Z-scores - Algebra II

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Question

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?

Answer

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.

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

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Question

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)?

Answer

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%.

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Question

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.

Answer

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:

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

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Question

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?

Answer

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 .

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Question

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?

Use the normal distribution table to calculate the probabilities. Round your answer to the nearest thousandth.

Answer

Let X represent the salaries of employees at XYZ Corporation.

We want to determine the probability that X is between 69,000 and 78,000:

To approximate this probability, we convert 69,000 and 78,000 to standardized values (z-scores).

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

$\frac{69000-60000}{7500}=1.2$

$\frac{78000-60000}{7500}=2.4$

We then want to determine the probability that z is between 1.2 and 2.4

The proportion of employees who earn between 69,000 and 78,000 is 0.107.

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Question

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?

Answer

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.

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

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Question

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

Answer

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.

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

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Question

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?

Answer

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 .

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

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

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Question

Suppose a student scored a on a test. The mean of the tests are , and the standard deviation is . What is the student's z-score?

Answer

Write the formula for z-score where is the data, $\mu$ is the population mean, and $\sigma$ is the population standard deviation.

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

Substitute the variables.

$z=\frac{65-72}{8} =- \frac{7}{8} = -0.875$

The z-score is:

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Question

Suppose Bob's test score is 50. Determine the z-score if the standard deviation is 3, and the mean is 75.

Answer

Write the formula for z-scores. This tells how many standard deviations above the below the mean.

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

Substitute the known values into the equation.

$z=\frac{50-75}{3} =-\frac{25}{3}$

The z-score is: $-\frac{25}{3}$

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Question

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

Answer

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:

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Question

Determine the z-score of a test result is 45, the mean is 60, and the standard deviation is 8.

Answer

Write the formula for z-scores. Z-scores determine the number of standard deviations below or above the mean.

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

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

$z=\frac{45-60}{8}= -\frac{15}{8}$

The z-score is $-\frac{15}{8}$ .

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Question

Suppose Billy scored a 54 on his exam. The mean of the exam grades is 66 and the standard deviation is 3. What is the z-score?

Answer

Write the formula for z-scores. Z scores determine how many standard deviations a score is above or below the mean.

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

Substitute the known values in order to determine the z-score.

$z=\frac{X-\mu}{\sigma} = \frac{54-66}{3} = -4$

The answer is:

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Question

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?

Answer

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:

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Question

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)?

Answer

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%.

Compare your answer with the correct one above

Question

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.

Answer

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:

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

Compare your answer with the correct one above

Question

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?

Answer

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 .

Compare your answer with the correct one above

Question

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?

Use the normal distribution table to calculate the probabilities. Round your answer to the nearest thousandth.

Answer

Let X represent the salaries of employees at XYZ Corporation.

We want to determine the probability that X is between 69,000 and 78,000:

To approximate this probability, we convert 69,000 and 78,000 to standardized values (z-scores).

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

$\frac{69000-60000}{7500}=1.2$

$\frac{78000-60000}{7500}=2.4$

We then want to determine the probability that z is between 1.2 and 2.4

The proportion of employees who earn between 69,000 and 78,000 is 0.107.

Compare your answer with the correct one above

Question

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?

Answer

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.

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

Compare your answer with the correct one above

Question

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

Answer

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.

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

Compare your answer with the correct one above

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