Question

urgent!!!! The scores of a high school entrance exam are approximately normally distributed with a given mean Mu = 82.4 and standard deviation Sigma = 3.3. What percentage of the scores are between 75.8 and 89?

68%

95%

99.7%

100%

Answer 1

Answer:95%

Step-by-step explanation:

Answer 2

The percentage of the scores between 75.8 and 89 is 95%.

What is the percentage of scores between 75.8 and 89 by using the Z-score formula?

The z-score formula is a formula that is applied to find the association between a value and a dataset of a mean.

$\mathbf{Z = \dfrac{x - \mu}{\sigma}}$

when;

• x = 75.8

$\mathbf{Z = \dfrac{75.8 - 82.4}{3.3}}$

Z = - 2

when;

• x = 82.4

$\mathbf{Z = \dfrac{89 - 82.4}{3.3}}$

Z = 2

Now, the percentage of the scores between 75.8 and 89 is:

P(75.8 < X < 89) = P(-2< Z< 2)

P(75.8 < X < 89) = P(Z< 2) - P(Z< -2)

Using the z-tables,

P(75.8 < X < 89) = 0.9772 - 0.0227

P(75.8 < X < 89) = 0.9545

P(75.8 < X < 89) ≅ 95%

Learn more about z-scores here:

https://brainly.com/question/25638875

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