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 ≈ |
|
≈ |
|
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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