Statistics Probability

Question

from 1985 to 2006 the average height of an NBA player was 6 foot 7 (79 inches) suppose the distribution is approximately normal and the standard deviation is 2.8 inches. what is the probability that of 38 NBA players selected at random, their average height is more that 78 inches?

Answer 1

So to find the probability for a group you need to use the formula z = (x - μ)/(σ/(√n) which translates to z = (sample mean - population mean) / (Standard deviation / √sample size). In this case we know:

sample mean = 78 in.

population mean = 79 in.

standard deviation = 2.8 in.

and sample size = 38

So from this we get the equation z = (78 - 79)/(2.8/√38) = -2.20

If we then use a z-score table, we can see that a z-score of -2.20 correlates with a percentile of 0.01390. So what does this percentile mean. This means that a proportion of 0.01390 out of 1 (or 1.39%) are less, or shorter than that sample mean of 78 in. But the question is what is the probability that the mean is GREATER than 78 in. Getting that is really simple from here. knowing that proportion of 0.01390 means we know the opposite proportion too (1-0.01390 = 0.9861). Converting this again to a percentage, we can get a 98.61% probability that their average height will be > 78 in.