Question

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

Answer 1

mean, m=82.4
standard deviation, sigma=3.3
Need proportion of sample between 75.8 and 89.

first calculate Z-scores of 75.8 and 89
Z(75.8)=(75.8-82.4)/3.3=-2
Z(89)=(89-82.4)/3.3=2

So
P(75.8<X<89)
=P(-2<Z<2)
=P(Z<2)-P(Z<-2)
=0.9772-0.0227
=0.9545

Answer 2

The percentage of the scores are between 75.8 and 89 can be calculated using z-scores.

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

Given:

The mean is $m=82.4$.

The standard deviation is $\sigma =3.3$.

Calculate the Z- score for 75.8.

$Z(75.8)=\dfrac{(75.8-82.4)}{3.3} \\Z(75.8) = -2$

Calculate the Z- score for 89.

$Z(89)=\dfrac{(89-82.4)}{3.3}\\Z(89)=2$

Calculate the percentage of the scores are between 75.8 and 89.

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

Refer the z-table, put the value,

$P(75.8<X<89)=0.9772-0.0227\\P(75.8<X<89)=0.9545$

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

Learn more about z-score here:

https://brainly.com/question/13299273