Question

Suppose, household color TVs are replaced at an average age of μ = 8.2 years after purchase, and the (95% of data) range was from 5.6 to 10.8 years. Thus, the range was 10.8 − 5.6 = 5.2 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 μ − 2σ to μ + 2σ is often used for "commonly occurring" data values. Note that the interval from μ − 2σ to μ + 2σ is 4σ in length. This leads to a "rule of thumb" for estimating the standard deviation from a 95% range of data values.Estimating the standard deviation

For a symmetric, bell-shaped distribution,

standard deviation ≈

range 4

≈

high value − low value 4

where it is estimated that about 95% of the commonly occurring data values fall into this range.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.)
yrs

(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 8.2 years with a standard deviation of 1.3 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 14% of the TVs under guarantee. For how long should the guarantee be made (rounded to the nearest tenth of a year)?
yrs.

Answer 1

approximate the standard deviation =5.2/4 =1.3

b)

probability =P(X>5)=P(Z>(5-8.2)/1.3)=P(Z>-2.46)=1-P(Z<-2.46)=1-0.0069=0.9931

c)

probability =P(X<10)=(Z<(10-8.2)/1.3)=P(Z<1.38)=0.9162

d)

for 14th percentile critical value of z=		-1.08
therefore corresponding value=mean+z*std deviation=			6.8 years