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 40 players will be less than 69.5 in.? c. What is the probability that a sample mean of 40 players will be more than 71 in.? d. What is the probability that a sample mean of 40 players will be between 70 and 71.5 in.? i Player heights Full data set Height (in.) 80 73 69 79 73 68 77 73 68 77 72 68 76 72 67 76 71 67 75 71 67 75 71 67 ?Click the icon to view the data table of heights a. The population mean is (Round to one decimal place as needed.) 757167 75 71 66 74 71 65 747065 74 70 65 747065 74 70 65 74 70 64 7470 62 7469 62 73 69 62 73 6961 Enter your answer in the answer box and then click Check Answer. PrintDone

Answer 1

Solution:- Given that values 80,79,77,77,76,76,75,75,75,75,74,74,74,74,74,74,74,74,73,73,73,73,73,72,72,71,71,71,71,71,71,70,70,70, 70, 70,70,69,69,69,69,68,68,68,67,67,67,67,67,66,65,65,65,65,65,64,62,62,62,61

a. The population mean is 70.5

= > ? = 70.5

standard deviation is 4.4

=>  ? = 4.4

b. the probability that a sample mean of 40 players will be less than 69.5 in : 0.0749
P(X < 69.5) = P(Z < (X - ?)/(?/sqrt(n)))
= P(z < (69.5 - 70.5)/(4.4/sqrt(40))
= P(Z < -1.4374)
= 0.0749
c. The probability that a sample mean of 40 players will be more than 71 in :
P(X > 71) = P(Z > (71 - 70.5)/(4.4/sqrt(40))
= P(Z > 0.7187)
= 0.2358

d. The probability that sample mean of 40 players will be between 70 and 71.5
P(70 < X < 71.5) = P((70 - 70.5)/(4.4/sqrt(40) < Z < (71.5 - 70.5)/(4.4/sqrt(40))
= P(-0.7187 < Z < 1.4374)
= 0.6893