Measures of Central Tendency ( Mean , Median , Quartiles and Mode ) Concise Solutions Chapter-24 Class -10. Solutions of Chapter-24 Concise Maths for Mean , Median , Quartiles and Mode . Solutions of Exercise – 24 (A), Exercise – 24 (B), Exercise – 24 (C) Exercise – 24 (D), Exercise – 24 (E) for Concise Selina Maths of ICSE Board Class 10th. Selina Publisher Solutions of Concise Maths  Measures of Central Tendency Concise Chapter-24 for ICSE Maths Class 10 is available here. All Solutionsof Concise Selina Maths of Measures of Central Tendency ( Mean , Median , Quartiles and Mode ) Concise Chapter-24 has been solved according instruction given by council.

## Measures of Central Tendency ( Mean , Median , Quartiles and Mode )Concise Solutions Chapter-24 Class -10

This Post is the  Solutions of Concise Maths Measures of Central Tendency  Chapter-24 for ICSE Class 10th. ICSE Maths text book of Concise is In series of famous ICSE writer in maths publications. Concise is most famous among students

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(Mean) Exercise – 24 (A)

(Mean ) Exercise – 24 (B),

(Median) Exercise – 24 (C)

(Mode) Exercise – 24 (D),

(Ogive) Exercise – 24 (E),

### How to Solve Concise Maths Selina Publications Chapter-24 Measures of Central Tendency ( Mean , Median , Quartiles and Mode ) Class 10

Note:- Before viewing Solutions of Chapter-24 Measures of Central Tendency of Concise Selina Maths . Read the Chapter-24 Measures of Central Tendency Carefully and then solve all example of your text bookThe Chapter-24 Measures of Central Tendency is main Chapter in ICSE board.

Measures of Central Tendency EXERCISE – 24 A Concise Maths Solution Chapter – 24 ICSE Class – 10

#### Question 1.

Find the mean of following set of numbers:
(i) 6, 9, 11, 12 and 7
(ii) 11, 14, 23, 26, 10, 12, 18 and 6.

#### Question 2.

Marks obtained (in mathematics) by a students are given below :
60, 67, 52, 76, 50, 51, 74, 45 and 56
(a) Find the arithmetic mean
(b) If marks of each student be increased by 4;
what will be the new value of arithmetic mean.

(a) Hence x = 9

(b) If marks of each students be increased by 4 then new mean will be = 59 + 4 = 63

#### Question 3.

Find the mean of natural numbers from 3 to 12.

Numbers betweeen 3 to 12 are 3, 4, 5, 6, 7, 8, 9, 10, 11, 12.
Here n = 10

#### Question 4.

(a) Find the means of 7, 11, 6, 5 and 6. (b) If each number given in (a) is diminished by 2; find the new value of mean.

(a) The mean of 7, 11, 6, 5 and 6.

(b) If we subtract 2 from each number, then the mean will be 7 – 2 = 5

#### Question 5.

If the mean of 6, 4, 7, a and 10 is 8. Find the value of ‘a’

No. of terms = 5
Mean = 8
∴ Sum of number (Σxi) = 5 x 8 = 40 …(i)
But Σxi = 6 + 4 + 7 + a+10 = 27 + a ….(ii)
From (i) and (ii)
27 + a = 40 ⇒ a = 40 – 27
∴ a = 13

#### Question 6.

The mean of the number 6, y, 7, x and 14 is 8. Express y in terms of x.

No. of terms = 5 and
mean = 8
∴ Sum of numbers (Σxi) = 5 x 8 = 40 ….(i)
But sum of numbers given = 6 + y + 7 + x + 14
= 21 + y + x + ….(ii)
From (i) and (ii)
27 + y + x = 40
⇒ y = 40 – 27 – x
⇒ y= 13 – x

#### Question 7.

The ages of 40 students are given in the following table
Find the arithmetic mean.

#### Question 8.

If 69.5 is the mean of 72, 70, x, 62, 50, 71, 90, 64, 58 and 82, find the value of x.

No. of terms = 10
Mean = 69.5
∴ Sum of numbers = 69,5 x 10 = 695 ….(i)
But sum of given number = 72 + 70 + x + 62 + 50 + 71 + 90 + 64 + 58 + 82 = 619+x ….(ii)
From (i) and (ii)
619 + x = 695
⇒ x = 695 – 619 = 76

#### Question 9.

The following table gives the heights of plants in centimeter. If the mean height of plants is 60.95 cm; find the value of ‘f’.

#### Question 10.

From the data given below, calculate the mean wage, correct to the nearest rupee.
(i) If the number of workers in each category is doubled, what would be the new mean wage? [1995]
(ii) If the wages per day in each category are increased by 60%; what is the new mean wage?

(iii) If the number of workers in each category is doubled and the wages per day per worker are reduced by 40%. What would be the new mean wage?

(i) Mean remains the same if the number of workers in each catagory is doubled.
∴ Mean = 80.
(ii) Mean will be increased by 60% if the wages per day per worker is increased by 60%.
∴ New mean = 80 x $\frac { 160 }{ 100 }$= 128
(iii) No change in the mean if the number of worker is doubled but if wages per worker is reduced by 40%, then
New mean = 80 x $\frac { 60 }{ 100 }$= 48

#### Question 11.

The contents of 100 match boxes were checked to determine the number of matches they contained.
(i) Calculate, correct to one decimal place, the mean number of matches per box.
(ii) Determine how many extra matches would have to be added to the total contents of the 100 boxes to bring the mean up to exactly 39 matches. [1997]

(ii) In the second case,
New mean = 39 matches
∴ Total contents = 39 x 100 = 3900
But total no of matches already given = 3813
∴ Number of new matches to be added = 3900 – 3813 = 87

#### Question 12.

If the mean of the following distribution is 3, find the value of p.

Mean = 3

#### Question 13.

In the following table, Σf= 200 and mean = 73. Find the missing frequencies f1 and f2.

#### Question 14.

Find the arithmetic mean (correct to the nearest whole-number) by using step-deviation method.

Let the Assumed mean = 30

#### Question 15.

Find the mean (correct to one place of decimal) by using short-cut method.

Let the Assumed mean A = 45

### Solutions of Exe-24(B) Measures of Central Tendency ( Mean , Median , Quartiles and Mode ) Chapter-24 Concise Maths

#### Question 1.

The following table gives the ages of 50 students of a class. Find the arithmetic mean of their ages.

Question 2.

The following table gives the weekly wages of workers in a factory.
Calculate the mean by using :
(i) Direct Method
(ii) Short-Cut Method

(i) Direct Method:

#### Question 3.

The following are the marks obtained by 70 boys in a class test :
Calculate the mean by :
(i) Short-Cut Method
(ii) Step-Deviation Method.

(i) Short cut Method :

#### Question 4.

Find mean by ‘step-deviation method :

#### Question 5.

The mean of following frequency distribution is 21$\frac { 1 }{ 7 }$ Find the value of ‘f ‘.

#### Question 6.

Using step-deviation method, calculate the mean marks of the following distribution.

Let Assumed mean = 72.5

#### Question 7.

Using the information given in the adjoining histogram; calculate the mean.

#### Question 8.

If the mean of the following observations is 54, find the value of p.

Mean = 54

⇒ 2106 + 54p = 2370 + 30p
⇒ 54p – 30p = 2370 – 2106 ⇒ 24p = 264
p = $\frac { 264 }{ 24 }$ = 11
Hence p = 11

#### Question 9.

The mean of the following distribution is 62.8 and the sum of all the frequencies is 50. Find the missing frequencies f1 and f2.

Mean = 62.8
and sum of frequencies = 50

#### Question 10.

Calculate the mean of the distribution given below using the short cut method.

#### Question 11.

Calculate the mean of the following distribution :

### Concise Maths Solutions Exe – 24 (C)Measures of Central Tendency ( Mean , Median , Quartiles and Mode ) Chapter-24

#### Question 1.

A student got the following marks in 9 questions of a question paper : 3, 5, 7, 3, 8, 0, 1, 4 and 6.
Find the median of these marks.

Arranging the given data in descending order, we get:
8, 7, 6, 5,4,3, 3, 1,0
The middle term is 4 which is 5th terms
∴ Median = 4

#### Question 2.

The weights (in kg) of 10 students of a class are given below :
21, 28.5, 20.5, 24, 25.5, 22, 27.5, 28, 21, 24 Find the median of their weights.

Arranging the given data in descending order.
We get 23.5, 28, 27.5, 25.5, 24, 24, 22, 21, 21, 20.5
the middle terms are 24 and 24, 5th and 6th terms.

#### Question 3.

The marks obtained by 19 students of a class are given below :
27, 36, 22, 31, 25, 26, 33, 24, 37, 32, 29, 28, 36, 35, 27, 26, 32, 35, 28. Find :
(i) Media
(ii) Lower quartile
(iii) Upper quartile
(iv) Inter – quartle range

(i) Arranging in order say ascending, we get
22, 24, 25, 26, 26, 27, 27, 28, 28, 29, 31, 32, 32, 33, 35, 35,36, 36, 37
Middle term is 10th term i.e. 29
∴ Median = 29

#### Question 4.

From the following data, find :
(i) Median
(ii) Upper quartile
(iii) Inter-quartile range
25,10, 40, 88, 45, 60, 77, 36,18, 95, 56, 65, 7, 0, 38 and 83.

(i) Arrange in ascending order, we get
0,7, 10, 18, 25, 36, 38, 40, 45, 56, 60, 65 ,77, 83, 88, 95

#### Question 5.

The ages of 37 students in a class are given in the following table :

Find the median.

#### Question 6.

The weight of 60 boys are given in the following distribution table :

Find : (i) Median
(ii) Lower quartile
(iii) Upper quartile
(iv) Inter-quartile range

#### Question 7.

Estimate the median for the given data by drawing ogive :

Through mark of 25.5 on the y-axis, draw a line parallel to x-axis which meets the curve at A. From A, draw a perpendicular to x-axis, which meets x-axis at B.
∴ The value of B is the median which is 28.

#### Question 8.

By drawing an ogive; estimate the median for the following frequency distribution :

Through mark of 28th on the y- axis, draw a line parallel to x-axis which meets the curve at A. From A, draw a perpendicular line segment to x- axis. Which meets it at B.

∴ The value of B is the median which is 18.4

#### Question 9.

From the following cumulative frequency table, draw ogive and then use it to find :
(i) Median,
(ii) Lower quartile,
(iii) Upper quartile.

No. of terms = 80
Median = 40th term Through mark of 40 draw a line parallel to x-axis which meets the curve at A. From A, draw a perpendicular to x-axis which meets it at B
(ii) Lower quartile (Q1) = $\frac { n }{ 4 }$ th term
$\frac { 80 }{ 4 }$ th term (Here n = 80 which is even)
= 20th term =18
(iii) Upper quartile (Q1) = $\frac { 3 }{ 4 }$ nth term =$\frac { 3 x80 }{ 4 }$ = 60th term = 66 .

∴ Value of B is the median which is 40.

#### Question 10.

In a school 100 pupils have heights as tabulated below:

Find the median height by drawing an ogive.

Through mark 50, draw a line parallel to x- axis which meets the curve at A. From A, draw perpendicular to x-axis which meets x-axis at B is the median which is 148 cm.

### Exe – 24 (D) Selina Concise Maths Solutions of Chapter-24 Measures of Central Tendency ( Mean , Median , Quartiles and Mode )

#### Question 1.

Find the mode of the following data:
(i) 7,9,8,7,7,6,8,10,7 and 6
(ii) 9,11,8,11,16,9,11,5,3,11,17 and 8

(i) Mode = 7
because it occurs 4 times
(ii) Mode =11
because it occurs 4 times

#### Question 2.

The following table shows the frequency distribution of heights of 50 boys:
Find the mode of heights.

Mode is 122 because it occurs maximum times i.e its., frequency is 18.

#### Question 3.

Find the mode of following data, using a histogram:

Mode class = 20 – 30
Mode = 24

We see in the histogram that line AD and CB intersect at P. Draw perpendicular Q to the horizontal x-axis. Which is the value of the mode = 24

#### Question 4.

The following table shows the expenditure of 60 boys on books. Find the mode of their expenditure:

Model class is = 30 – 35
and Mode = 34

We see in the histogram that line AD and CB intersect at P. Draw perpendicular Q to the horizontal axis. Which is the value of the mode.

#### Question 5.

Find the median and mode for the set of numbers 2,2,3,5,5,5,6,8 and 9.

Median = $\frac { 9 +1 }{ 2 }$ = 5th term which is 5
Mode = 5, because it occurs in maximum times.

#### Question 6.

A boy scored following marks in various class tests during a term, each test being marked out of 20.
15,17,16,7,10,12,14,16,19,12,16.
(i) What are his modal marks ?
(ii) What are his median marks ?
(iii) What are his total marks ?
(iv) What are his mean marks ?

Arranging the given data in ascending order : 7, 10,12, 12,14, 15,16,16, 16, 17,19.
(i) Mode = 16 as it occurs in maximum times.
(ii) Median= $\frac { 11 +1 }{ 2 }$ = 6th term which is 15
(iii) Total marks = 7 + 10+ 12+ 12+ 14+ 15+ 16 + 16+ 16+ 17+ 19= 154

#### Question 7.

Find the mean, median and mode of the following marks obtained by 16 students in a class test marked out of 10 marks.
0,0,2,2,3,3,3,4,5,5,5,5,6, 6,7,8

(iii) Mode = 5 as it occurs in maximum times.

#### Question 8.

At a shooting competition the score of a com-petitor were as given below :

(i) What was his modal score ?
(ii) What was his median score ?
(iii) What was his total score ?
(iv) What was his mean score ?

### Chapter-24 Measures of Central Tendency ( Mean , Median , Quartiles and Mode ) Selina Concise Maths Solutions of Exe – 24(E)

#### Question 1.

The following distribution represents the height of 160 students of a school
Draw an ogive for the given distribution taking 2 cm = 5 cm of height on one axis and 2 cm = 20 students on the other axis. Using the graph, determine :
(i) The median height.
(ii) The inter quartile range.
(iii) The number of students whose height is above 172

The cumulative frequency table may be prepared as follows :

Now, we take height along x-axis and number of students along the y-axis. Now, plot the point (145, 12), (150, 32), (155, 62), (160, 100), (165, 124), (170, 140), (175, 152) and (180, 160). On the graph paper and join them with free hand.
(i) Here N = 160 ⇒ $\frac { N }{ 2 }$ = 80
Which is even now take a point A on the y-axis representing 80. Through A draw horizontal line meeting the ogive at B. From B, draw BC ⊥ x-axis, meeting the x-axis at C.
The abscissa of C is 157.5 So, median = 157.5 cm
(ii) Proceeding in the same way as we have done in above, we have, Q1 = 152 and Q3 = 164
So, inter quartile range = Q3 – Q1 = 164 – 152 = 12 cm

(iii) From the ogive, we see that the number of students whose height is less than 172 is 145.
No. of students whose height is above 172 cm = 160 – 145 = 15

##### PQ

The following table gives the weekly wages of workers in a factory.

Calculate : (i) the mean, (ii) the modal class, (iii) the number of workers getting weekly w ages below Rs. 80 and (iv) the number of workers get¬ting Rs 65 or more but less than Rs. 85 as weekly wages. [2002]

(ii) Modal class 55-60 (It has maximum frequency)
(iii) No. of workers getting wages below Rs. 80 = 60
(iv) No. of worker getting Rs. 65 is more but less than 85 as weekly wages = 37

#### Question 2.

Draw an ogive for the data given below and from the graph determine :
(i) the median marks,
(ii) the number of students who obtained more than 75% marks ?

Through 60.5 th marks, draw a line segment parallel to x-axis which meets the curve at A.
From A, draw a line segment perpendicular to, x-axis meeting at B.
∴ B is the median = 43 (approx.)
No. of students who obtained upto 75% marks in the test =111
∴ No. of students who obtained more than 75% = 120- 111 =9

#### Question 3.

The mean of 1, 7, 5, 3, 4, and 4 is m. The numbers 3,2,4,2,3,3 and p have mean m-1 and median q. Find p and q.

Mean of 1,7, 5, 3,4, and 4 = $\frac { 24 }{ 6 }$ =4
∴ m = 4.
Now mean of 3,2,4,2,3,3 and p = m- 1= 4- 1 = 3
i.e. 17+p = 3xn when n =7
17 + p = 3×7 = 21
⇒ p = 21 – 17 = 4
Median of 3, 2,4,2, 3, 3 and 3 is q
Arranging in ascending order, 2,2, 3,3,3,3,4,4
Mean = 4th terms is 3.
∴ q = 3

#### Question 4.

In a malaria epidemic, the number of cases diagnosed were as follows :
Date (July) 1 2 3 4 5 6 7 8 9 10 11 12 Number 5 12 20 27 46 30 31 18 11 5 0 1 On what days does the mode, the upper and lower quartiles occur ?

(i) Mode = 5th July (because it has the maximum frequencies i,e. 46)

#### Question 5.

The incomes of the parents of 100 students in a class in a certain university are tabulated below :

(i) Draw a cumulative frequency curve to estimate the median income.
(ii) If 15 % of the students ae given freeships on the basis of the income of their parents, find the annual income of parents, below which the freeships will be awarded.
(iii) Calculate the Arithmetic mean.

(Cummulative  Frequancy table)

(i) No. of terms= 100
∴ Mean = $\frac { 100 }{ 2 }$ = 50th term
Through 50 mark, draw a line segment parallel to x-axis which meets the curve at A. From A, draw a perpendicular to x-axis meeting it at B, B is the median.
∴ B = 17.6 thousands
(ii) Upper quartile = 100 x $\frac { 3 }{ 4 }$ = 75th term
From the Curve Q3 = 23.2
(iii) Lower-quartile = 100 x $\frac { 1 }{ 4 }$=25th term
From the curve Q1 = 12.8
∴ Inter-quartile range = Q3– Q1 = 23.2 – 12.8
= 10.4 thousands
(iv) 15% of 100 students = $\frac { 100 x 15 }{ 100 }$ = 15
From C.F. 15, draw a horizontal line which intersects the curve at P. From P, draw a perpendicular to x-axis meeting it at 11.2
∴ Freeship to parents = Rs. 11.2 thousands upto
the income of Rs. 11.2 thousands

#### Question 6.

The marks of 20 students in a test were as follows : 2, 6, 8, 9, 10, 11, 11, 12, 13, 13, 14, 14, 15, 15, 15, 16, 16, 18, 19 and 20. Calculate-
(i) the mean
(ii) the median
(iii) the mode. [2002]

Arranging in ascending order,
2, 6, 8, 9, 10, 11, 11, 12, 13, 13, 14, 14, 15, 15, 15, 15, 16, 16, 18, 19, 20
No. of terms = 20
Ix = 2 + 6 + 8 + 9+ 10+ 11 + 11 + 12+ 13 + 13 + 14 + 14 + 14 + 15 + 15 + 15 + 16 + 16 + 18 +
19 + 20 = 257

(iii) Mode = 15 (as it has maximum frequency i.e. it has 3)

#### Question 7.

The marks obtained by 120 students in a Mathematics test are given below:

Draw an ogive for the given distribution on a graph sheet. Use a suitable scale for your ogive. Use your ogive to estimate :
(i) the median
(ii) the number of students who obtained more than 75% marks in a test ?
(iii) the number of students who did not pass in the test if the pass percentage was 40. [2002]
(iv) the lower quartile

From a point B (30) on v-axis, draw a line parallel to x- axis meeting the curve at Q and from Q Draw a line parallel to .Y-axis meeting it at 30.
∴ Lower quartile = 30
Through 60.5th marks, draw a line segment parallel to x-axis which meets the curve at A. From A, draw a line segment perpendicular to, x-axis meeting at B.
∴ B is the median = 43 (approx.)
No. of students who obtained upto 75% marks in the test = 110
∴ No. of students who obtained more than 75% = 120- 110 = 10
No. of students who obtained less than 40% marks in the test = 52 (∴ in the graph x = 40, y = 52)

Question 8.

Using a graph paper, draw an Ogive for the following distribution which shows a record of the weight in kilograms of 200 students.

Use your Ogive to estimate the following :
(i) The percentage of students weighing 55 kg or more,
(ii) The weight above which the heaviest 30% of the students fall.
(iii) The number of students who are (a) under-weight and (b) over-weight, if 55.70 kg is considered as standard weight. (2005)

∴ Heaviest 60 students in weight = 9 + 20 + 31 = 60 (From the graph, the required weight is 65 kg or more but less than 80 kg)

(iii) Total number of students who are (i) under weight = 47 and (ii) over weight = 152 (∴ Standard weight is 55.70 kg)

Question 9.

The distribution given below shows the marks obtained by 25 students in an aptitude test. Find the mean, median and mode of the distribution.

Mode = Marks with maximum frequency is 6 ∴ Mode = 6

#### Question 10.

The mean of the following distribution is 52 and the frequency of class interval 30-40 is Find ‘f’ .Find ‘ f ‘.

#### Question 11.

The monthly income of a group of 320 employees in a company is given below:

Draw an ogive of the given distribution on a graph sheet taking 2 cm = Rs. 1000 on one axis and 2 cm= 50 employees on the other axis.
From the graph determine
(i)the median wage
(ii)the number of employees whose income is below Rs. 8500.
(iii)If the salary of a senior employee is above Rs. 11,500, find the number of senior employees in the company.
(iv) the upper quartile.

(i)For median wage, Take OP =$\frac { 320 }{ 2 }$ =160 on y-axis, Draw a line PQ || x-axis and from Q,
draw QM ⊥ x-axis, abcissa of M point is 9400 ⇒ Median = Rs. 9400
(ii) Take OM’ = 8500 on.t-axis. Draw Q’M’|| toy-axis and P’Q’ || X-axis
Where ordinate of P’ is 92.5
There are approximately 93 employees whose monthly wage is below Rs. 8500
(iii) There are approximately 18 employees whose salary is above Rs. 11500.
(iv) Upper quartile
Mark a point A ony-axis on $\frac { 3N }{ 4 }$ = $\frac { 3 x 320 }{ 4 }$= 240 and draw a line AB || X-axis, then draw BB’
⊥ x-axis abscissa of B’ is upper quartile i.e., Rs. 10250.

#### Question 12.

A Mathematics aptitude test of 50 students was recorded as follows :

Draw a histogram for the above data using a graph paper and locate the mode.

1. Draw the histogram with given data.
2. Inside the highest rectangle which represents the maximum frequency (or modal class), draw two lines AC and BD diagonally from the upper comer C and D or adjacent rectangle which intersect at K.
3. Draw KL ⊥ X-axis.
Value ofL is the mode which is 82.5 (approx).

#### Question 13.

Marks obtained by 200 students in an examination are given below :

Draw an ogive for the given distribution taking 2 cm = 10 marks on one axis and 2 cm = 20 students on the other axis. Using the graph, determine
(i) The median marks.
(ii) The number of students who failed if minimum marks required to pass is 40.
(iii) If scoring 85 and more marks is considered as grade one, find the number of students who secured grade one in the examination.

(i) Median is 57.
(ii)44 students failed.
(iii) No. of students who secured grade one = 200 – 188 = 12

#### Question 14.

Marks obtained by 40 students in a short assessment are given below, where a and ft are two missing data.

If the mean of the distribution is 7.2, find a and b.

#### Question 15.

Find the mode and median of the following frequency distribution :

#### Question 16.

The median of the following observations 11, 12, 14, (x – 2), (x + 4), (x + 9), 32, 38, 47 arranged in ascending order is 24. Find the value of x and hence find the mean.

#### Question 17.

The numbers 6, 8, 10, 12, 13 and x are arranged in an ascending order. If the mean of the observations is equal to the median, find the value of x. (2014)

#### Question 18.

(Use a graph paper for this question.) The daily pocket expenses of 200 students in a school are given below:

Draw a histogram representing the above distribution and estimate the mode from the graph. (2014)

Steps of construction :
(i) Draw a line BC = 6.5 cm.
(ii) Centre B and C draw arcs AB = 5.5 cm and AC = 5 cm
(iii) Join AB and AC, ABC is the required triangle,
(iv) Draw the angle bisetors of B and C. Let these bisectors meet at O.
(v) Taking O as centre. Draw a incircle which touches all the sides of the ∆ ABC.
(vi) From O draw a perpendicular to side BC which cut at N.
(vii) Measure ON which is required radius of the incircle.
ON = 1.5 cm

#### Question 19.

The marks obtained by 100 students in a Mathematics test are given below :

Draw an ogive for the given distribution on a graph sheet.
Use a scale of 2 cm 10 units on both axes).
Use the ogive to estimate the:
(i) median.
(ii) lower quartile.
(iii) number of students who obtained more than 85% marks in the test.
(iv) number of students failed, If the pass percentage was 35. (2014)

N= 100
Median = $\frac { 100 }{ 2 }$ = 50th term Median = 45
(ii) Lower quartile : (Q1)
N = 100
⇒ $\frac { N }{ 4 }$ = $\frac { 100 }{ 4 }$ = 25
∴ Q1 = 32
(iii) Mo. of students with 85% less = 94
∴ More than 85% marks = 100 94 6
(iv) Number of students who did not pass = 30

##### PQ.

The marks obtained by 30 students in a class assessment of 5 marks is given below :

Calculate the mean, median and mode of the above distribution. (2015)

Which is between 10 and 20
∴ Median = 3
(ii) Mode frequency of 3 is the greatest
∴ Mode = 3

##### PQ.

The weight of 50 workers is given below :

Draw an ogive of the given distribution using a graph sheet. Take 2 cm = 10 kg on one axis and 2 cm = 5 workers along the other axis. Use the ogive drawn to estimate the following:
(i) the upper and lower quartiles.
(ii) if weighing 95 Kg and above is considered overweight, find the number of workers who are overweight. (2015)

Plot the points (60, 4), (70, 11), (80, 22), (90, 36), (100, 42) (110, 47) and (120, 50) on the graph and join them in order with free hand.
This is the required ogive

(i) Upper Quartile = $\frac { 50 x 3 }{ 4 }$ th term = $\frac { 150 }{ 4 }$ th = -37.5th term
Lower Quartile = $\frac { 50 }{ 4 }$ th = 12.5th term
Upper quartile is 42 kg and lower quartile is 72 kg.
(ii) 95 kg and above are over weight
∴ No. of over weight students are 50 – 39 = 11 students.

#### Question 20.

The mean of following number is 68. Find the value of ‘x’. 45, 52, 60, x, 69, 70, 26, 81 and 94. Hence estimate the median. (2016)

##### PQ.

The table shows the distribution of the scores obtained by 160 shooters in a shooting competition. Use a graph sheet and draw an ogive for the distribution. (Take 2 cm = 10 scores on the X-axis and 2 cm = 20 shooters on the Y-axis). (2016)

Use your graph to estimate the following :
(i) The median.
(ii) The interquartile range.
(iii) The number of shooters who obtained a score of more than 85%.

Through mark 80 on y-axis, draw a horizontal line which meets the ogive drawn at point Q.
Through Q, draw a vertical line which meets the x-axis at the mark of 43(app.).
∴ Median = 43
(ii) Since the number of terms = 160

(iii) Since 85% scores = 85% of 100 = 85
Through mark for 85 on x-axis, draw a vertical line which meets the ogive drawn at point B.
Through the point B, draw a horizontal line which meets thej-axis at the mark of 148 = 160- 148= 12
So, the number of shooters who obtained more than 85% score is 12.

#### Question 21

The marks of 10 students of a class in an examination arranged in ascending order is as follows:

13, 35, 43, 46, x, x + 4, 55, 61, 71, 80

If the median marks is 48, find the value of x. Hence, find the mode of the given data.

Here the number of observations i. e is 10, which is even.’

So, the given data is 13, 35, 43, 46, 46, 50, 55, 61, 71, 80.

In the given data, 46 occurs most frequently.

∴ Mode = 46

#### Question 22

The daily wages of 80 workers in a project are given below.

 Wages 400- 450 450- 500 500- 550 550-600 600-650 650-700 700- 750 No.of workers 2 6 12 18 24 13 5

Use a graph paper to draw an ogive for the above distribution. (Use a scale of 2 cm = Rs. 50 on x – axis and 2 cm = 10 workers on y – axis). Use your ogive to estimate.

1. i. the median wages of the workers.
2. thelower quartile wage of workers.

iii. the number of workers who earn more than Rs. 625 daily.

The cumulative frequency table of the given distribution is as follows:

 Wages (Rs.) Upper limit No. of workers C.f. 400-450 450 2 2 450-500 500 6 8 500-550 550 12 20 550-600 600 18 38 600-650 650 24 62 650-700 700 13 75 700-750 750 5 80

The ogive is as follows:

Number of workers = n = 80

1) Median = term = 40th term, draw a horizontal line which meets the curve at point A.

Draw vertical line parallel to y axis from A to meet x axis at B.

The value of point B is 605.

2) Lower quartile (Q1)= term=20th term = 550

3) Through mark of point 625 on x axis draw a vertical line which meets the graph at point C Then through point C, draw a horizontal line which meets the y axis at the mark of 50.

Thus, the number of workers that earn more than Rs. 625 daily = 80 – 50 = 30

#### Question 23

The histogram below represents the scores obtained by 25 students in a Mathematics mental test. Use the data to:

1. Frame a frequency distribution table.
2. To calculate mean.

3 To determine the Modal class.

1. The frequency distribution table is as follows:
 Class interval Frequency 0-10 2 10- 20 5 20-30 8 30-40 4 40-50 6

ii.

 Class interval Frequency (f) Mean value (x) fx 0-10 2 5 10 10- 20 5 15 75 20-30 8 25 200 30-40 4 35 140 40-50 6 45 270 Sf = 25 Sf = 695

iii. Here the maximum frequency is 8 which is corresponding to class 20 – 30.

Hence, the modal class is 20 – 30.

Thanks