# Frequency Distribution, Mean, Median

## Assignment on Frequency distribution, Mean, Median, and Standard Deviation

PADM 6700Homework                                                                                         Chapters 4,5, and 6                                                              Summer 2021
Homework assignments are designed to reinforce the course material covered in the class lectures and the textbook. To ensure students are grasping what is being covered, homework assignments are given for each chapter covered in the course lecture. All homework points are extra points. Late homework is not accepted unless the assignments are accompanied by documentation or a situation preventing the submission of homework was discussed before the assignment's due date.
To receive credit all responses must be circled or clearly indicated. Chapters 4-5 are due on Tuesday,

#### Question 1 (2 points)

A doctor’s office staff studied the waiting times for patients who arrive at the office with a request for emergency service. The following data with waiting times in minutes were collected over one month

 2 5 10 12 4 4 5 17 11 8 12 21 6 21 6 7 13 18 3

Use classes of 0-4, 5-9, and so on the following:

a. Show the frequency distribution.

0-4 = 4

5-9 = 8

10 – 14 = 5

15 – 19 = 2

20 + = 1

b. Show the relative frequency distribution.

 Bin Frequency Rel frequency 0-4 4 0.2 5-9 8 0.4 10-14 5 0.25 15-19 2 0.1 20+ 1 0.05 Total 20 1

c. Show the cumulative frequency distribution.

 Bin Frequency Cumfrequency 0-4 4 4 5-9 8 12 10-14 5 17 15-19 2 19 20+ 1 20 Total 20

d. Show the cumulative relative frequency distribution.

 Bin Frequency Rel frequency CumRelfrequency 0-4 4 0.2 0.2 5-9 8 0.4 0.6 10-14 5 0.25 0.85 15-19 2 0.1 0.95 20+ 1 0.05 1 Total 20 1

e. What proportion of the patients needing emergency service wait nine (9) minutes or less?

The cumulative relative frequency for 5-9 shows the patients needing emergency service wait nine (9) minutes or lessi.e 0.6. Thus, patients needing emergency service wait nine (9) minutes or less is 60%.

Question 2 5.2 Page 104 (1 Point)

27+31+33+35+40+64+65 = 295/7 = 42.142 (Mean)

Median = (n+1)/2 th value = 4th value

Which is 35 (Median)

The department should send the median age.

Question 3 5.4 Page 104 (1Point)

26+35+51+63+147= 322/5= 64.4 (Mean)

Median = (n+1)/2 th value = 3rd value

Which is 51 (Median)

The army should report the median.

Question 4 5.12 Pages 106 – 107 (2 Points)

a) The level of measurement are on ordinal scale.

b) The percentage distribution is as follows:

 Rating Frequency % To a very great extent 42 0.3784 To a great extent 31 0.2793 To a moderate extent 19 0.1712 To some extent 12 0.1081 Not at all 7 0.0631 Total 111 1

Question 5 5.14 Page 108 (2 Points)

a) The level of measurement is on an ordinal scale.
b) The percentage distribution is as follows:
 Rating Frequency % Far above expectations 15 0.1145 Above expectations 22 0.1679 Meets expectations 77 0.5878 Below expectations 9 0.0687 Below expectations 8 0.0611 Total 131 1

Measures of central tendency:

Median = (n+1)/2 th value = 66th value

The 66th value is “meets expectations” and is the median.

c) Since about 50% of the employees have rated “meets expectations”, “below expectations” or “far below expectations”. Thus, the feeling of the head of the Department that at least 90% of them fall into the top two categories is not correct.

Question 6 6.2 Page 118 (1 point)

Median: Since n is even, the formula for the median is:

Average of n/2 th value and (n/2)+th value

Since n=14, median is average of 7th and 8th value

Arranging the data in ascending order,

7th value is 7

And 8th value is 8

Thus, the median is an average of 7 and 8 which is 7.5

Mean = Σxi/ n = 99/14 = 7.07

Standard deviation= Xi Xi-Xmean (Xi-Xmean)^2 0 -7.07 49.9849 3 -4.07 j 3 -4.07 16.5649 4 -3.07 9.4249 5 -2.07 4.2849 6 -1.07 1.1449 7 -0.07 0.0049 8 0.93 0.8649 9 1.93 3.7249 9 1.93 3.7249 10 2.93 8.5849 11 3.93 15.4449 12 1.93 24.3049 12 4.93 24.3049 Σ=99 Σ=178.9286
Question 7 6.4 Page 118 (2 Points)

Question 8 6.10 Page 118 (2 Points)

Mean = Σxi/ n = 595/10 = 59.5

Median = Since n is even, the formula for the median is:

Average of n/2 the value and (n/2)+th value

Since n=10, median is average of 5th and 6th value

Arranging the data in ascending order,

7th value is 40

And 8th value is 75
Thus, the median is an average of 40 and 75 which is 57.5

Standard deviation = Xi Xi-Xmean (Xi-Xmean)^2 75 15.5 240.25 20 -39.5 1560.25 15 -44.5 1980.25 95 35.5 1260.25 30 -29.5 870.25 100 40.5 1640.25 40 -19.5 380.25 10 --49.5 2450.25 90 30.5 930.25 120 60.5 3660.25 595 0 14972.5

Range = Maximum-Minimum = 120-10 = 110