Class 11 RD Sharma Solutions – Chapter 32 Statistics – Exercise 32.1
Question 1. Calculate the mean deviation about the median of the following observation :
(i) 3011, 2780, 3020, 2354, 3541, 4150, 5000
(ii) 38, 70, 48, 34, 42, 55, 63, 46, 54, 44
(iii) 34, 66, 30, 38, 44, 50, 40, 60, 42, 51
(iv) 22, 24, 30, 27, 29, 31, 25, 28, 41, 42
(v) 38, 70, 48, 34, 63, 42, 55, 44, 53, 47
Solution:
(i) 3011, 2780, 3020, 2354, 3541, 4150, 5000
Calculating Median (M) of the following observation:
Arranging numbers in ascending order,
2354, 2780, 3011, 3020, 3541, 4150, 5000
Median is the middle number of all the observations.
Therefore, Median = 3020 and n = 7
x_{i} |d_{i}| = |x_{i }– 3020| 3011 9 2780 240 3020 0 2354 666 3541 521 4150 1130 5000 1980 Total 4546 Calculating Mean Deviation:
= 1/7 × 4546
= 649.42
Hence, Mean Deviation is 649.42.
(ii) 38, 70, 48, 34, 42, 55, 63, 46, 54, 44
Calculating Median (M) of the following observation:
Arranging numbers in ascending order,
34, 38, 42, 44, 46, 48, 54, 55, 63, 70
Median is the middle number of all the observations.
Here, the number of observations are even,
therefore the Median = (46 + 48)/2 = 47
Median = 47 and n = 10
x_{i} |d_{i}| = |x_{i} – 47| 38 9 70 23 48 1 34 13 42 5 55 8 63 16 46 1 54 7 44 3 Total 86 Calculating Mean Deviation:
= 1/10 × 86
= 8.6
Hence, Mean Deviation is 8.6.
(iii) 34, 66, 30, 38, 44, 50, 40, 60, 42, 51
Calculating Median (M) of the following observation:
Arranging numbers in ascending order,
30, 34, 38, 40, 42, 44, 50, 51, 60, 66
Median is the middle number of all the observations.
Here, the number of observations are even,
therefore the Median = (42 + 44)/2 = 43
Median = 43 and n = 10
x_{i} |d_{i}| = |x_{i} – 43| 30 13 34 9 38 5 40 3 42 1 44 1 50 7 51 8 60 17 66 23 Total 87 Calculating Mean Deviation:
= 1/10 × 87
= 8.7
Hence, Mean Deviation is 8.7.
(iv) 22, 24, 30, 27, 29, 31, 25, 28, 41, 42
Calculating Median (M) of the following observation:
Arranging numbers in ascending order,
22, 24, 25, 27, 28, 29, 30, 31, 41, 42
Median is the middle number of all the observations.
Here, the number of observations are even,
therefore the Median = (28 + 29)/2 = 28.5
Median = 28.5 and n = 10
x_{i} |d_{i}| = |x_{i} – 28.5| 22 6.5 24 4.5 30 1.5 27 1.5 29 0.5 31 2.5 25 3.5 28 0.5 41 12.5 42 13.5 Total 47 Calculating Mean Deviation:
= 1/10 × 47
= 4.7
Hence, Mean Deviation is 4.7.
(v) 38, 70, 48, 34, 63, 42, 55, 44, 53, 47
Calculating Median (M) of the following observation:
Arranging numbers in ascending order,
34, 38, 43, 44, 47, 48, 53, 55, 63, 70
Median is the middle number of all the observation.
Here, the number of observations are even,
therefore the Median = (47 + 48)/2 = 47.5
Median = 47.5 and n = 10
x_{i} |d_{i}| = |x_{i} – 47.5| 38 9.5 70 22.5 48 0.5 34 13.5 63 15.5 42 5.5 55 7.5 44 3.5 53 5.5 47 0.5 Total 84 Calculating Mean Deviation:
= 1/10 × 84
= 8.4
∴ The Mean Deviation is 8.4.
Question 2. Calculate the mean deviation from the mean for the following data :
(i) 4, 7, 8, 9, 10, 12, 13, 17
(ii) 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17
(iii) 38, 70, 48, 40, 42, 55, 63, 46, 54, 44
(iv) 36, 72, 46, 42, 60, 45, 53, 46, 51, 49
(v) 57, 64, 43, 67, 49, 59, 44, 47, 61, 59
Solution:
(i) 4, 7, 8, 9, 10, 12, 13, 17
We know, Mean Deviation,
Where, |d_{i}| = |x_{i} – x|
So, let us assume x to be the mean of the given observation.
Now, x = [4 + 7 + 8 + 9 + 10 + 12 + 13 + 17]/8
= 80/8
= 10
Number of observations, n = 8
x_{i} |d_{i}| = |x_{i} – 10| 4 6 7 3 8 2 9 1 10 0 12 2 13 3 17 7 Total 24 MD = 1/8 * 24
= 3
(ii) 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17
Since,
Mean Deviation,
Where, |d_{i}| = |x_{i} – x|
So, let us assume x to be the mean of the given observation.
x = [13 + 17 + 16 + 14 + 11 + 13 + 10 + 16 + 11 + 18 + 12 + 17]/12
= 168/12
= 14
Number of observations, n = 12
x_{i} |d_{i}| = |x_{i} – 14| 13 1 17 3 16 2 14 0 11 3 13 1 10 4 16 2 11 3 18 4 12 2 17 3 Total 28 Now,
MD = 1/12 × 28
= 2.33
(iii) 38, 70, 48, 40, 42, 55, 63, 46, 54, 44
We know that,
Mean Deviation,
Where, |d_{i}| = |x_{i} – x|
So, let us assume x to be the mean of the given observation.
x = [38 + 70 + 48 + 40 + 42 + 55 + 63 + 46 + 54 + 44]/10
= 500/10
= 50
Number of observations, n = 10
x_{i} |d_{i}| = |x_{i} – 50| 38 12 70 20 48 2 40 10 42 8 55 5 63 13 46 4 54 4 44 6 Total 84 MD = 1/10 × 84
= 8.4
(iv) 36, 72, 46, 42, 60, 45, 53, 46, 51, 49
Mean Deviation,
Where, |d_{i}| = |x_{i} – x|
So, let us assume x to be the mean of the given observation.
x = [36 + 72 + 46 + 42 + 60 + 45 + 53 + 46 + 51 + 49]/10
= 500/10
= 50
Number of observations, n = 10
x_{i} |d_{i}| = |x_{i} – 50| 36 14 72 22 46 4 42 8 60 10 45 5 53 3 46 4 51 1 49 1 Total 72 MD = 1/10 × 72
= 7.2
(v) 57, 64, 43, 67, 49, 59, 44, 47, 61, 59
Mean Deviation,
Where, |d_{i}| = |x_{i} – x|
So, let us assume x to be the mean of the given observation.
x = [57 + 64 + 43 + 67 + 49 + 59 + 44 + 47 + 61 + 59]/10
= 550/10
= 55
Number of observations, n = 10
x_{i} |d_{i}| = |x_{i} – 55| 57 2 64 9 43 12 67 12 49 6 59 4 44 11 47 8 61 6 59 4 Total 74 MD = 1/10 × 74
= 7.4
Question 3. Calculate the mean deviation of the following income groups of five and seven members from their medians:
I Income in ₹ | II Income in ₹ |
4000 | 3800 |
4200 | 4000 |
4400 | 4200 |
4600 | 4400 |
4800 | 4600 |
4800 | |
5800 |
Solution:
Dataset I :
Since the data is arranged in ascending order,
4000, 4200, 4400, 4600, 4800
Median (Middle of ascending order observation) = 4400
Total observations, n = 5
Now, Mean Deviation,
x_{i} |d_{i}| = |x_{i} – 4400| 4000 400 4200 200 4400 0 4600 200 4800 400 Total 1200 MD(I) = 1/5 × 1200
= 240
Dataset II :
Since the data is arranged in ascending order,
3800, 4000, 4200, 4400, 4600, 4800, 5800
Median (Middle of ascending order observation) = 4400
Total observations, n = 7
Now, Mean Deviation,
x_{i} |d_{i}| = |x_{i} – 4400| 3800 600 4000 400 4200 200 4400 0 4600 200 4800 400 5800 1400 Total 3200 MD(II) = 1/7 × 3200
= 457.14
Therefore, the Mean Deviation of set 1, MD(I) is 240 and set 2, MD(II) is 457.14
Question 4. The lengths (in cm) of 10 rods in a shop are given below:
40.0, 52.3, 55.2, 72.9, 52.8, 79.0, 32.5, 15.2, 27.9, 30.2
(i) Find the mean deviation from the median.
(ii) Find the mean deviation from the mean also.
Solution:
(i) The mean deviation from the median
Arranging the data in ascending order,
15.2, 27.9, 30.2, 32.5, 40.0, 52.3, 52.8, 55.2, 72.9, 79.0
We know that,
Since, the number of observations are even,
therefore Median = (40 + 52.3)/2 = 46.15
Median = 46.15
Also, number of observations, n = 10
x_{i} |d_{i}| = |x_{i} – 46.15| 40.0 6.15 52.3 6.15 55.2 9.05 72.9 26.75 52.8 6.65 79.0 32.85 32.5 13.65 15.2 30.95 27.9 19.25 30.2 15.95 Total 167.4 MD = 1/10 * 167.4
=16.74
(ii) Mean deviation from the mean also.
Where, |d_{i}| = |x_{i} – x|
So, let us assume x to be the mean of the given observation.
Now, x = [40.0 + 52.3 + 55.2 + 72.9 + 52.8 + 79.0 + 32.5 + 15.2 + 27.9 + 30.2]/10
= 458/10
= 45.8
And, number of observations, n = 10
x_{i} |d_{i}| = |x_{i} – 45.8| 40.0 5.8 52.3 6.5 55.2 9.4 72.9 27.1 52.8 7 79.0 33.2 32.5 13.3 15.2 30.6 27.9 17.9 30.2 15.6 Total 166.4 MD = 1/10 * 166.4
= 16.64
Question 5. In question 1(iii), (iv), (v) find the number of observations lying betweenand , where M.D. is the mean deviation from the mean.
Solution:
(iii) 34, 66, 30, 38, 44, 50, 40, 60, 42, 51
We know that,
Where, |d_{i}| = |x_{i} – x|
So, let us assume x to be the mean of the given observation.
x = [34 + 66 + 30 + 38 + 44 + 50 + 40 + 60 + 42 + 51]/10
= 455/10
= 45.5
And, number of observations, n = 10
x_{i} |d_{i}| = |x_{i} – 45.5| 34 11.5 66 20.5 30 15.5 38 7.5 44 1.5 50 4.5 40 5.5 60 14.5 42 3.5 51 5.5 Total 90 MD = 1/10 × 90
= 9
Now,
So, There are total 6 observation between and
(iv) 22, 24, 30, 27, 29, 31, 25, 28, 41, 42
We know that,
Where, |d_{i}| = |x_{i} – x|
So, let us assume x to be the mean of the given observation.
x = [22 + 24 + 30 + 27 + 29 + 31 + 25 + 28 + 41 + 42]/10
= 299/10
= 29.9
Also, number of observations, n = 10
x_{i} |d_{i}| = |x_{i} – 29.9| 22 7.9 24 5.9 30 0.1 27 2.9 29 0.9 31 1.1 25 4.9 28 1.9 41 11.1 42 12.1 Total 48.8 MD = 1/10 × 48.8
= 4.88
And,
So, there are 5 observations in between.
(v) 38, 70, 48, 34, 63, 42, 55, 44, 53, 47
We know that,
Where, |d_{i}| = |x_{i} – x|
So, let us assume x to be the mean of the given observation.
x = [38 + 70 + 48 + 34 + 63 + 42 + 55 + 44 + 53 + 47]/10
= 494/10
= 49.4
Number of observations, n = 10
x_{i} |d_{i}| = |x_{i} – 49.4| 38 11.4 70 20.6 48 1.4 34 15.4 63 13.6 42 7.4 55 5.6 44 5.4 53 3.6 47 2.4 Total 86.8 MD = = 1/10 × 86.8
= 8.68
Also,
There are 6 observations in between.
Question 6. Show that the two formulae for the standard deviation of the ungrouped data and are equivalent, where
Solution:
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