1. (10 points) Assume that the heights of NBA players are normally distributed w

Question

1. (10 points) Assume that the heights of NBA players are normally distributed with mean height 73.5 inches and standard deviation 2.2 inches. Show what you did (such as sketches, calculation set-ups, calculator entries) to find these answers: a. What is the probability of randomly selecting an NBA player that is between 72 and 73 inches tall? Ans: b. How tall would an NBA player be whose height was at the 67th percentile? (ie, 67% of all NBA players are shorter than this player) Ans: Ans: in this interval c. Find a value h such that only 6.3% of all NBA players have heights more than h inches. d Find a symmetric interval about the mean such that 72% of all heights of NBA players lie Ans: e. Consider a randomi selection of n = 10 players from the NBA. What is the probability that average height of these players is greater than 75 inches? ns

Explanation / Answer

Answer:

a).

z value for 72, z = (72-73.5)/2.2 = -0.68

z value for 73, z = (73-73.5)/2.2 = -0.23

P( 72<x<73) = P( -0.68<z<-0.23) = P( z < -0.23) –P( z < -0.68)

=0.409 -0.2483

=0.1607

b).

z value for67th percentile = 0.44

x = mean+z*sd =73.5+0.44*2.2 = 74.468

c).

z value for top 6.3% =1.53

x = mean+z*sd =73.5+1.53*2.2 = 76.866

d).

z values for middle 72% = (-1.08, 1.08)

lower limit   =73.5-1.08*2.2 = 71.124

upper limit =73.5+1.08*2.2 = 75.876

e).

when n=10, standard error =sd/sqrt(n) =2.2/sqrt(10) = 0.6957

z value for 75, z = (75-73.5)/0.6957 = 2.16

P( mean x >75) = P( z > 2.16) =0.0154

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