The A.C. Nielsen Company reported in the Nilsen Report on Television that the mean weekly television viewing time for children aged $2-6$years is $24.85$hours. Assume that the weekly television viewing times of such children are normally distributed with a standard deviation of $6.23$hours and apply the empirical rule to fill in the blanks.

Part (a): Approximately role="math" localid="1652510212472" $68\%$of all such children watch between ___ and ___.

Part (b): Approximately role="math" localid="1652510208999" $95\%$of all such children watch between ___ and ___.

Part (c): Approximately $99.7\%$of all such children watch between ___ and ___.

Part (d): Draw graphs similar to those in Fig. $6.27$on page $272$to portray your results.

Consider the given question,

The weekly television viewing time of children aged $2-6$years is approximately normal with mean$24.85$hours and standard deviation$6.23hours$.

By property of empirical rule, approximately $68\%$of values lie in the interval,

$$\begin{gathered} \left(\mu -\sigma ,\mu +\sigma \right)=\left(24.85-6.23,24.85+6.23\right) \\ =\left(18.62,31.08\right) \end{gathered}$$

Thus, approximately $68\%$of all children watch between role="math" localid="1652510519610" $18.62,31.08$hoursof TV per week.

By property of empirical rule, approximately $95\%$of values lie in the interval,

$$\begin{gathered} \left(\mu -2\sigma ,\mu +2\sigma \right)=\left(24.85-\left(2\right)6.23,24.85+\left(2\right)6.23\right) \\ =\left(12.39,37.31\right) \end{gathered}$$

Thus, approximately $95\%$of all children watch between$12.39,37.31$hours.

By property of empirical rule, approximately $99.7\%$of values lie in the interval,

$$\begin{gathered} \left(\mu -3\sigma ,\mu +3\sigma \right)=\left(24.85-\left(3\right)6.23,24.85+\left(3\right)6.23\right) \\ =\left(6.16,43.54\right) \end{gathered}$$

Thus, approximately $99.7\%$of all children watch between$6.16,43.54$hours.

On drawing graph representing approximately $68\%$of all children,

On drawing graph representing approximately $95\%$of all children,

On drawing graph representing approximately $99.7\%$of all children,