Chapter 11.10, Problem 30PPSE

Solution

To explain: If there are one or more possibilities for mean and standard deviation with respect to the given information.

There are two possible values of mean and standard deviation with respect to the given information. The possible values of mean and standard deviation are

  $(\text{mean, standard deviation)}=(206,42)\text{ and }(269,21)$

Given:Data point 332 is 3 standard deviations away from the mean. Data point 248 is 1 standard deviation away from the mean.

Concept used:

The number of standard deviations away from the mean is known as the Z score.

Z score is calculated using the formula $Z=\frac{X-\mu }{\sigma }$

Calculation:

The first case is where 332 is 3 standard deviations more than the mean and 248 is 1 standard deviation more than the mean. It can be written in mathematical form as shown below

  $\begin{array}{l}3=\frac{332-\mu }{\sigma }\to \mu +3\sigma =332\\ 1=\frac{248-\mu }{\sigma }\to \mu +\sigma =248\end{array}$

The difference between these two data points is two standard deviations and the standard deviation is solved as shown below

  $\begin{array}{l}\mu +3\sigma =332---(1)\\ \mu +1\sigma =248---(2)\\ (1)-(2)\\ \mu +3\sigma -\mu -\sigma =332-248\\ 2\sigma =84\\ \sigma =42\\ \mu +3\sigma =332\\ \mu +3\times 42=332\\ \mu =332-126=206\end{array}$

The mean and standard deviation in this case are $(206,42)$ respectively.

The second case is where 332 is 3 standard deviations more than the mean and 248 is 1 standard deviation below than the mean. It can be written in mathematical form as shown below

  $\begin{array}{l}3=\frac{332-\mu }{\sigma }\to \mu +3\sigma =332\\ -1=\frac{248-\mu }{\sigma }\to \mu -\sigma =248\end{array}$

The difference between these two data points is four standard deviations and the standard deviation is solved as shown below

  $\begin{array}{l}\mu +3\sigma =332---(1)\\ \mu -1\sigma =248---(2)\\ (1)-(2)\\ \mu +3\sigma -\mu +\sigma =332-248\\ 4\sigma =84\\ \sigma =21\\ \mu +3\sigma =332\\ \mu +3\times 21=332\\ \mu =332-63=269\end{array}$

The mean and standard deviation in this case are $(269,21)$ respectively.

Conclusion:

There are two possible values of mean and standard deviation possible when 332 is three standard deviations from the mean and 248 is 1 standard deviation from the mean. The possible values of mean and standard deviation are (206,42) and (269,21) respectively.