Question

# Suppose, household color TVs are replaced at an average age of μ = 8.6 years after...

Suppose, household color TVs are replaced at an average age of μ = 8.6 years after purchase, and the (95% of data) range was from 6.0 to 11.2 years. Thus, the range was 11.2 – 6.0 = 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 symmetrical 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 8.6 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 7% of the TVs under guarantee. For how long should the guarantee be made (rounded to the nearest tenth of a year)?
yrs

a)

standard deviation ≈ range/4 =5.2/4 =1.3

b)

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

 probability = P(X>5) = P(Z>-2.77)= 1-P(Z<-2.77)= 1-0.0028= 0.9972

c)

probability that someone will keep a color TV fewer than 10 years before replacement :

 probability = P(X<10) = P(Z<1.08)= 0.8599

(try 0.8592 if this comes wrong)

d)

 for 7th percentile critical value of z= -1.48 therefore corresponding value=mean+z*std deviation= 6.7

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