Q. 6.106 - Elementary Statistics

Question

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 $68\%$ of all such children watch between ___ and  ___.

Part (b): Approximately $95\%$ 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.

Step-by-Step Solution

Answer

Part (a): Approximately $68\%$of all such children watch between $18.62$hours and $31.08$ hours.

Part (b): Approximately $95\%$ of all such children watch between $12.39$ hours and $37.31$ hours.

Part (c): Approximately $99.7\%$ of all such children watch between $6.16$ hours and $43.54$ hours.

Part (d): Graph of approximately $68\%$of all such children,

Graph of approximately $95\%$ of all such children,

Graph of approximately $99.7\%$ of all such children,

Part (a) Step 1. Given information.

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 $6.23hours$.

Part (a) Step 2. Determine the approximate of all such children within 68 % .

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 $18.62,31.08$ hours of TV per week.

Part (b) Step 1. Determine the approximate of all such children within 95 % .

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.

Part (c) Step 1. Determine the approximate of all such children within 99 . 7 % .

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.

Part (d) Step 1. Drawing graph representing approximately 68 % , 95 % , 99 . 7 % of all children.

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,