Question

A book publisher knows that it takes an average of nine business days from when the material for the book is finalized until the first edition is printed and ready to sell. Suppose the exact amount of time has a standard deviation of four days. Use an appropriate normal transformation to answer the following questions. a. Suppose the publisher examines the printing time for a sample of 36 books. What is the probability that the sample mean time is shorter than eight days? 0.0668 b. Suppose the publisher examines the printing time for a sample of 36 books. What is the probability that the sample mean time is between 7 and 10 days? 0.9973 c. Suppose the publisher signs a contract for the printer to print 100 books. If the average printing time for the 100 books is longer than 9.3 days, the printer must pay a penalty. What is the probability the penalty clause will be activated? 0.2266 d. Suppose the publisher signs a contract for the printer to print 10 books. If the average printing time for the 10 books is longer than 9.7 days, the printer must pay a penalty. What is the probability the penalty clause will be activated?

Answer 1

a)

for normal distribution z score =(X-μ)/σ
here mean=       μ=	9
std deviation   =σ=	4.000
sample size       =n=	36
std error=σx̅=σ/√n=	0.66667

probability =P(X<8)=(Z<(8-9)/0.667)=P(Z<-1.5)=0.0668

b)

probability =P(7<X<10)=P((7-9)/0.667)<Z<(10-9)/0.667)=P(-3<Z<1.5)=0.9332-0.0013=0.9318

c)

sample size       =n=	100
std error=σx̅=σ/√n=	0.40000

probability =P(X>9.3)=P(Z>(9.3-9)/0.4)=P(Z>0.75)=1-P(Z<0.75)=1-0.7734=0.2266

d)

sample size       =n=	10
std error=σx̅=σ/√n=	1.2649

probability =P(X>9.7)=P(Z>(9.7-9)/1.265)=P(Z>0.55)=1-P(Z<0.55)=1-0.7088=0.2912

(please try 0.2900 if this comes wrong)