Question

The amount of time (minutes) spent at a Smart Centre by 40 shoppers are given below.

40	59	34	25	38	28	28	40	24	41
14	59	52	9	53	27	59	4	22	42
30	43	51	37	36	54	32	44	12	51
44	12	38	40	41	42	35	17	50	39

a. Complete the following frequency distribution for the data.

Class	Frequency
0 to < 10
10 to < 20
20 to < 30
30 to < 40
40 to < 50
50 to < 60
Total	40

b. What is the class midpoint for the fourth class?

c. What proportion of the shoppers spent less than 30 minutes at the Smart Centre?

d. Is the distribution symmetric, skewed left, or skewed right?

(click to select)Symmetric or Skewed Left or Skewed Right

Answer 1

(a)

The table is given by

Class interval	Class boundary	Midpoint	Frequency
0-9	0-9.5	4.5	2
10-19	9.5-19.5	14.5	4
20-29	19.5-29.5	24.5	6
30-39	29.5-39.5	34.5	9
40-49	39.5-49.5	44.5	10
50-59	49.5-59.5	54.5	9
Total	N/A	N/A	40

(b)

From the table, it is observed that the class midpoint of the 4th class is 34.5

(c)

Class interval	Class boundary	Midpoint	Frequency	Cumulative frequency()
0-9	0-9.5	4.5	2	2
10-19	9.5-19.5	14.5	4	6
20-29	19.5-29.5	24.5	6	12
30-39	29.5-39.5	34.5	9	21
40-49	39.5-49.5	44.5	10	31
50-59	49.5-59.5	54.5	9	40
Total	N/A	N/A	40	N/A

The cumulative frequency of less than type of the 3rd class is 12. Hence the proportion is given by

Hence the required proportion is 30%

(d)

We are to find the mean, median and the SD to determine the Pearson's coefficient of skewness.

Class interval	Class boundary	Midpoint(x)	Frequency(f)	xf
0-9	0-9.5	4.5	2	9
10-19	9.5-19.5	14.5	4	58
20-29	19.5-29.5	24.5	6	147
30-39	29.5-39.5	34.5	9	310.5
40-49	39.5-49.5	44.5	10	445
50-59	49.5-59.5	54.5	9	490.5
Total	N/A	N/A	40	1460

Hence the mean is given by

Here n=40; n/2=20; It lies in the 4th class(observe it from the table in (b))

The median is given by

As the median is greater than the mean, the distribution is skewed to the left.

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