Question

I want to select average students . If the mean score on the qualifying test is 43.2 and the standard deviation is 8.6. find the score that cutt off the middle 50 percent of all scores

Answer 1

Answer:

The score that cutt off the middle 50 percent of all scores is 43.2

Explanations:

For a normal distribution, the normal value is the mean

It divides the curve in 50%-50%

Mean score, μ = 43.2

Standard Deviation, σ = 8.6

P(X≤a)=0.5

Use Z =(X-μ)/δ ~N(0;1)

b =(a-μ)/σ

P(Z≤b)=0.5

The value is b=0 (the mean for Z)

a=(b*σ)+μ

a=(0*8.6)+43.2

a=43.2

Therefore, the score that cutt off the middle 50 percent of all scores is 43.2