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

Exercise 24(A)

1. Find the mean of the following set of numbers:

(i) 6, 9, 11, 12 and 7

Ans:

(i) By definition of mean we can say that 

Mean = ∑x/ n

Here n = 5

Thus now Mean = $\dfrac{~\left( 6~+~9~+~11~+~12~+~7 \right)}{5}$= 9

(ii) 11, 14, 23, 26, 10, 12, 18 and 6

Ans: The formula for mean is, 

Mean = ∑x/ n

Here n = 8 

Similar to 1 we have, 

Thus now Mean = $\dfrac{\left( 11~+~14~+~23~+~26~+~10~+~12~+~18~+~6 \right)}{8}$= $\dfrac{120}{8}$=15 

2. Marks obtained (in mathematics) by 9 students are given below:

60, 67, 52, 76, 50, 51, 74, 45 and 56

1. Find the arithmetic mean

Ans: It is given that in mathematics 9 students have obtained the following marks 60, 67, 52, 76, 50, 51, 74, 45 and 56.

Now by definition of Mean we have, 

Mean = ∑x/ n

Here we know that n = 9 

Thus, 

Mean = $\dfrac{\left( 60~+~67~+~52~+~76~+~50~+~51~+~74~+~45~+~56 \right)}{9}$= 59 

2. If marks of each student are increased by 4; what will the new value of the arithmetic mean?

Ans: To find the value of the arithmetic mean. If the marks of each student are increased by 4.

Now if the marks are increased by 4 then the new arithmetic mean is = 59+4 = 63

3. Find the mean of the natural numbers from 3 to 12.

Ans: We have to find the mean of the natural numbers from 3 to 12.

The natural numbers between 3 to 12 are 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12.

Here n = 10 

Mean = ∑x/ n

Mean = $\dfrac{\left( 3~+~4~+~5~+~6~+~7~+~8~+~9~+~10~+~11~+~12 \right)}{10}$= 7.5

4. 

1. Find the mean of 7, 11, 6, 5, and 6

Ans: The formula of Mean = ∑x/ n

Now we have to find the mean of 7, 11, 6, 5, and 6

hence  n = 5.

Mean = $\dfrac{\left( 7~+~11~+~6~+~5~+~6 \right)}{5}$= 7. 

2. If each number given in (a) is diminished by 2, find the new value of mean.

Ans: Now if  2 is subtracted from each number, then the new mean will be 7 - 2 = 5. 

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

Ans: It is given that, 

The number of terms (n) = 5 

Mean value = 8 

Sum of all the terms = 8 x 5 = 40 …… (i)

Sum of numbers = 6 + 4 + 7 + a + 10 = 27 + a ….. (ii)

Now equating  (i) and (ii), we get

27 + a = 40

Hence, a = 13. 

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

Ans:

It is given that, 

No. of terms (n) = 5 and mean = 8

Now sum of all the terms = 5 x 8 = 40 ……. (i)

Now sum of all the numbers = 6 + y + 7 + x + 14 = 27 + y + x ……. (ii)

Now let us just equate (i) and (ii), we get

27 + y + x = 40

x + y = 13

Hence, y = 13 – x

7. The ages of 40 students are given in the following table:

Find the arithmetic mean.

Ans: 

Now the arithmetic mean by definition is, 

Mean = $\dfrac{\sum fi~xi}{\sum fi}$= $\dfrac{614}{40~}$= 15.35

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

Ans:  From the question, we can say that, 

The number of terms = 10 

Mean value = 69.5

∴ Sum of numbers = 69.5 $\times 10=695$….(i)

The sum of the given number = 72 + 70 + x + 62 + 50 + 71 + 90 + 64 + 58 + 82 = 619+x ….(ii)

Now let us equate (i) and (ii)

619 + x = 695

⇒ x = 695 – 619 = 76

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

Ans:

Now Mean = 60.95

$\dfrac{1718+60f}{28+f}$= 60.95

$\Rightarrow ~$1718+60f = 60.95 (28+f)

$\Rightarrow $ 1718+ 60f = 1706.6 + 60.95f

$\Rightarrow $ 0.95f = 11.4   

f = $\dfrac{11.4}{0.95}$= 12

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? 

Ans:

Now mean is given by, 

$\underline{x~}$= $\dfrac{\sum fi~xi}{\sum fi}$= $\dfrac{3360}{42}$= 80 

(i) The mean would remain the same even if the number of workers in each category is doubled.

Hence the mean would be 80.

(ii) If the wages per day in each category are increased by 60%, what is the new mean wage?

Ans: To find the new mean wage, if the wages per day in each category are increased by 60%

Now the mean will be increased by 60% if the wages per day per worker is increased by 60%

Therefore the New mean = 80 $\times \dfrac{160}{100}$= 128. 

(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?

Ans: Even if the number of workers is doubled then there would be no change in the mean but if the wages per worker is reduced by 40%, then

Now new mean would be =  80 $\times \dfrac{60}{100}$ = 48

11. The contents of 100 matchboxes were checked to determine the number of matches they contained.

(i) Calculate, correct to one decimal place, the mean number of matches per box.

Ans:

(i) We have to calculate correctly to one decimal place, the mean number of matches per box.

Mean is given by, $\underline{x~}$= $\dfrac{\sum fi~xi}{\sum fi}$= $\dfrac{3813}{100}$= 38.13 = 38.1

(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.

Ans: Now in the second case the new mean = 39 matches 

∴ Total contents = 39 x 100 = 3900

But we know that the number of matches = 3813 

∴ Number of new matches to be added = 3900 – 3813 = 87

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

Ans:

It is given that mean = 3 

The mean by the formula is given by, 

$\underline{x~}$= $\dfrac{\sum fi~xi}{\sum fi}$

3 = $\dfrac{6p+87}{33}$

6p= 12

$\Rightarrow $p = 2

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

Ans: Now we have, 

It is given that ∑f= 200

86+f1+f2 = 200

F1+f2 =114 ……(1)

Now mean is given by ,  $\underline{x~}$= $\dfrac{\sum fi~xi}{\sum fi}$

73 = $\dfrac{~7000~+~50f1~+100f2}{200}$

$\Rightarrow $f1+2f2 =152 …..(2) 

Subtract (2) from (1) we get, 

f2= 38

Now , f1 = 114 -38=76

Hence, f1 = 114 and f2 = 38

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

Ans:  

Now we know that Mean formula= A + $\dfrac{\sum ft}{\sum fi}\times i$

= $\dfrac{30+\left( 598 \right)5}{384}$

= 30 - 7.79 = 22.21= 22 approximately.

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

Ans: 

Now let us assume that the mean value is A =45.

Now mean = A + $\dfrac{\sum fd}{\sum f}$

= 45 + $\dfrac{\left( -88 \right)}{200}$= 45 - 0.44

= 44.56 

Exercise 24(B) 

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

Ans:

By formula we know that arithmetic mean is given by, 

$\underline{x~}$= $\dfrac{\sum fi~xi}{\sum fi}$= $\dfrac{1074}{50}$= 21.48

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

Calculate the mean by using:

(i) Direct Method

Ans: Mean by the direct method 

By formula we know that arithmetic mean is given by, 

$\underline{x~}$= $\dfrac{\sum fi~xi}{\sum fi}$= $\dfrac{5520}{80}$

(ii) Short – Cut Method

Ans: Finding meaning by short-cut method. 

Here we know that A = 72.5

Now by formula we know that mean is given by, 

$\underline{x~}$=A+ $\dfrac{\sum fi~di}{\sum fi}$= 72.5+($\dfrac{-280}{80}$) = 72.5 - 3.5 = 69.

3. The following are the marks obtained by 70 boys in a class test:

Calculate the mean by:

(i) Short – cut method

Ans:

Finding the mean by, 

(i) Short–Cut Method

Now we know that, A = 65 

Now by formula, we know that mean is given by, 

$\underline{x~}$=A+ $\dfrac{\sum fi~di}{\sum fi}$= 65 + ($\dfrac{-270}{70}$) = 65 - 3.86 =61.14. 

(ii) Step – Deviation Method

Ans: Step – deviation method

Now we know that, A = 65 

Now by formula, we know that mean is given by, 

$\underline{x~}$=A+h $\times $ $\dfrac{\sum fi~ui}{\sum fi}$= 65 + 10 $\times $($\dfrac{-27}{70}$) = 65 - 3.86 =61.14. 

4. Find mean by step – deviation method:

Ans: Now we have to find the mean by using the step-division method.

Here we know that A = 87.50 and h = 7 

Now by formula, we know that mean is given by, 

$\underline{x~}$=A+h $\times $ $\dfrac{\sum fi~ui}{\sum fi}$= 87.5+7$\times \dfrac{29}{150}$= 87.5 $\times 1.35=88.85$

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

Ans:

It is given that $\underline{x~}$= 21 $\dfrac{1}{7}$= $\dfrac{148}{7}$

By formula we know that arithmetic mean is given by, 

$\underline{x~}$= $\dfrac{\sum fi~xi}{\sum fi}$= $\dfrac{1235+35f}{63+f}$

$\Rightarrow \dfrac{148}{7}$=$\dfrac{1235+35f}{63+f}$

$\Rightarrow 9324~+148f~=8645+245f$

$\Rightarrow f=~\dfrac{679}{97}$= 7

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

Ans: Let the assumed mean be 72.5

By formula we know that arithmetic mean is given by, 

$\underline{x~}$=A+ $\dfrac{\sum fi~di}{\sum fi}$= 72.5 + $\dfrac{-280}{80}$= 72.5 -3.5 =69.

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

Ans:

By formula we know that arithmetic mean is given by, 

$\underline{x~}$=$\dfrac{\sum fx}{\sum f}$=$\dfrac{2850}{75}$= 38

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

Ans:

It is given that the mean is 54.

By formula we know that arithmetic mean is given by, 

$\underline{x~}$=$\dfrac{\sum fixi}{\sum fi}$

54 = $\dfrac{2370+30p}{39+p}$

⇒ 2106 + 54p = 2370 + 30p

⇒ 54p – 30p = 2370 – 2106 

⇒ 24p = 264

p = $\dfrac{264}{24}$=11

Hence p =11

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.

Ans: It is given that mean = 62.8 and the sum of frequencies = 50

Now sum of frequencies = 50 

30 +${{f}_{1}}+{{f}_{2}}=50~$

${{f}_{1}}+{{f}_{2}}=50~$-30 =20 …. (1)

Mean = $\underline{x~}$=$\dfrac{\sum fixi}{\sum fi}$

62.8 = $\dfrac{2060+30{{f}_{1}}+70{{f}_{2}}}{50}$

$\Rightarrow $2060+30${{f}_{1}}+70{{f}_{2}}$=3140

$\Rightarrow $30${{f}_{1}}+70{{f}_{2}}$=3140-2060

$\Rightarrow $3${{f}_{1}}+7{{f}_{2}}$=108……(2)

Now multiplying (1) by 7 and (2) by 1 we have, 

7${{f}_{1}}+7{{f}_{2}}=140$

3${{f}_{1}}+7{{f}_{2}}=108$

Subtracting (1) and (2) we have, 

4${{f}_{1}}=32$

${{f}_{1}}=~8$

But ${{f}_{1}}+{{f}_{2}}=20$

Now ${{f}_{2}}=12$

Hence ${{f}_{1}}=~8$and ${{f}_{2}}=12$

10. Calculate the mean of the distribution given below using the shortcut method.

Ans:

A =45.5

Mean =A+$\dfrac{\sum fiui}{\sum fi}\times 10$

= 45.5 + $\dfrac{7}{50}\times 10$

= 45.5 +1.4 = 46.9.

Exercise 24(C)

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.

Ans:  Now the first step is to arrange the given data in the descending order, 

8, 7, 6, 5, 4, 3, 3, 1, 0

We have to find the median from the given numbers and hence the middle term is 4 which is nothing but the 5th term and hence the median = 4 

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 and 24.

Find the median of their weights.

Ans: Now the first step is to start arranging the given data in the descending order, which would be.

28.5, 28, 27.5, 25.5, 24, 24, 22, 21, 21, 20.5

After arranging the numbers the middle terms are 24 and 24 which are nothing but the 5th and 6th terms.

Thus now the median = $\dfrac{\left( 24+24 \right)}{2}$= 24

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 and 28. 

1. Find median 

Ans: It is given that the marks obtained by 19 students of a class are 27, 36, 22, 31, 25, 26, 33, 24, 37, 32, 29, 28, 36, 35, 27, 26, 32, 35 and 28.

Now the first step is to arrange the numbers in ascending order. 

22, 24, 25, 26, 26, 27, 27, 28, 28, 29, 21, 32, 32, 33, 35, 35, 36, 36, 37

2.  To Find the Median.

The middle term is the 10th term which is 29.

Hence the median = 29. 

3. Find Lower Quartile

Ans: Lower quartile

As the number of students is 19, hence n = 19. 

${{q}_{1}}$= $\dfrac{n+1th}{4}$term 

${{q}_{1}}$= $\dfrac{19+1th}{4}$term 

${{q}_{1}}$=  5th term.

4. Find Upper Quartile 

Ans: Upper quartile 

${{q}_{3}}={{\left[ \dfrac{3\left( n+1 \right)}{4} \right]}^{th}}$term 

${{q}_{3}}={{\left[ \dfrac{3\left( 19+1 \right)}{4} \right]}^{th}}$term 

${{q}_{3}}={{15}^{th~}}term~=~35~$

5. Find Interquartile Range

Ans: Interquartile range is given by =${{q}_{3}}$ -${{q}_{1}}$= 35 -26 = 9 

4. From the following data, find:

1. Median

25, 10, 40, 88, 45, 60, 77, 36, 18, 95, 56, 65, 7, 0, 38 and 83

Ans: The first step is to arrange the given numbers in ascending order, so on we have.

0, 7, 10, 18, 25, 36, 38, 40, 45, 56, 60, 65, 77, 83, 88, 95

2. To Find the Median 

The median is the mean of the 8th and the 9th term. 

Thus now the median would be = $\dfrac{\left( 40+45 \right)}{2}$= $\dfrac{85}{2}$= 42.5.

3. Find Upper Quartile

Ans: Upper quartile is given by, 

${{q}_{3}}={{\left[ \dfrac{3\left( n+1 \right)}{4} \right]}^{th}}$term 

${{q}_{3}}$= terms = ${{12}^{th~}}$term = 65.

4. Find Interquartile Range

Ans: Interquartile range is given by = ${{q}_{3}}$ -${{q}_{1}}$

As the number of terms is 16, hence n =16. 

Lower quartile ${{q}_{1}}$= $\dfrac{n~}{4}$th term  = $\dfrac{16~}{4}$= 4th term = 18. 

∴  ${{q}_{1}}$= 18

Interquartile range =  ${{q}_{3}}$ -${{q}_{1}}$ = 65-18=47

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

Ans:

From the question, we know that the number of terms (n) = 37 

Median = $\dfrac{37+1th}{2}$term = 19th term 

The 19th term is 14,

Therefore the median = 14

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

Find : (i) Median

Ans:

It is given that the number of terms = 60

1. Median: 

The mean of the 30th and the 31st term is the median.

∴ Median = $\dfrac{39+39}{2}$= 39 

2. Find Lower Quartile

Ans: Lower quartile ${{q}_{1}}$is given by = $\dfrac{1\times n}{4}$th term 

= $\dfrac{1\times 60~th~}{4}$term, as n = 60 

 ${{q}_{1}}$= 15th term = 38. 

3. Find Upper Quartile

Ans: Upper quartile (${{q}_{3}}$)= $\dfrac{3n}{4}$th term 

= $\dfrac{3}{4}\times 60=~45th~term=~40~$

4. Find Interquartile Range

Ans: Interquartile range is given by =   ${{q}_{3}}$- ${{q}_{1}}$ = 40- 38= 2

7. Estimate the median for the given data by drawing ogive :

Ans:

Number of terms = 50

∴ Median = $\dfrac{25th~+26th~}{2}$= 25.5 = 25.5 th term. 

Now through the mark of 25.5 on the y-axis, draw a line which is parallel to x-axi which then meets the curve at A. Now from A dra a perpendicular to x-axis at B. 

Therefore the value of B is the median which is 28.

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

Ans:

Number of terms = 55

∴ Median = $\dfrac{55+1}{2}$=28th term 

Draw a line parallel to the x-axis that meets the curve at A through the 28th mark on the y-axis. Draw a perpendicular line segment from A to the x-axis. Which it meets at B.

9. From the following cumulative frequency table, draw ogive and then use it to find :

(i) Median

Ans:

From the tabular column we can figure out that the number of terms = 80 

(i) Median = Now through the 40th term mark draw a  line which is parallel to the x-axis which then meets the curve at A.Now from A draw a perpendicular to x-axis which meets it at B.

Hence the value of B is the median = 40. 

(ii) Find Lower Quartile.

Ans: Lower quartile(${{q}_{1}}$)= $\dfrac{n}{4}$th term.= 80th/4 = 20th term 

The 20th term is 18.

(iii) Find Upper Quartile

Ans: Upper quartile(${{q}_{3}}$) = $\dfrac{3}{4}\times nth~term~$= $\dfrac{3\times 80}{4}$= 60th term = 66 

10.In a school 100 pupils have heights as tabulated below:

Find the median height by drawing an ogive.

Ans:

From the question we can say that the number of terms are 100

Therefore median = n/2 = $\dfrac{100}{2}$ = 50th term 

Now through the mark 50, draw a line that is parallel to the x-axis which then meets the curve at A. Now from A  draw, a perpendicular to the x-axis which then meets the x-axis at B is the median which is nothing but 148 cm. 

Exercise 24 (D)

1. Find the mode of the following data:

(i) 7, 9, 8, 7, 7, 6, 8, 10, 7 and 6

Ans: We have to find the mode for the given sequence of numbers

(i) 7, 9, 8, 7, 7, 6, 8, 10, 7 and 6

It is seen that 7 occurs 4 times i the given data 

Hence we could say that mode = 7.

(ii) 9, 11, 8, 11, 16, 9, 11, 5, 3, 11, 17 and 8

Ans: 9, 11, 8, 11, 16, 9, 11, 5, 3, 11, 17 and 8

As we could figure out that 11 occurs 4 times in the given data.

We can say that mode =11.

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

Find the mode of heights.

Ans: From definition we can say that mode is the value that appears most often in a set of data values. From the given data given to us we could say that mode is 122 cm as it has occurred most number of times which is that its frequency is 18 

3. Find the mode of following data, using a histogram:

Ans: First let us sketch a histogram, 

The intersection of lines AD and CB at P can be seen in the histogram. Draw a Q parallel to the horizontal x-axis. Clearly it is evident from the histogram we could say that mode is in 20-30, because in that class there are 20 frequencies.

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

Ans: The first step is to sketch a histogram from the given table provided. 

Clearly it is evident that the mode is in 30-35 as it has the highest frequency which is 23. 

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

Ans: The set of numbers given are 2,2,3,5,5,5,6,8 and 9

1. Median = $\dfrac{n+1}{2}$= $\dfrac{9+1}{2}$= 5th term. 

The 5th term is the mode which is 5. 

2. Mode: From definition we can say that mode is the value that appears most often in a set of data values. 

Mode is 5 as it repeats a maximum number of times. 

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 ?

Ans: It is given that the boy scored the following marks out of 20.

15,17,16,7,10,12,14,16,19,12,16.

The first step is to arrange the given data in the ascending order: 7, 10,12, 12,14, 15,16,16, 16, 17,19.

(i) Mode is the value that appears most often in a set of data values. 

From the definition we could easily say that mode is 16 as it occurs a maximum number of times. 

(ii) What are his median marks ?

Ans: Median = $\dfrac{n+1}{2}= \dfrac{11+1}{2}$= 6th term which is 15. 

(iii) What are his total marks ?

Ans: Now the total marks = 7 + 10+ 12+ 12+ 14+ 15+ 16 + 16+ 16+ 17+ 19= 154

(iv) What are his mean marks ?

Ans:Mean($\underset{\scriptscriptstyle-}{x}$)= $\dfrac{{{x}_{1}}+~{{x}_{2}}+.....+{{x}_{n}}}{x}$= $\dfrac{154}{11}$=14

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

Ans: It is given that the marks obtained by 16 students in a class test marked out of 10 marks are 0,0,2,2,3,3,3,4,5,5,5,5,6, 6,7,8

(i) Mean =  $\dfrac{{{x}_{1}}+~{{x}_{2}}+.....+{{x}_{n}}}{x}$= $\dfrac{64}{16}$= 4

(ii) Median = Mean of 8th and the 9th term = $\dfrac{4+5}{2}$= $\dfrac{9}{2}$= 4.5

(iii) Mode is the value that appears most often in a set of data values. 

From the definition we could easily say that mode is 5 as it occurs a maximum number of times. 

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

(i) What was his modal score ?

Ans:

Modal score = 4 as its frequency is 7, which is maximum.

(ii) What was his median score ?

Ans: Median = $\dfrac{n+1}{2}$= $\dfrac{25+1}{2}$=13th term which is 3

(iii) What was his total score ?

Ans: Total score = 80

(iv) What was his mean score ?

Ans: Mean = $\underline{x~}$= $\dfrac{\sum fi~xi}{\sum fi}$= $\dfrac{80}{25}$=3.2

Exercise 24(E)

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:

1. The median height.

Ans:

Now let us first draw an ogive taking into consideration the height of the student along the x-axis and the cumulative frequency along the y-axis. 

Median = 160/2 = 80th term 

Now from the mark for 80, we have to draw a parallel line to the x-axis which then meets the curve, then from the curve, we should draw a vertical line that meets the x-axis at the mark of 157.5.

2. Determine the interquartile range.

Ans: To find the interquartile range.

We know that the number of terms = 160 

Lower quartile (q1)= ($\dfrac{160}{4}$)= 4oth term = 152 

Upper quartile (q3) = (3$\times \dfrac{160}{4}$) = 120th term = 164 

Inner Quartile range = q3 – q1= 164-152=12.

3. The number of students whose height is above 172 cm.

Ans: To find the number of students whose height is above 172 cm. Draw a vertical line that meets the curve through the mark for 172 on the x-axis; then, from the curve, draw a horizontal line that meets the y-axis at the mark for 145.

Now,  the number of students whose height is above 172cm =160-144=16.

2. Draw ogive for the data given below and from the graph determine: 

1. The median marks.

Ans:

Here 1cm = 10 marks on X axis

1cm = 20 students on Y axis

(i) To Find the median marks.

So now the median = 120/2 = 60th term 

Now through mark 60, draw a parallel line to the x-axis which meets the curve at A. From A, then draw a perpendicular to the x-axis meeting it at B. 

Hence the value of point B is the median = 43.

(ii) Determine the Number of students who obtained more than 75% marks.

Ans: Now total marks = 100

75% of total marks = $\dfrac{75}{100}\times 100=75~marks~$

Hence now the number of students achieving more than 75% marks =120-111= 9 students. 

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.

Ans: From the data given 

Mean of 1, 7, 5, 3, 4 and 4 = $\dfrac{~\left( 1~+~7~+~5~+~3~+~4~+~4 \right)}{6}$= $\dfrac{24}{6}$= 4

Hence, m = 4. 

Now mean of 3, 2, 4, 2, 3, 3 and p = m-1 = 4-1=3 

Thus, 17+p = 3$\times n......,$where n = 7 

17+p =21

p = 4

Now arrange the terms in an ascending order we get: 

2, 2, 3, 3, 3, 3, 4, 4

Mean = 4th term = 3 

Hence q =3. 

4.  In a malaria epidemic, the number of cases diagnosed were as follows:

On what days do the mode and upper and lower quartiles occur?

Ans:

(i) Mode is calculated with the number which has the highest frequency and so the mode is 5th July as it has the maximum frequency.

(ii) Total number of terms = 206 

Upper quartile = 206 $\times \dfrac{3}{4}$= 154.5th = 7th July 

Lower quartile = 206 $\times \dfrac{1}{4}$= 51.5th = 4th July 

5. The income of the parents of 100 students in a class in a certain university is tabulated below.

(i) Draw a cumulative frequency curve to estimate the median income.

Ans: To draw the cumulative frequency curve. 

We have to plot the points (8, 8), (16, 43), (24, 78), (32, 92) and (40, 100) to get the curve as follows. 

Here, N = 100

$\dfrac{N}{2}$= 50 

At y= 50, affix A. 

Draw a horizontal line which then meets the curve at B through A.

Now through B draw a vertical line which then meets OX at M. 

OM = 17.6 units.

Hence, median income = 17.6 thousands.

(ii) If 15% of the students are given free ships on the basis of the income of their parents, find the annual income of parents, below which the freeships will be awarded.

Ans: Now 15% of 100 students = $\dfrac{\left( 15\times 100 \right)}{100}$=15

From the cumulative frequency 15 , draw a horizontal line which then intersects the curve at P. 

Now from P, draw a perpendicular to the x-axis which meets it at Q which is equal to 9.6.

Hence, freeship would be provided to the students providing the annual income of their parents is close to 9.6 thousands. 

(iii) Calculate the Arithmetic mean

Ans: Arithmetic mean = $\dfrac{\sum ~fx}{~\sum ~f}$= $\dfrac{1832}{100}$= 18.32

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 

Ans: Now the first step is to arrange the terms in an ascending order:

2, 6, 8, 9, 10, 11, 11, 12, 13, 13, 14, 14, 15, 15, 15, 16, 16, 18, 19, 20

Number of terms = 20 

∑ x = 2 + 6 + 8 + 9 + 11 + 11 + 12 + 13 + 13 + 14 + 14 + 15 + 15 + 15 + 15 + 16 + 16 + 18 + 19 + 20 = 257

(i) Mean = $\dfrac{\sum x}{\sum n}$= $\dfrac{257}{20}$= 12.85 

(ii) Calculate the median 

Ans: Median = $\dfrac{\left( 10th~term~+~11th~term \right)}{2}$= $\dfrac{\left( 13+14 \right)}{2}$= 13.5

(iii) Calculate the mode

Ans: Mode = The mode is 15 as it has the maximum frequency which is 3. 

7. The marks obtained by 120 students in a mathematics test is 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

Ans:

(i)  From the graph, the median = $\dfrac{\left( 120+1 \right)}{2}$= 60.5 th term 

Now through mark 60.5, draw a parallel line to the x-axis which then meets the curve at A. Then from A draw a perpendicular to x-axis which meets it at B

The value of point B is the median = 43 

(ii) the number of students who obtained more than 75% in the test.

Ans: The number of students who obtained more than 75% in the test?

Number of students who obtained up to 75% marks in the test = 110.

Number of students who obtained more than 75% marks in the test = 120 – 110 = 10.

(iii) Use your ogive to estimate the number of students who did not pass in the test if the pass percentage was 40.

Ans: The number of students who obtained less than 40% marks in the test = 52 as from the graph x=40 and y = 52. 

(iv) Use your ogive to estimate the lower quartile

Ans: Lower quartile = q1= 120 $\times $$\left( \dfrac{1}{4} \right)$ = 30th term = 30 

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

Use your ogive to estimate the following:

(i) The percentage of students weighing 55 kg or more

Ans:

(i) The number of students weighing more than 55 kg = 200 - 44 = 156

Thus, the percentage of students weighing 55 kg or more = 200 - 44 = 156 

Now the percentage of students weighing 55 kilograms or more=$\dfrac{156}{200}\times 100=78%$

(ii) Use your ogive to estimate the weight above which the heaviest 30% of the student fall

Ans: 30% of students = $\dfrac{\left( 30\times 200 \right)}{100}$= 60 

Heaviest 60 students in weight = 9+21+30=60 

Weight = 65 kilogram 

(iii) Use your ogive to estimate the number of students who are (a) underweight (b) overweight, if 55.70 kg is considered as standard weight.

Ans: The number of students who are (a) underweight (b) overweight, if 55.70 kg is considered as standard weight? 

1. Underweight students when 55.70 kg is standard = 46 from the graph.

2. Overweight students when 55.70 kg is standard = 200 - 55.70 = 154 from the graph. 

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.

Ans:

Now we know that the number of terms = 25

(i) Mean = $\dfrac{171}{25}$= 6.85

(ii) Median = $\dfrac{\left( 25+1 \right)}{2}$= 13th term = 7

(iii) Mode = 6 is the mode as it has the maximum frequency which is 6.

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

Ans: 

Mean = $\dfrac{\sum ~fx}{\sum ~f}$= $\dfrac{\left( 1940~+~35f \right)}{\left( 36~+~f \right)}$…… (i)

Now it is given that the mean = 52 …….(ii)

Equating (i) and (ii) we have 

$\dfrac{\left( 1940~+~35f \right)}{\left( 36~+~f \right)}$=52

1940+35f = 1872+52f

17f = 68

Thus, f =4 

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 paper 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.

Ans:

It is given that the number of employees = 320

(i) Median = 320/2 = 160^th term

From the mark 160, draw a parallel line to the x-axis which then meets the curve at A.Now from A draw a perpendicular to x-axis meeting it at B. 

From here the value of point B is the median = rupees 9.3 thousands

(ii) From the graph determine the number of employees whose income is below Rs 8500.

Ans: The number of employees with the income below Rs 8,500 = 95 (approximately from the graph).

(iii) From the graph determine if the salary of a senior employee is above Rs 11,500, find the number of senior employees in the company.

Ans: Number of employees with income below Rs 11,500 = 305 ((approximately from the graph).

Hence now from the graph, the number of senior employees = 320 - 305 = 15 

(iv) From the graph determine the upper quartile.

Ans: The upper quartile = Q3= 320 $\times \dfrac{3}{4}$=240th term = Rupees 10,300.

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