Question

A farmer believes that the yields of his tomato plants have a normal distribution with an average yield of 10 lbs. and a standard deviation of 2 lbs. The farmer would like to identify the plants which yield the highest 5% and save them for breeding purposes. Calculate the yield which separates the highest 5% of yields from the lowest 95% of yields.

MAJOR HELP IF ABLE TO DO ON EXCEL

Answer 1

Let X be the yield of a tomato plant

$X\sim N(10, 2^{2})$

Then

$z= \frac{X-10}{2}\sim N(0,1)$

To find c such that

P( X < c) = 0.95

$\Rightarrow P(z<\frac{c-10}{2})=0.95$

$\Rightarrow \frac{c-10}{2}=1.65$ (from z table or using excel "=NORM.S.INV(0.95)"gives 1.65 )

$\Rightarrow c=13.3$

Using Excel

"=NORM.INV(0.95,10,2)" gives the value  13.3

this formula in excel  returns the value of X for which the cumulative probability is 0.95 with mean =10 and

standard deviation = 2

Thus the yield that separates highest 5% yields is 13.3