Question

The height measurements of 600 adult males are arranged in ascending order and it is observed that 180th and 450th entries are 46.3 and 67.8 inch respectively. If the measurements are normally distributed. i. Find the mean and the standard deviation of the distribution.

Answer 1

The mean and the standard deviation of the distribution is 55.70 inches and 17.93 inches respectively.

Step-by-step explanation:

We are given that the height measurements of 600 adult males are arranged in ascending order and it is observed that 180th and 450th entries are 46.3 and 67.8 inches respectively.

Also, the measurements are normally distributed.

Let X = height measurements of adult males

The z-score probability distribution for the normal distribution is given by;

                             Z  =  $\frac{X-\mu}{\sigma}$  ~ N(0,1)

where, $\mu$ = population mean

           $\sigma$ = standard deviation

The total number of adult males = 600 and the 180th entry is 46.3 inches.

The z-score = $\frac{X-\mu}{\sigma}$  

Here the z-score will be 180 is how much % of 600, i.e;

z-score = $\frac{180}{600}\times 100$ = 30%

In the z table, the critical value of x which has an area of 30% or below is -0.5244, this means that;

       $-0.5244 = \frac{X-\mu}{\sigma}$

        $-0.5244 = \frac{46.3-\mu}{\sigma}$

        $-0.5244\sigma = 46.3-\mu$

         $\mu = 46.3+0.5244\sigma$     ----------- [equation 1]

Similarly, it is given that 450th entry is 67.8 inches.

The z-score = $\frac{X-\mu}{\sigma}$  

Here the z-score will be 450 is how much % of 600, i.e;

z-score = $\frac{450}{600}\times 100$ = 75%

In the z table, the critical value of x which has an area of 75% or below is 0.6745, this means that;

       $0.6745 = \frac{X-\mu}{\sigma}$

        $0.6745 = \frac{67.8-\mu}{\sigma}$

        $0.6745\sigma = 67.8-\mu$

        $\mu = 67.8-0.6745\sigma$    ----------- [equation 2]

From equation 1 and 2 we get;

 $46.3+0.5244\sigma = 67.8-0.6745\sigma$

 $0.5244\sigma+0.6745\sigma = 67.8-46.3$

 $1.1989\sigma = 21.5$

     $\sigma = \frac{21.5}{1.1989}$ = 17.93

Now, putting the value of $\sigma$ in equation 1, we get;

     $\mu = 46.3+0.5244\sigma$

     $\mu = 46.3+(0.5244\times 17.93)$

     $\mu = 55.70$

Hence, the mean and the standard deviation of the distribution is 55.70 inches and 17.93 inches respectively.