The data table contains frequency distribution of the heights of the players in
ID: 3324333 • Letter: T
Question
The data table contains frequency distribution of the heights of the players in a basketball league.
a. Calculate the mean and standard deviation of this population.
b. What is the probability that a sample mean of 45
players will be less than 69.5
in.?
c. What is the probability that a sample mean of 45
players will be more than 71
in.?
d. What is the probability that a sample mean of 45
players will be between 70
and 71.5
in.?
Explanation / Answer
Solution:
(All probablities are calculated by using Z-table or normal table.)
Part a
From the given data, we have
Population size = N = 60
Population mean = µ = 69.96667
Population standard deviation = = 4.554628
Part b
We have to find P(Xbar < 69.5)
Z = (Xbar - µ) / [/sqrt(n)]
We are given
n = 45
µ = 69.96667
= 4.554628
Xbar = 69.5
Z = (69.5 - 69.96667) / (4.554628/sqrt(45))
Z = -0.68733
P(Z<-0.68733) = P(Xbar < 69.5) = 0.245938
Required probability = 0.245938
Part c
We have to find P(Xbar>71)
P(Xbar>71) = 1 - P(Xbar<71)
Z = (71 - 69.96667) / (4.554628/sqrt(45))
Z = 1.521922
P(Z<1.521922) = 0.935986
P(Xbar>71) = 1 - P(Xbar<71)
P(Xbar>71) = 1 - 0.935986
P(Xbar>71) = 0.064014
Required probability = 0.064014
Part d
We have to find P(70<Xbar<71.5)
P(70<Xbar<71.5) = P(Xbar<71.5) – P(Xbar<70)
For Xbar = 71.5
Z = (71.5 - 69.96667) / (4.554628/sqrt(45))
Z = 2.258338186
P(Z<2.258338186) = P(Xbar<71.5) = 0.988037708
For Xbar = 70
Z = (70 - 69.96667) / (4.554628/sqrt(45))
Z = 0.049089506
P(Z<0.049089506) = P(Xbar<70) = 0.519576017
P(70<Xbar<71.5) = P(Xbar<71.5) – P(Xbar<70)
P(70<Xbar<71.5) = 0.988037708 - 0.519576017
P(70<Xbar<71.5) = 0.468462
Required probability = 0.468462
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