Question

NEED HELP ASAP FOR 70PTS

Answer 1

Solution: We are given:

$\mu=515, \sigma=110$

(a) What percentage of standardized test scores is between 185 and 845​?

In order to find the percentage of scores that fall between 185 and 845, we use the z score formula first:

When x = 185, we have:

$z=\frac{x-\mu}{\sigma}$

        $=\frac{185-515}{110} =-3$

When x=845, we have:

$z=\frac{x-\mu}{\sigma}$

        $=\frac{845-515}{110} =3$

Therefore, we have to find $P(-3\leq z \leq 3)$.

From the empirical rule of normal distribution 99.7% of data falls within 3 standard deviation's from mean.

Therefore, 99.7% of standardized test scores is between 185 and 845​.

​(b) What percentage of standardized test scores is less than 185 or greater than 845​?

In order to find the percentage of scores that is less than 185 or greater than 845, we use the z score formula first:

When x = 185, we have:

$z=\frac{x-\mu}{\sigma}$

        $=\frac{185-515}{110} =-3$

When x=845, we have:

$z=\frac{x-\mu}{\sigma}$

        $=\frac{845-515}{110} =3$

Therefore, we have to find $P(z3)$.

From the empirical rule of normal distribution 0.15% of data falls 3 standard deviation's below mean and 0.15% of data falls 3 standard deviation's above mean.

Therefore $P(z3)$ = 0.15% +0.15% =0.3%

Therefore, 0.3% of standardized test scores is less than 185 or greater than 845.

​(c) What percentage of standardized test scores is greater than 735​?

Answer: In order to find the percentage of scores that is greater than 735, we use the z score formula first:

When x = 735, we have:

$z=\frac{x-\mu}{\sigma}$

        $=\frac{735-515}{110} =2$

Therefore, we have to find $P(z>2)$.

From the empirical rule of normal distribution 2.5% of data falls 2 standard deviation's above mean.

Therefore, 2.5% of standardized test scores is greater than 735.