   Chapter 5, Problem 5.5P

Chapter
Section
Textbook Problem

At St. Algebra College, the 200 freshmen enrolled in introductory biology took a final exam on which their mean score was 72 and their standard deviation was 6. The following table presents the grades of 10 students. Convert each into a Z score and determine the number of people who scored higher or lower than each of the 10 students. (HINT: Multiply the appropriate proportion by N and round the result.) X i Z score Number of Students Above Number of students Below 60 57 55 67 70 72 78 82 90 95

To determine

To find:

The Z score, number of students above and below the score.

Explanation

Given:

The grades of 200 freshmen with a mean of 72 and standard deviation of 6 are-

60,57,55,67,70,72,78,82,90,95.

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 72 and standard deviation is 6.

For the score of 60,

Substitute 60 for Xi, 72 for X¯, and 6 for s in the above mentioned formula,

Z=60726=126=2

From the normal distribution table the area below 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 above the score of 2 is,

0.4772+0.5=0.9772

The number of students below 60 is,

200×2.28%=4.565

The number of students above 60 is,

200×97.72%=195.44195

For the score of 57,

Substitute 57 for Xi, 72 for X¯, and 6 for s in the above mentioned formula,

Z=57726=156=2.5

From the normal distribution table the area below the score of 2.5 is 0.0062.

The area between the mean and the score 2.5 is 0.4938.

Use the concept of symmetry, the area above the score of 2.5 is,

0.4938+0.5=0.9938

The number of students below 57 is,

200×0.0062=1.241

The number of students above 57 is,

200×0.9938=198.76199

For the score of 55,

Substitute 55 for Xi, 72 for X¯, and 6 for s in the above mentioned formula,

Z=55726=176=2.83

From the normal distribution table the area below the score of 2.83 is 0.0023.

The area between the mean and the score 2.83 is 0.4977.

Use the concept of symmetry, the area above the score of 2.83 is,

0.4977+0.5=0.9977

The number of students below 55 is,

200×0.0023=0.460

The number of students above 55 is,

200×0.9977=199.54200

For the score of 67,

Substitute 67 for Xi, 72 for X¯, and 6 for s in the above mentioned formula,

Z=67726=56=0.83

From the normal distribution table the area below the score of 0.83 is 0.2033.

The area between the mean and the score 0.83 is 0.2967.

Use the concept of symmetry, the area above the score of 0.83 is,

0.2967+0.5=0.7967

The number of students below 67 is,

200×0.2033=40.6641

The number of students above 67 is,

200×0.7967=159.34159

For the score of 70,

Substitute 70 for Xi, 72 for X¯, and 6 for s in the above mentioned formula,

Z=70726=26=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 number of students below 70 is,

200×0.3707=74

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