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Nady [450]
2 years ago
14

The scores of a high school entrance exam are approximately normally distributed with a given mean mu = 82.4 and standard deviat

ion sigma = 3.3. what percentage of the scores are between 75.8 and 89? 68% 95% 99.7% 100%
Mathematics
1 answer:
Slav-nsk [51]2 years ago
6 0

The percentage of the scores that are between 75.8 and 89 is; 95%

<h3>How to find the percentage from z-score?</h3>

We are given;

Population mean; μ

Standard deviation; σ = 3.3

Thus;

z-score for a mean score of 75.8 is;

z = (75.8 - 82.4)/3.3

z = -2

z-score for a mean score of 89 is;

z = (89 - 82.4)/3.3

z = 2

From online p-value from z-score calculator, the p-value between both z-scores is;

p-value = 0.9545 = 95.45%

Approximating to the nearest percent = 95%

Read more about z-score at; brainly.com/question/25638875

#SPJ1

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Let \left(1+\frac{1}{\tan^{2}A} \right)\cdot \left(1+\frac{1}{\cot^{2}A} \right), we proceed to prove the trigonometric expression by trigonometric identity:

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