Question

"Durable press" cotton fabrics are treated to improve their recovery from wrinkles after washing. Unfortunately, the treatment also reduces the strength of the fabric. The breaking strength of untreated fabric is normally distributed with mean 51 pounds and standard deviation 2.3 pounds. The same type of fabric after treatment has normally distributed breaking strength with mean 18.9 pounds and standard deviation 1.6 pounds. A clothing manufacturer tests 5 specimens of each fabric. All 10 strength measurements are independent. (Round your answers to four decimal places.)

(a) What is the probability that the mean breaking strength of the 5 untreated specimens exceeds 50 pounds?

(b) What is the probability that the mean breaking strength of the 5 untreated specimens is at least 25 pounds greater than the mean strength of the 5 treated specimens?

Answer 1

a)

for normal distribution z score =(X-μ)/σ
here mean=       μ=	51
std deviation   =σ=	2.3000
sample size       =n=	5
std error=σx̅=σ/√n=	1.0286

probability =

P(X>50)

=

P(Z>-0.97)=

1-P(Z<-0.97)=

1-0.1660=

0.8340

b)

let difference between mean =51-18.9 =32.1

and std deviation =sqrt(2.3²/5+1.6²/5)=1.253

hence probability that the mean breaking strength of the 5 untreated specimens is at least 25 pounds greater than the mean strength of the 5 treated specimens =P(X>25)=P(Z>(25-32.1)/1.253)=P(Z>-5.67)=1.0000