Suppose, household color TVs are replaced at an average age of p 8.0 years after

Question

Suppose, household color TVs are replaced at an average age of p 8.0 years after purchase, 10.4-5.64.8 years. Let x be the age (in years) andthe 95% of data ) range was from 5.6 to 10.4 years. Thus, the range was ) 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% range of data values extending from 20 to + 2a is often used for standard deviation from a -commonly occurring" data values. Note that the interval frem -20 to . 2is 4 in length. This leads to a "rule of thumb" for estimating the 95% range of data values. Estimating the standard deviation For a symmetric, bell-shaped distribution, standard deviation Fange, high valuelow value 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 decimat ) 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 s.o years with a standard deviation of 1.2 years before it breaks. Suppose that a company guarantees color TVs and vwill replace a yrs TV that breaks while under guarantee with a new one. However, the company does not want to replace more than n% or the rs ander guarantee for her leg told the uarantee be made (rounded to the nearest tenth of a year)? yrs

Explanation / Answer

a) std deviaiton =(10.4-5.6)/4=1.2

b)

probability to keep it for more then 5 years =P(X>5)=1-P(X<5)=1-P(Z<(5-8)/1.2)=1-P(Z<-2.5)=1-0.0062=0.9938

c)probability to keep it for less then 5 years=P(X<10)=P(Z<(10-8)/1.2)=P(Z<1.6667)=0.9522

d)

at 11th percentlie ; from standard normal distribution z=-1.2265

hence gaurantee period =mean +z*Std deviaiton =6.5 Years

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