Question

a) The following data relates to daily bill (in Kshs) on consumption of a certain commodity for 60 households

Daily bills (Kshs) No. of households
10 - 20 6
20 - 30 7
30 - 40 11
40 - 50 10
50 - 60 6
60 - 70 5
70 - 80 9
80 - 90 3
90 - 100 3


i) Calculate the mean
ii) Calculate the median
iii) Calculate the mode
iv) Calculate the standard deviation
v) Calculate the coefficient of skewness
vi) Comment on the skewness of this distribution
Calculate the coefficient of variation

Answer 1

Answer on Question #63447 – Math – Statistics and Probability

The following data relates to daily bill (in Kshs) on consumption of a certain commodity for 60 households

Daily bills (Kshs) No. of households

i) Calculate the mean

ii) Calculate the median

iii) Calculate the mode

iv) Calculate the standard deviation

v) Calculate the coefficient of skewness

vi) Comment on the skewness of this distribution

vii) Calculate the coefficient of variation

Solution (i)

The mean is

$$\begin{array}{l} \bar{x} = \frac{\sum_{i=1}^{n} x_i f_i}{N} \\ = \frac{15 \cdot 6 + 25 \cdot 7 + 35 \cdot 11 + 45 \cdot 10 + 55 \cdot 6 + 65 \cdot 5 + 75 \cdot 9 + 85 \cdot 3 + 95 \cdot 3}{60} = 49.5 \end{array}$$

Answer (i): The mean is $\bar{x} = 49.5$

Solution (ii)

The median is in the class where the cumulative frequency reaches half the sum of the absolute frequencies. That is to say, the median is within the class $\frac{N}{2}$

$$M_e = L_i + \frac{\frac{N}{2} - F_{i-1}}{f_i} a_i$$

$L_i$ is the lower limit of the median class.

$\frac{N}{2}$ is half the sum of the absolute frequency.

$F_{i-1}$ is the absolute frequency immediately below the median class.

$a_{i}$ is the width of the class containing the median class.

$$a _ {i} = 1 0$$ $$\frac {N}{2} = \frac {6 0}{2} = 3 0$$

Median class: 40-50

$$L _ {i} = 4 0$$ $$f _ {i} = 1 0$$ $$F _ {i - 1} = 2 4$$ $$M _ {e} = 4 0 + \frac {3 0 - 2 4}{1 0} \cdot 1 0 = 4 0 + 6 = 4 6$$

Answer (ii): The median is $M_e = 46$ .

Solution (iii)

The mode is the most repeated value in a distribution.

Mode for Grouped Data Formula

$$M _ {o} = L _ {i} + \frac {f _ {i} - f _ {i - 1}}{\left(f _ {i} - f _ {i - 1}\right) + \left(f _ {i} - f _ {i + 1}\right)} \cdot a _ {i}$$

$L_{i}$ is the lower limit of the modal class.

$f_{i}$ is the absolute frequency of the modal class.

$f_{i-1}$ is the absolute frequency immediately below the modal class.

$f_{i + 1}$ is the absolute frequency immediately after the modal class.

$a_{i}$ is the width of the class containing the modal class.

10-20 6

20-30 7

30-40 11

40-50 10

50-60 6

60-70 5

70-80 9

80-90 3

90-100 3

Modal class 30-40

$$\begin{array}{l} M _ {o} = L _ {i} + \frac {f _ {i} - f _ {i - 1}}{\left(f _ {i} - f _ {i - 1}\right) + \left(f _ {i} - f _ {i + 1}\right)} \cdot a _ {i} = 3 0 + \frac {1 1 - 7}{(1 1 - 7) + (1 1 - 1 0)} \cdot 1 0 = \\ = 3 0 + \frac {4}{5} \cdot 1 0 = 3 8 \\ \end{array}$$

Answer(iii): the mode is $M_{o} = 38$ .

Solution (iv)

To calculate the standard deviation,

use the following formula:

$$\sigma = \sqrt {\frac {\sum_ {i = 1} ^ {n} x _ {i} ^ {2} f _ {i}}{N} - \bar {x} ^ {2}}$$$$\sigma = \sqrt {\frac {\sum_ {i = 1} ^ {n} x _ {i} ^ {2} f _ {i}}{N} - \bar {x} ^ {2}} = \sqrt {2 9 6 8 . 3 3 3 - (4 9 . 5) ^ {2}} = 2 2. 7 6$$

Answer: The standard deviation is $\sigma = 22.76$

Solution (v)

The coefficient of skewness we calculate using the following formula

$$K_s = \frac{\mu_3}{\sigma^3} = \frac{\sum_{i=1}^{n} (x_i - \bar{x})^3 f_i}{N \sigma^3}$$

$\sigma = 22.76$ and $\bar{x} = 49.5$ we found before, find the sum

So $K_s = \frac{\mu_3}{\sigma^3} = \frac{\sum_{i=1}^{n} (x_i - \bar{x})^3 f_i}{N \sigma^3} = 0.285$.

Answer (v): the coefficient of skewness is $K_s = 0.285$

Solution (vi)

Positive skew: The right tail is longer; the mass of the distribution is concentrated on the left of the figure. The distribution is said to be right-skewed, right-tailed, or skewed to the right.

Answer (vi): right-skewed.

Solution (vii)

The coefficient of variation is defined as the ratio of the standard deviation $\sigma$ to the mean $\bar{x}$

$$c_v = \frac{\sigma}{\bar{x}} = \frac{22.76}{49.5} = 0.46$$

Answer (vii): The coefficient of variation is $c_v = 0.46$.

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