Z-scores

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Algebra II › Z-scores

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Your teacher tells you that the mean score for a test was a and that the standard deviation was for your class.

You are given that the -score for your test was . What did you score on your test?

CORRECT

Explanation

The formula for a z-score is

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

where $\mu$ = mean and $\sigma$ = standard deviation and =your test grade.

Plugging in your z-score, mean, and standard deviation that was originally given in the question we get the following.

$2.5=\frac{x-70}{7}$

Now to find the grade you got on the test we will solve for .

$2.5=\frac{x-70}{7}$

$2.5\cdot 7={x-70}$

The mean grade on a science test was 79 and there was a standard deviation of 6. If your sister scored an 88, what is her z-score?

CORRECT

Explanation

Use the formula for z-score:

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

Where is her test score, $\mu$ is the mean, and $\sigma$ is the standard deviation.

$z=\frac{88-79}{6}=1.5$

On a statistics exam, the mean score was and there was a standard deviation of . If a student's actual score of , what is his/her z-score?

CORRECT

Explanation

The z-score is a measure of an actual score's distance from the mean in terms of the standard deviation. The formula is:

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

Where $\large \mu, \sigma$ are the mean and standard deviation, respectively. is the actual score.

If we plug in the values we have from the original problem we have

${z = \frac{92-80}{7}}$

which is approximately .

A large group of test scores is normally distributed with mean 78.2 and standard deviation 4.3. What percent of the students scored 85 or better (nearest whole percent)?

CORRECT

Explanation

If the mean of a normally distributed set of scores is $\mu = 78.2$ and the standard deviation is $\sigma = 4.3$ , then the -score corresponding to a test score of is

$\frac{X-\mu}{\sigma} =\frac{85-78.2}{4.3} = \frac{6.8}{4.3} \approx 1.58$

From a -score table, in a normal distribution,

We want the percent of students whose test score is 85 or better, so we want $P (z\geq 1.58)$ . This is

$P (z\geq 1.58) = 1- P (z< 1.58) = 1- 0.9429 = 0.0571$

or about 5.7 % The correct choice is 6%.

Sarah scored an 8.5 out of ten on her gymnastics floor routine. If the mean of the scores is 9.2 and the standard deviation is 1.3, what is her z-score?

CORRECT

Explanation

Write the formula for z-scores. Z-scores are indicators of how many standard deviations above or below the mean.

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

$x=\textup{score}$

$\mu = \textup{mean}$

$\sigma =\textup{ standard deviation}$

Substitute the known values.

$z=\frac{8.5-9.2}{1.3} = \frac{-0.7}{1.3} = -0.538$

The answer is:

Find the z-score if the result of a test score is 6, the mean is 8, the standard deviation is 2.

CORRECT

$\textup{The answer is not given.}$

Explanation

Write the formula to determine the z-scores.

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

Substitute all the known values into the formula to determine the z-score.

$z=\frac{6-8}{2}$

Simplify this equation.

$z=\frac{-2}{2} = -1$

The answer is:

In a normal distribution, if the mean score is 8 in a gymnastics competition and the student scores a 9.3, what is the z-score if the standard deviation is 2.5?

CORRECT

Explanation

Write the formula to find the z-score. Z-scores are defined as the number of standard deviations from the given mean.

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

Substitute the values into the formula and solve for the z-score.

$z=\frac{x-\mu}{\sigma } = \frac{9.3-8}{2.5} =0.52$

You just took your ACT. The mean score was a with a standard deviation of . If you scored a , what is your z-score?

CORRECT

Explanation

Use the formula for z-score:

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

Where is your score, $\mu$ is the mean, and $\sigma$ is the standard deviation.

$z=\frac{26-22}{3}=1.33$

A distributor manufactures a product that has an average weight of pounds.

If the standard deviation is pounds, determine the z-score of a product that has a weight of pounds.

CORRECT

$-\frac{1}{2}$

Explanation

The z-score can be expressed as

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

where

$\mu= \text{average }=20$

$\sigma= \text{standard: deviation}=2$

Therefore the z-score is:

$z=\frac{x-\mu }{\sigma }=\frac{18-20}{2}=\frac{-2}{2}=-1$

The salaries of employees at XYZ Corporation follow a normal distribution with mean 60,000 and standard deviation 7,500. What proportion of employees earn approximately between 69,000 and 78,000?

Normal-distribution