# 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
645 views

## 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

### Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Get Solutions

### Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Get Solutions

Find more solutions based on key concepts
Show solutions
In Exercises 516, evaluate the given quantity. log416

Finite Mathematics and Applied Calculus (MindTap Course List)

Find the limit (if possible): limx25x+2.

Calculus: An Applied Approach (MindTap Course List)

Add or Subtract Perform the addition or subtraction and simplify. 49. 1x2+1x2+x

Precalculus: Mathematics for Calculus (Standalone Book)

Fill in the blanks. 3.The distance between two points P1(a, b) and P2(c, d) is _________.

Applied Calculus for the Managerial, Life, and Social Sciences: A Brief Approach

Fill in each blank: 1g=mg

Elementary Technical Mathematics

Evaluate using partial fractions. 2 ln |x + 1| − 2 ln |x + 2| + C ln |x + 1| + 3 ln |x + 2| + C 3 ln |x + 1| −...

Study Guide for Stewart's Single Variable Calculus: Early Transcendentals, 8th

The normal plane to at t = 1 has equation:

Study Guide for Stewart's Multivariable Calculus, 8th

Explain how individual differences threaten the internal validity of a nonequivalent group design.

Research Methods for the Behavioral Sciences (MindTap Course List)

Is it Proportional? In Exercises S17 through S25, determine whether or not the relationship described is a prop...

Functions and Change: A Modeling Approach to College Algebra (MindTap Course List)

5. The distance from Potsdam to larger markets and limited air service have hindered the town in attracting new...

Modern Business Statistics with Microsoft Office Excel (with XLSTAT Education Edition Printed Access Card) (MindTap Course List)

Radioactive Decay Series The following system of differential equations is encountered in the study of the deca...

A First Course in Differential Equations with Modeling Applications (MindTap Course List)