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

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  =    ~ N(0,1)

where, = population mean = 50

            = standard deviation = 7

So, X ~ Normal()

Now, the 9th percentile is calculated as;

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

            P( < ) = 0.09

            P(Z < ) = 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;

                      x = 40.52

Hence, the 9th percentile is 40.52.