Class 11: Statistics – Exercise 32.3 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Question 1: Compute the mean deviation from the median of the following distribution:

Class	0-10	10-20	20-30	30-40	40-50
Frequency	5	10	20	5	10

Answer:

Calculation of Mean Deviation from Median

$\displaystyle \begin{array} {|c | c| c| c| c | c | } \hline \text{Class} & x_i & f_i & \text{Cum. Freq.} & |d_i| = |x_i - 25 | & f_i |d_i| \\ \hline 0-10 & 5 & 5 & 5 & 20 & 100 \\ \hline 10-20 & 15 & 10 & 15 & 10 & 100 \\ \hline 20-30 & 25 & 20 & 35 & 0 & 0 \\ \hline 30-40 & 35 & 5 & 40 & 10 & 50 \\ \hline 40-50 & 45 & 10 & 50 & 20 & 200 \\ \hline & & N = \sum f_i = 50 & & & \sum_{i=1}^{5} f_i |d_i|= 450 \\ \hline \end{array}$

$\displaystyle \text{Clearly, } N = 50 \Rightarrow \frac{N}{2} = \frac{50}{2} = 25$

$\displaystyle \text{The cumulative frequency just greater than } \frac{N}{2} = 25 \text{ is } 35.$

$\displaystyle \text{The corresponding value of class } \text{ is } 20-30$

$\displaystyle \text{Therefore the median class } \text{ is } 20-30$

$\displaystyle \therefore l = 20, \hspace{0.5cm} f = 20, \hspace{0.5cm} F = 15, \hspace{0.5cm} N = 50, \hspace{0.5cm} h = 10$

$\displaystyle \therefore \text{ Median } = l + \frac{\Big( \frac{N}{2} - F \Big) }{f} \times h$

$\displaystyle = 20 + \frac{\Big( \frac{50}{2} - 15 \Big) }{20} \times 10$

$\displaystyle = 20 + \frac{\Big( 25 - 15 \Big) }{20} \times 10$

$\displaystyle = 25$

$\displaystyle \text{Mean deviation from the median } = \frac{\sum_{i=1}^5 f_i|d_i|}{N} = \frac{450}{50} = 9$

$\\$

Question 2: Find the mean deviation from the mean from the following data:

(i)

Classes	0-100	100-200	200-300	300-400	400-500	500-600	600-700	700-800
Frequency	4	8	9	10	7	5	4	3

(ii)

Classes	95-105	105-115	115-125	125-135	135-145	145-155
Frequency	9	13	16	26	30	12

(iii)

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

Answer:

(i)

Calculation of Mean Deviation from Mean

$\displaystyle \begin{array} {|c | c| c| c| c | c | } \hline \text{Class} & x_i & f_i & f_ix_i & |x_i - \overline{X}| & f_i|x_i - \overline{X}| \\ & & & & = |x_i - 358 |& \\ \hline 0-100 & 50 & 4 & 200 & 308 & 1232 \\ \hline 100-200 & 150 & 8 & 1200 & 208 & 1664 \\ \hline 200-300 & 250 & 9 & 2250 & 108 & 972 \\ \hline 300-400 & 350 & 10 & 3500 & 8 & 80 \\ \hline 400-500 & 450 & 7 & 3150 & 92 & 644 \\ \hline 500-600 & 550 & 5 & 2750 & 192 & 960 \\ \hline 600-700 & 650 & 4 & 2600 & 292 & 1168 \\ \hline 700-800 & 750 & 3 & 2250 & 392 & 1176 \\ \hline & N= & & \sum_{i=1}^{8} f_ix_i & & \sum_{i=1}^{8} f_i |x_i - \overline{X}| \\ & \sum f_i = 50 & & = 17900 & & = 7896 \\ \hline \end{array}$

$\displaystyle \text{Clearly, } N = 50 \text{ and } \sum f_ix_i = 17900$

$\displaystyle \therefore \overline{X} = \frac{\sum f_ix_i}{N} = \frac{17900}{50} = 358$

From the above table we get

$\displaystyle \sum f_i|x_i- \overline{X}| = 7896 \text{ and } N = \sum f_i = 50$

$\displaystyle \therefore M.D. = \frac{1}{N} \sum f_i |x_1 - \overline{X}| = \frac{7896}{50} = 157.92$

$\\$

(ii)

Calculation of Mean Deviation from Mean

$\displaystyle \begin{array} {|c | c| c| c| c | c | } \hline \text{Class} & x_i & f_i & f_ix_i & |x_i - \overline{X}| & f_i|x_i - \overline{X}| \\ & & & & = |x_i - 128.58 | & \\ \hline 95-105 & 100 & 9 & 900 & 28.58 & 257.22 \\ \hline 105-115 & 110 & 13 & 1430 & 18.58 & 241.54 \\ \hline 115-125 & 120 & 16 & 1920 & 8.58 & 137.28 \\ \hline 125-135 & 130 & 26 & 3380 & 1.42 & 36.92 \\ \hline 135-145 & 140 & 30 & 4200 & 11.42 & 342.6 \\ \hline 145-155 & 150 & 12 & 1800 & 21.42 & 257.04 \\ \hline & N= & & \sum_{i=1}^{6} f_ix_i & & \sum_{i=1}^{6} f_i |x_i - \overline{X}| \\ & \sum f_i = 106 & & = 13630 & & = 1272.6 \\ \hline \end{array}$

$\displaystyle \text{Clearly, } N = 106 \text{ and } \sum f_ix_i = 13630$

$\displaystyle \therefore \overline{X} = \frac{\sum f_ix_i}{N} = \frac{13630}{106} = 128.58$

From the above table we get

$\displaystyle \sum f_i|x_i- \overline{X}| = 1272.6 \text{ and } N = \sum f_i = 106$

$\displaystyle \therefore M.D. = \frac{1}{N} \sum f_i |x_1 - \overline{X}| = \frac{1272.6}{106} = 12.005$

$\\$

(iii)

Calculation of Mean Deviation from Mean

$\displaystyle \begin{array} {|c | c| c| c| c | c | } \hline \text{Class} & x_i & f_i & f_ix_i & |x_i - \overline{X}| & f_i|x_i - \overline{X}| \\ & & & & = |x_i - 358 | & \\ \hline 0-10 & 5 & 6 & 30 & 22 & 132\\ \hline 10-20 & 15 & 8 & 120 & 12 & 96 \\ \hline 20-30 & 25 & 14 & 350 & 2 & 28\\ \hline 30-40 & 35 & 16 & 560 & 8 & 128 \\ \hline 40-50 & 45 & 4 & 180 & 18 & 72 \\ \hline 50-60 & 55 & 2 & 110 & 28 & 56 \\ \hline & N= & & \sum_{i=1}^{6} f_ix_i & & \sum_{i=1}^{6} f_i |x_i - \overline{X}| \\ & \sum f_i = 50 & & = 1350 & & = 512 \\ \hline \end{array}$

$\displaystyle \text{Clearly, } N = 50 \text{ and } \sum f_ix_i = 1350$

$\displaystyle \therefore \overline{X} = \frac{\sum f_ix_i}{N} = \frac{1350}{50} = 27$

From the above table we get

$\displaystyle \sum f_i|x_i- \overline{X}| = 512 \text{ and } N = \sum f_i = 27$

$\displaystyle \therefore M.D. = \frac{1}{N} \sum f_i |x_1 - \overline{X}| = \frac{512}{50} = 10.24$

$\\$

Question 3: Compute the mean deviation from the mean of the following distribution:

Class	10-20	20-30	30-40	40-50	50-60	60-70	70-80	80-90
No. of Students	8	10	15	25	20	18	9	5

Answer:

Calculation of Mean Deviation from Mean

$\displaystyle \begin{array} {|c | c| c| c| c | c | } \hline \text{Class} & x_i & f_i & f_ix_i & |x_i - \overline{X}| = |x_i - 49 | & f_i|x_i - \overline{X}| \\ \hline 10-20 & 15 & 8 & 120 & 34 & 272\\ \hline 20-30 & 25 & 10 & 250 & 24 & 240 \\ \hline 30-40 & 35 & 15 & 525 & 14 & 210 \\ \hline 40-50 & 45 & 25 & 1125 & 4 & 100 \\ \hline 50-60 & 55 & 20 & 1100 & 6 & 120 \\ \hline 60-70 & 65 & 18 & 1170 & 16 & 288 \\ \hline 70-80 & 75 & 9 & 675 & 26 & 234 \\ \hline 80-90 & 85 & 5 & 425 & 36 & 180 \\ \hline & N= & & \sum_{i=1}^{8} f_ix_i & & \sum_{i=1}^{8} f_i |x_i - \overline{X}| \\ & \sum f_i = 110 & & = 5390 & & = 1644 \\ \hline \end{array}$

$\displaystyle \text{Clearly, } N = 110 \text{ and } \sum f_ix_i = 5390$

$\displaystyle \therefore \overline{X} = \frac{\sum f_ix_i}{N} = \frac{5390}{110} = 49$

From the above table we get

$\displaystyle \sum f_i|x_i- \overline{X}| = 1644 \text{ and } N = \sum f_i = 110$

$\displaystyle \therefore M.D. = \frac{1}{N} \sum f_i |x_1 - \overline{X}| = \frac{1644}{110} = 14.95$

$\\$

Question 4: The age distribution of 100 life insurance policy holders is as follows:

Age ( on nearest birthday)	17-19.5	20-25.5	26-35.5	36-40.5	41-50.5	51-55.5	56-60.5	61-70.5
No. of Persons	5	16	12	26	14	12	6	5

Calculate the mean deviation from the median age.

Answer:

To make the function continuous, we need to subtract 0.25 from the lower limit and add 0.25 to the upper limit of the class.

Calculation of Mean Deviation from Median

$\displaystyle \begin{array} {|c | c| c| c| c | c | } \hline \text{Class} & x_i & f_i & \text{Cum. Freq.} & |d_i| & f_i |d_i| \\ & & & & = |x_i - 38.63 | & \\ \hline 16.75 - 19.75 & 18.25 & 5 & 5 & 20.38 & 101.9 \\ \hline 19.75 - 25.75 & 22.75 & 16 & 21 & 15.88 & 254.08 \\ \hline 25.75 - 35.75 & 30.75 & 12 & 33 & 7.88 & 94.56 \\ \hline 35.75 - 40.75 & 38.25 & 26 & 59 & 0.38 & 9.88 \\ \hline 40.75-50.75 & 45.75 & 14 & 73 & 7.12 & 99.68 \\ \hline 50.75-55.75 & 53.25 & 12 & 85 & 14.62 & 175.44 \\ \hline 55.75-60.75 & 58.25 & 6 & 91 & 19.62 & 117.72 \\ \hline 60.75-70.75 & 65.75 & 5 & 96 & 27.12 & 135.6 \\ \hline & & N = \sum f_i = 96 & & & \sum_{i=1}^{8} f_i |d_i|= 988.68 \\ \hline \end{array}$

$\displaystyle \text{Clearly, } N = 96 \Rightarrow \frac{N}{2} = \frac{96}{2} = 48$

$\displaystyle \text{The cumulative frequency just greater than } \frac{N}{2} = 48 \text{ is } 59.$

$\displaystyle \text{The corresponding value of class } \text{ is } 35.75-40.75$

$\displaystyle \text{Therefore the median class } \text{ is } 35.75-40.75$

$\displaystyle \therefore l = 35.75, \hspace{0.5cm} f = 26, \hspace{0.5cm} F = 33, \hspace{0.5cm} N = 96, \hspace{0.5cm} h = 5$

$\displaystyle \therefore \text{ Median } = l + \frac{\Big( \frac{N}{2} - F \Big) }{f} \times h$

$\displaystyle = 35.75 + \frac{\Big( \frac{96}{2} - 33 \Big) }{26} \times 5$

$\displaystyle = 35.75 + \frac{\Big( 48 - 33 \Big) }{26} \times 5$

$\displaystyle = 38.63$

$\displaystyle \text{Mean deviation from the median } = \frac{\sum_{i=1}^8 f_i|d_i|}{N} = \frac{988.86}{96} = 10.30$

$\\$

Question 5: Find the mean deviation from the mean and from median of the following distribution:

Marks	0-10	10-20	20-30	30-40	40-50
No. of Students	5	8	15	16	6

Answer:

Calculation of Mean Deviation from Median

$\displaystyle \begin{array} {|c | c| c| c| c | c | } \hline \text{Class} & x_i & f_i & \text{Cum. Freq.} & |d_i| = |x_i - 28 | & f_i |d_i| \\ \hline 0-10 & 5 & 5 & 5 & 23 & 115 \\ \hline 10-20 & 15 & 8 & 13 & 13 & 104 \\ \hline 20-30 & 25 & 15 & 28 & 3 & 45 \\\hline 30-40 & 35 & 16 & 44 & 7 & 112 \\ \hline 40-50 & 45 & 6 & 50 & 17 & 102 \\ \hline & & N = \sum f_i = 50 & & & \sum_{i=1}^{5} f_i |d_i|= 478 \\ \hline \end{array}$

$\displaystyle \text{Clearly, } N = 50 \Rightarrow \frac{N}{2} = \frac{50}{2} = 25$

$\displaystyle \text{The cumulative frequency just greater than } \frac{N}{2} = 25 \text{ is } 28.$

$\displaystyle \text{The corresponding value of class } \text{ is } 20-30$

$\displaystyle \text{Therefore the median class } \text{ is } 20-30$

$\displaystyle \therefore l = 20, \hspace{0.5cm} f = 15, \hspace{0.5cm} F = 13, \hspace{0.5cm} N = 50, \hspace{0.5cm} h = 10$

$\displaystyle \therefore \text{ Median } = l + \frac{\Big( \frac{N}{2} - F \Big) }{f} \times h$

$\displaystyle = 20 + \frac{\Big( \frac{50}{2} - 13 \Big) }{15} \times 10$

$\displaystyle = 20 + \frac{\Big( 25 - 13 \Big) }{15} \times 10$

$\displaystyle = 28$

$\displaystyle \text{Mean deviation from the median } = \frac{\sum_{i=1}^5 f_i|d_i|}{N} = \frac{478}{50} = 9.56$

Calculation of Mean Deviation from Mean

$\displaystyle \begin{array} {|c | c| c| c| c | c | } \hline \text{Class} & x_i & f_i & f_ix_i & |x_i - \overline{X}| = |x_i - 27 | & f_i|x_i - \overline{X}| \\ \hline 0-10 & 5 & 5 & 25 & 22 & 110\\ \hline 10-20 & 15 & 8 & 120 & 12 & 96\\ \hline 20-30 & 25 & 15 & 375 & 2 & 30 \\ \hline 30-40 & 35 & 16 & 560 & 8 & 128 \\ \hline 40-50 & 45 & 6 & 270 & 18 & 108 \\ \hline & N= & & \sum_{i=1}^{5} f_ix_i & & \sum_{i=1}^{5} f_i |x_i - \overline{X}| \\ & \sum f_i = 50 & & = 1350 & & = 472 \\ \hline \end{array}$

$\displaystyle \text{Clearly, } N = 50 \text{ and } \sum f_ix_i = 1350$

$\displaystyle \therefore \overline{X} = \frac{\sum f_ix_i}{N} = \frac{1350}{50} = 27$

From the above table we get

$\displaystyle \sum f_i|x_i- \overline{X}| = 472 \text{ and } N = \sum f_i = 50$

$\displaystyle \therefore M.D. = \frac{1}{N} \sum f_i |x_1 - \overline{X}| = \frac{472}{50} = 9.44$

$\\$

Question 6: Calculate mean deviation about median age for the age distribution of 100 persons given below:

Age	16-20	21-25	26-30	31-35	36-40	41-45	46-50	51-55
No. of Persons	5	6	12	14	26	12	16	9

Answer:

Since the function is not continuous, we subtract 0.5 from the lower limit of the class and add 0.5 to the upper limit of the class so that the class interval remains the same, while the function becomes continuous.

Calculation of Mean Deviation from Median

$\displaystyle \begin{array} {|c | c| c| c| c | c | } \hline \text{Class} & x_i & f_i & \text{Cum. Freq.} & |d_i| = |x_i - 38 | & f_i |d_i| \\ \hline 15.5-20.5 & 18 & 5 & 5 & 20 & 100 \\ \hline 20.5-25.5 & 23 & 6 & 11 & 15 & 90 \\ \hline 25.5-30.5 & 28 & 12 & 23 & 10 & 120 \\ \hline 30.5-35.5 & 33 & 14 & 37 & 5 & 70 \\ \hline 35.5-40.5 & 38 & 26 & 63 & 0 & 0 \\ \hline 40.5-45.5 & 43 & 12 & 75 & 5 & 60 \\ \hline 45.5-50.5 & 48 & 16 & 91 & 10 & 160 \\ \hline 50.5-55.5 & 53 & 9 & 100 & 15 & 135 \\ \hline & & N = \sum f_i = 100 & & & \sum_{i=1}^{8} f_i |d_i|= 735 \\ \hline \end{array}$

$\displaystyle \text{Clearly, } N = 100 \Rightarrow \frac{N}{2} = \frac{100}{2} =50$

$\displaystyle \text{The cumulative frequency just greater than } \frac{N}{2} = 50 \text{ is } 63.$

$\displaystyle \text{The corresponding value of class } \text{ is } 35.5-40.5$

$\displaystyle \text{Therefore the median class } \text{ is } 35.5-40.5$

$\displaystyle \therefore l = 35.5, \hspace{0.5cm} f = 26, \hspace{0.5cm} F = 37, \hspace{0.5cm} N = 100, \hspace{0.5cm} h = 5$

$\displaystyle \therefore \text{ Median } = l + \frac{\Big( \frac{N}{2} - F \Big) }{f} \times h$

$\displaystyle = 35.5 + \frac{\Big( \frac{100}{2} - 37 \Big) }{26} \times 5$

$\displaystyle = 35.5 + \frac{\Big( 50 - 37 \Big) }{26} \times 5$

$\displaystyle = 38$

$\displaystyle \text{Mean deviation from the median } = \frac{\sum_{i=1}^8 f_i|d_i|}{N} = \frac{735}{100} = 7.35$

Thus the mean deviation from the median age is 7.35 years

$\\$

Question 7: Calculate the mean deviation about the mean for the following frequency distribution:

Class Interval	0-4	4-8	8-12	12-16	16-20
Frequency	4	6	8	5	2

Answer:

In order to avoid tedious calculations of computing mean $\overline{X}$ , let us compute $\overline{X}$ by step-deviation method. The formula for the same is :

$\displaystyle \overline{X} = a + h \Bigg( \frac{1}{N} \sum_{i=1}^{n} f_id_i \Bigg), \text{ where } d_i = \frac{x_i - a}{h}, a = \text{ assumed mean and } \\ \\ h = \text{ common factor. }$

Let us take the assumed mean $a = 10$ and $h = 4$ and form the following table.

Calculation of Mean Deviation from Mean

$\displaystyle \begin{array} {|c |c | c| c| c| c | c | } \hline \text{Class} & x_i & f_i & d_i = \frac{x_i - 10}{4} & f_ix_i & |x_i - \overline{X}| & f_i|x_i - \overline{X}| \\ & & & & & = |x_i - 9.2 | & \\ \hline 0-4 & 2 & 4 & -2 & -8 & 7.2 & 28.8\\ \hline 4-8 & 6 & 6 & -1 & -6 & 3.2 & 19.2\\ \hline 8-12 & 10 & 8 & 0 & 0 & 0.8 & 6.4 \\ \hline 12-16 & 14 & 5 & 1 & 5 & 4.8 & 24 \\ \hline 16-20 & 18 & 2 &2 & 4 & 8.8 & 17.6 \\ \hline & & N= & & \sum_{i=1}^{5} f_ix_i & & \sum_{i=1}^{5} f_i |x_i - \overline{X}| \\ & & \sum f_i = 25 & & = -5 & & = 96 \\ \hline \end{array}$

$\displaystyle \text{Clearly, } N = 25 \text{ and } \sum f_ix_i = -5$

$\displaystyle \therefore \overline{X} = a + h \Bigg( \frac{1}{N} \sum_{i=1}^{n} f_id_i \Bigg)$

$\displaystyle = 10 + 4 \Bigg( \frac{1}{25} \times ( - 5) \Bigg) = 10 - \frac{20}{25} = 10 - 0.8 = 9.2$

From the above table we get

$\displaystyle \sum f_i|x_i- \overline{X}| = 96 \text{ and } N = \sum f_i = 25$

$\displaystyle \therefore M.D. = \frac{1}{N} \sum f_i |x_1 - \overline{X}| = \frac{96}{25} = 3.84$

$\\$

Question 8: Calculate the mean deviation from the median of the following data:

Class Interval	0-6	6-12	12-18	18-24	24-30
Frequency	4	5	3	6	2

Answer:

Calculation of Mean Deviation from Median

$\displaystyle \begin{array} {|c | c| c| c| c | c | } \hline \text{Class} & x_i & f_i & \text{Cum. Freq.} & |d_i| = |x_i - 14 | & f_i |d_i| \\ \hline 0-6 & 3 & 4 & 4 & 11 & 44 \\ \hline 6-12 & 9 & 5 & 9 & 5 & 25 \\ \hline 12-18 & 15 & 3 & 12 & 1 & 3 \\ \hline 18-24 & 21 & 6 & 18 & 7 & 42 \\ \hline 24-30 & 27 & 2 & 20 & 13 & 26 \\ \hline & & N = \sum f_i = 20 & & & \sum_{i=1}^{5} f_i |d_i|= 140 \\ \hline \end{array}$