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Algebra II : Basic Statistics

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Example Questions

Example Question #1 : Z Scores

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?

Possible Answers:

-2.475

-0.875

-0.75

-0.85

-1.143

Correct answer:

-0.875

Explanation:

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: -0.875

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Example Question #2 : Z Scores

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

Possible Answers:

$-\frac{25}{3}$

$-\frac{75}{47}$

$-\frac{25}{6}$

$-\frac{2}{3}$

Correct answer:

$-\frac{25}{3}$

Explanation:

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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Example Question #2 : Z Scores

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

Possible Answers:

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

Correct answer:

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:

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Example Question #11 : Z Scores

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

Possible Answers:

$-\frac{15}{64}$

$-\frac{15}{16}$

$\frac{1}{8}$

$\textup{There is not enough information to determine the z-score.}$

$-\frac{15}{8}$

Correct answer:

$-\frac{15}{8}$

Explanation:

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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Example Question #12 : Z Scores

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?

Possible Answers:

-3.75

Correct answer:

Explanation:

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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Example Question #13 : Z Scores

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?

Possible Answers:

-1.268

-0.574

-0.642

0.338

-0.538

Correct answer:

-0.538

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: -0.538

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Example Question #14 : Z Scores

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

Possible Answers:

1.55

1.25

0.75

1.33

Correct answer:

1.33

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.

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

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Example Question #1 : How To Find Direct Variation

varies directly with the square root of . If x = 25 , then y = 48 . What is the value of if x = 81 ?

Possible Answers:

491.52

86.4

155.52

$26\tfrac{2}{3}$

None of these answers are correct.

Correct answer:

86.4

Explanation:

If varies directly with the square root of , then for some constant of variation ,

$\small y = K \sqrt{x}$

If x = 25 , then y = 48 ; therefore, the equation becomes

$\small \small 48 = K \sqrt{25}$ ,

5K = 48 .

Divide by 5 to get K = 9.6 , making the equation

$\small \small y = 9.6 \sqrt{x}$ .

If x = 81 , then $\small \small \small y = 9.6 \sqrt{81} = 9.6 \cdot 9 = 86.4$ .

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Example Question #2 : Variable Relationships

If varies directly with $x^{3}$ and when x=5 , y=61.125 due to the effect of a constant, what is the value of when x=11 ?

Possible Answers:

y=43.331

y=650.859

y=134.475

y=1,331

y=672.375

Correct answer:

y=650.859

Explanation:

Since varies directly with $x^{3}$ , $y=Kx^{3}$ where is a constant.

1. Solve for when x=5 and y=61.125 .

$y=Kx^{3}$

$61.125=K(5^{3})$

$K=\frac{61.125}{125}=0.489$

2. Use your equation to solve for when x=11 .

$y=0.489x^{3}$

$y=0.489(11^{3})=650.859$

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Example Question #433 : Basic Statistics

If varies indirectly with $\sqrt{x}$ and when x=4 , y=12.3 due to the effect of a constant, what is the value of when x=17 ?

Possible Answers:

y=5.97

y=25.36

y=50.71

y=16.49

y=0.24

Correct answer:

y=5.97

Explanation:

Since varies indirectly with $\sqrt{x}$ , $y=\frac{K}{\sqrt{x}}$

1. Solve for when y=12.3 and x=4 .

$y=\frac{K}{\sqrt{x}}$

$12.3=\frac{K}{\sqrt{4}}$

$12.3= \frac{K}{2}$

K=24.6

2. Use the equation you found to solve for when x=17 .

$y=\frac{24.6}{\sqrt{x}}$

$y=\frac{24.6}{\sqrt{17}}=5.97$

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Algebra II : Basic Statistics

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #1 : Z Scores

Example Question #2 : Z Scores

Example Question #2 : Z Scores

Example Question #11 : Z Scores

Example Question #12 : Z Scores

Example Question #13 : Z Scores

Example Question #14 : Z Scores

Example Question #1 : How To Find Direct Variation

Example Question #2 : Variable Relationships

Example Question #433 : Basic Statistics

All Algebra II Resources

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