Question

Assume the random variable X is normally​ distributed, with mean u=50 and standard deviation SD=6. Find the 15 th percentile.

Answer 1

44.85.

Explanation:

We have been given that the random variable X is normally​ distributed, with mean u=50 and standard deviation SD=6. We are asked to find the 15th percentile.

We will use normal distribution table to find the z-score corresponding to 15th percentile or 0.15.

Using normal distribution table, we will get z-score of
`-1.03`.

Now, we will use z-score formula to find corresponding x-value to z-score of
`-1.03`.

`z=(x-\mu)/(\sigma)`, where,

z = Z-score,

x = Sample score,

`\mu` = Mean,

`\sigma` = Standard deviation.

Upon substituting our given values, we will get:

`-1.03=(x-50)/(6)`

`-1.03*5=(x-50)/(6)*6`

`-5.15=x-50`

`-5.15+50=x-50+50`

`44.85=x`

Therefore, the 15th percentile is 44.85.