On a standardized aptitude test, scores are normally distributed with a mean of 100 and a standard deviation of 10. Find the percent of scores that are: a) Between 90 and 100

34.134%

Step-by-step explanation:

Given :

Normal distribution ;

Mean, m = 100 ; Standard deviation, s = 10

Percentage score between 90 and 100

P(90 ≤ x ≤ 100)

Recall : Zscore = (x - m) / s

P[(90 - 100) / 10 ; (100 - 100) / 10]

P[(10/10 ; 0 / 10)]

P(Z ≤ 1) - P(Z≤0) (Using Z probability calculator)

0.84134 - 0.5

= 0.34134

= 0.34134 * 100%

= 34.134%

4.

step-by-step explanation:

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