Question

Suppose, household color TVs are replaced at an average age of μ = 8.8 years after purchase, and the (95% of data) range was from 5.8 to 11.8 years. Thus, the range was 11.8 - 5.8 = 6.0 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 sysmmetric 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.) _____years

(b) What is the probability that someone will keep a color TV more than 5 years before repalcement? (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 8.8 years with a standard deviation of 1.5 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 13% of the TVs under guarantee. For how long should the guarantee be made (rounded to the nearest tenth of a year)? ____ years

Answer 1

(a) The empirical rule indicates that for a sysmmetric 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.

Here standard deviation = Range/4 = 6/4 = 1.5 years

(b) What is the probability that someone will keep a color TV more than 5 years before repalcement?

Pr(x > 5) = 1 - Pr(x < 5 years ; 8.8 years ; 1.5 years)

Z = (5 - 8.8)/1.5 = -2.5333

Pr(x > 5) = 1 - Pr(x < 5 years ; 8.8 years ; 1.5 years) = 1 - Pr(Z < -2.5333) = 1 - 0.0057 = 0.9943

(c) Pr(x < 10 years)

Z = (10 - 8.8)/1.5 = 0.8

Pr(x < 10 years) = Pr(x < 10 years ; 8.8 years ; 1.5 years) = Pr(Z < 0.8) = 0.7881

(d) Here lets say the number of years the guarantee would be made is x₀

Pr(x < x₀ ; 8.8 ; 1.5) = 0.13

Taking Z table

Z = -1.1264 = (x₀ - 8.8)/1.5

x₀ = 8.8 - 1.5 * 1.1264 = 7.1 years