College of Doctoral Studies
DBA-830:Module 2 Problem Set
Problem1:
Findthe mean, median, and mode for the scores in the following frequencydistribution table:
(Seepp. 99-100 for directions in your text book)
X f fx
8 1 8
7 4 28
6 2 12
5 2 10
4 2 8
3 1 3
Solution
X F FX Cum.FX
3 1 3 3
4 2 8 11
5 2 10 21
6 2 12 33
7 4 28 61
8 1 8 69
6 12 69
Mean=â fx/f =69/12 = 5.75
Median= âfx/f =69/12 = 5.75
Therefore,the median is at 6.
Modeis the number with the highest frequency 7 with 4 frequencies.
Problem2:
Compute the variance and standard deviation for the following data:
3, 4, 4, 5, 5, 6, 6, 7, 7, 7, 7, 8 These are data are for a population.
Mean,X = 3+4+4+5+5+6+6+7+7+7+7+8/12 = 69/12 = 5.75
X (X-M) (X-M)2
3 -2.75 7.5625
4 -1.75 3.0625
4 -1.75 3.0625
5 -0.75 0.5625
5 -0.75 0.5625
6 0.25 0.0625
6 0.25 0.0625
7 1.25 1.5625
7 1.25 1.5625
7 1.25 1.5625
7 1.25 1.5625
8 2.25 5.0625
26.25
Variance,S2 = 26.25/12 = 2.1875
Std.Dev., S = â2.1875 = 1.4790
Compute the variance and standard deviation for the same data, except now consider that the data are for a sample from a population.
(Seepp. 109 and 116for directions intextbook)
X (X-M) (X-M)2
3 -2.75 7.5625
4 -1.75 3.0625
4 -1.75 3.0625
5 -0.75 0.5625
5 -0.75 0.5625
6 0.25 0.0625
6 0.25 0.0625
7 1.25 1.5625
7 1.25 1.5625
7 1.25 1.5625
7 1.25 1.5625
8 2.25 5.0625
26.25
Variance,S2 = 26.25/12-1 = 26.25/11 = 2.3864
Std.Dev., S = â2.3864 = 1.5448
Problem3:
Fourstudents each received a raw score of 22 on a different test. Computethe z-score for the raw score value of 22 for each student, giveninformation about the distributions of X values for each test. Whichstudent (a, b, c, or d) had the highest score value, compared to thenormative group who took the same test? Which student did the poorestin this group, compared to the normative group who took the sametest? Why?
X = 22, when M = 20, SD = 5, Z = ?
X = 22, when M = 30, SD = 5, Z = ?
X = 22, when M = 16, SD = 2, Z = ?
X = 22, when M = 10, SD = 8, Z = ?
(Seep. 160 in textbook)
Z = X-M/SD = (22-20)/5 = 2/5 = 0.4
Z = X-M/SD = (22-30)/5 = -8/5 = -1.6
Z = X-M/SD = (22-16)/2 = 6/2 = 3.0
Z = X-M/SD = (22-10)/8 = 12/8 = 1.5
StudentC had the highest score of the four students because his score was 3standard deviations above the mean while on the other hand, student Bperformance was the poorest as he was 1.6 standard deviations belowthe mean. That meant that compared to the mean of the class, studentC performed three times as the class average of 16 while student Bperformed 1.6 times less than the class average of 30.