# Measure of central Tendency Homework Help By Experienced Statisticians

If you need measure of central tendency homework help, there is no better platform that you can visit than Statistics Homework Helper. Students like you have sought the help of our experienced statistics homework tutors and gone ahead to secure decent grades in their projects. You too can join the list of students who ace their homework by taking our measure of central tendency homework help. Our tutors will put their best foot forward to make sure that impeccable solutions are sent your way before your deadline.

## Mean, Mode and Median

1. The following is a hypothetical list of exam scores. Calculate the mode, median, and mean, and discuss how these three scores compare to each other.
100
78
89
92
55
90
80
65
90
88
95
90
62
Solution
Mode
Mode = the most occurring value = 90
Median
Median = the middle number after arranging the value ascending or descending order
Median = 69
Mean
Mean= (∑▒x)/n=(100+78+89+92+55+90+80+65+90+88+95+90+62)/13
Mean=1074/13=82.6154
The mean is the average score of the student will the median is the 50% score of the student while the mode is the most occur scores of the student in the examination.
Hypothetical Frequency Distribution of Exam Scores.
a. Calculate the cumulative frequencies (Cf), proportions (Prop), percentages (Pctg), and cumulative percentages (Cpctg) and place them in the table (you can copy and paste the table into your answer box).
b. Calculate the mean of the distribution. Note: There are multiple frequencies for each score.

 Grade f Cf Prop Pctg Cpctg 100 2 2 0.0377 3.77 3.77 93 2 4 0.0377 3.77 7.55 90 4 8 0.0755 7.55 15.09 88 3 11 0.0566 5.66 20.75 87 7 18 0.1321 13.21 33.96 85 6 24 0.1132 11.32 45.28 79 8 32 0.1509 15.09 60.38 76 5 37 0.0943 9.43 69.81 75 7 44 0.1321 13.21 83.02 65 9 53 0.1698 16.98 100.00
N=53

## Deviation Score

3. Compute deviation scores for each of the hypothetical grades listed below. What is the sum of the deviation scores? Note: Remember that you must first calculate the mean.
100 16.8
78 -5.2
89 5.8
92 8.8
55 -28.2
90 6.8
80 -3.2
65 -18.2
88 4.8
95 11.8
Mean= (∑▒x)/n=(100+78+89+92+55+90+80+65+88+95)/10
Mean= 832/10=83.2
Sum of the deviation = Zero (0)
4.Using the grades listed in question #3, calculate the range, variance, and standard deviation of the grade distribution. Briefly explain why it is important to square deviation scores in calculating the variance and standard deviation.

 Grade (X−X‾) {(X -\overline{X})}" role="presentation" style="position: relative;">(X−Xˉ${(X -\overline{X})}$[Math Processing Error] (X−X‾)2 {(X -\overline{X})^2}" role="presentation" style="position: relative;">(X−Xˉ${(X -\overline{X})^2}$[Math Processing Error] 100 16.8 282.24 78 -5.2 27.04 89 5.8 33.64 92 8.8 77.44 55 -28.2 759.24 90 6.8 46.24 80 -3.2 10.24 65 -18.2 331.24 88 4.8 23.04 95 11.8 139.24

Range
Range = Largest value – smallest value
=100- 55 = 45
Variance
Variance = (∑▒(X-X ̅ )^2 )/(n-1)=1765.60/9=196.1778

## Standard deviation

Standard deviation = √((∑▒(X-X ̅ )^2 )/(n-1))=√196.1778=14.006
Briefly explain why it is important to square deviation scores in calculating the variance and standard deviation
The reason for squaring the deviation scores in calculating the variance and standard deviation score is because squaring the deviation score always gives a positive value and so that the sum will not be equals to zero. In this exercise, the sum of the deviation is zero, therefore there is need to find the square of the deviation in calculating the variance and the standard deviation.
Related Topics