A particular variety of watermelon weighs on average 20.2 pounds with a standard

Question

A particular variety of watermelon weighs on average 20.2 pounds with a standard deviation of 1.38 pounds. Consider the sample mean weight of 80 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 20.24? 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 0.9? Give your answer to four decimal places. Use the standard deviation as you entered it above to answer this question.

Explanation / Answer

Mean ( u ) =20.2
Standard Deviation ( sd )= 1.38/ Sqrt(n) = 0.1543
Number ( n ) = 80
Normal Distribution = Z= X- u / (sd/Sqrt(n) ~ N(0,1)                  
P ( Z < x ) = 0.9
Value of z to the cumulative probability of 0.9 from normal table is 1.282
P( x-u/s.d < x - 20.2/0.15429 ) = 0.9
That is, ( x - 20.2/0.15429 ) = 1.28
--> x = 1.28 * 0.15429 + 20.2 = 20.3977                  

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