Suppose, household color TVs are replaced at an average age of ? = 7.6 years aft

Question

Suppose, household color TVs are replaced at an average age of ? = 7.6 years after purchase, and the (95% of data) range was from 3.8 to 11.4 years. Thus, the range was 11.4 ? 3.8 = 7.6 years. Let x be the age (in years) at which a color TV is replaced. Assume that x has a distribution that is approximately normal.

(a) The empirical rule indicates that for a symmetric and bell-shaped distribution, approximately 95% of the data lies within two standard deviations of the mean. Therefore, a 95% range of data values extending from ? ? 2? to ? + 2? is often used for "commonly occurring" data values. Note that the interval from ? ? 2? to ? + 2? is 4? in length. This leads to a "rule of thumb" for estimating the standard deviation from a 95% range of data values.

Use this "rule of thumb" to approximate the standard deviation of x values, where x is the age (in years) at which a color TV is replaced. (Round your answer to one decimal place.)

(b) What is the probability that someone will keep a color TV more than 5 years before replacement? (Round your answer to four decimal places.)


(c) What is the probability that someone will keep a color TV fewer than 10 years before replacement? (Round your answer to four decimal places.)


(d) Assume that the average life of a color TV is 7.6 years with a standard deviation of 1.9 years before it breaks. Suppose that a company guarantees color TVs and will replace a TV that breaks while under guarantee with a new one. However, the company does not want to replace more than 7% of the TVs under guarantee. For how long should the guarantee be made (rounded to the nearest tenth of a year)?

Explanation / Answer

a) std deviation=range/4 =7.6/4=1.9

b) probability that someone will keep a color TV more than 5 years before replacement

=P(X>5)=P(Z>(5-7.6)/1.9)=P(Z>-1.37)=0.9147

c)

P(X<10)=P(Z<(10-7.6)/1.9)=P(Z<1.26)=0.8962

d)

at 7th percentile ; critical value z =-1.48

hence corresponding value =mean +z*std deviation=7.6-1.48*1.9=4.8 Years

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