3 Marks Question

Question 13 Marks

A student noted the number of cars passing through a spot on a road for $100$ periods each of $3$ minutes and summarized it in the table given below. Find the mode of the data:

Number of cars	0-10	10-20	20-30	30-40	40-50	50-60	60-70	70-80
Frequency	7	14	13	12	20	11	15	8

Answer

Here, the maximum class frequency is 20, and the class corresponding to this frequency is $40-50$. So, the modal class is $40-50.$
Therefore $h = 10, l = 40, f_1 = 20, f_0 = 12 , f_2 = 11$
Mode = l + $\left[ {\frac{{{f_1}\; - {f_0}}}{{2{f_1} - \;{f_0} - {f_2}}}} \right] \times$ h = 40 + $\left[ {\frac{{20 - 12}}{{2(20) - \;12 - 11}}} \right] \times$ $10$
$= 40 +$ $\frac{{80}}{{17}}$ = 40 + 4.7 = 44.7
Hence the mode of the data is $44.7$ cars.

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Question 23 Marks

The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.

Runs scored	Number of batsmen
3000-4000	4
4000-5000	18
5000-6000	9
6000-7000	7
7000-8000	6
8000-9000	3
9000-10000	1
10000-11000	1

Find the mode of the data.

Answer

Since the maximum number of batsman have their runs scored in the interval 4000-5000, the modal class is $4000-5000.$
Therefore, $l = 4000, h = 1000, f_1 =18, f_0= 4, f_2=$
Mode = l +$ \left[ {\frac{{f1\; - f0}}{{2f1 - \;f0 - f2}}} \right] \times $ h = 4000 + $\left[ {\frac{{18 - 4}}{{2(18) - \;4 - 9}}} \right] \times$ $1000$
$= 4000 +$ $\frac{{14000}}{{23}} $ $= 4000 + 608.7 = 4608.7$
Hence, the mode of the data is $4608.7$

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Question 33 Marks

The following data gives the information on the observed lifetimes (in hours) of $225$ electrical components:

Lifetimes (in hours)	0-20	20-40	40-60	60-80	80-100	100-120
Frequency	10	35	52	61	38	29

Determine the modal lifetimes of the components.

Answer

Here, the maximum class frequency is 61, and the class corresponding to this frequency is $60-80$. So, the modal class is $60-80.$
Therefore $h = 20, l = 60, f_1 = 61, f_0 = 52 , f_2 = 38$
${Mode = \;l\; + \left[ {\frac{{{f_1}\; - {f_0}}}{{2{f_1} - \;{f_0} - {f_2}}}} \right] \times h = \;60 + \left[ {\frac{{61 - 52}}{{2(61) - \;52 - 38}}} \right] \times 20 = }$ $\begin{array}{*{20}{l}} {\;60 + \left[ {\frac{9}{{122 - 90}}} \right] \times 20 = 60 + \;\frac{{180}}{{32}} = 60\; + \;5.625 = 65.625} \end{array}$
Therefore, the modal lifetime of the components is $65.625$ hours.

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Question 43 Marks

The following table gives the literacy rate (in percentage) of $35$ cities. Find the mean literacy rate.

Literacy rate (in %)	45-55	55-65	65-75	75-85	85-95
Number of cities	3	10	11	8	3

Answer

Take a = 70, h = 10

Literacy rate (in %)	Number of cities ($f_i$)	Class mark ($x_i$)	$d_i = x_i –70$	${u_i} = \frac{{{x_i} - 70}}{{10}}$	$f_iu_i$
45-55 55-65 65-75 75-85 85-95	3 10 11 8 3	50 60 70 80 90	–20 –10 0 10 20	–2 –1 0 1 2	–6 –10 0 8 6
Total	$\sum$ $f_i_= 35$				$\sum$ $f_iu_i$ = -2

sing the step-deviation method,
$\overline x $ = a + $\left( {\frac{{\sum {{f_i}{u_i}} }}{{\sum {{f_i}} }}} \right) \times$ h = 70 +$ \left( {\frac{{ - 2}}{{35}}} \right) \times$ 10
= 70 - $\frac{4}{7}$ = 70 - 0.57 = 69.43%
Hence, the mean literacy rate is 69.43%

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Question 53 Marks

A class teacher has the following absentee record of $40$ students of a class for the whole term. Find the mean number of days a student was absent.

Number of days	0-6	6-10	10-14	14-20	20-28	28-38	38-40
Number of students	11	10	7	4	4	3	1

Answer

Number of days	Number of students ($f_i$)	Class mark ($x_i$)	$f_ix_i$
0-6	11	3	33
6-10	10	8	80
10-14	7	12	84
14-20	4	17	68
20-28	4	24	96
28-38	3	33	99
38-40	1	39	39
Total	$\sum$ $f_i = 40$		$\sum$ $f_ix_i = 499$

Using the direct method,
$\overline x = \frac{{\sum {{f_i}{x_i}} }}{{\sum {{f_i}} }} = \frac{{499}}{{40}}$ = 12.475
Hence, the mean number of days a student was absent is 12.48

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Question 63 Marks

To find out the concentration of $SO_2$ in the air (in parts per million, i.e. ppm), the data was collected for 30 localities in a certain city and is presented below:

Concentration of $SO_2 $(in ppm)	Frequency
0.00-0.04	4
0.04-0.08	9
0.08-0.12	9
0.12-0.16	2
0.16-0.20	4
0.20-0.24	2

Find the mean concentration of $SO_2$ in the air.

Answer

Take $a = 0.14, h = 0.04$

Concentration of $SO_2$ (in ppm)	Frequency $(f_i)$	Class Mark $(x_i)$	$d_i = x_i – 0.14$	${u_i} = \frac{{{x_i} - 0.14}}{{0.04}}$	$f_iu_i$
0.00-0.04 0.04-0.08 0.08-0.12 0.12-0.16 0.16-0.20 0.20-0.24	4 9 9 2 4 2	0.02 0.06 0.10 0.14 0.18 0.22	–0.12 –0.08 0.04 0 0.04 0.08	–3 –2 –1 0 1 2	–12 –18 –9 0 4 4
Total	$\sum {{f_i}}$ = 30				$\sum$ $f_iu_i= -31$

Using the step-deviation method,
$\overline x$ = a +$ \left( {\frac{{\sum {{f_i}{u_i}} }}{{\sum {{f_i}} }}} \right) \times$ h = 0.14 +$\left( {\frac{{ - 31}}{{30}}} \right) \times$ (0.04)
$= 0.14 – 0.041 = 0.0999$ ppm.
Therefore, the mean concentration of $SO_2$ in the air is $0.099$

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Question 73 Marks

The table below shows the daily expenditure on food of $25$ households in a locality.

Daily expenditure (in ₹)	100-150	150-200	200-250	250-300	300-350
Number of households	4	5	12	2	2

Find the mean daily expenditure on food by a suitable method.

Answer

Daily expenditure	Frequency $f_i$	Mid value $x_i$	$d_i= x_{i }– 225$	$u_i = \frac{(x_i – 225)} {50}$	$f_iu_i$
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
	$\sum f_i$ = 25				$\sum f_iu_i$= -7

assumed mean $(a) = 225,$
$h = 50$
Mean = $\overline {x}$ = a + $\left( \frac { \sum f_iu_i} {\sum f_i } \right) \times$ h
$= 225 + 50$$\left( {\frac{{ - 7}}{{25}}} \right)$
$= 225 - 14 = 211$

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Question 83 Marks

Thirty women were examined in a hospital by a doctor and the number of heartbeats per minute was recorded and summarized as follows. Find the mean heartbeats per minute for these women, choosing a suitable method.

Number of heartbeats per minute	65-68	68-71	71-74	74-77	77-80	80-83	83-86
Number of women	2	4	3	8	7	4	2

Answer

Take a = 75.5, h = 3

Number of heart beats per minute	Number of women $(f_i)$	Class mark $(x_i)$	$d_i = x_i - 75.5$	$u_i$ = $\frac{x_i\;-75.5}3$	$f_iu_i$
65-68 68-71 71-74 74-77 77-80 80-83 83-86	2 4 3 8 7 4 2	66.5 69.5 72.5 75.5 78.5 81.5 84.5	–9 –6 –3 0 3 6 9	–3 –2 –1 0 1 2 3	–6 –8 –3 0 7 8 6
Total	$\sum f_i$ = 30				$\sum f_iu_i$ = 4

Using the step-deviation method,
$\overline x$ = a + $\frac{\displaystyle\overset{}{\underset{}{\sum f_ix_i}}}{\displaystyle\sum_{}^{}f_i}\times$ h = 75.5 + $\left[\frac4{30}\right]\times$ 3 = 75.5 + 0.4 = 75.9
Hence, the mean heart beats per minute are 75.9

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Question 93 Marks

Consider the following distribution of daily wages of $50$ workers of a factory:

Daily wages (in Rs.)	500-520	520-540	540-560	560-580	580-600
Number of workers	12	14	8	6	10

Find the mean of the daily wages of the workers of the factory by using an appropriate method.

Answer

The distribution of daily wages of 50 workers of a factory is as follows,

Daily wages of workers(Class interval)	Number of workers $(f_i)$	Class mark $(x_i)$	$d_i = x_i−550$	$f_id_i$
500-520	12	510	−40 − 40	1 2 $\times$ (−40) = −480
520-540	14	530	−20 − 20	14 $\times$ −20 = −280
540-560	8	550	0	8 $\times$ 0 = 0
560-580	6	570	20	6 $\times$ 20 = 120
580-600	10	590	40	10 $\times$ 40 = 400
Total	50			-240

Using the assumed mean method, the mean of daily wages of the workers is,
$\bar{x}=a+\frac{\sum f_{i} d_{i}}{\sum f_{i}}$
let a = 550,Substitute values in the above formula,
$\bar{x}=550+\frac{(-240)}{50}$
$= 545.2$
Thus, the mean of daily wages of the 50 workers of the factory is Rs 545.20

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Question 103 Marks

A survey was conducted by a group of students as a part of their environmental awareness programme, in which they collected the following data regarding the number of plants in $20$ houses in a locality. Find the mean number of plants per house.

Number of plants	0-2	2-4	4-6	6-8	8-10	10-12	12-14
Number of houses	1	2	1	5	6	2	3

Which method did you use for finding the mean, and why?

Answer

Number of plants	Number of houses$(f_i)$	Class mark$(x_i)$	$f_{i}x_i$
0-2 2-4 4-6 6-8 8-10 10-12 12-14	1 2 1 5 6 2 3	1 3 5 7 9 11 13	1 6 5 35 54 22 39
Total	${\sum f_i}$ = 20		${\sum f_ix_i}$ = 162

$\therefore \overline x = \frac{\displaystyle\overset{}{\underset{}{\sum f_ix_i}}}{\displaystyle\sum_{}^{}f_i}$ ... Using direct method because numerical values of x_i and $f_i$ are small
= $\frac{162}{20}$
= 8.1 plants
We have used direct method for finding the mean because numerical values of $x_i$ and $f_i$ are small.

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Question 113 Marks

A survey regarding the heights (in cm) of 51 girls of Class X of a school was conducted and the following data was obtained. Find the median height.

Height (in cm)	No. of girls
Less than 140	4
Less than 145	11
Less than 150	29
Less than 155	40
Less than160	46
Less than 165	51

Answer

We have,

Class Intervals	Frequency (f)	C.F
Below 140	4	4
140-145	7	11
145-150	18	29
150-155	11	40
155-160	6	46
160-165	5	51
	$N = \sum f = 51$

Here, $\frac{N}{2} = \frac{{51}}{2} = 25.5$ which is in the class 145-150
Here, $l_1 = 145, h = 5, N = 51, C = 11, F = 18$
$\therefore $ Median $ = {l_1} + \frac{{\frac{N}{2} - C}}{f} \times h$
$ = 145 + \frac{{25.5 - 11}}{{18}} \times 5$
$ = 145 + \frac{{72.5}}{{18}} \Rightarrow 149.03$
$\therefore $ Median height of the girls = 149.03

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Question 123 Marks

The marks distribution of $30$ students in a mathematics examination. Find the mode of this data. Also, compare and interpret the mode and the mean.

Class interval	Number of students $(f_i)$	Classmark $(x_i)$	$f_ix_i$
10 - 25	2	17.5	35.0
25 - 40	3	32.5	97.5
40 - 55	7	47.5	332.5
55 - 70	6	62.5	375.0
70 - 85	6	77.5	465.0
85 - 100	6	92.5	555.0
Total	$\sum f_{i}$ = 30		$\sum f_{i}x_i$ = 1860.0

Answer

Since the maximum number of students (i.e., 7) have got marks in the interval 40 - 55, the modal class is 40 - 55
Therefore, the lower limit (l) of the modal class = 40
the class size ( h) = 15
the frequency $(f_1)$ of modal class = 7
the frequency $( f_0)$ of the class preceding the modal class = 3
the frequency $( f_2)$ of the class succeeding in the modal class = 6
Now, using the formula:
Mode $=l+\left(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right) \times h$
we get Mode = $40+\left(\frac{7-3}{14-6-3}\right) \times 15=52$
So, the mode mark is 52
Now $\bar{x}=\frac{\sum f_{i} x_{i}}{\sum f_{i}}$
$\bar{x}=\frac{1860.0}{30}=62$
So, the maximum number of students obtained 52 marks, while on average a student obtained 62 marks.

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Question 133 Marks

The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is ₹18. Find the missing frequency $f$.

Daily pocket allowance (in Rs.)	11-13	13-15	15-17	17-19	19-21	21-23	23-25
Number of children	7	6	9	13	$f$	5	4

Answer

The frequency distribution table from the given data can be drawn as :

Daily allowance (in ₹)	Class mark $\left(x_i\right)$	Number of children $\left(f_i\right)$	$f_i x_i$
11-13	12	7	84
13-15	14	6	84
15-17	16	9	144
17-19	18	13	234
19-21	20	$f$	20$f$
21-23	22	5	110
23-25	24	4	96
		$\sum f_i=f+44$	$\Sigma f_i x_i=20 f+752$

Now, mean, $\bar{x}=\frac{\sum f_i x_i}{\sum f_i} \Rightarrow 18=\frac{20 f+752}{f+44}$
$\Rightarrow 20 f+752=18 f+792 \Rightarrow 20 f-18 f=792-752$
$\Rightarrow 2 f=40 \Rightarrow f=\frac{40}{2}=20$
Hence, the fourth frequency, $f$ is 20.

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Question 143 Marks

If mean of the following distribution is 7.5, then find the value of P.

x	3	5	7	9	11	13
f	6	8	15	P	8	4

Answer

$\sum f_i x_i=(3 \times 6)+(5 \times 8)+(7 \times 15)+(9 \times p)+(11 \times 8)+(13 \times 4) $
$\sum f_i x_i=18+40+105+9 p+88+52 $
$\sum f_i x_i=303+9 p$
$\sum f_i=6+8+15+p+8+4 $
$\sum f_i=41+p$
$\bar{x}=\frac{\sum f_i x_i}{\sum f_i} $
$7.5=\frac{303+9 p}{41+p}$
$7.5(41+p)=303+9 p $
$307.5+7.5 p=303+9 p $
$307.5-303=9 p-7.5 p $
$4.5=1.5 p \\ p=\frac{4.5}{1.5} \\ p=3$
The value of $p$ is 3.

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