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.

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.

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= 6^{th} = 13

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

= 7^{th} =14

Median = (13 + 14)/2

= 27/2

= 13.5

So, the absolutevalues 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

So, the mean deviationabout the median for the given data is 2.33.

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= 5^{th} = 46

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

= 6^{th} =49

Median = (46 + 49)/2

= 95

= 47.5

So, the absolutevalues 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 deviationabout the median for the given data is 7.

x | 5 | 10 | 15 | 20 | 25 |

f | 7 | 4 | 6 | 3 | 5 |

**Answer
5** :

Let us make the tableof the given data and append other columns after calculations.

X | f | f | |x | f |

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., |x_{i} – x̅|, as shown in thetable.

x | 10 | 30 | 50 | 70 | 90 |

f | 4 | 24 | 28 | 16 | 8 |

**Answer
6** :
Let us make the table of the given data and append other columns after calculations.

X | f | f | |x | f |

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 |

x | 5 | 7 | 9 | 10 | 12 | 15 |

f | 8 | 6 | 2 | 2 | 2 | 6 |

**Answer
7** :
Let us make the table of the given data and append other columns after calculations.

X | f | c.f. | |x | f |

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 13^{th} and 14^{th} observations. Both of theseobservations lie in the cumulative frequency 14, for which the correspondingobservation is 7.

Then,

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

= (7 + 7)/2

= 14/2

= 7

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

x | 15 | 21 | 27 | 30 | 35 |

f | 3 | 5 | 6 | 7 | 8 |

**Answer
8** :

Let us make the tableof the given data and append other columns after calculations.

X | f | c.f. | |x | f |

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 = (15^{th} observation+ 16^{th }observation)/2

= (30 + 30)/2

= 60/2

= 30

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

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 f | Mid – points x | f | |x | f |

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 |

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 f | Mid – points x | f | |x | f |

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 |

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