raw score is below the mean. The z score is given by:

$z=\frac{x-\mu}{\sigma}\\ \\For\ a\ sample \ n, the\ z\ score\ is:\\ \\z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n} } }$

For x > 76 and a sample (n) = 49:

$z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n} } }\\\\z=\frac{76-75}{\frac{3.5}{\sqrt{49} } }=2$

From the normal distribution table, the probability that the average score of the 49 players exceeded 76 = P(x > 76) = P(z > 2) = 1 - P(z < 2) = 1 - 0.9772 = 0.0228