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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.?


67 67 67 66 66 66 66 66 66 65 65 65 65 6 6 63 63 62 61 -ght 72 72 72 71 71 71 71 70 70 70 70 69 69 69 68 68 68 68 68 65555433333322 77 7 8777

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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