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 mu = 14 years with a (95% of data) range from 9 to 19 years. Let x = age at which a refrigerator is replaced. Assume that x has a distribution that is approximately normal. 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 mu - 2sigma to mu + 2sigma is often used for "commonly occurring" data values. Note that the interval from mu - 2sigma to mu + 2sigma is 4sigma 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, 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.) What is the probability that someone will keep a refrigerator fewer than 11 years before replacement? (Round your answer to four decimal places.) What is the probability that someone will keep a refrigerator more than 18 years before replacement? (Round your answer to four decimal places.) Assume that the average life of a refrigerator is 14 years, with the standard deviation given In part 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 4% of the refrigerators under guarantee. For how long should the guarantee be made (rounded to the nearest tenth of a year)?

Explanation / Answer

SD= Range/4= ( 19-9)/4 = 2.5

P( x <11)

Z value for 11, z=( 11-14)/2.5 =-1.2

P( x <11) = P( z < -1.2) = 0.1151

P( x >18)

Z value for 18, z=( 18-14)/2.5 =1.6

P( x >18) = P( z > 1.6) = 0.0548

Mean =14

SD= 2.5

Z value for top 4% = 1.751

The required value = mean + z*sd = 14+1.751*2.5 =18.3775

The required value =18.4

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