A particular variety of watermelon weighs on average 273 pou

A particular variety of watermelon weighs on average 27.3 pounds with a standard deviation of 1.53 pounds. Consider the sample mean weight of 72 watermelons of this variety. Assume the individual watermelon weights are independent.

a. What is the expected value of the sample mean weight? Give an exact answer.  

b. What is the standard deviation of the sample mean weight? Give your answer to four decimal places.  

c. What is the approximate probability the sample mean weight will be less than 27.38? Give your answer to four decimal places. Use the standard deviation as you entered it above to answer this question.  

d. What is the value c such that the approximate probability the sample mean will be less than c is 0.98? Give your answer to four decimal places. Use the standard deviation as you entered it above to answer this question.  

Solution

a)
Mean ( u ) =27.3

b)
Standard Deviation ( sd )=1.53/Sqrt(72) = 0.18031
Number ( n ) = 72
Normal Distribution = Z= X- u / (sd/Sqrt(n) ~ N(0,1)                  

c)
P(X < 27.38) = (27.38-27.3)/1.53/ Sqrt ( 72 )
= P ( Z <0.4437) From Standard NOrmal Table
= 0.6714                  
P ( Z < x ) = 0.98

d)
Value of z to the cumulative probability of 0.98 from normal table is 2.054
P( x-u/s.d < x - 27.3/1.53 ) = 0.98
That is, ( x - 27.3/1.53 ) = 2.05
--> x = 2.05 * 1.53 + 27.3 = 30.4426