Answer to Question #157280 in Statistics and Probability for Vincent Miyato

Question

A client company of your firm is a horticultural shop selling a wide variety of products to its customers. The analysis of weekly sales of plants throughout the year is summarized in the following frequency distribution:

Weekly sales of plants ($'000) Number of weeks

12.55 but less than 12.80 9

12.80 but less than 13.05 19

13.05 but less than 13.30 10

13.30 but less than 13.55 8

13.55 but less than 13.80 6

Required:

(a) Construct a cumulative frequency distribution curve (5 marks)

(b) Calculate the mean,modal and median weekly sales of plants for the horticultural shop. (8 marks)

(c) Determine the variance and standard deviation of the weekly sales of plants for the horticultural shop. (6 marks)

(d) Calculate the coefficient of variation and coefficient of skewness for the weekly sales of plants for the horticultural shop. ( 6 marks)

(Total: 25 marks)

Answer 1

(a)

$$\begin{matrix} Class\ interval & Frequency & Cumulative\ frequency \\ 12.55\leq x<12.80 & 9 & 9 \\ 12.80\leq x<13.05& 19 & 28 \\ 13.05\leq x<13.30 & 10 & 38 \\ 13.30\leq x<13.55 & 8 & 46 \\ 13.55\leq x<13.80 & 6 & 52 \\ \end{matrix}$$

(b)

$$12.675(9)+12.925(19)+13.175(10)+13.425(8)$$

$$+13.675(6)=680.85$$

$$mean=\dfrac{680.85}{52}\approx13.093$$

26th and 27th values lie in the class $12.80\leq x<13.05$

$$median=12.80+\dfrac{\dfrac{52}{2}-9}{19}\times0.25\approx13.024$$

The modal class is $12.80\leq x<13.05$ as it contains the most values.

$$mode=12.80+\dfrac{19-9}{2\cdot19-9-10}\times0.25\approx12.932$$

(c)

$$(12.675)^2(9)+(12.925)^2(19)+(13.175)^2(10)$$

$$+(13.425)^2(8)+(13.675)^2(6)=8919.6425$$

$$s^2=\dfrac{1}{52-1}(8919.6425-\dfrac{(680.85)^2}{52})=0.0998$$

$$s=\sqrt{s^2}=\sqrt{0.0998}=0.3159$$

(d)

$$C.V.=\dfrac{s}{mean}\times100=\dfrac{0.3159}{13.093}\times100=24.127$$

$$Sk=\dfrac{mean-median}{3s}=\dfrac{13.093-13.024}{3(0.3159)}=0.0728$$