given mean=495

standard deviation=121

a) probability that monthly expenditure is less than 300$

z=(300-495)/121=-1.6115

p(z<-1.6115)=0.5-0.4452=0.0548

b)probability that monthly expenditure lies between 300 and 600 is

for 300 to 495

p(300<x<495)=0.4452

for 495 to 600

z=(600-495)/121=0.86

p(z<0.86)=0.3051

total p(300<x<600)=0.4452+0.3051=0.7503

c)top 8% means finding x having probability 0.08

for 0.5-0.08=0.42

z value corresponding to p value 0.42 is 1.28

(X-495)/121=1.28

X=650

so, for discount he has to spent atleast 649.88$ or 650$

good

Secure Information

Content will be erased after question is completed.

Enter the email address associated with your account, and we will email you a link to reset your password.

Forgot your password?

Sign Up