Question

Suppose that yield level, represented by Y, in a randomly selected farm is normally distributed with a mean level of 70ppm (parts per million) and standard deviation of 13 ppm.

a) What is probability that randomly selected field will have yield level less than 60ppm?

b) What is the probability yield level being greater than 90 ppm?

c) What is the probability yield level will be between 65-95 ppm?

d) Find the value of y such that P (Y > y) = 40%.

Answer 1

A) Given data

Mean value (Y')=70

Standard deviation (S)=13

We need to determine the probability for the following cases

a) Probability that randomly selected field less than 60 is

P(Y<60)

First determine the z score Value

Z score=(Y-Y')/S

=(60-70)/13

=-0.769=-0.77

P(Y<60)=P(Z<-0.77)

=0.2206 (From normal area tables)

b) Probability that randomly selected yield is greater than 90 is

P(Y>90)

Z score=(90-70)/13=1.538=1.54

P(Y>90)=P(Z>1.54)

=0.0618 (From normal area tables)

c) Probability that randomly selected yield is between 65-95 is

P(65<Y<95)

For y=65

Z score=(65-70)/13=-0.38

For y=95

Z score=(95-70)/13=1.92

P(65<Y<95)=P(-0.38<Z<1.92)

=0.6206 (From normal area tables)

d) We need to determine y Value for which P(Y>y)=40%=0.4

Z score=(y-y')/S

P(Y>y)=P(Z>y)=0.4

=(y-70)/13=0.4

y Value=0.4×13+70

=5.2+70=75.2