Suppose, household color TVs are replaced at an average age of μ=8.2years after purchase, and the (95% of data) range was from 6.2 to 10.2 years. Thus, the range was 10.2−6.2=4.0 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 → μ+2σ is often used for "commonly occurring" data values. Note that the interval from μ–2σ → μ+2σis4σ 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≈ran≥ 4≈high value−low 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 decimal place.) (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.)

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Suppose, household color TVs are replaced at an average age of μ=8.2years after purchase, and the (95% of data) range was from 6.2 to 10.2 years. Thus, the range was 10.2−6.2=4.0 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 → μ+2σ is often used for &quot;commonly occurring&quot; data values. Note that the interval from μ–2σ → μ+2σis4σ in length. This leads to a &quot;rule of thumb&quot; for estimating the standard deviation from a 95% range of data values.Estimating the standard deviation  For a symmetric, bell-shaped distribution,  standard deviation≈ran≥ 4≈high value−low value  where it is estimated that about 95% of the commonly occurring data values fall into this range. Use this &quot;rule of thumb&quot; 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.)  (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.)

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Step 1   A normal distribution follows a bell shaped curve which is centred at the mean and it is also called Gaussian distribution. The normal distribution with 0 mean and 1 standard deviation is called standard normal distribution.   The relation between a normal random variable x and z-score is z=x−μσ, where μ is mean and σ is standard deviation of x. The probability that x lies in an interval is equal to the area of that interval.  Step 2   Answer (a): From the rule of thumb, the range is equal to 4σ, where σ is standard deviation. So to find the standard deviation of x divide the range by 4 and the range is 4.  σ=ran≥4  =44  =1  So the standard deviation of TV age is 1.   Answer (b): It is asked to find the probability that the age of a TV is more than 5 years that is x&amp;gt;5. First find the standard score corresponding to x=5 by substituting μ=8.2 and σ=1 ∈ z=x−μσ.  z=5−8.21  =−3.2  Using the table of standard normal distribution the area on the right of z=−3.2 is 0.9993. So the probability that the age of a TV is more than 5 year is 0.9993 that is very close to 1.   Step 3   Answer (c): It is asked to find the probability that the age of a TV is less than 10 years that is x&amp;lt;10. First find the standard score corresponding to x=10 by substituting μ=8.2 and σ=1 ∈ z=x−μσ.  z=10−8.21  =1.8  Using the table of standard normal distribution the area on the left of z=1.8 is 0.9641. So the probability that the age of a TV is less than 10 year is 09641.

Suppose, household color TVs are replaced at an average age of \mu

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