2 Marks Questions

Question 12 Marks

In a class test, $50$ students obtained marks are as follows. Find the modal class and the median class.

Marks	$0-20$	$20-40$	$40-60$	$60-80$	$80-100$
Numbers	$4$	$6$	$25$	$10$	$5$

Answer

Modal class is the class with highest frequency.
Here, $25$ students got their marks between
Thus, Modal class $=40-60$

Class interval	Frequency	Cumulative frequency
$0-20$	$4$	$4$
$20-40$	$6$	$4 + 6 = 10$
$40-60$	$25$	$10 + 25 = 35$
$60 – 80$	$10$	$35 + 10 = 45$
$80 - 100$	$5$	$45 + 5 = 50$
	Total$(n) = 50$

We have,
$N=50$
$\frac{N}{2}=\frac{50}{2}=25$
The cumulative frequency with just greater $25$ and this belongs to the class $40-60$.
Hence, median class $40-60$.
Thus both the modal class and the median class are $40-60$.

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Question 22 Marks

Given below is a frequency distribution table showing daily income of 100 workers of a factory:

Daily income of workers (in Rs)	200-300	300-400	400-500	500-600	600-700
Number of workers	12	18	35	20	15

Convert this table to a cumulative frequency distribution table of more than type'.

Answer

The table can be re-written in 'more than type' as:

Daily income of the workers (In Rs)	Cumulative Frequency (C.F.)
More than 200	12+ (18+35+ 20+ 15) = 100
More than 300	18+ (35+ 20+ 15) = 88
More than 400	35+ (20 + 15) = 70
More than 500	20+ 15 = 35
More than 600	15

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Question 32 Marks

Find the mode of the following frequency distribution:

Class	0-10	10-20	20-30	30-40	40-50
Frequency	3	8	9	10	3

Answer

To find the mean of the given data, we can find the midpoint of the given class.

Weight (in Kg)	Number of students (f)	Mid-point of the class w	Product wf
50-52	18	51	918
52-54	21	53	1113
54-56	17	55	935
56-58	28	57	1596
58-60	16	59	944
60-62	35	61	2135
62-64	15	63	945

Here total number of students
$
=18+21+17+28+16+35+15=150
$
Total weight $=\sum w f=8586 kg$
$
\begin{aligned}
\text { Mean } & =\frac{\text { Total weight }}{\text { Number of students }} \\
& =\frac{8586}{150}=57.24
\end{aligned}
$
Mean weight of the students is 57.24 kg .

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Question 42 Marks

The following are the ages of 300 patients getting medical treatment in a hospital on a particular day

Age(in years)	10-20	20-30	30-40	40-50	50-60	60-70
Number of patients	60	42	55	70	53	20

Form'less than type' cumulative frequency distribution.

Answer

The following is less than type cumulative frequency distribution.

Age (in years)	Cumulative frequency
Less than 20	60
Less than 30	60+42=102
Less than 40	102+55 157
Less than 50	157+70=227
Less than 60	227+53=280
Less than 70	280+20= 300

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Question 52 Marks

Convert the following distribution to a 'more than type' cumulative frequency distribution:

Class	10-20	20-30	30-40	40-50	50-60
Frequency	4	8	10	12	10

Answer

The following is more than type cumulative frequency distribution.

More than type	Cumulative Frequency
More than or equal to 10	4 + 40 = 44
More than or equal to 20	32+8 = 40
More than or equal to 30	22+10=32
More than or equal to 40	10+12=22
More than or equal to 50	10

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Question 62 Marks

Find the mode of the following frequency distribution:

Class	$0-10$	$10-20$	$20-30$	$30-40$	$40-50$
Frequency	$3$	$8$	$9$	$10$	$3$

Answer

Here, the maximum frequency is $10$ and the corresponding class is $30-40$.
So, $30-40$ is the modal class such that
Lower limit of the modal class, $l=30$
Width of the modal class, $h=10$
Frequency of the modal class, $f=10$
Frequency of the class preceeding the modal class, $f_1=9$
Frequency of the class following the modal class,
$f_2=3$
We know that,
$\text { Mode }=l+\frac{f-f_1}{2 f-f_1-f_2} \times h$
Substituting values in the formula, we get,
$\text { Mode }=30+\frac{10-9}{20-9-3} \times 10$
$\Rightarrow \text { Mode }=30+\frac{10}{8}$
$\Rightarrow \text { Mode }=30+1.25=31.25$
Therefore, the mode of given frequency distribution is $31.25$

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Question 72 Marks

Find the mode of the following distribution of marks obtained by 50 students.

Marks	0-10	10-20	20-30	30-40	40-50
Number of students	4	8	10	20	8

Answer

$
\text { Mode }=l+\left(\frac{f_1-f_0}{2 f_1-f_0-f_2}\right) \times h
$
The modal class with highest frequency $=30-40$ where $l=$ Lower limit of modal class $=30$
$f_1=$ Frequency of modal class $=20$
$f_0=$ Frequency of class before modal class $=10$
$f_2=$ Frequency of class after modal class $=8$
$h =$ Class Interval $=40-30=10$
Substituting the above values in formula of mode,
Mode $=30+\left(\frac{20-10}{2(20)-10-8}\right) \times 10$
Mode $=30+\left(\frac{10}{22}\right) \times 10=30+4.545 \approx 34.55$
Hence, the mode is 34.55 .

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Question 82 Marks

If mode of the following frequency distribution is $55,$ then find the value of $x,$

Class:	$0-15$	$15-30$	$30-45$	$45-60$	$60-75$	$75-90$
Frequency:	$10$	$7$	$X$	$15$	$10$	$12$

Answer

Mode of the frequency distributions is $55$
Modal class is $45-60$
lower $\operatorname{limit}(t)=45$
Class height $(h)=15$
$f_0=15$
$f_1=x$
$f_2=10$
$\text {Mode }=I=\left(\frac{f_0-f_1}{2 f_0-f_1-f_2}\right) \times h$
$55=45+\left(\frac{15-x}{2 \times 15-x-10}\right) \times 15$
$55-45=\left(\frac{15-x}{2 \times 15-x-10}\right) \times 15$
$10(30-x-10)=(15-x) \times 15$
$10(20-x)=215-15 x$
$5 x=25$
$x=5$
So, the value of $x$ for the frequency distribution is $5 .$

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Question 92 Marks

Find the mode of the following frequency distribution:

Class	$15-20$	$20-25$	$25-30$	$30-35$	$35-40$	$40-45$
Frequency	$3$	$8$	$9$	$10$	$3$	$2$

Answer

Class	Frequency
$15-20$	$3$
$20-25$	$8$
$25-30$	$9$
$30-35$	$10$
$35-40$	$3$
$40-45$	$2$

Highest frequency is $10$
$\therefore \text { Modal class is } 30-35$
$\Rightarrow l =30, h=5, f_1=10, f_0=9, f_2=3$
$ M_0=l+\left[\frac{f_1-f_0}{2 f_1-f_0-f_2}\right] \times h$
$ =30+\left[\frac{10-9}{2 \times 10-9-3}\right] \times 5$
$ =30+\frac{1}{20-12} \times 5$
$ =\frac{240+\frac{5}{8}}{8}$
$ =\frac{245}{8}$
$=3.625$

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Question 102 Marks

The frequency distribution table of agriculture holding in a village is given below :

Area of Land (in hectares)	1-3	3-5	5-7	7-9	9-11	11-13
Number of Families	20	45	80	55	40	12

Find the modal agriculture holding per family.

Answer

Here, highest frequency is 80 . Therefore, model class is $5-7$.
Now, lower limit of model class, $1=5$
class size of modal class, $h =2$
frequency of class preceding the model class, $f_0=45$
frequency of model class, $f 1=80$
frequency of class succeeding the model class, $f_2=55$
$
\begin{aligned}
\Rightarrow \text { Mode } & =1+\left(\frac{f_1-f_0}{2 f_1-f_0-f_2}\right) \times h \\
& =5+\left(\frac{80-45}{2 \times 80-45-55}\right) \times 2 \\
& =5+\left(\frac{35}{160-100}\right) \times 2 \\
& =5+\frac{35 \times 2}{60} \\
& =5+1.167 \\
& =6.167
\end{aligned}
$

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Question 112 Marks

Using the empirical relationship between the three measures of central tendency, find the median of a distribution, whose mean is $169$ and mode is $175.$

Answer

Given, mean $= 169$ and mode $=175$
We know that,
$\therefore 3 \text { Median }=2 \text { Mean }+ \text { Mode }$
$\text { Median } =\frac{2 \text { Mean }+ \text { Mode }}{3}$
$ =\frac{2 \times 169+175}{3}$
$ =\frac{338+175}{3}$
$=\frac{513}{3}$
$=171$

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Question 122 Marks

The following distribution shows the transport expenditure of $100$ employees:

Expenditure $($in Rs.$):$	$200-400$	$400-600$	$600-800$	$800-1000$	$1000-1200$
Number of employees:	$21$	$25$	$19$	$23$	$12$

Find the mode of the distribution

Answer

Expenditure $($in Rs.$)$	$fi$
$200-400$	$21$
$400-600$	$25$
$600-800$	$19$
$800-1000$	$23$
$1000-1200$	$12$

Highest $f =25$
$ \Rightarrow$ Modal class $=400-600$
$\Rightarrow l= 400, h=200, f_1=25, f_0=21, f_2=19$
$M _0=l+\left[\frac{f_1-f_0}{2 f_1-f_0-f_2}\right] \times h$
$=400+\left[\frac{25-21}{2 \times 25-21-19}\right] \times 200$
$=400+\left[\frac{4}{50-40}\right] \times 200$
$=400+\frac{4}{10} \times 200=400+80$
$M _0=480$

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Question 132 Marks

Fine the mean for the following distribution :

Classes	$5-15$	$15-25$	$25-35$	$35-45$
Frequency	$2$	$4$	$3$	$1$

Answer

Classes	$F_i$	$X_i$	$f_ix_i$
$5-15$	$2$	$10$	$20$
$15-25$	$4$	$20$	$80$
$25-35$	$3$	$30$	$90$
$35-45$	$1$	$40$	$40$
	$\Sigma f_i=10$		$\Sigma f_ix_i = 230$

$\bar{x}=\frac{\sum f i x i}{\sum f i}$
$=\frac{230}{10}$
$\bar{x}=23$

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