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# Class 10 RD SHARMA Solutions Maths Chapter 15 - Statistics

## Statistics Exercise Ex. 15.3

### Solution 2

Let us find class marks (xi) for each interval by using the relation.

Now we may compute xi and fixias following

 Number of plants Number of houses (fi) xi fixi 0 - 2 1 1 1 x 1 = 1 2 - 4 2 3 2 x 3 = 6 4 - 6 1 5 1 x 5 = 5 6 - 8 5 7 5 x 7 = 35 8 - 10 6 9 6 x 9 = 54 10 - 12 2 11 2 x 11 = 22 12 - 14 3 13 3 x 13 = 39 Total 20 162

From the table we may observe that

So, mean number of plants per house is 8.1.
We have used here direct method as values of class marks (xi) and fi are small.

### Solution 4

We may find class mark of each interval (xi) by using the relation.

Class size h of this data = 3
Now taking 75.5 as assumed mean (a) we may calculate di, ui, fiui as following.

 Number of heart beats per minute Number of women fi xi di = xi -75.5 fiui 65 - 68 2 66.5 - 9 - 3 - 6 68 - 71 4 69.5 - 6 - 2 - 8 71 - 74 3 72.5 - 3 - 1 - 3 74 - 77 8 75.5 0 0 0 77 - 80 7 78.5 3 1 7 80 - 83 4 81.5 6 2 8 83 - 86 2 84.5 9 3 6 Total 30 4

Now we may observe from table that

So mean hear beats per minute for these women are 75.9 beats per minute.

### Solution 21

 Number of mangoes Number of boxes fi 50 - 52 15 53 - 55 110 56 - 58 135 59 - 61 115 62 - 64 25

We may observe that class intervals are not continuous. There is a gap of 1 between two class intervals. So we have to add to upper class limit and subtract from lower class limit of each interval.
And class mark (xi) may be obtained by using the relation

Class size (h) of this data = 3

Now taking 57 as assumed mean (a) we may calculate di, ui, fiui as follows:

 Class interval fi xi di = xi - 57 fiui 49.5 - 52.5 15 51 -6 -2 -30 52.5 - 55.5 110 54 -3 -1 -110 55.5 - 58.5 135 57 0 0 0 58.5 - 61.5 115 60 3 1 115 61.5 - 64.5 25 63 6 2 50 Total 400 25

Now, we have:

Clearly mean number of mangoes kept in a packing box is 57.19.

Note: We have chosen step deviation method here as values of fi, di are big and also there is a common multiple between all di.

### Solution 22

We may calculate class mark (xi) for each interval by using the relation

Class size = 50

Now taking 225 as assumed mean (a) we may calculate di, ui, fiui as follows:

 Daily expenditure (in Rs) fi xi di = xi - 225 fiui 100 - 150 4 125 -100 -2 -8 150 - 200 5 175 -50 -1 -5 200 - 250 12 225 0 0 0 250 - 300 2 275 50 1 2 300 - 350 2 325 100 2 4 Total -7

Now we may observe that -

### Solution 23

 Concentration of SO2 (in ppm) Frequency Class mark xi di = xi - 0.14 fiui 0.00 - 0.04 4 0.02 -0.12 -3 -12 0.04 - 0.08 9 0.06 -0.08 -2 -18 0.08 - 0.12 9 0.10 -0.04 -1 -9 0.12 - 0.16 2 0.14 0 0 0 0.16 - 0.20 4 0.18 0.04 1 4 0.20 - 0.24 2 0.22 0.08 2 4 Total 30 -31

### Solution 25

We may find class marks by using the relation

Class size (h) for this data = 10

Now taking 70 as assumed mean (a) we may calculate di, ui, and fiui as follows:

 Literacy rate (in %) Number of cities fi xi di= xi - 70 ui =di/10 fiui 45 - 55 3 50 -20 -2 -6 55 - 65 10 60 -10 -1 -10 65 - 75 11 70 0 0 0 75 - 85 8 80 10 1 8 85 - 95 3 90 20 2 6 Total 35 -2

Now we may observe that

So, mean literacy rate is 69.43%.

### Solution 26

Here, we have cumulative frequency distribution less than type. First we convert it into an ordinary frequency distribution.

### Solution 5

Let the assumed mean be A = 8

Here, h = 2

 Class interval Mid value xi di = xi - 8 fi fiui 3 - 5 5 - 7 7 - 9 9 - 11 11 - 13 4 6 8 10 12 -4 -2 0 2 4 -2 -1 0 1 2 5 10 10 7 8 -10 -10 0 7 16 N = 40

### Solution 24 (i)

We may find class mark of each interval by using the relation

 Number of days Number of students fi xi di = xi - 21 fidi 0 - 6 6 - 12 12 - 18 18 - 24 24 - 30 30 - 36 36 - 42 10 11 7 4 4 3 1 3 9 15 21 27 33 39 -18 -12 -6 0 6 12 18 -180 -132 -42 0 24 36 18 40

Assumed mean A = 21

Hence, the number of days a student was absent is 14.1.

### Solution 30

We may find class mark of each interval by using the relation

 Marks Number of students fi xi di = xi - 47.5 fidi 30 - 35 35 - 40 40 - 45 45 - 50 50 - 55 55 - 60 60 - 65 14 16 28 23 18 8 3 32.5 37.5 42.5 47.5 52.5 57.5 62.5 -15 -10 -5 0 5 10 15 -210 -160 -140 0 90 80 45 110

Assumed mean A = 47.5

Hence, mean marks of the students is 44.81.

## Statistics Exercise Ex. 15.4

### Solution 2

The median height of the students is Rs 167.13.

### Solution 8

We can find cumulative frequencies with their respective class intervals as below -

 Life time Number of lamps (fi) Cumulative frequency 1500 - 2000 14 14 2000 - 2500 56 14 + 56 = 70 2500 - 3000 60 70 + 60 = 130 3000 - 3500 86 130 + 86 = 216 3500 - 4000 74 216 + 74 = 290 4000 - 4500 62 290 + 62 = 352 4500 - 5000 48 352 + 48 = 400 Total (n) 400

Now we may observe that cumulative frequency just greater than is 216 belonging to class interval 3000 - 3500.
Median class = 3000 - 3500
Lower limit (l) of median class = 3000
Frequency (f) of median class = 86
Cumulative frequency (cf) of class preceding median class = 130
Class size (h) = 500

So, median life time of lamps is 3406.98 hours.

### Solution 9

We may find cumulative frequencies with their respective class intervals as below

 Weight (in kg) 40 - 45 45 - 50 50 - 55 55 - 60 60 - 65 65 - 70 70 - 75 Number of students (f) 2 3 8 6 6 3 2 c.f. 2 5 13 19 25 28 30

Cumulative frequency just greater than   is 19, belonging to class interval 55 - 60.
Median class = 55 - 60
Lower limit (l) of median class = 55
Frequency (f) of median class = 6
Cumulative frequency (cf) of median class = 13
Class size (h) = 5

So, median weight is 56.67 kg.

### Solution 17

Here class width is not same. There is no need to adjust the frequencies according to class intervals. Now given frequency table is of less than type represented with upper class limits. As policies were given only to persons having age 18 years onwards but less than 60 years, we can define class intervals with their respective cumulative frequency as below

 Age (in years) Number of policy holders (fi) Cumulative frequency (cf) 18 - 20 2 2 20 - 25 6 - 2 = 4 6 25 - 30 24 - 6 = 18 24 30 - 35 45 - 24 = 21 45 35 - 40 78 - 45 = 33 78 40 - 45 89 - 78 = 11 89 45 - 50 92 - 89 = 3 92 50 - 55 98 - 92 = 6 98 55 - 60 100 - 98 = 2 100 Total (n)

Now from table we may observe that n = 100.

Cumulative frequency (cf) just greater than  is 78 belonging to interval 35 - 40
So, median class = 35 - 40
Lower limit (l) of median class = 35
Class size (h) = 5
Frequency (f) of median class = 33
Cumulative frequency (cf) of class preceding median class = 45

So, median age is 35.76 years.

### Solution 18

The given data is not having continuous class intervals. We can observe that difference between two class intervals is 1. So, we have to add and subtract

to upper class limits and lower class limits.
Now continuous class intervals with respective cumulative frequencies can be represented as below:

 Length (in mm) Number or leaves fi Cumulative frequency 117.5 - 126.5 3 3 126.5 - 135.5 5 3 + 5 = 8 135.5 - 144.5 9 8 + 9 = 17 144.5 - 153.5 12 17 + 12 = 29 153.5 - 162.5 5 29 + 5 = 34 162.5 - 171.5 4 34 + 4 = 38 171.5 - 180.5 2 38 + 2 = 40

From the table we may observe that cumulative frequency just greater then
is 29, belonging to class interval 144.5 - 153.5.
Median class = 144.5 - 153.5
Lower limit (l) of median class = 144.5
Class size (h) = 9
Frequency (f) of median class = 12
Cumulative frequency (cf) of class preceding median class = 17

So, median length of leaves is 146.75 mm.

### Solution 4

The cumulative frequency table is

 Salary Frequency fi Cumulative frequency (c.f.) 5 - 10 10 - 15 15 - 20 20 - 25 25 - 30 30 - 35 35 - 40 40 - 45 45 - 50 49 133 63 15 6 7 4 2 1 49 182 245 260 266 273 277 279 280

The cumulative frequency greater than and nearest to 140 is 182.

So, median class is 10 - 15

Lower limit (l) = 10, class size (h) = 5, c.f. = 49

Frequency of median class = 133

Hence, mean median salary is Rs. 13421.

## Statistics Exercise Ex. 15.5

### Solution 7

We may compute class marks (xi) as per the relation

Now taking 30 as assumed mean (a) we may calculate di and fidi as follows.

 Age (in years) Number of patients fi class mark xi di= xi - 30 fidi 5 - 15 6 10 -20 -120 15 - 25 11 20 -10 -110 25 - 35 21 30 0 0 35 - 45 23 40 10 230 45 - 55 14 50 20 280 55 - 65 5 60 30 150 Total 80 430

From the table we may observe that

Clearly, mean of this data is 35.38. It represents that on an average the age of a patient admitted to hospital was 35.38 years.
As we may observe that maximum class frequency is 23 belonging to class interval 35 - 45.
So, modal class = 35 - 45
Lower limit (l) of modal class = 35
Frequency (f1) of modal class = 23
Class size (h) = 10
Frequency (f0) of class preceding the modal class = 21
Frequency (f2) of class succeeding the modal class = 14

Clearly mode is 36.8.It represents that maximum number of patients admitted in hospital were of 36.8 years.

### Solution 8

From the data given as above we may observe that maximum class frequency is 61 belonging to class interval 60 - 80.
So, modal class = 60 - 80
Lower class limit (l) of modal class = 60
Frequency (f1) of modal class = 61
Frequency (f0) of class preceding the modal class = 52
Frequency (f2) of class succeeding the modal class = 38
Class size (h) = 20

So, modal lifetime of electrical components is 65.625 hours.

### Solution 10

We may observe from the given data that maximum class frequency is 10 belonging to class interval 30 - 35.
So, modal class = 30 - 35
Class size (h) = 5
Lower limit (l) of modal class = 30
Frequency (f1) of modal class = 10
Frequency (f0) of class preceding modal class = 9
Frequency (f2) of class succeeding modal class = 3

It represents that most of states/U.T have a teacher - student ratio as 30.6

Now we may find class marks by using the relation

Now taking 32.5 as assumed mean (a) we may calculate di, ui and fiui as following.

 Number of students per teacher Number of states/U.T (fi) xi di = xi - 32.5 ui fiui 15 - 20 3 17.5 -15 -3 -9 20 - 25 8 22.5 -10 -2 -16 25 - 30 9 27.5 -5 -1 -9 30 - 35 10 32.5 0 0 0 35 - 40 3 37.5 5 1 3 40 - 45 0 42.5 10 2 0 45 - 50 0 47.5 15 3 0 50 - 55 2 52.5 20 4 8 Total 35 -23

So mean of data is 29.2
It represents that on an average teacher - student ratio was 29.2.

### Solution 12

From the given data we may observe that maximum class frequency is 20 belonging to 40 - 50 class intervals.
So, modal class = 40 - 50
Lower limit (l) of modal class = 40
Frequency (f1) of modal class = 20
Frequency (f0) of class preceding modal class = 12
Frequency (f2) of class succeeding modal class = 11
Class size = 10

So mode of this data is 44.7 cars.

### Solution 16

We may observe from the given data that maximum class frequency is 40 belonging to 1500 - 2000 intervals.
So, modal class = 1500 - 2000
Lower limit (l) of modal class = 1500
Frequency (f1) of modal class = 40
Frequency (f0) of class preceding modal class = 24
Frequency (f2) of class succeeding modal class = 33
Class size (h) = 500

So modal monthly expenditure was Rs. 1847.83
Now we may find class mark as

Class size (h) of give data = 500
Now taking 2750 as assumed mean (a) we may calculate di, ui and fiui as follows:

 Expenditure (in Rs) Number of families fi xi di = xi - 2750 ui fiui 1000 - 1500 24 1250 -1500 -3 -72 1500 - 2000 40 1750 -1000 -2 -80 2000 - 2500 33 2250 -500 -1 -33 2500 - 3000 28 2750 0 0 0 3000 - 3500 30 3250 500 1 30 3500 - 4000 22 3750 1000 2 44 4000 - 4500 16 4250 1500 3 48 4500 - 5000 7 4750 2000 4 28 Total 200 -35

Now from table may observe that

So, mean monthly expenditure was Rs. 2662.50.

### Solution 17

From the given data we may observe that maximum class frequency is 18 belonging to class interval 4000 - 5000.
So, modal class = 4000 - 5000
Lower limit (l) of modal class = 4000
Frequency (f1) of modal class = 18
Frequency (f0) of class preceding modal class = 4
Frequency (f2) of class succeeding modal class = 9
Class size (h) = 1000

So mode of given data is 4608.7 runs.

### Solution 3 (iv)

The cumulative frequency table is

 Salary Frequency fi 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 60 - 70 8 10 10 16 12 6 7

Modal class is 30 - 40

Lower limit (l) = 30, class size (h) = 10, f = 16, f1 = 10 and f2 = 12

Hence, the mode is 36.

## Statistics Exercise Ex. 15.6

### Solution 8

Less Than Series:

 Class interval Cumulative Frequency Less than 10 22 Less than 20 32 Less than 30 40 Less than 40 55 Less than 50 60 Less than 60 66

We plot the points (10, 22), (20, 32), (30, 40), (40, 55), (50, 60) and (60, 66) to get 'less than type' ogive.

More Than Series:

 Class interval Frequency More than 0 66 More than 10 44 More than 20 34 More than 30 26 More than 40 11 More than 50 6

We plot the points (0, 66), (10, 44), (20, 24), (30, 26), (40, 11), and (50, 6) to get more than ogive.

From the graph, median = 21.25 cm

### Solution 9

More than type frequency distribution is given by

 Class interval Frequency fi Cumulative frequency More than 20 More than 30 More than 40 More than 50 More than 60 More than 70 More than 80 10 8 12 24 6 25 15 100 90 82 70 46 40 15

So, more than type frequency curve is

### Solution 10

Less than type frequency distribution is given by

 Class interval Frequency fi Cumulative frequency Less than 220 Less than 240 Less than 260 Less than 280 Less than 300 12 14 8 6 10 12 26 34 40 50

So, less than type frequency curve is

## Statistics Exercise 15.66

### Solution 1

There are three main measure of central tendency the mode, the median and the mean.

Each of these measures describes a different indication of the typical or central value in the distribution.

The mode is the most commonly occuring value in a distribution.

Median is middle value of distribution.

While standard deviation is a measure of dispersion of a set of data from its mean.

So, the correct option is (d).

### Solution 4

It is well known that relation between mean, median and mode 1 s

3 median = mode + 2 mean

Mode = 3 median - 2 mean

So, the correct option is (a).

### Solution 5

Mean is an average value of any given data which cannot be determined by a graph.

Value of median and mode can easily be calculated by graph.

Median is middle value of a distribution and mode is highest frequent value of a given distribution.

So, the correct option is (a).

### Solution 6

The median of a series may be determined through the graphical presentation of data in the forms of Ogives.

Ogive is a curve showing the cummulative frequency for a given set of data.

To get the median we present the data graphically in the form of 'less than' ogive  or 'more than' ogive

Then the point of intersection of the two graphs gives the value of the median.

So, the correct option is (d).

### Solution 7

Histogram is used to plot the distribution of numerical data or frequency of occurrences of data.

Mode is the most commonly occurring value in the data.

So in distribution or Histogram, the value of the x-coordinate corresponding to the peak value on y - axis, is the mode.

So, the correct option is (a).

### Solution 8

Mode is the most frequent value in the data.

Mode is the value which occurs the most number of times.

So, the correct option is (c).

### Solution 10

We know that the relation between mean, median & mode is

3 Median = Mode + 2 Mean

Hence, Mode = 3 Median - 2 Mean

So, the correct option is (c).

## Statistics Exercise 15.67

### Solution 12

We know that the relation between mean, median & mode is

3 Median = mode + 2 Mean

Hence, mode = 3 Median - 2 Mean

So, the correct option is (d).

### Solution 18

Mode is the number in observation data is that which repeats most number of time

In the given data 48 comes twice and 43 comes twice but mode is 43.

Hence if x = 43 then 43 comes thrice.

So x + 3 = 43 + 3 = 46

So, the correct option is (c).

### Solution 19

In the given data 15, 16, 17 comes twice but given 15 is mode.

Hence 15 comes more than 16, 17.

This is only possible if x = 15.

So, the correct option is (a).

## Statistics Exercise 15.68

### Solution 25

We know, 3 Median = Mode + 2 Mean

mean = 24

mode = 12

3 median = 12 + 2 × 24

= 12 + 48

= 60

median = 20

So, the correct option is (c).

### Solution 36

While computing the mean of the grouped data, we assume that the frequencies are centred at the class marks of the classes.

Hence, correct option is (b).

## Statistics Exercise 15.69

### Solution 41

di's are the deviations from a of mid-points of classes.

Hence, correct option is (c).

### Solution 42

The abscissa of the point of intersection of less than type and of the more than type cumulative frequency curves of a grouped data given its median.

Hence, correct option is (b).