Question

Assume the random variable X is normally distributed with mean 53 and standard deviation of 6. Find the 9th percentile

Answer 1

Answer:

The 9th percentile of X is 44.96

Step-by-step explanation:

Percentiles for a Normal Distribution

The standard normal distribution can be used for computing percentiles. For example, the median is the 50th percentile being the center value, the first quartile is the 25th percentile, and so on.

To compute percentiles of a normal distribution we can use the formula

$X=\mu+z\sigma$

Where $\mu$ is the mean, $\sigma$ is the standard deviation of the variable X, and z is the z-score value from the standard tables

The value of X can also be directly obtained from digital tables included in math packages and tools like Excel.

The Excel NORM.INV Function calculates the inverse of the cumulative Normal Distribution Function for a given value of p, $\mu$ and $\sigma$.

We'll use the values

$p=9\%=0.09,\ \mu=53, \ \sigma=6$

NORM.INV(0.09,53,6) results in 44.96 which means the 9th percentile of X is 44.96

We could have used the standard normal distribution, which only needs the value of p

$NORM.S.INV( 0.09 )=-1.34$

That is the value of z, we now apply the formula

$X=\mu+z\sigma$

$X=53+(-1.34)\cdot 6=44.96$

We get the very same result as before