NCERT Solutions Class 11th Maths Chapter – 15 – Statistics Exercise 15.1
| Textbook | NCERT |
| class | Class – 11th |
| Subject | Mathematics |
| Chapter | Chapter – 15 |
| Chapter Name | Statistics |
| grade | Class 11th Maths solution |
| Medium | English |
| Source | last doubt |
NCERT Solutions Class 11th Maths Chapter – 15 – Statistics Exercise 15.1
?Chapter – 15?
✍Statistics✍
?Exercise 15.1?
- Find the mean deviation about the mean for the data in Exercises 1 and 2.
1. 4, 7, 8, 9, 10, 12, 13, 17
?♂️Solution: 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.
2. 38, 70, 48, 40, 42, 55, 63, 46, 54, 44
?♂️Solution: 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.
Find the mean deviation about the median for the data in Exercises 3 and 4.
3. 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17
?♂️Solution: First we have to arrange the given observations into ascending order,
10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18.
The number of observations 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 absolute values 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 deviation about the median for the given data is 2.33.
4. 36, 72, 46, 42, 60, 45, 53, 46, 51, 49
?♂️Solution: First we have to arrange the given observations into ascending order,
36, 42, 45, 46, 46, 49, 51, 53, 60, 72.
The number of observations 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 absolute values 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 deviation about the median for the given data is 7.
Find the mean deviation about the mean for the data in Exercises 5 and 6.
5.
| xi | 5 | 10 | 15 | 20 | 25 |
| fi | 7 | 4 | 6 | 3 | 5 |
?♂️Solution: Let us make the table of 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 calculated data,
The absolute values of the deviations from the mean, i.e., |xi – x̅|, as shown in the table.
so, the mean deviation about the mean for the given data is 6.32
6.
| xi | 10 | 30 | 50 | 70 | 90 |
| fi | 4 | 24 | 28 | 16 | 8 |
?♂️Solution: 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 |
so, the mean deviation about the mean for the given data is 16
7. Find the mean deviation about the median for the data in Exercises 7 and 8.
| xi | 5 | 7 | 9 | 10 | 12 | 15 |
| fi | 8 | 6 | 2 | 2 | 2 | 6 |
?♂️Solution: 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 is even.
Median is the mean of the 13th and 14th observations. Both of these observations lie in the cumulative frequency 14, for which the corresponding observation is 7.
Then,
Median = (13th observation + 14th observation)/2
= (7 + 7)/2
= 14/2
= 7
So, the absolute values of the respective deviations from the median, i.e., |xi – M| are shown in the table.
hence, the mean deviation about the median for the given data 3.23
8.
| xi | 15 | 21 | 27 | 30 | 35 |
| fi | 3 | 5 | 6 | 7 | 8 |
?♂️Solution: Let us make the table of 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 is odd.
So 29/2 = 14.5
The cumulative frequency greater than 14.5 is 21, for which the corresponding observation is 30.
Then,
Median = (15th observation + 16th observation)/2
= (30 + 30)/2
= 60/2
= 30
So, the absolute values of the respective deviations from the median, i.e., |xi – M| are shown in the table.
hence the mean deviation about the median for the given data 5.1
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 |
?♂️Solution: Let us make the table of the given data and append other columns after calculations.
| Income per day in ₹ | Number of persons fi | Mid – pointsxi | 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 |
the sum of calculated data,
x‾ = a + (∑fidi/N) × h
= 350+4/50×100
= 358
Mean deviation = ∑f i |x i – x ̄/N
= 7856/50
= 157.92
hence , the mean deviation about the mean for given data is 157.92
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 |
?♂️Solution: Let us make the table of the given data and append other columns after calculations.
| Height in cms | Number of boys fi | Mid – pointsxi | 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 |
the sum of the calculated data,
Mean x‾ = a + (∑fidi/∑fi) × h
= 130+( – 47/100)×10
= 130 − 4.7
= 125.3
mean deviation = ∑f i | x i – x ̄ /N
= 1128.8/100
= 11.288
11. Find the mean deviation about median for the following data:
| Marks | 0 -10 | 10 -20 | 20 – 30 | 30 – 40 | 40 – 50 | 50 – 60 |
| Number of girls | 6 | 8 | 14 | 16 | 4 | 2 |
?♂️Solution: Let us make the table of the given data and append other columns after calculations.
| Marks | Number of Girls fi | Cumulative frequency (c.f.) | Mid – pointsxi | |xi – Med| | fi|xi – Med| |
| 0 – 10 | 6 | 6 | 5 | 22.85 | 137.1 |
| 10 – 20 | 8 | 14 | 15 | 12.85 | 102.8 |
| 20 – 30 | 14 | 28 | 25 | 2.85 | 39.9 |
| 30 – 40 | 16 | 44 | 35 | 7.15 | 114.4 |
| 40 – 50 | 4 | 48 | 45 | 17.15 | 68.6 |
| 50 – 60 | 2 | 50 | 55 | 27.15 | 54.3 |
| 50 | 517.1 |
The class interval containing Nth/2 or 25th item is 20-30
So, 20-30 is the median class.
Then,
Median = l + (((N/2) – c)/f) × h
Where, l = 20, c = 14, f = 14, h = 10 and n = 50
Median = 20 + (((25 – 14))/14) × 10
= 20 + 7.85
= 27.85
The absolute values of the deviations from the median, i.e., [x-Med], as shown in the table.
So ∑6i = 1 fi|xi – Med. = 517.1
And M.D. (M)=1/N∑6i fi|xi – Med.
= (1/50) x 517.1
= 10.34
hence the mean deviation about the median for the given data is 10.34
12. Calculate the mean deviation about median age for the age distribution of 100 persons given below:
| Age(in years) | 16 – 20 | 21 – 25 | 26 – 30 | 31 – 35 | 36 – 40 | 41 – 45 | 46 – 50 | 51 – 55 |
| Number | 5 | 6 | 12 | 14 | 26 | 12 | 16 | 9 |
[Hint Convert the given data into continuous frequency distribution by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of each class interval]
?♂️Solution: The given data is converted into continuous frequency distribution by subtracting 0.5 from the lower limit and adding the 0.5 to the upper limit of each class intervals and append other columns after calculations.
| Age | Number fi | Cumulative frequency (c.f.) | Mid – pointsxi | |xi – Med| | fi|xi – Med| |
| 15.5 – 20.5 | 5 | 5 | 18 | 20 | 100 |
| 20.5 – 25.5 | 6 | 11 | 23 | 15 | 90 |
| 25.5 – 30.5 | 12 | 23 | 28 | 10 | 120 |
| 30.5 – 35.5 | 14 | 37 | 33 | 5 | 70 |
| 35.5 – 40.5 | 26 | 63 | 38 | 0 | 0 |
| 40.5 – 45.5 | 12 | 75 | 43 | 5 | 60 |
| 45.5 – 50.5 | 16 | 91 | 48 | 10 | 160 |
| 50.5 – 55.5 | 9 | 100 | 53 | 15 | 135 |
| 100 | 735 |
N=100
⇒N/2=50
Thus, the cumulative frequency slightly greater than 50 is 63 and falls in the median class 35.5−40.5.
Median Median =l+N/2−F/f×h
=35.5+(50−37)26×5
=35.5+2.5
=38
Mean deviation about the median age Mean deviation about the median age =∑8i =1 fi/di/N
=735100
=7.35
Thus, the mean deviation from the median age is 7.35 years