**Find the mean deviation about the mean for the data 4, 7, 8, 9, 10, 12, 13, 17**

Solution:

Given data is 4, 7, 8, 9, 10, 12, 13, 17

**Mean** of the data

**Find the mean deviation about the mean for the data 38, 70, 48, 40, 42, 55, 63, 46, 54, 44**

Solution:

Given data is 38, 70, 48, 40, 42, 55, 63, 46, 54, 44

Mean of the given data

**3.Find the mean deviation about the median for the data ****13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17**

Solution:

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

**Find the mean deviation about the median for the data**

**36, 72, 46, 42, 60, 45, 53, 46, 51, 49**

Solution:

Given data is 36, 72, 46, 42, 60, 45, 53, 46, 51, 49

The number of observations is 10, which is even.

Arranging the data in ascending order, we obtain

36, 42, 45, 46, 46, 49, 51, 53, 60, 72

**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} | |x_{i}–x| | f_{i}|x_{i}–x| |

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 |

**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} | |x_{i} – x| | f_{i}|x_{i} – x| |

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 |

**Find the mean deviation about the mean for the data**

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

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

Solution:

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

5 | 8 | 8 |

7 | 6 | 14 |

9 | 2 | 16 |

10 | 2 | 18 |

12 | 2 | 20 |

15 | 6 | 26 |

N = 26, which is even.

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

|x_{i} – M| | 2 | 0 | 2 | 3 | 5 | 8 |

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

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

**8. Find the absolute value of the derivations from median, for the data**

x_{i} | 15 | 21 | 27 | 30 | 35 |

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

Solution:

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

15 | 3 | 3 |

21 | 5 | 8 |

27 | 6 | 14 |

30 | 7 | 21 |

35 | 8 | 29 |

N = 29, which is an odd number.

**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:

Income per day | Number of persons | x_{i} | f_{i}x_{i} | |x_{i} –x| | f_{i}|x_{i} –x| |

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 |

**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:

Height in cms | Number of boys | x_{i} | f_{i}x_{i} | |x_{i} – x| | f_{i}| x_{i} – x| |

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 |

12530 | 1128.8 |

**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:

Marks | Number of girls | Cumulative frequency(c.f) | x_{i} | |x_{i} – median| | f_{i}|x_{i} –x| |

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 the (^{N}/_{2}) or 25^{th} item is 20-30

Thus, 20-30 is the median class

**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:

We need to convert the given dat into continuous frequency distribution by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of each interval in order to make it continuous.

age | Number | c.f | x_{i} | |x_{i} – x| | f_{i}|x_{i}–x| |

16-20 | 5 | 5 | 18 | 20 | 100 |

21-25 | 6 | 11 | 23 | 15 | 90 |

26-30 | 12 | 23 | 28 | 10 | 120 |

31-35 | 14 | 37 | 33 | 5 | 70 |

36-40 | 26 | 63 | 38 | 0 | 0 |

41-45 | 12 | 75 | 43 | 5 | 60 |

46-50 | 16 | 91 | 48 | 10 | 160 |

51-55 | 9 | 100 | 53 | 15 | 135 |

100 | 735 |

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

Therefore, 35.5 – 40.5 is the median class.

We know that,