 # Scores on a quiz were normally distributed and had a mean of 10 and standard deviation of 3. For each of the following scores, find the Z score and the percentage of area above and below the score. X i Z score % Area Above % Area Below 5 6 7 8 9 11 12 14 15 16 18 ### Essentials Of Statistics

4th Edition
HEALEY + 1 other
Publisher: Cengage Learning,
ISBN: 9781305093836 ### Essentials Of Statistics

4th Edition
HEALEY + 1 other
Publisher: Cengage Learning,
ISBN: 9781305093836

#### Solutions

Chapter
Section
Chapter 5, Problem 5.1P
Textbook Problem
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## Scores on a quiz were normally distributed and had a mean of 10 and standard deviation of 3. For each of the following scores, find the Z score and the percentage of area above and below the score. X i Z score % Area Above % Area Below 5 6 7 8 9 11 12 14 15 16 18

Expert Solution
To determine

To find:

The Z score, percentage of area above and below the score.

Solution:

The Z score and the percentage of area above and below the score is given in the table below,

 Xi Z score % Area Above % Area Below 5 −1.67 95.25 4.75 6 −1.33 90.82 9.18 7 −1 84.13 15.87 8 −0.67 74.86 25.14 9 −0.33 62.93 37.07 11 0.33 37.07 62.93 12 0.67 25.14 74.86 14 1.33 9.18 90.82 15 1.67 4.75 95.25 16 2 2.28 97.72 18 2.67 0.38 99.62

### Explanation of Solution

Given:

The scores of the quiz are normally distributed with a mean of 10 and standard deviation of 3.

The scores of the quiz are-

5,6,7,8,9,11,12,14,15,16,18

Description:

The normal curve is symmetrical and its mean is equal to the median. The area below and above the mean is 50% or 0.05.

Formula used:

Let the data values be denoted by Xi.

The formula to calculate the Z score is given by,

Z=XiX¯s

Where, X¯ is the mean and s is the standard deviation of the distribution.

Calculation:

Given that the mean is 10 and standard deviation is 3.

For the score of 5,

Substitute 5 for Xi, 10 for X¯, and 3 for s in the above mentioned formula,

Z=5103=53=1.67

From the normal distribution table the area below the score of 1.67 is 0.0475.

The area between the mean and the score 1.67 is 0.4525.

Use the concept of symmetry, the area above the score of 1.67 is, 0.4525+0.5=0.9525

The percentage of area below and above the score of 1.67 is 4.75% and 95.25% respectively.

For the score of 6,

Substitute 6 for Xi, 10 for X¯, and 3 for s in the above mentioned formula,

Z=6103=43=1.33

From the normal distribution table the area below the score of 1.33 is 0.0918.

The area between the mean and the score 1.33 is 0.4082.

Use the concept of symmetry, the area above the score of 1.33 is, 0.4082+0.5=0.9082

The percentage of area below and above the score of 1.33 is 9.18% and 90.82%.

For the score of 7,

Substitute 7 for Xi, 10 for X¯, and 3 for s in the above mentioned formula,

Z=7103=33=1

From the normal distribution table the area below the score of 1 is 0.1587.

The area between the mean and the score 1 is 0.3413.

Use the concept of symmetry, the area above the score of 1 is, 0.3413+0.5=0.8413

The percentage of area below and above the score of 1 is 15.87% and 84.13%.

For the score of 8,

Substitute 8 for Xi, 10 for X¯, and 3 for s in the above mentioned formula,

Z=8103=23=0.67

From the normal distribution table the area below the score of 0.67 is 0.2514.

The area between the mean and the score 0.67 is 0.2486.

Use the concept of symmetry, the area above the score of 0.67 is, 0.2486+0.5=0.7486

The percentage of area below and above the score of 0.67 is 25.14% and 74.86%.

For the score of 9,

Substitute 9 for Xi, 10 for X¯, and 3 for s in the above mentioned formula,

Z=9103=13=0.33

From the normal distribution table the area below the score of 0.33 is 0.3707.

The area between the mean and the score 0.33 is 0.1293.

Use the concept of symmetry, the area above the score of 0.33 is, 0.1293+0.5=0.6293

The percentage of area below and above the score of 1 is 37.07% and 62.93%.

For the score of 11,

Substitute 11 for Xi, 10 for X¯, and 3 for s in the above mentioned formula,

Z=11103=13=0.33

From the normal distribution table the area above the score of 0.33 is 0.3707.

The area between the mean and the score 0.33 is 0.1293.

Use the concept of symmetry, the area below the score of 0.33 is, 0.1293+0.5=0.6293

The percentage of area above and below the score of 0.33 is 37.07% and 62.93%.

For the score of 12,

Substitute 12 for Xi, 10 for X¯, and 3 for s in the above mentioned formula,

Z=12103=23=0.67

From the normal distribution table the area above the score of 0.67 is 0.2514.

The area between the mean and the score 0.67 is 0.2486.

Use the concept of symmetry, the area below the score of 0.67 is, 0.2486+0.5=0.7486.

The percentage of area above and below the score of 0.67 is 25.14% and 74.86%.

For the score of 14,

Substitute 14 for Xi, 10 for X¯, and 3 for s in the above mentioned formula,

Z=14103=43=1.33

From the normal distribution table the area above the score of 1.33 is 0.0918.

The area between the mean and the score 1.33 is 0.4082.

Use the concept of symmetry, the area below the score of 1.33 is, 0.4082+0.5=0.9082.

The percentage of area above and below the score of 1.33 is 9.18% and 90.82%.

For the score of 15,

Substitute 15 for Xi, 10 for X¯, and 3 for s in the above mentioned formula,

Z=15103=53=1.67

From the normal distribution table the area above the score of 1.67 is 0.0475.

The area between the mean and the score 1.67 is 0.4525.

Use the concept of symmetry, the area below the score of 1.67 is, 0.4525+0.5=0.9525

The percentage of area above and below the score of 1.67 is 4.75% and 95.25% respectively.

For the score of 16,

Substitute 16 for Xi, 10 for X¯, and 3 for s in the above mentioned formula,

Z=16103=63=2

From the normal distribution table the area above the score of 2 is 0.0228.

The area between the mean and the score 2 is 0.4772.

Use the concept of symmetry, the area below the score of 2 is, 0.4772+0.5=0.9772.

The percentage of area above and below the score of 2 is 2.28% and 97.72% respectively.

For the score of 18,

Substitute 18 for Xi, 10 for X¯, and 3 for s in the above mentioned formula,

Z=18103=83=2.67

From the normal distribution table the area above the score of 2.67 is 0.0038.

The area between the mean and the score 2.67 is 0.4962.

Use the concept of symmetry, the area below the score of 2.67 is, 0.4962+0.5=0.9962.

The percentage of area above and below the score of 2.67 is 0.38% and 99.62% respectively.

Conclusion:

Therefore, the Z score and the percentage of area above and below the score is given in the table below,

 Xi Z score % Area Above % Area Below 5 −1.67 95.25 4.75 6 −1.33 90.82 9.18 7 −1 84.13 15.87 8 −0.67 74.86 25.14 9 −0.33 62.93 37.07 11 0.33 37.07 62.93 12 0.67 25.14 74.86 14 1.33 9.18 90.82 15 1.67 4.75 95.25 16 2 2.28 97.72 18 2.67 0.38 99.62

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