The diameter of a brand of tennis balls is approximately normally distributed, w

Question

The diameter of a brand of tennis balls is approximately normally distributed, with a mean of 2.63 inches and a standard deviation of 0.03 inch. If you select a random sample of 9 tennis balls, a. What is the sampling distribution of the mean? b. What is the probability that the sample mean is less than 2.61 inches? c. What is the probability that the sample mean is between 2.62 and 2.64 inches? d. The probability is 6-% that the sample mean will be between what two values symmetrically distributed around the population mean?

Explanation / Answer

Mean ( u ) =2.63
Standard Deviation ( sd )=0.03
Number ( n ) = 9
Normal Distribution = Z= X- u / (sd/Sqrt(n) ~ N(0,1)                                  
a)
P(X < 2.61) = (2.61-2.63)/0.03/ Sqrt ( 9 )
= -0.02/0.01= -2
= P ( Z <-2) From Standard NOrmal Table
= 0.0228                                      

b)
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 2.62) = (2.62-2.63)/0.03/ Sqrt ( 9 )
= -0.01/0.01
= -1
= P ( Z <-1) From Standard Normal Table
= 0.15866
P(X < 2.64) = (2.64-2.63)/0.03/ Sqrt ( 9 )
= 0.01/0.01 = 1
= P ( Z <1) From Standard Normal Table
= 0.84134
P(2.62 < X < 2.64) = 0.84134-0.15866 = 0.6827

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