Chapter 32: Statistics – Exercise 32.3

Statistics – Exercise – 32.3 – Q.1

We have to calculate mean deviation from the median. So, first we calculate the median.

CI	x	F	cf	d = (x – med)	fd
0 – 10	5	5	5	20	100
10 – 20	15	10	15	10	100
20 – 30	25	20	35	0	0
30 – 40	35	5	91	10	50
40 – 50	45	10	101	20	200
		50			450

 

Statistics – Exercise – 32.3 – Q.2(i)

CI	x	f	xf	d = (x-mean)	fd
0 – 100	50	4	200	308	1232
100 – 200	150	8	1200	208	1664
200 – 300	250	9	2250	108	972
300 – 400	350	10	3500	8	80
400 – 500	450	7	3150	92	644
500 – 600	550	5	2750	192	960
600 – 700	650	4	2600	292	1168
700 – 800	750	3	2250	392	1176
		50	17900		7896

 

Statistics – Exercise – 32.3 – Q.2(ii)

Classes	fi	xi	di	fidi
95 – 105	9	100	-3	-27	28.58	257.22
105 – 115	13	110	-2	-26	18.58	241.54
115 – 125	16	120	-1	-16	8.58	137.28
125 – 135	26	130	0	0	1.42	36.92
135 – 145	30	140	1	30	11.42	342.6
145 – 155	12	150	2	24	21.42	257.04
	N = 106			Total = -15		Total = 1272.60

 

Statistics – Exercise – 32.3 – Q.2(iii)

CI	x	f	xf	d = (x - mean)	fd
0 – 10	5	6	30	22	132
10 – 20	15	8	120	12	96
20 – 30	25	14	350	2	28
30 – 40	35	16	560	8	128
40 – 50	45	4	180	18	72
50 – 60	55	2	110	28	56
		50	1350		512
	Mean		27
	Deviation		10.24

 

Statistics – Exercise – 32.3 – Q.3

Find the mean deviation from the mean for the data:

Classes	0 -10	10 – 20	20 – 30	30 – 40	40 – 50	50 – 60
Frequencies	6	8	14	16	4	2

 

Statistics – Exercise – 32.3 – Q.4

We have to calculate mean deviation from the median. So, first we calculate the median.

CI	x	f	cf	D = (x – med)	fd
17 – 19.5	18.25	5	5	20	100
20 – 25.5	22.75	16	21	15.5	248
26 – 35.5	30.75	12	33	7.5	90
36 – 40.5	38.25	26	59	0	0
41 – 50.5	45.75	14	73	7.5	105
51 – 55.5	53.25	12	85	15	180
56 – 60.5	58.25	6	91	20	120
61 – 70.5	65.75	5	96	27.5	137.5
		96			980.5

We have N = 96 ⟹ N/2 = 48

The cumulative frequency just greater than N/2 is 59 and the corresponding value of x is 38.25.

Hence, median = 38.25

 

Statistics – Exercise – 32.3 – Q.5

M.D from Median

Marks	Students	xi	Cum. Freq		fidi
0 – 10	5	5	5	55/3	275/3
10 – 20	8	15	13	25/3	200/3
20 – 30	15	25	28	3-May	75/3
30 – 40	16	35	44	35/3	560/3
40 – 50	6	45	50	65/3	390/3
	N = 50				Total = 500

M.D from mean

Marks	Students	xi		fidi	|xi - 27|	fi|xi - 27|
0 – 10	5	5	-3	-15	22	110
10 – 20	8	15	-2	-16	12	96
20 – 30	15	25	-1	-15	2	30
30 – 40	16	35	0	0	8	128
40 – 50	6	45	1	6	18	108
	N = 50			Total = 40		Total = 472

 

Statistics – Exercise – 32.3 – Q.6

Converting the given data into continuous frequency distribution by sub tracing 0.5 from the lower limit and adding 0.5 to the upper limit of each class interval.

Age	x:	f:	Cumulative frequency	|d:|=|x:-38|	f:|d:|
15.5 – 20.5	18	5	5	20	100
20.5 – 25.5	23	6	11	15	90
25.5 – 30.5	28	12	23	10	120
30.5 – 35.5	33	14	37	5	70
35.5 – 40.5	38	26	63	0	0
40.5 – 45.5	43	12	75	5	60
45.5 – 50.5	48	16	91	10	160
50.5 – 55.5	53	9	100	15	135
		N = Σf: = 100			Σf:|d:| = 735

Clearly, N = 100 ⟹ N/2 = 50.

Cumulative frequency is just greater than N/2 is 63 and the corresponding class is 35.5 - 40.5.

I = 35.5, f = 26, h = 5, F = 37

 

Statistics – Exercise – 32.3 – Q.7

Classes	fi	xi	fixi	|xi - 9.2|	fi|xi - 9.2|
0 – 4	4	2	8	7.2	28.8
4 – 8	6	6	35	3.2	19.2
8 – 12	8	10	80	0.8	6.4
12 – 16	5	14	70	4.8	24
16 – 20	2	18	36	8.8	17.6
	N = 25		Total = 230		Total = 96.0

 

Statistics – Exercise – 32.3 – Q.8

Classes	fi	xi	fixi	|xi - 14.1|	fi|xi - 14.1|
0 – 6	4	3	12	11.1	44.4
6 – 12	5	9	45	5.1	25.5
12 – 18	3	15	45	0.9	2.7
18 – 24	6	21	126	6.9	41.4
24 – 30	2	27	54	12.9	25.8
	N = 20		Total = 282		Total = 139.8