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 = 12.5^th number

Hence, the median value will be the 12^th and 13^th 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

X_i	µ	X_i– µ	(X_i– 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 X_i– µ and square the answers to get (X_i– µ)² to make them all positive.

Summation of the squares, Σ (X_i– µ)²will 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.