Question

If the random variable X is normally distributed with mean of 50 and standard deviation of 7, find the 9th percentile.

Answer 1

Answer:

The 9th percentile is 40.52.

Step-by-step explanation:

We are given that the random variable X is normally distributed with a mean of 50 and a standard deviation of 7.

Let X = the random variable

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

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

where, $\mu$ = population mean = 50

           $\sigma$ = standard deviation = 7

So, X ~ Normal($\mu=50, \sigma^{2} = 7^{2}$)

Now, the 9th percentile is calculated as;

            P(X < x) = 0.09         {where x is the required value}

            P( $\frac{X-\mu}{\sigma}$ < $\frac{x-50}{7}$ ) = 0.09

            P(Z < $\frac{x-50}{7}$ ) = 0.09

Now, in the z table the critical value of x that represents the below 9% of the area is given as -1.3543, i.e;

                     $\frac{x-50}{7}=-1.3543$

                     $x-50=-1.3543 \times 7$

                     $x=50 -9.48$

                      x = 40.52

Hence, the 9th percentile is 40.52.