CENTRAL TENDENCY AND STANDARD DEVIATION

CENTRAL TENDENCY AND STANDARD DEVIATION

Answer:  CENTRAL TENDENCY AND STANDARD DEVIATION 5

Question 1

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

1. 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

1. 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

1. 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

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.
Scroll to Top