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Maths 3 Marks Question

Statistics · Gujarat Board (GSEB) STD 10 (GSEB - English Medium). 9 questions.

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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 $\sum f_i=35$       $\sum f_i u_i=-2$
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 $(x_i)$ $d_i= x_i-550$ $f_id_i$
$500-520$ $12$ $510$ $−40 − 40$ $1 2 \times (−40) = −480$
$520-540$ $14$ $530$ $−20 − 20$ $14 \times −20 = −280$
$540-560$ $8$ $550$ $0$ $8 \times 0 = 0$
$560-580$ $6$ $570$ $20$ $6 \times 20 = 120$
$580-600$ $10$ $590$ $40$ $10 \times 40 = 400$
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 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 $140$ $4$ $4$
$140-145$ $7$ $11$
  $18$ $29$
$150-155$ $11$ $40$
$155-160$ $6$ $46$
$160-165$ $5$ $51$
  $N = \sum f = 51$  
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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