The heights of WNBA players are normally distributed with a mean of 72\" and a s

Question

The heights of WNBA players are normally distributed with a mean of 72" and a standard deviation of 2", and the NBA players' heights have a mean of 79" and a standard deviation of 3".

You are in a gym watching a very good player practicing. This player is 5 feet 10 inches. Is it more likely that this is a WNBA player or a NBA player?

Part 1 of 2

First find the z-score of this player for each league. (Round your answers to two decimal places.)

WNBA z=

NBA z=

By using the Standard Normal Table, find the area to the left of the z-score for each league. (Round your answers to one decimal place.)

WNBAA Area=

NBA Area=

Explanation / Answer

Please leave a comment in case you don't understand any part of the question.

Given in question are normal distribution params for WNBA and NBA:

WNBA -

Mean = 72inches = 6 ft
Stdev = 2inches = .167 ft

NBA

Mean= 79inhces = 6.583 ft
Stdev = 3inches = .25 ft

1. Z score of player who is 5 ft and 10inches = 5.833

So, P(Z = 5.833 for WNBA ) = (5.833-6)/.167 = -1
wnba Z = -1

So, P(Z = 5.833 for NBA ) = (5.833-6.583)/.25 = -3
nba Z = -3

It is more likely that the player is from WNBA as Z=-3 for NBA means that there is very low probability ( i.e. .0013 ) to be from NBA compared to higher probability of .1587 for WNBA.

Hence, this player is more likely a WNBA player

2.

P( left of Z=-1) = 0.1587 = .16 ( 1 decimal place)

WNBAA Area= .16

P( left of Z=-3) = 0.0013 = 0.0 ( 1 decimal place)

NBA Area= 0.0

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