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.

3 0

2 years ago

Timed test please help

insens350 [35]

Last option: − ∞ < y < 5

8 0

3 years ago

65% of what number is 39?

aleksklad [387]

You have to turn 65 into a decimal. That would be .65
They do 39/.65
You should get 60 as your final answer.

6 0

3 years ago

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