Chapter 14.4, Problem 10E

(a)

To determine

To find:

The values are within one standard deviation of the mean .

Answer to Problem 10E

The one standard deviation of the mean $=136$ .

Explanation of Solution

Given:

The normally distributed $=200$ .

Concept used:

Normal distribution:- A normal distribution is a frequency distribution that often occurs

There is a large number of values in a set of data.

The graph of this distribution is a symmetric ,bell-shaped curve.

Calculation:

The normally distributed $=200$ .

  $68\%$ of the data fall in with in one standard deviation of the mean.

  $\begin{array}{l}=\frac{68}{100}\\ =0.68\\ =\left(0.68\right)\left(200\right)\\ =136\end{array}$

Therefore ,

The one standard deviation of the mean $=136$ .

(b)

To determine

To find:

The values are within two standard deviation of the mean .

Answer to Problem 10E

The $2$ standard deviation of the mean $=190$ .

Explanation of Solution

Given:

The normally distributed $=200$ .

Concept used:

Normal distribution:- A normal distribution is a frequency distribution that often occurs

There is a large number of values in a set of data.

The graph of this distribution is a symmetric ,bell-shaped curve.

Calculation:

The normally distributed $=200$ .

  $95\%$ of the data fall in with in $2$ standard deviation of the mean.

  $\begin{array}{l}=\frac{95}{100}\\ =0.95\\ =\left(0.95\right)\left(200\right)\\ =190\end{array}$

Therefore ,

The $2$ standard deviation of the mean $=190$ .

(c)

To determine

To find:

The values fall in the interval between the mean and standard deviation of the mean .

Answer to Problem 10E

  $107.52\text{to}292.48$

Explanation of Solution

Given:

The normally distributed $=200$ .

Concept used:

Normal distribution:- A normal distribution is a frequency distribution that often occurs

There is a large number of values in a set of data.

The graph of this distribution is a symmetric ,bell-shaped curve.

Calculation:

The normally distributed $=200$ .

  $68\%$ of the data fall in with in one standard deviation of the mean.

  $\begin{array}{l}=\frac{68}{100}\\ =0.68\\ =\left(0.68\right)\left(200\right)\\ =136\end{array}$

Using the formula

  $\begin{array}{l}=\overline{X}\pm t\sigma \\ =200\pm 0.6\left(136\right)\\ =\left(200+92.48\right),\left(200-92.48\right)\\ =292.48,107.52\end{array}$

Therefore ,

The interval about the mean that includes of the data $\left(107.52\text{to}292.48\right)$ .