3 Marks Question

Question 13 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 23 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 33 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 43 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 53 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 63 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 $(x_i)$	$di = xi−550$	$f_id_i$
$500-520$	$12$	$510$	$−40 − 40$	$1 2 \times (−40) = −480$
$520-540$	$14$	$530$	$−20 − 20$	$14 \times −20 = −280$
$540-560$	$8$	$550$	$0$	$8 \times 0 = 0$
$560-580$	$6$	$570$	$20$	$6 \times 20 = 120$
$580-600$	$10$	$590$	$40$	$10 \times 40 = 400$
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 73 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 $(x_i)$	$f_ix_i$
$0-2$ $2-4$ $4-6$ $6-8$ $8-10$ $10-12$ $12-14$	$1$ $2$ $1$ $5$ $6$ $2$ $3$	$1$ $3$ $5$ $7$ $9$ $11$ $13$	$1$ $6$ $5$ $35$ $54$ $22$ $39$
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 83 Marks

The distribution below gives the weights of $30$ students of a class. Find the median weight of the students.

Weight(in kg)	Number of students
$40-45$	$2$
$45-50$	$3$
$50-55$	$8$
$55-60$	$6$
$60-65$	$6$
$65-70$	$3$
$70-75$	$2$

Answer

Weight (in kg)	Number of students	Cumulative frequency
$40-45$ $45-50$ $50-55$ $55-60$ $60-65$ $65-70$ $70-75$	$2$ $3$ $8$ $6$ $6$ $3$ $2$	$2$ $5$ $13$ $19$ $25$ $28$ $30$

Now, $n = 30$
So, $\frac{n}{2} = \frac{{30}}{2} = 15$
This observation lies in the class $55-60,$
So, $55-60$ is the median class.
Therefore,
$l = 55$
$h = 5$
$f = 6$
$cf = 13$
$\therefore $ Median $ = l + \left( {\frac{{\frac{n}{2} - cf}}{f}} \right) \times h = 55 + \left( {\frac{{15 - 13}}{{6}}} \right) \times 5$
$ = 55 + \frac{{10}}{6} = 55 + \frac{5}{3}$
$= 55 + 1.67 = 56.67$
Hence, the median weight of the students is $56.67\ kg.$

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Question 93 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 $150$	$29$
	$40$
Less than $160$	$46$
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 103 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, f1 = 61, f0 = 52 , f2 = 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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