Question

The annual coffee expenditures for households are approximately normally distributed with a mean of $ 48.57 and a standard deviation of $ 10.00 . a. Find the probability that a household spent less than ​$25.00. b. Find the probability that a household spent more than ​$55.00. c. What proportion of the households spent between ​$30.00 and ​$40​.00? d. 95​% of the households spent less than what​ amount? a. The probability that a household spent less than ​$25.00 is?

Answer 1

Given The annual coffee expenditures for households are approximately normally distributed

Mean = $ 48.57

Standard deviation = $ 10.00

standard normal variate Z =

a) P(X < $25.00 )

for x = $25.00 = = -2.35

P(X < $25.00 ) = P(Z < -2.35) = 0.0094 = 0.94%

b) P(X> $55.00 )

Z = 0.643

P(X> $55.00 ) = P(Z > 0.643) = 1- P(Z<0.643) = 1 - 0.7398 = 0.26011 = 26.01%

c) P(30<x<40)

for x=30 Z = -1.857

for x = 40, Z = -0.857

P(30<x<40) = P(-1.857 < Z <-0.857) = P(Z<-0.857) - P(Z<-1.857) = 0.19572 - 0.03166 = 0.1641 = 16.41%

d) P(Z) = 0.95

=> Z = 1.645

X = 16.45+48.57 = 65.02

95% of households spent less than $65.02.