Question

8. The weight of coffee in a can is normally distributed with a mean of 16.1 oz. and standard deviation of 0.5 oz (a) According to the label, a can contains 16 oz. of coffee. What is the probability that there is less than 16 oz. of coffee in a can? (b) What is the probability that a can contains 16 to 16.5 oz. of coffee? (c) What is the lower 10th percentile of the distribution of the coffee weight? (Ten percent of the cans contain coffee less than this weight.)

Answer 1

a)

for normal distribution z score =(X-μ)/σ
here mean=       μ=	16.1
std deviation   =σ=	0.5000

P(it contains less then 16 oz.):

probability =

P(X<16)

=

P(Z<-0.2)=

0.4207

b)

probability =

P(16<X<16.5)

=

P(-0.2<Z<0.8)=

0.7881-0.4207=

0.3674

c)

for 10th percentile critical value of z=		-1.28
therefore corresponding value=mean+z*std deviation=			15.460