Question

Suppose, household color TVs are replaced at an average age of μ = 8.4 years after purchase, and the (95% of data) range was from 5.0 to 11.8 years. Thus, the range was 11.8 – 5.0 = 6.8 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 8.4 years with a standard deviation of 1.7 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 6% of the TVs under guarantee. For how long should the guarantee be made (rounded to the nearest tenth of a year)? yrs

Answer 1

a)accordingly explain in this question we precede further,

Here, 95% confidence interval is given that is (5, 11.8)

And Let X~N(mue, sigma)

So for normal distribution,

95% CI is 4Sigma,

That is,

b)

Let X is someone will keep TV for this much year (means X year)

X~N(8.4,1.7)

From question given

Probability of someone will keep TV for more than 5 year=P(X>5)

Let make it standard normal,

=P(Z>-2)=0.9772

Using standard normal table probbilprob=0.9772

c)

Let X is someone will keep TV for this much year (means X year)

X~N(8.4,1.7)

From question given

Probability of someone will keep TV for fewer than 10 year=P(X<10)

Let make it standard normal,

Similarly above,

=P(Z<0.94)=0.8264

Using stadard normal Z table proaprobabi value =0.8264

So, Here X is someone will keep TV for this much year (means X year)

X~N(8.4,1.7)

Probability of someone will keep TV for less than X year and company will change it is not more than 6%

So, maximum 0.06 probability that company will change the TV.

Mathamatically,

So, using standard normal table Z=(-1.555)

And transform z to X,

X=(-1.555*1.7)+8.4=5.757

here we round X=5 because if we round upper than probability will increase form 0.06

So year=5