Question

A random variable is normally distributed with a mean of 25 and a standard deviation of 5. If an observation is randomly selected from the distribution, what value will be exceeded 85% of the time

Answer 1

Answer:

Value of 26.93 will be exceeded 85% of the time.

Step-by-step explanation:

We are given that a random variable is normally distributed with a mean of 25 and a standard deviation of 5.

Let X = a random variable

So, X ~ N($\mu=25,\sigma^{2} =5^{2}$)

The z score probability distribution is given by;

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

where, $\mu$ = population mean

            $\sigma$ = standard deviation

Now, we have to find that value which will be exceeded 85% of the time, i.e.;

        P(X > $x$) = 0.85

    1 - P(X $\leq x$) = 0.85

         P(X $\leq x$) = 0.15

         P( $\frac{X-\mu}{\sigma}$ $\leq \frac{x-25}{5}$ ) = 0.15

         P(Z $\leq \frac{x-25}{5}$ ) = 0.15

Now, in the z table the critical value of $x$ whose less than area is 0.15 is given as 0.3853, i.e;

                 $\frac{x-25}{5}$ = 0.3853

                 $x-25=0.3853 \times 5$

                    $x$ = 25 + 1.9265 = 26.93

Therefore, value of 26.93 will be exceeded 85% of the time.