1. The life of a certain type of automobile tire is normally distributed with me

Question

1. The life of a certain type of automobile tire is normally distributed with mean 34,000 miles and standard

deviation 4000 miles.

(a) What is the probability that such a tire lasts between 30,000 and 35,000 miles?

(b) The manufacturer of the tire refunds a customer’s money if a tire does not last for a certain guaranteed mileage. How should the manufacturer set the guaranteed mileage so that he only has to refund money for 3% of the tires that he sells?

2. A film-coating process produces films whose thicknesses are normally distributed with a mean of ?

microns and a standard deviation of 10 microns. For a certain application, the minimum acceptable

thickness is 90 microns. To what value should the mean be set so that only 1% of the films will be too

thin?

3. Find the exact value of P( | Y - µ | < 1.75? ) for an exponential random variable with parameter ? .

Compare with the corresponding lower bound given by Tchebysheff’s Theorem.

Explanation / Answer

Ans:

1)mean=34000,standard deviation=4000

a)

z(30000)=(30000-34000)/4000=-1

z(35000)=(35000-34000)/4000=0.25

P(-1<z<0.25)=P(z<0.25)-P(z<-1)=0.5987-0.1587=0.44

b)P(Z<=z)=0.03

z=normsinv(0.03)=-1.88

Gauranteed mileage=34000-1.88*4000=26480

2)P(Z<=z)=0.01

z=normsinv(0.01)=-2.33

z=-2.33=(90-mean)/10

mean=90+2.33*10=113.3 microns

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