Question Solved1 Answer Question 5 (12 pts) The number of television viewing hours per household and the prime viewing times are two factors that affect television advertising income. A random sample of 25 households in a partic- ular viewing area produced the following estimates of viewing hours per household 3.0, 6.0, 7.5, 15.0, 12.0, 6.5, 8.0, 4.0, 5.5, 6.0 5.0 12.0, 1.0, 3.5, 3.0, 7.5, 5.0, 100, 8.0, 3.5, 9.0, 2.0, 6.5, 1.0, 5.0 1 a. Scan the data and use the range to find an approximate value for s. b. Calculate the sample mean i and the sample standard deviation s. Compare s with the approximate value obtained in part a, c. Find the percentage of the viewing hours per household that falls into the interval : +2s. Compare with the corresponding percentage given by the Empirical Rule (assuming data is relatively mound-shaped).

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Transcribed Image Text: Question 5 (12 pts) The number of television viewing hours per household and the prime viewing times are two factors that affect television advertising income. A random sample of 25 households in a partic- ular viewing area produced the following estimates of viewing hours per household 3.0, 6.0, 7.5, 15.0, 12.0, 6.5, 8.0, 4.0, 5.5, 6.0 5.0 12.0, 1.0, 3.5, 3.0, 7.5, 5.0, 100, 8.0, 3.5, 9.0, 2.0, 6.5, 1.0, 5.0 1 a. Scan the data and use the range to find an approximate value for s. b. Calculate the sample mean i and the sample standard deviation s. Compare s with the approximate value obtained in part a, c. Find the percentage of the viewing hours per household that falls into the interval : +2s. Compare with the corresponding percentage given by the Empirical Rule (assuming data is relatively mound-shaped).
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Transcribed Image Text: Question 5 (12 pts) The number of television viewing hours per household and the prime viewing times are two factors that affect television advertising income. A random sample of 25 households in a partic- ular viewing area produced the following estimates of viewing hours per household 3.0, 6.0, 7.5, 15.0, 12.0, 6.5, 8.0, 4.0, 5.5, 6.0 5.0 12.0, 1.0, 3.5, 3.0, 7.5, 5.0, 100, 8.0, 3.5, 9.0, 2.0, 6.5, 1.0, 5.0 1 a. Scan the data and use the range to find an approximate value for s. b. Calculate the sample mean i and the sample standard deviation s. Compare s with the approximate value obtained in part a, c. Find the percentage of the viewing hours per household that falls into the interval : +2s. Compare with the corresponding percentage given by the Empirical Rule (assuming data is relatively mound-shaped).
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(a) In the given data of number of television viewing hours per houshold, the minimum value is =1& the maximum value is =15.Therefore, range of the given sample is -{:[" maximum value - minimum value "],[=15-1],[=14]:}Now, the range rule tells us-sd=(" Range ")/(4)quad[[" sd is the standard "],[" deviation of the data set "]]we should remember that, this formula is very rough for calculating sd.Then, sd=(14)/(4)=3.5(b) The formula of the sample mean and sample standard deviation is -Samplemean= bar(x)=(1)/(n)sum_(i=1)^(n)x_(i)quad&quad sample sd=sqrt((1)/(n-1)sum_(i=1)^(n)(x_(i)-( bar(x)))^(2))From the table above,{:[" sample mean "=(155.5)/(25)=6.22],[" sample sd "=sqrt(((1)/(25-1)xx293.54))],[=3.49726]:}In (a) the ... See the full answer