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### NCERT Solutions for Class-11 Maths Exercise 15.1

**Question****1. Find the mean deviation about the mean for the data**

**4, 7, 8, 9, 10, 12, 13, 17**

**Solution :**

The given data is

4, 7, 8, 9, 10, 12, 13, 17

**Question****2. Find the mean deviation about the mean for the data**

**38, 70, 48, 40, 42, 55, 63, 46, 54, 44**

**Solution :**

The given data is

38, 70, 48, 40, 42, 55, 63, 46, 54, 44

Mean of the given data,

The deviations of the respective observations from the mean are

–12, 20, –2, –10, –8, 5, 13, –4, 4, –6

The absolute values of the deviations, i.e. , are

12, 20, 2, 10, 8, 5, 13, 4, 4, 6

The required mean deviation about the mean is

**Question****3. Find the mean deviation about the median for the data.**

**13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17**

**Solution :**

The given data is

13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17

Here, the numbers of observations are 12, which is even.

Arranging the data in ascending order, we obtain

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 = 6^{th} = 13

(12/2)+ 1)^{th} observation = 6 + 1

= 7^{th} = 14

Median = (13 + 14)/2

= 27/2

= 13.5

So, the absolute values of the respective deviations from the median, i.e., |x_{i} – 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

**Question****4. Find the mean deviation about the median for the data**

**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 = 5^{th} = 46

(10/2)+ 1)^{th} observation = 5 + 1

= 6^{th} = 49

Median = (46 + 49)/2

= 95

= 47.5

So, the absolute values of the respective deviations from the median, i.e., |x_{i} – 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.

**Question****5. Find the mean deviation about the mean for the data.**

x_{i} | 5 | 10 | 15 | 20 | 25 |

f_{i} | 7 | 4 | 6 | 3 | 5 |

**Solution :**

x_{i} | f_{i} | f_{i} x_{i} | ||

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 |

**Question****6. Find the mean deviation about the mean for the data**

x_{i} | 10 | 30 | 50 | 70 | 90 |

f_{i} | 4 | 24 | 28 | 16 | 8 |

**Solution :**

x_{i} | f_{i} | f_{i }x_{i} | ||

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 |

**Question****7. Find the mean deviation about the median for the data.**

x_{i} | 5 | 7 | 9 | 10 | 12 | 15 |

f_{i} | 8 | 6 | 2 | 2 | 2 | 6 |

**Solution :**

The given observations are already in ascending order.

Adding a column corresponding to cumulative frequencies of the given data, we obtain the following table.

x_{i} | f_{i} | c.f. |

5 | 8 | 8 |

7 | 6 | 14 |

9 | 2 | 16 |

10 | 2 | 18 |

12 | 2 | 20 |

15 | 6 | 26 |

|x_{i }– M| | 2 | 0 | 2 | 3 | 5 | 8 |

fi | 8 | 6 | 2 | 2 | 2 | 6 |

f_{i} |x_{i} – M| | 16 | 0 | 4 | 6 | 10 | 48 |

Now, N = 26, which is even.

Median is the mean of the 13^{th} and 14^{th} observations. Both of these observations lie in the cumulative frequency 14, for which the corresponding observation is 7.

Then,

Median = (13^{th} observation + 14^{th }observation)/2

= (7 + 7)/2

= 14/2

= 7

So, the absolute values of the respective deviations from the median, i.e., |x_{i} – M| are shown in the table.

.

**Question****8. Find the mean deviation about the median for the data**

xi | 15 | 21 | 27 | 30 | 35 |

fi | 3 | 5 | 6 | 7 | 8 |

**Solution :**

The given observations are already in ascending order.

Adding a column corresponding to cumulative frequencies of the given data, we obtain the following table.

x_{i} | f_{i } | c.f. |

15 | 3 | 3 |

21 | 5 | 8 |

27 | 6 | 14 |

30 | 7 | 21 |

35 | 8 | 29 |

Here, N = 29, which is odd.

|x_{i} – M| | 15 | 9 | 3 | 0 | 5 |

f_{i} | 3 | 5 | 6 | 7 | 8 |

f_{i} |x_{i} – M| | 45 | 45 | 18 | 0 | 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 = (15^{th} observation + 16^{th }observation)/2

= (30 + 30)/2

= 60/2

= 30

So, the absolute values of the respective deviations from the median, i.e., |x_{i} – M| are shown in the table.

**Question****9. Find the mean deviation about the mean for the data.**

Income per day | Number of persons |

0-100 | 4 |

100-200 | 8 |

200-300 | 9 |

300-400 | 10 |

400-500 | 7 |

500-600 | 5 |

600-700 | 4 |

700-800 | 3 |

**Solution :**

The following table is formed.

Income per day | Number of persons fi | Mid-point x_{i} | f_{i} x_{i} | ||

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 | 1168 |

700 – 800 | 3 | 750 | 2250 | 392 | 1176 |

50 | 17900 | 7896 |

Here,

**Question****10. Find the mean deviation about the mean for the data**

Height in cms | Number of boys |

95-105 | 9 |

105-115 | 13 |

115-125 | 26 |

125-135 | 30 |

135-145 | 12 |

145-155 | 10 |

**Solution :**

The following table is formed.

Height in cms | Number of boys f_{i} | Mid-point x_{i} | f_{i }x_{i} | ||

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 |

**Question****11. Find the mean deviation about median for the following data:**

Marks | Number of girls |

0-10 | 6 |

10-20 | 8 |

20-30 | 14 |

30-40 | 16 |

40-50 | 4 |

50-60 | 2 |

**Solution :**

The following table is formed.

Marks | Number of girls f_{i} | Cumulative frequency (c.f.) | Mid-point x_{i} | |x_{i} – Med.| | f_{i }|x_{i} – 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 N^{th}/2 or 25^{th} 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

**Question****12. Calculate the mean deviation about median age for the age distribution of 100 persons given below:**

Age | Number |

16-20 | 5 |

21-25 | 6 |

26-30 | 12 |

31-35 | 14 |

36-40 | 26 |

41-45 | 12 |

46-50 | 16 |

51-55 | 9 |

**Solution :**

The given data is not continuous. Therefore, it has to be converted into continuous frequency distribution by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of each class interval.

The table is formed as follows.

Age | Number f_{i} | Cumulative frequency (c.f.) | Mid-point x_{i} | |x_{i} – Med.| | fi |x_{i }– 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 |

The class interval containing N^{th}/2 or 50^{th} item is 35.5 – 40.5

So, 35.5 – 40.5 is the median class.

Then,

Median = l + (((N/2) – c)/f) × h

Where, l = 35.5, c = 37, f = 26, h = 5 and N = 100

Median = 35.5 + (((50 – 37))/26) × 5

= 35.5 + 2.5

= 38

## Conclusiosn for NCERT SOLUTIONS FOR CLASS 11 MATHS CHAPTER 15-STATISTICS EXERCISE 15.1

An academic team of knowledgeable members of SWC has produced and published the NCERT Solutions for class 11’s maths chapter for your use as a reference. You can get answers to all of the chapters of the NCERT maths class 11 here at SWC. Please make use of the following NCERT answers that were created by SWC as a reference for this chapter. In addition to that, study the chapter’s theory before attempting to solve the NCERT problems.