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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

2510124451711812216
21671318
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.

BinFrequencyRel frequency
0-440.2
5-980.4
10-1450.25
15-1920.1
20+10.05
Total201

c. Show the cumulative frequency distribution.

BinFrequencyCumfrequency
0-444
5-9812
10-14517
15-19219
20+120
Total20

d. Show the cumulative relative frequency distribution.

BinFrequencyRel frequencyCumRelfrequency
0-440.20.2
5-980.40.6
10-1450.250.85
15-1920.10.95
20+10.051
Total201

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:

RatingFrequency%
To a very great extent420.3784
To a great extent310.2793
To a moderate extent190.1712
To some extent120.1081
Not at all70.0631
Total1111

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:
RatingFrequency%
Far above expectations150.1145
Above expectations220.1679
Meets expectations770.5878
Below expectations90.0687
Below expectations80.0611
Total1311

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=

XiXi-Xmean(Xi-Xmean)^2
0-7.0749.9849
3-4.07j
3-4.0716.5649
4-3.079.4249
5-2.074.2849
6-1.071.1449
7-0.070.0049
80.930.8649
91.933.7249
91.933.7249
102.938.5849
113.9315.4449
121.9324.3049
124.9324.3049
Σ=99Σ=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 =

XiXi-Xmean(Xi-Xmean)^2
7515.5240.25
20-39.51560.25
15-44.51980.25
9535.51260.25
30-29.5870.25
10040.51640.25
40-19.5 380.25
10--49.52450.25
9030.5930.25
12060.53660.25
595014972.5

Range = Maximum-Minimum = 120-10 = 110