CENTRAL TENDENCY AND STANDARD DEVIATION
Answer: CENTRAL TENDENCY AND STANDARD DEVIATION 5
Question 1
- Mean – is a measure of average for a given set of values. It is found by adding all the values then dividing by the number of values and is given by the formula;
Mean = ……where n is the number of values.
The mean sales for Presto Printing Company in February is obtained as follows:
Feb2 – feb10 = 4794+5954+3309+3106+7124+2349+3123+4128
=33887
Feb11 – feb19 = 3198+2198+7287+1323+4598+3987+3099+3098
=28788
Feb20 – feb28 = 5950+3209+6531+3098+4598+4873+9976+5878
= 44113
Hence, the mean =
= $4, 449.50 mean sales
- Median – is the most middle value in a list of values, obtained by first rewriting the list in a numerical order then finding the middle value depending on the average of the total number of values.
Let’s rearrange the values first;
1323, 2198, 2349, 3098, 3098, 3099, 3106, 3123, 3198, 3209, 3309, 3987, 4128, 4598, 4598, 4794, 4873, 5878, 5950, 5954, 6531, 7124, 7287, 9976
The median index is;
(24 + 1) / 2 = 12.5th number
Hence, the median value will be the 12th and 13th divided by 2;
Median =
= $4, 057.50 median sales
- Mode – is the value that occurs or is repeated most in a set of values.
From our arranged set of values, the occurrences of the values are;
1323 – 1 2198 – 1
2349 – 1 3098 – 2
3099 – 1 3106 – 1
3123 – 1 3198 – 1
3209 – 1 3309 – 1
3987 – 1 4128 – 1
4598 – 2 4794 – 1
4873 – 1 5878 – 1
5950 – 1 5954 – 1
6531 – 1 7124 – 1
7287 – 1 9976 – 1
It then appears then two values are most repeated; therefore, the set values have two modes which are: $4, 598 and $3, 098 of the monthly sales.
Question 2
- setting up a frequency distribution table and calculating the relative frequencies
Let’s begin by sorting the values in a numerical order;
60, 61, 69, 76, 79, 79, 79, 81, 85, 87, 87, 88, 92, 92, 98, 99, 100, 100, 100, 1006
The Relative frequencies are calculated by diving the frequency of a given interval then dividing by the number of values given in the set and they sum up-to 1 (Mertler, 2016).
Then using the frequency intervals given;
Scores interval | frequency | Relative frequency |
60 – 70 | 3 | 3/20 = 0.15 |
70 – 80 | 4 | 4/20 = 0.20 |
80 – 90 | 5 | 5/20 = 0.25 |
90 – 100 | 8 | 8/20 =0.40 |
Total | 20 | 1 |
- b) The variance and standard deviation of Matt’s quiz scores
Xi | µ | Xi – µ | (Xi – µ)2 |
60 | -25.6 | 655.36 | |
61 | -24.6 | 605.16 | |
69 | -16.6 | 275.56 | |
76 | -9.6 | 92.16 | |
79 | -6.6 | 43.56 | |
79 | -6.6 | 43.56 | |
79 | -6.6 | 43.56 | |
81 | -4.6 | 21.16 | |
85 | -0.6 | 0.36 | |
87 | 1.4 | 1.96 | |
87 | 1.4 | 1.96 | |
88 | 2.4 | 5.76 | |
92 | 6.4 | 40.96 | |
92 | 6.4 | 40.96 | |
98 | 12.4 | 153.76 | |
99 | 13.4 | 179.56 | |
100 | 14.4 | 207.36 | |
100 | 14.4 | 207.36 | |
100 | 14.4 | 207.36 | |
100 | 14.4 | 207.36 | |
Σ = 1712 | 1712/20 = 85.6 | Σ = 3034.8 |
We first begin by getting the mean, µ = ; where n is the number of scores
= 1712 and n = 20
Therefore, µ = = 85.6
Then we subtract the mean from each score to get Xi – µ and square the answers to get (Xi – µ)2 to make them all positive.
Summation of the squares, Σ (Xi – µ)2 will be : 3034.8
Therefore, the variance will be:
S2 = = = 159.7263
And the standard deviation is the square root of the variance,
Standard deviation, σ = = 12.6383
Question 3
Average daily sales =
= $5, 953.14
7.5% of the average weekly sales = = 446.4855
For the week; 446.4855 * 7 days = $3, 125.40
Therefore, it would be better for ben to earn a flat of $ 4, 000 for the week because the 7.5% amount ($ 3, 125.40) is much less.
References
Mertler, C. A., & Reinhart, R. V. (2016). Advanced and multivariate statistical methods: Practical application and interpretation. Routledge.
Question: CENTRAL TENDENCY AND STANDARD DEVIATION
Calculate and apply measures of central tendency and standard deviation to business applications. |
