Question

1. Accrotime is a manufacturer of quartz crystal watches. Accrotime researchers have shown that the watches...

1. Accrotime is a manufacturer of quartz crystal watches. Accrotime researchers have shown that the watches have an average life of 29 months before certain electronic components deteriorate, causing the watch to become unreliable. The standard deviation of watch lifetimes is 5 months, and the distribution of lifetimes is normal.

(a) If Accrotime guarantees a full refund on any defective watch for 2 years after purchase, what percentage of total production will the company expect to replace? (Round your answer to two decimal places.)
%

(b) If Accrotime does not want to make refunds on more than 12% of the watches it makes, how long should the guarantee period be (to the nearest month)?

2.

Suppose x has a distribution with μ = 56 and σ = 14.

(a) If random samples of size n = 16 are selected, can we say anything about the x distribution of sample means?

Yes, the x distribution is normal with mean μx = 56 and σx = 0.9.Yes, the x distribution is normal with mean μx = 56 and σx = 14.     Yes, the x distribution is normal with mean μx = 56 and σx = 3.5.No, the sample size is too small.


(b) If the original x distribution is normal, can we say anything about the x distribution of random samples of size 16?

Yes, the x distribution is normal with mean μx = 56 and σx = 14.No, the sample size is too small.     Yes, the x distribution is normal with mean μx = 56 and σx = 0.9.Yes, the x distribution is normal with mean μx = 56 and σx = 3.5.


Find P(52 ≤ x ≤ 57). (Round your answer to four decimal places.)

Homework Answers

Answer #1
for normal distribution z score =(X-μ)/σx
mean μ= 29
standard deviation σ= 5

a)

percentage of total production will the company expect to replace :

probability =P(X<24)=(Z<(24-29)/5)=P(Z<-1)=0.1587 ~ 15.87 %

b)

for 12th percentile critical value of z=-1.17
therefore corresponding value=mean+z*std deviation=23 months

2)a)

No, the sample size is too small.

b)

Yes, the x distribution is normal with mean μx = 56 and σx = 3.5

probability =P(52<X<57)=P((52-56)/3.5)<Z<(57-56)/3.5)=P(-1.14<Z<0.29)=0.6141-0.1271=0.4870

(please try 0.4859 if this comes wrong)

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