Question

For a normal distribution with μ=500 and σ=100, what is the minimum score necessary to be in the top 60% of the distribution?

Answer 1

Answer:

475.

Step-by-step explanation:

We have been given that for a normal distribution with μ=500 and σ=100. We are asked to find the minimum score that is necessary to be in the top 60% of the distribution.

We will use z-score formula and normal distribution table to solve our given problem.

$z=\frac{x-\mu}{\sigma}$

Top 60% means greater than 40%.

Let us find z-score corresponding to normal score to 40% or 0.40.

Using normal distribution table, we got a z-score of $-0.25$.

Upon substituting our given values in z-score formula, we will get:

$-0.25=\frac{x-500}{100}$

$-0.25*100=\frac{x-500}{100}*100$

$-25=x-500$

$-25+500=x-500+500$

$x=475$

Therefore, the minimum score necessary to be in the top 60% of the distribution is 475.