2 Marks Questions

Question 12 Marks

A survey conducted on 20 households in a locality by a group of students resulted in the following frequency table for the number of family members in a household:

Family size	1-3	3-5	5-7	7-9	9-11
Number of families	7	8	2	2	1

Find the mode of this data.

Answer

The frequency distribution table is given as:

Family size$(x_i)$	1-3	3-5	5-7	7-9	9-11
No. of families$(f_i)$	$f_0= 7$	$f_1= 8$	$f_2= 2$	2	1

From the given frequency table, the maximum class frequency is 8, and the class corresponding to this frequency is 3 – 5. So, the modal class is 3 - 5.
Now modal class = 3 - 5, lower limit (l) of modal class = 3, class size (h) = 2
frequency $(f_1)$ of the modal class = 8,
frequency $(f_0)$ of class preceding the modal class = 7
frequency $(f_2)$ of class succeeding the modal class = 2
Now, let us substitute these values in the formula :
$\text { Mode }=l+\left(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right) \times h$
$=3+\left(\frac{8-7}{2 \times 8-7-2}\right) \times 2=3+\frac{2}{7}=3.286$
Therefore, the mode of the data above is 3.286

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Question 22 Marks

The wickets taken by a bowler in 10 cricket matches are as follows :
2, 6, 4, 5, 0, 2,1, 3, 2, 3
Find the mode of the above data.

Answer

At first, we arrange the given data in ascending order:- 0, 1, 2, 2, 2, 3, 3, 4, 5, 6
Now find the most occurring number i.e. 2
Hence, mode = 2

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Question 32 Marks

The distribution given below shows the number of wickets taken by bowlers in one-day cricket matches. Find the mean number of wickets by choosing a suitable method. What does the mean signify?

Number of Wickets	20-60	60-100	100-150	150-250	250-350	350-450
Number of Bowlers	7	5	16	12	2	3

Answer

Here, the class size varies, and hence $x_i$, so $f_ix_i$ will be large. Let us apply the step deviation method with a = 200 and h = 20. Then, we obtain the data as in Table

Number of tickets taken	Number of bowlers $(f_i)$	$x_i$	$d_i = x_i – 200$	$u_{i}=\frac{d_{i}}{20}$	$u_if_i$
20-60	7	40	-160	-8	-56
60-100	5	80	-120	-6	-30
100-150	16	125	-75	-3.75	-60
150-250	12	200	0	0	0
250-350	2	300	100	5	10
350-450	3	400	200	10	30
Total	45				-106

Therefore, $\bar{x}=200+20\left(\frac{-106}{45}\right)=200-47.11=152.89$
This tells us that, on average, the number of wickets taken by these 45 bowlers in one-day cricket is 152.89

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Question 42 Marks

The table below gives the percentage distribution of female teachers in the primary schools of rural areas of various states and union territories (U.T.) of India. Find the mean percentage of female teachers by all the three methods discussed in this section.

Percentage of female teachers	15 - 25	25 - 35	35 - 45	45 - 55	55 - 65	65 - 75	75 - 85
Number of states/U.T.	6	11	7	4	4	2	1

Answer

Let, $a = 50$

C.I.	Number of states/ U.T. $(f_i)$	$x_i$	$d_i = x_i - 50$	$f_id_i$
15 - 25	6	20	-30	-180
25 - 35	11	30	-20	-220
35 - 45	7	40	-10	-70
45 - 55	4	50	0	0
55 - 65	4	60	10	40
65 - 75	2	70	20	40
75 - 85	1	80	30	30

From table, $\Sigma f_id_i= -360 , \Sigma f_i=36$
we know that, mean=$\overline { x } = a + \frac { \Sigma f _ { i } d _ { i } } { \Sigma f _ { i } } $
$= 50 + \frac { - 360 } { 35 } $
$= 39.71$

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Question 52 Marks

The marks obtained by 30 students of Class X of a certain school in a mathematics paper consisting of 100 marks are presented in the table given below. Find the mean of the marks obtained by the students in mathematics paper.

Marks obtained $(x_i)$	10	20	36	40	50	56	60	70	72	80	88	92	95
Number of Students $(f_i)$	1	1	3	4	3	2	4	4	1	1	2	3	1

Answer

Recall that to find the mean marks, we require the product of each $x_i$ with the corresponding frequency $f_i$. So, let us put them in a column as shown in Table given below:

Marks obtained $(x_i)$	Number of students $(f_i)$	$f_ix_i$
10	1	10
20	1	20
36	3	108
40	4	160
50	3	150
56	2	112
60	4	240
70	4	280
72	1	72
80	1	80
88	2	176
92	3	276
95	1	95
Total	$\sum f_i $= 30	$\sum f_ix_i$ = 1779

Now, $\bar{x}=\frac{\Sigma f_{i} x_{i}}{\Sigma f_{i}}=\frac{1779}{30}=59.3$
Therefore, the mean marks obtained by students in mathematics paper is 59.3

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