Statistics and probability

Question

It is known that the average height of all athletes playing for a certain private school is 181.42 centimeters with a standard deviation of 11.54 centimeters. If 21 players are selected at random, what is the probability that the mean height of the sample is at most 182.70 centimeters? Assume a normal distribution. Round your answer to four decimal places. Do not place any unit.

Answer 1

Let x = mean height. Want to compute P(x < 182.7)

z score = (sample mean - population mean)/(population variance/sqrt(sample size))

convert to z score P(z < (182,7 - 181.42)/(11.54/sqrt(21)) = P(z < 0.508) which from standard normal probability table is 0.7190.

Answer 2

We want to find P[ Xbar ≤ 182.7 ]

We can take this inequality, and do the same thing on the left hand side and the right hand side, but on the left side I'll use symbols and on the right side i'll use the corresponding numbers that we were given.

P[ (Xbar-μ)/(σ/√n) ≤ (182.7-181.42)/(11.54/√21) ]

The symbols on the left hand side make up a formula for "z," which should look familiar. The right hand side is just a number we can calculate, it happens to be about 0.51.

So now we have P[ z ≤ 0.51 ]. There are many ways we can find this. We can use a graphing calculator, R, wolfram alpha, or a standard normal table (a z-table) such as this one: http://www.z-table.com.

The table shows the area to the left of a given z-score, which is what we want because the inequality we want has a "less than" sign. So we look up a z-score of 0.51 in the table and we see that the associated probability is 0.695.