Chapter 15 Statistics Ex 15.1 Solutions
Question - 1 : - Find the mean deviation about the mean for the data in Exercises 1 and 2.4, 7, 8, 9, 10, 12, 13, 17
Answer - 1 : -
First we have to find (x̅) of the given data
So, the respective values of the deviations from mean,
i.e., xi – x̅ are, 10 – 4 = 6, 10 – 7 = 3, 10 – 8 = 2, 10 – 9 = 1, 10 – 10 = 0,
10 – 12 = – 2, 10 – 13 = – 3, 10 – 17 = – 7
6, 3, 2, 1, 0, -2, -3, -7
Now absolute values of the deviations,
6, 3, 2, 1, 0, 2, 3, 7
MD = sum of deviations/ number of observations
= 24/8
= 3
So, the mean deviation for the given data is 3.
Question - 2 : - Find the mean deviation about the mean for the data in Exercises 1 and 2.38, 70, 48, 40, 42, 55, 63, 46, 54, 44
Answer - 2 : -
First we have to find (x̅) of the given data
So, the respective values of the deviations from mean,
i.e., xi – x̅ are, 50 – 38 = -12, 50 -70 = -20, 50 – 48 = 2, 50 – 40 = 10, 50 – 42 = 8,
50 – 55 = – 5, 50 – 63 = – 13, 50 – 46 = 4, 50 – 54 = -4, 50 – 44 = 6
-12, 20, -2, -10, -8, 5, 13, -4, 4, -6
Now absolute values of the deviations,
12, 20, 2, 10, 8, 5, 13, 4, 4, 6
MD = sum of deviations/ number of observations
= 84/10
= 8.4
So, the mean deviation for the given data is 8.4.
Question - 3 : - Find the mean deviation about the median for the data in Exercises 3 and 4.13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17
Answer - 3 : -
First we have toarrange the given observations into ascending order,
10, 11, 11, 12, 13,13, 14, 16, 16, 17, 17, 18.
The number ofobservations is 12
Then,
Median = ((12/2)th observation+ ((12/2)+ 1)th observation)/2
(12/2)th observation= 6th = 13
(12/2)+ 1)th observation= 6 + 1
= 7th =14
Median = (13 + 14)/2
= 27/2
= 13.5
So, the absolutevalues of the respective deviations from the median, i.e., |xi –M| are
3.5, 2.5, 2.5, 1.5,0.5, 0.5, 0.5, 2.5, 2.5, 3.5, 3.5, 4.5
Mean Deviation,
= (1/12) × 28
= 2.33
So, the mean deviationabout the median for the given data is 2.33.
Question - 4 : - Find the mean deviation about the median for the data in Exercises 3 and 4.36, 72, 46, 42, 60, 45, 53, 46, 51, 49
Answer - 4 : -
First we have toarrange the given observations into ascending order,
36, 42, 45, 46, 46,49, 51, 53, 60, 72.
The number ofobservations is 10
Then,
Median = ((10/2)th observation+ ((10/2)+ 1)th observation)/2
(10/2)th observation= 5th = 46
(10/2)+ 1)th observation= 5 + 1
= 6th =49
Median = (46 + 49)/2
= 95
= 47.5
So, the absolutevalues of the respective deviations from the median, i.e., |xi –M| are
11.5, 5.5, 2.5, 1.5,1.5, 1.5, 3.5, 5.5, 12.5, 24.5
Mean Deviation,
= (1/10) × 70
= 7
So, the mean deviationabout the median for the given data is 7.
Question - 5 : - Find the mean deviation about the mean for the data in Exercises 5 and 6. | xi | 5 | 10 | 15 | 20 | 25 |
| fi | 7 | 4 | 6 | 3 | 5 |
Answer - 5 : -
Let us make the tableof the given data and append other columns after calculations.
| Xi | fi | fixi | |xi – x̅| | fi |xi – x̅| |
| 5 | 7 | 35 | 9 | 63 |
| 10 | 4 | 40 | 4 | 16 |
| 15 | 6 | 90 | 1 | 6 |
| 20 | 3 | 60 | 6 | 18 |
| 25 | 5 | 125 | 11 | 55 |
| 25 | 350 | | 158 |
The sum of calculateddata,
The absolute values ofthe deviations from the mean, i.e., |xi – x̅|, as shown in thetable.
Question - 6 : - Find the mean deviation about the mean for the data in Exercises 5 and 6. | xi | 10 | 30 | 50 | 70 | 90 |
| fi | 4 | 24 | 28 | 16 | 8 |
Answer - 6 : - Let us make the table of the given data and append other columns after calculations.
| Xi | fi | fixi | |xi – x̅| | fi |xi – x̅| |
| 10 | 4 | 40 | 40 | 160 |
| 30 | 24 | 720 | 20 | 480 |
| 50 | 28 | 1400 | 0 | 0 |
| 70 | 16 | 1120 | 20 | 320 |
| 90 | 8 | 720 | 40 | 320 |
| 80 | 4000 | | 1280 |
| xi | 5 | 7 | 9 | 10 | 12 | 15 |
| fi | 8 | 6 | 2 | 2 | 2 | 6 |
Answer - 7 : - Let us make the table of the given data and append other columns after calculations.
| Xi | fi | c.f. | |xi – M| | fi |xi – M| |
| 5 | 8 | 8 | 2 | 16 |
| 7 | 6 | 14 | 0 | 0 |
| 9 | 2 | 16 | 2 | 4 |
| 10 | 2 | 18 | 3 | 6 |
| 12 | 2 | 20 | 5 | 10 |
| 15 | 6 | 26 | 8 | 48 |
Now, N = 26, which iseven.
Median is the mean ofthe 13th and 14th observations. Both of theseobservations lie in the cumulative frequency 14, for which the correspondingobservation is 7.
Then,
Median = (13th observation+ 14th observation)/2
= (7 + 7)/2
= 14/2
= 7
So, the absolutevalues of the respective deviations from the median, i.e., |xi –M| are shown in the table.
| xi | 15 | 21 | 27 | 30 | 35 |
| fi | 3 | 5 | 6 | 7 | 8 |
Answer - 8 : -
Let us make the tableof the given data and append other columns after calculations.
| Xi | fi | c.f. | |xi – M| | fi |xi – M| |
| 15 | 3 | 3 | 15 | 45 |
| 21 | 5 | 8 | 9 | 45 |
| 27 | 6 | 14 | 3 | 18 |
| 30 | 7 | 21 | 0 | 0 |
| 35 | 8 | 29 | 5 | 40 |
Now, N = 29, which isodd.
So 29/2 = 14.5
The cumulativefrequency greater than 14.5 is 21, for which the corresponding observation is30.
Then,
Median = (15th observation+ 16th observation)/2
= (30 + 30)/2
= 60/2
= 30
So, the absolutevalues of the respective deviations from the median, i.e., |xi –M| are shown in the table.
Question - 9 : - Find the mean deviation about the mean for the data in Exercises 9 and 10. | Income per day in ₹ | 0 – 100 | 100 – 200 | 200 – 300 | 300 – 400 | 400 – 500 | 500 – 600 | 600 – 700 | 700 – 800 |
| Number of persons | 4 | 8 | 9 | 10 | 7 | 5 | 4 | 3 |
Answer - 9 : -
Let us make the tableof the given data and append other columns after calculations.
| Income per day in ₹ | Number of persons fi | Mid – points xi | fixi | |xi – x̅| | fi|xi – x̅| |
| 0 – 100 | 4 | 50 | 200 | 308 | 1232 |
| 100 – 200 | 8 | 150 | 1200 | 208 | 1664 |
| 200 – 300 | 9 | 250 | 2250 | 108 | 972 |
| 300 – 400 | 10 | 350 | 3500 | 8 | 80 |
| 400 – 500 | 7 | 450 | 3150 | 92 | 644 |
| 500 – 600 | 5 | 550 | 2750 | 192 | 960 |
| 600 – 700 | 4 | 650 | 2600 | 292 | 1160 |
| 700 – 800 | 3 | 750 | 2250 | 392 | 1176 |
| 50 | | 17900 | | 7896 |
Question - 10 : - Find the mean deviation about the mean for the data in Exercises 9 and 10. | Height in cms | 95 – 105 | 105 – 115 | 115 – 125 | 125 – 135 | 135 – 145 | 145 – 155 |
| Number of boys | 9 | 13 | 26 | 30 | 12 | 10 |
Answer - 10 : -
Let us make the tableof the given data and append other columns after calculations.
| Height in cms | Number of boys fi | Mid – points xi | fixi | |xi – x̅| | fi|xi – x̅| |
| 95 – 105 | 9 | 100 | 900 | 25.3 | 227.7 |
| 105 – 115 | 13 | 110 | 1430 | 15.3 | 198.9 |
| 115 – 125 | 26 | 120 | 3120 | 5.3 | 137.8 |
| 125 – 135 | 30 | 130 | 3900 | 4.7 | 141 |
| 135 – 145 | 12 | 140 | 1680 | 14.7 | 176.4 |
| 145 – 155 | 10 | 150 | 1500 | 24.7 | 247 |
| 100 | | 12530 | | 1128.8 |
