Suppose, household color TVs are replaced at an average age of μ = 7.4 years after purchase, and the (95% of data) range was from 5.0 to 9.8 years. Thus, the range was 9.8 − 5.0 = 4.8 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 symmetric and
bellshaped 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, bellshaped distribution,
standard deviation ≈ 

≈ 

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 7.4 years with a
standard deviation of 1.2 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)?
yrs
a) standard deviation ≈ =4.8/4 =1.2
b)
for normal distribution z score =(Xμ)/σ  
here mean= μ=  7.4 
std deviation =σ=  1.2000 
probability that someone will keep a color TV more than 5 years before replacement:
probability =  P(X>5)  =  P(Z>2)=  1P(Z<2)=  10.0228=  0.9772 
c)
probability that someone will keep a color TV fewer than 10 years before replacement :
probability =  P(X<10)  =  P(Z<2.17)=  0.9850 
d)
for 13th percentile critical value of z=  1.13  
therefore corresponding value=mean+z*std deviation=  6.0 
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