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#2 Find the mean of the data summarized in the given frequency distribution. Com

ID: 3045544 • Letter: #

Question

#2

Find the mean of the data summarized in the given frequency distribution. Compare the computed mean to the actual mean of 51.1 degrees. Low Temperature o Frequency 40-44 45-49 50-54 55-59 60-64 degrees. The mean of the frequency distribution is Round to the nearest tenth as needed.) Which of the following best describes the relationship between the computed mean and the actual mean? A. B. O c. ( D. The computed mean is close to the actual mean because the difference between the means is more than 5%. The computed mean is close to the actual mean because the difference between the means is less than 5%. The computed mean is not close to the actual mean because the difference between the means is less than 5% The computed mean is not close to the actual mean because the difference between the means is more than 5%.

Explanation / Answer

Solution:

Here, we have to find mean for frequency distribution.

Mean = XF / F

Where, X is midpoint of the class and F is the frequency of class.

Calculations are given as below:

Midpoint X

F

X*F

42

3

126

47

7

329

52

12

624

57

5

285

62

1

62

Total

28

1426

Mean = XF / F

Mean = 1426/28 = 50.92857

Mean = 50.9

The mean of the frequency distribution is 50.9 degrees.

Which of the following best describes the relationship between the computed mean and the actual mean?

Answer: B. The computed mean is close to the actual mean because the difference between the means is less than 5%.

Question 2

Here, we have to compute range, variance, and standard deviation for the sample data.

Range = Maximum – minimum

Variance = (X - mean)^2/(n – 1)

SD = sqrt(Variance)

Table for calculations is given as below:

No.

X

(X - mean)

(X - mean)^2

1

148

7.5

56.25

2

147

6.5

42.25

3

148

7.5

56.25

4

140

-0.5

0.25

5

136

-4.5

20.25

6

124

-16.5

272.25

Total

843

0

447.5

Mean

140.5

Maximum

148

Minimum

124

Range

24

Variance

89.5

SD

9.460443964

Range = 148 – 124 = 24 mmHg

Sample variance = 447.5/(6 - 1) = 447.5/5 = 89.5 mmHg^2

Sample standard deviation = sqrt(89.5) = 9.5 mmHg

What should be the value of the standard deviation?

Answer: D. There is no way to tell what the standard deviation should be.

Midpoint X

F

X*F

42

3

126

47

7

329

52

12

624

57

5

285

62

1

62

Total

28

1426