Question

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

Answer 1

The 9th percentile is 40.52.

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 =
`(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(
`(X-\mu)/(\sigma)` <
`(x-50)/(7)` ) = 0.09

P(Z <
`(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;

`(x-50)/(7)=-1.3543`

`x-50=-1.3543 * 7`

`x=50 -9.48`

x = 40.52

Hence, the 9th percentile is 40.52.