# Class 11 NCERT Solutions- Chapter 15 Statistics – Exercise 15.1

### Find the mean deviation about the mean for the data in questions 1 and 2.

### Question 1. 4, 7, 8, 9, 10, 12, 13, 17

**Solution:**

Given observations are = 4, 7, 8, 9, 10, 12, 13, 17

Therefore, Total Number of observations = n = 8

Step 1:Calculating mean(a) for given data about which we have to find the mean deviation.Mean(a) = ∑(all data points) / Total number of data points

= (4 + 7 + 8 + 9 + 10 + 12 + 13 + 17) / 8

= 80 / 8

a = 10

Step 2:Finding deviation of each data point from mean(x_{i }– a)

x_{i}4 7 8 9 10 12 13 17 x_{i}– a4 – 10 = -6 7 – 10 = -3 8 – 10 = -2 9 – 10 = -1 10 – 10 = 0 12 – 10 = 2 13 – 10 = 3 17 – 10 = 7

Step 3:Taking absolute value of deviations, we get|x

_{i}– a| = 6, 3, 2, 1, 0, 2, 3, 7

Step 4:Required mean deviation about the mean isM.D (a) = ∑ (|x

_{i}– a|) / n= (6 + 3 + 2 + 1 + 0 + 2 + 3 + 7) / 8

= 24 / 8

= 3

So, mean deviation for given observations is3

### Question 2. 38, 70, 48, 40, 42, 55, 63, 46, 54, 44

**Solution:**

Given observations are = 38, 70, 48, 40, 42, 55, 63, 46, 54, 44

Therefore, Total Number of observations = n = 10

Step 1:Calculating mean(a) for given data about which we have to find the mean deviation.Mean(a) = ∑(all data points)/ Total number of data points

= (38 + 70 + 48 + 40 + 42 + 55 + 63 + 46 + 54 + 44)/ 10

= 500 / 10

a = 50

Step2 :Finding deviation of each data point from mean(x_{i }– a)

x_{i}38 70 48 40 42 55 63 46 54 44 x_{i}– a38 – 50 = -1270 – 50 = 20 48 – 50 = -2 40 – 50 = -10 42 – 50 = -8 55 – 50 = 5 63 – 50 = 13 46 – 50 = -4 54 – 50 = 4 44 – 50 = -6

Step 3:Taking absolute value of deviations, we get|x

_{i}– a| = 12, 20, 2, 10, 8, 5, 13, 4, 4, 6

Step 4:Required mean deviation about the mean isM.D(a)

= ∑(|x_{i}– a|) / n= (12 + 20 + 2 + 10 + 8 + 5 + 13 + 4 + 4 + 6) / 10

= 84 / 10

= 8.4

So, mean deviation for given observations is8.4

### Find the mean deviation about the median for the data in questions 3 and 4.

### Question 3. 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17

**Solution:**

Given observations are = 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17

Therefore, Total Number of observations = n = 12

Step 1:Calculating median(M) for given data about which we have to find the mean deviation.To calculate median for given data we have to arrange observations in ascending order

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

As number of observations are even median is given by following formula

Median = [(n/2)

^{th }observation+ ((n/2) + 1)^{th }observation] / 2(n/2)

^{th}observation = 12 / 2 = 6^{th }observation= 13((n/2)+1)

^{th}observation = (12/2) + 1= 7^{th}observation = 14Therefore, Median = (13 + 10) / 2 = 13.5

Step 2:Finding deviation of each data point from median(x_{i}– M)

x_{i}10 11 11 12 13 13 14 16 16 17 17 18 x_{i}-M10 – 13.5 = -3.5 11 – 13.5 = -2.5 11 – 13.5 = -2.5 12 – 13.5 = -1.5 13 – 13.5 = -0.5 13 – 13.5 = 0.5 14 – 13.5 = 0.5 16 – 13.5 = 2.5 16 – 13.5 = 2.5 17 – 13.5 = 3.5 17 – 13.5 = 3.5 18 – 13.5 = 4.5

Step 3:Taking absolute values of deviations we get|x

_{i}– M| = 3.5, 2.5, 2.5, 1.5, 0.5, 0.5, 0.5, 2.5, 2.5, 3.5, 3.5, 4.5

Step 4:Required mean deviation about median is given byM.D (M) = ∑(|x

_{i}– M|) / n= (3.5 + 2.5 + 2.5 + 1.5 + 0.5 + 0.5 + 0.5 + 2.5 + 2.5 + 3.5 + 3.5 + 4.5) / 12

= 28 / 12

= 2.33

So, mean deviation about median for given observations is2.33

### Question 4. 36, 72, 46, 42, 60, 45, 53, 46, 51, 49

**Solution:**

Given observations are = 36, 72, 46, 42, 60, 45, 53, 46, 51, 49

Therefore, Total Number of observations = n = 10

Step 1:Calculating median(M) for given data about which we have to find the mean deviation.To calculate median for given data we have to arrange observations in ascending order

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

As number of observations are even median is given by following formula

Median = [(n/2)

^{th}observation+ ((n/2) + 1)^{th}observation] / 2(n/2)

^{th }observation = 10 / 2 = 5^{th}observation= 46((n/2) + 1)

^{th }observation = (10/2) + 1 = 6^{th}observation = 49Therefore, Median = (46 + 49) / 2 = 47.5

Step 2:Finding deviation of each data point from median(x_{i }-M)

x_{i}36 42 45 46 46 49 51 53 60 72 x_{i}– M-11.5 -5.5 -2.5 -1.5 -1.5 1.5 3.5 5.5 12.5 24.5

Step 3:Taking absolute values of deviations we get|x

_{i }– M| = 11.5, 5.5, 2.5, 1.5, 1.5, 1.5, 3.5, 5.5, 12.5, 24.5

Step 4:Required mean deviation about median is given byM.D (M) = ∑(|x

_{i}– M|) / n= (11.5 + 5.5 + 2.5 + 1.5 + 1.5 + 1.5 + 3.5 + 5.5 + 12.5 + 24.5) / 10

= 70 / 10

= 7

So, mean deviation about median for given observations is7

### Find the mean deviation about the mean for the data in questions 5 and 6.

### Question 5.

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

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

**Solution: **

Given table data is of discrete frequency distribution as

we have n = 5 distinct values(x

_{i}) along with their frequencies(f_{i}).

Step 1:Let us make a table of given data and append other columns after calculation

x_{i}f_{i}x_{i }* f_{i}|x_{i }– a|f_{i }* |x_{i }– a|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 Total25 350 158

Step2 :Now to find the mean we have to first calculate the sum of given dataN = ∑ f

_{i}= (7 + 4 + 6 + 3 + 5) = 25∑ x

_{i }* f_{i}= (35 + 40 + 90 + 60 + 125) = 350

Step 3:Find the mean using following formulaMean (a) = ∑(x

_{i}* f_{i})/ N = 350 / 25 = 14

Step 4:Using above mean find absolute values of deviationsfrom mean i.e |x

_{i }– a| and also find values of f_{i}*|xi -a| column.Now,

∑ f

_{i}*|x_{i}– a| = (63 + 16 + 6 + 18 + 55) = 158Using formula, M.D (a) = ∑(f

_{i}* |x_{i}– a|)/ N= 158 / 25

= 6.32

So, mean deviation for given observations is6.32

### Question 6.

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

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

**Solution:**

Given table data is of discrete frequency distribution as

we have n = 5 distinct values(x

_{i}) along with their frequencies(f_{i}).

Step 1:Let us make a table of given data and append other columns after calculation

x_{i}f_{i}x_{i }* f_{i}|x_{i }– a|f_{i }* |x_{i }– a|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 Total80 4000 1280

Step 2:Now to find the mean we have to first calculate the sum of given data. From the above table,N = ∑(f

_{i}) = 80 & ∑(x_{i}* f_{i}) = 4000

Step 3:Find the mean using following formulaMean (a) = ∑ (x

_{i }* f_{i})/ N = 4000 / 80 = 50

Step 4:Using above mean find absolute values of deviations from mean i.e |x_{i }– a| and also find valuesof f

_{i}* |x_{i}-a| column.Now, from above table,

∑ f

_{i}*|x_{i}– a| = (160 + 480 + 0 + 320 + 320) = 1280Using formula, M.D (a) = ∑ f

_{i }*|x_{i}-a| / N= 1280 / 80

= 16

So, mean deviation for given observations is16

### Find the mean deviation about the median for the data in question 7 and 8.

### Question 7.

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

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

**Solution: **

For the given discrete frequency distribution we have to find a mean deviation about the median.

Step 1:Let us make a table of given data and append another column of cumulative frequencies.The given observations are already in ascending order.

x_{i}f_{i}C.F|x_{i }– M|f_{i }* |x_{i }– 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 Total26 84

Step 2:Identifying the observation whose cumulative frequency is equalto or just greater than N / 2 and then Finding the median.

N = ∑f

_{i }= 26 is even. We divide N by 2. Thus, 26/2 = 13The cumulative frequency for greater than 13 is 14, for which corresponding observation is 7

Median = [(N/2)

^{th }observation + ((N/2) + 1)^{th}observation] / 2= (13

^{th}observation + 14^{th}observation) / 2= (7 + 7) / 2 =

7

Step 3:Now, find absolute values of the deviations from median,i.e., |x

_{i }– M| and f_{i }* |x_{i }– M| as shown in above table.∑ f

_{i}* |x_{i}– M| = (16 + 4 + 6 + 10 + 48) = 84Using formula, M.D (M) = ∑ (f

_{i }* |x_{i}– M|)/ N= 84 / 26

= 3.23

So, mean deviation about median for given observations is

3.23

### Question 8.

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

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

**Solution: **

For the given discrete frequency distribution we have to find a mean deviation about the median.

Step 1:Let us make a table of given data and append another column of cumulative frequencies.The given observations are already in ascending order.

x_{i}f_{i}C.F|x_{i }– M|f_{i }* |x_{i }– 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 Total29 148

Step 2:Identifying the observation whose cumulative frequency is equal to orjust greater than N / 2 and then Finding the median.

Here, N = ∑f

_{i}= 29 is odd we divide N by 2. Thus, 29 / 2 = 14.5The cumulative frequency for greater than 14.5 is 21, for which corresponding observation is 30

Therefore, Median = [(N/2)

^{th}observation+ ((N/2) + 1)^{th}observation] / 2= (15

^{th}observation + 16^{th}observation) / 2= (30 + 30)/2 = 30

Step 3:Now, find absolute values of the deviations from median,i.e., |x

_{i }– M| and f_{i}* |x_{i}– M| as shown in above table .∑f

_{i}* |x_{i}– M| = 148Using formula, M.D (M) = ∑(f

_{i }* |x_{i}– M|) / N= 148 / 29

= 5.1

So, mean deviation about median for given observations is

5.1

### Find the mean deviation about the mean for the following data in questions 9 and 10.

### Question 9.

| 0-100 | 100-200 | 200-300 | 300-400 | 400-500 | 500-600 | 600-700 | 700-800 |

| 4 | 8 | 9 | 10 | 7 | 5 | 4 | 3 |

**Solution: **

Given data is in continuous intervals along with their frequencies so it is in the

continuous frequency distribution. In this case, we assume that the frequency

in each class is centered at its mid-point.

Step 1:So, we find a midpoint for each interval.Then we append other columns similar to discrete frequency distribution.

Income per

day in Rs.

Number Of

Person f_{i}

Midpoints

x_{i}f_{i}x_{i}|x_{i }– a|f_{i }* |x_{i }– a|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 Total50 17900 7896

Step 2:Finding the sum of frequencies f_{i}‘s and sum of f_{i}x_{i}‘sN = ∑f

_{i }= 50∑f

_{i}x_{i }= 17900Then mean of given data is given by

a = ∑ (f

_{i}x_{i})/ N= 17900 / 50

a = 358

Step 3:Computing sum of column f_{i}* |x_{i}– a|∑f

_{i}* |x_{i}– a| = 7896Thus mean deviation about mean is given by

M.D (a) = ∑f

_{i }* |x_{i}– a| / N= 7896 / 50

= 157.92

Therefore, mean deviation about mean for given data is157.92

### Question 10.

| 95 – 105 | 105 – 115 | 115 – 125 | 125 – 135 | 135 – 145 | 145 – 155 |

| 9 | 13 | 26 | 30 | 12 | 10 |

**Solution:**

Given data is in the continuous frequency distribution.

Step 1:We find a midpoint for each interval and then append other columns.

Height in

cms

Number of

boys

Midpoints

x_{i}f_{i}x_{i}|x_{i }– a|f_{i }* |x_{i }– a|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 Total100 12530 1128.8

Step 2:Finding the sum of frequencies f_{i}‘s and the sum of f_{i}x_{i}‘sN = ∑f

_{i }= 100∑ f

_{i}x_{i}= 17900Then mean of given data is given by

a = ∑(f

_{i}x_{i}) / N= 12530 / 100

a = 125.3

Step 3:Computing sum of column f_{i }* |x_{i}– a|∑f

_{i }* |x_{i}– a| = 1128.8Thus mean deviation about mean is given by

M.D(a) = ∑(f

_{i}* |x_{i}– a|)/ N= 1128.8 / 100

= 11.288

Therefore, the mean deviation about mean for given data is11.288

### Find the mean deviation about the median for the following data in questions 11 and 12.

### Question 11.

Marks | 0 – 10 | 10 – 20 | 20 – 30 | 30 – 40 | 40 – 50 | 50 – 60 |

| 6 | 8 | 14 | 16 | 4 | 2 |

**Solution:**

Given data is in the continuous frequency distribution &

here the only difference is that we have to calculate the median.

Step 1:First we have to compute cumulative frequencies, then we find a midpoint for each interval and then append other columns.The data is already arranged in ascending order

Marks

Number of

Girls f_{i}C.F

Midpoints

x_{i}|x_{i }– M|f_{i }* |x_{i }– M|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 Total50 517.1

Step 2:First identifying the interval in which median lies and then applying the formula to compute medianThe class interval whose cumulative frequency is greater than equal to N/2 = 25 is 20 – 30.

So, 20 – 30 is the median class.

Then applying the formula

Median (M) = l + {[(N / 2) – C] / f} * hwhere, l = lower limit of the median class

h = width of median class

N = sum of frequencies

C = cumulative frequency of the class just preceding the median class

Therefore,

M = 20 + {[(25 – 14) / 14] * 10}

M = 27.85

Step 3:Finding absolute values of the deviations from median as shown in tablecomputing sum of column f

_{i}* |x_{i}– M|∑f

_{i}* |x_{i}– M| = 517.1Thus, mean deviation about median is given by

M.D(M) = ∑f

_{i}* |x_{i}– M| / N= 517.1 / 50

= 10.34

So, mean deviation about median for given observations is10.34

### Question12. Calculate the mean deviation about median age for the age distribution of 100 persons given below:

| 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 |

**Solution:**

Converting 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.

Check if data is arranged in ascending order.

Step 1:Finding midpoints and C.Fs and then appending other columns

Age

(in years)

Number

f_{i}C.F

Midpoints

x_{i}|x_{i }– M|f_{i }* |x_{i}– M|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 71 43 5 60 45.5-50.5 16 95 48 10 160 50.5-55.5 9 100 53 15 135 Total100 735

Step 2:First identifying the interval in which median lies and then applying the formula to compute medianThe class interval whose cumulative frequency is greater than equal to N/2 = 50 is 35.5-40.5.

So, 35.5-40.5 is the median class.

Then applying the formula

Median (M) = l + {[(N / 2) – C] / f} * hwhere l = lower limit of the median class = 35.5

h = width of median class = 5

N = sum of frequencies = 100

C = cumulative frequency of the class just preceding the median class = 37

f = frequency = 26

Therefore,

M = 35.5 + {[(50 – 37) / 26] * 5}

M = 38

Step 3:Finding absolute values of the deviations from median as shown in tablecomputing sum of column f

_{i}* |x_{i }– M|∑f

_{i}* |x_{i}– M| = 735Thus, mean deviation about median is given by

M.D(M) = ∑f

_{i}* |x_{i }– M| / N= 735 / 100

= 7.35

So, mean deviation about median for given observations is7.35

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