A consumer products magazine indicated that the average life of a refrigerator b

Question

A consumer products magazine indicated that the average life of a refrigerator before replacement is = 14 years with a (95% of data) range from 8 to 20 years. Let x = age at which a refrigerator 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 – 2 to + 2 is often used for "commonly occurring" data values. Note that the interval from – 2 to + 2 is 4 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. (Round your answer to one decimal place.) yr (b) What is the probability that someone will keep a refrigerator fewer than 11 years before replacement? (Round your answer to four decimal places.) (c) What is the probability that someone will keep a refrigerator more than 18 years before replacement? (Round your answer to four decimal places.) (d) Assume that the average life of a refrigerator is 14 years, with the standard deviation given in part (a) before it breaks. Suppose that a company guarantees refrigerators and will replace a refrigerator that breaks while under guarantee with a new one. However, the company does not want to replace more than 5% of the refrigerators under guarantee. For how long should the guarantee be made (rounded to the nearest tenth of a year)? yr

Explanation / Answer

a)

8 to 20 is 4 standard deviations.

Hence,

4*sigma = 20 - 8 = 12

Hence,

sigma = 3 [ANSWER]

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b)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    11      
u = mean =    14      
          
s = standard deviation =    3      
          
Thus,          
          
z = (x - u) / s =    -1      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -1   ) =    0.158655254 [ANSWER]

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c)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    18      
u = mean =    14      
          
s = standard deviation =    3      
          
Thus,          
          
z = (x - u) / s =    1.333333333      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   1.333333333   ) =    0.09121122 [ANSWER]

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d)

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.05      
          
Then, using table or technology,          
          
z =    -1.644853627      
          
As x = u + z * s,          
          
where          
          
u = mean =    14      
z = the critical z score =    -1.644853627      
s = standard deviation =    3      
          
Then          
          
x = critical value =    9.065439119 = 9.1 YEARS [ANSWER]  

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