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)^{2 }will be : 3034.8

Therefore, the **variance** will be:

S^{2 }= = = **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. |