Question

A particular variety of watermelon weighs on average 21.3 pounds with a standard deviation of 1.07 pounds. Consider the sample mean weight of 90 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 greater than 21? Give your answer to four decimal places. Hint: You can use R and the function pnorm to compute the cdf of the standard normal

d. Suppose a vender will only accept the batch of 90 watermelons if the total weight is greater than 1890 pounds. What is the probability it will be accepted? Give your answer to four decimal places

Answer 1

From the given data ,

Here,

a) The sample mean weight.

expected value=21.3

b) The standard deviation of sample mean weight.

standard deviation of sample mean weight=1.07/sqrt(90)=0.1128

c)  The approximate probability the sample mean weight will be greater than 21

z=(21-21.3)/0.1128

z=-2.66

P(z>-2.66)=1-normsdist(-2.66)=0.9961

d) The probability it will be accepted

sampling distribution of sample sums:

mean=90*21.3=1917

standard deviation=sqrt(90)*1.07=10.151

z=(1890-1917)/10.151

z=-2.66

P(z>-2.66)=0.9961