Question

The scores for a certain test of intelligence are normally distributed with mean 85 and standard deviation 11. Find the 90th percentile of these scores.

Answer 1

Answer:

99.1 is the 90th percentile of these scores.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 85

Standard Deviation, σ = 11

We are given that the distribution of score is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

We have to find the value of x such that the probability is 0.90

$P( X < x) = P( z < \displaystyle\frac{x - 85}{11})=0.90$  

Calculation the value from standard normal z table, we have,  

$P(z \leq 1.282) = 0.90$

Putting values, we get,

$\displaystyle\frac{x - 85}{11} = 1.282\\\\x = 99.1$

Thus, 99.1 is the 90th percentile of these scores.