Case study (4 Marks)

(i) Write the median class.
(ii) When first ball was picked up, what was the probability of calling out an even number?
(iii) (a) Find median of the given data.
(b) Find mode of the given data.

Answer

(i) The cumulative frequency distribution is as given below:

Numbers announced	0-15	15-30	30-45	45-60	60-75
Number of times (f)	8	9	10	12	9
Cumulative frequency	8	17	27	39	48 = N

We find that $N=\Sigma f_i=48$ and $\frac{N}{2}=24$. The cumulative frequency just greater than $\frac{N}{2}$ i.e. 24 is 27 and the corresponding class is 30-45. So, 30-45 is the median class.
(ii) There are 75 balls out of which 37 balls are even numbered.
∴ Probability of calling out an even number = $\frac{37}{75}$
(iii) (a) We find that 30-45 is the median class with l = 30, f = 10, h = 15, F = 17 and N = 48.
$\therefore \quad$ Median $=l+\frac{\frac{N}{2}-F}{f} \times h=30+\frac{24-17}{10} \times 15=30+\frac{7 \times 3}{2}=40.5$
(b) We observe that 45-60 is the modal class with l = 45, f = 12, $f_1=10, f_2=9$ and $h=15$.
$\therefore \quad$ Mode $=l+\frac{f-f_1}{2 f-f_1-f_2} \times h=45+\frac{12-10}{24-10-9} \times 15=45+\frac{2}{5} \times 15=51$

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Question 64 Marks

Activities like running or cycling reduce stress and the risk of ental disorder like depression. Running helps build endurance. Children develop stronger bones and uscles and are less prone to gain weight. The physical education teacher of a school has decided to conduct an inter school running tournaent in his school preises. The tie taken by a group of students to run 100 , was noted as follows:
Iage

Tie (in seconds)	0-20	20-40	40-60	60-80	80-100
Nuber of students	8	10	13	6	3

Based on the above, answer the following questions:
(i) What is the edian class of the above given data?
(ii) (a) Find the ean tie taken by the students to finish the race.
OR
(b) Find the ode of the above given data.
(iii) How any students took tie less than 60 seconds?

Answer

(i) The cumulative frequency distribution is

Time (in seconds)	0-20	20-40	20-40	60-80	80-100
Number of students	8	10	13	6	3
Cumulative frequency	8	18	31	37	40

We find that $N=\Sigma f_i=40$ and $\frac{N}{2}=20$ The cumulative frequency just greater $\frac{N}{2}$ i.e. 20 is 31 and the corresponding class is 40-60. So, median class is 40-60.
(ii) (a)

Computation of mean time

Time (in seconds) (Class)	Number of students $\left(f_i\right)$	Mid-values$x_i$	$u_i=\frac{x_i-50}{20}$	$f_i \mu_i$
0-20	8	10	-2	-16
20-40	10	30	-1	-10
40-60	13	50	0	0
60-80	6	70	1	6
80-100	3	90	2	6
	$\Sigma f_i=40$			$\Sigma f_1 u_i=-14$

We obtain,
$N=\Sigma f_i=40, a=50, h=20$ and $\Sigma f_i u_i=-14$
Mean time $=a+h\left(\frac{1}{N} \Sigma f_i u_i\right)=50+20 \times-\frac{14}{40}=43$
Hence, mean time taken by the students to finish the seconds.
OR
(b) We find that 40-60 is the modal class with $l=40, f=13, f_1=10, f_2=6$ and $h=20$.
$\therefore \quad$ Mode $=L+\frac{f-f_1}{2 f-f_1-f_2} \times h=40+\frac{13-10}{26-10-6} \times 20=46$
(iii) From the cumulative frequency table in (i), we find that the number of students who took less than 60 seconds to finish the race is 31.

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Question 74 Marks

Vocational training copleents traditional education by providing practical skills and hands-on experience. While education equips individuals with a broad knowledge base, vocational training focuses on job-specific skills, enhancing eployability thus aking the student self-reliant. Keeping this in view, a teacher ade the following table grving the frequency distribution of students/adults undergoing vocational training fro the training institute.
Iage

Age (in years)	15-19	20-24	25-29	30-34	35-39	40-44	45-49	50-54
Nuber of participants	62	132	96	37	13	11	10	4

Fro the above, answer the following questions:
(i) What is the lower liit of the odal class of the above data?
(ii) (a) Find the edian class of the above data.
OR
(b) Find the nuber of participants of age less than 50 years who undergo vocational training.
(iii) Give the epirical relationship between ean, edian and ode.

Answer

(i) Converting the frequency distribution into inclusive form, we obtain the following frequency distribution

Clearly, class 19.5-24.5 has the maximum frequency 132. So, 19.5-24.5 is the modal class.
(ii) (a) The cumulative frequency distribution is as given below:

Age (in years)	Number of participants	Cumulative frequency
14.5-19.5	62	62
19.5-24.5	132	194
24.5-29.5	96	290
29.5-34.5	37	327
34.5-39.5	13	340
39.5-44.5	11	351
44.5-49.5	10	361
49.5-54.5	4	365
		$N=\Sigma f_i=365$

We have, N= 365 $\Rightarrow \frac{ N }{2}=182.5$
The cumulative frequency just greater than $\frac{N}{2}$ =182.5 is 194 and the corresponding class is 19.5-24.5. So, 19.5-24.5 is the median class.
OR
(ii) (b) Clearly, number of participants of age less than 50 years is the cumulative frequency corresponding to the class 44.5-49.5 and is equal to 361. Thus, there are 361 participants of age less than 50 years.
(iii) The empirical ralation between mean, median and mode is:
Mode 3 Median -2 Mean.

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Question 84 Marks

India eteorological departent observes seasonal and rainfall every year in different sub-divisions of our country.

It helps the to copare and analyse the results. The table groen below shows sub-division wise seasonal (onsoon) rainfall () in 2018:

(i) Write the odal class.
(ii) Find the edian of the given data.
(iii) Find the ean rainfall in this season.
(iv) If sub-division having at least 1000 rainfall during onsoon season, is considered good rainfall sub. division, then how any sub-drvisions had good rainfall?

Answer

(i) 600-800 mm rainfall is in maximum number of sub-divisions. Therefore, 600-800 is the modal class.
(ii) The cumulative frequency distribution is as given below:

Rainfall (mm)	Number of sub-divisions (f)	Cumulative frequency
200-400	2	2
400-600	4	6
600-800	7	13
800-1000	4	17
1000-1200	2	19
1200-1400	3	22
1400-1600	1	23
1600-1800	1	24
		N = 24

We have, $N=\Sigma f_i=24 \Rightarrow \frac{N}{2}=12$
The cumulative frequency just greater than $\frac{N}{2}$ i.e. 12 is 13 and the corresponding class is 600-800. It is the median class with l = 600 f = 7 h = 200 and F = 6.
$\therefore \quad$ Median $=l+\frac{\frac{N}{2}-F}{f} \times h \Rightarrow$ Median $=600+\frac{12-6}{7} \times 200=600+171.4=771.4 mm$
(iii)

Computation of Mean

Rainfall (mm)	Mid values $\left(x_i\right)$	$\frac{x_i-1100}{200}=u_i$	Number of Sub-divisions $\left(f_i\right)$	$f_i u_i$
200-400	300	-4	2	-8
400-600	500	-3	4	-12
600-800	700	-2	7	-14
800-1000	900	-1	4	-4
1000-1200	1100	0	2	0
1200-1400	1300	1	3	3
1400-1600	1500	2	1	2
1600-1800	1700	3	1	3
				$\Sigma f_i u_i=-30$

We find that a = 1100 200, N = 24
Mean $=a+h \times\left(\frac{1}{N} \Sigma f_i u_i\right) \Rightarrow$ Mean $=1100+200 \times \frac{-30}{24}=1100-250=850 mm$
(IV) Number of sub-divisions having at least 1000 m rainfall = 2 + 3 + 1 + 1 = 7.
Hence, 7 sub-divisions have good rainfall

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Question 94 Marks

Electricity energy consumption is the form of energy. Global electricity consumption continues to increase faster than world population, leading to an increase in the average amount of electricity consumed per person (per capita electricity consumption)

Tariff	LT Residential	Bill Number	: 658473
Type of supply	Single phase	Connected load	: 3 kw
Meter Reading, date	31-11-2021	Meter reading	: 65789
Previous reading date	31-10-2021	Previous reading	: 65500
		Units Consumed	: 289

A survey is conducted for 56 families of a colony A and 80 families of colony B. The following table gives the weekly consumption of electricity of the families

Weekly consumption (in units):	0-10	10-20	20-30	30-40	40-50	50-60
Number of families in colony A:	16	12	18	6	4	0
Number of families in colony A:	0	5	10	20	40	5

(i)The median weekly consumption of electricity in colony A, is
(a) 12 units$\quad$ (b) 16 units$\quad$ (c) 20 units$\quad$ (d) 8 units
(ii) The mean weekly consumption of electricity in colony A, is
(a) 19.64 units$\quad$ (b) 22.5 units$\quad$ (c) 26 units$\quad$ (d) 28 units
(iii)The modal class in table for colony A is
(a) 0-10$\quad$ (b) 10-20$\quad$ (c) 20-30$\quad$ (d) 30-40
(iv) The modal weekly consumption of electricity in colony B is
(a) 38.2 units$\quad$ (b) 43.6 units$\quad$ (c) 26 units$\quad$ (d) 32 units

Answer

(i) : (c)

Computation of median

Weekly consumption	Number of families (f)	Cumulative frequency (c.f.)
0-10	16	16
10-20	12	28
20-30	18	46
30-40	6	52
40-50	4	56
50-60	0	56
		$\quad$$\quad$N = 56

We have, N = 56 The cumulative frequency just greater than N/2 = 28 is 46 and the corresponding class is 10-20. It is the median class with $l=10, N=56, f=12, f=16$ and $h=10$.
Median $=l+\frac{2^N-r}{f} \times h \Rightarrow$ Median $=10+\frac{28-16}{12} \times 10=20$

Weekly consumption	Number of families$\left(f_i\right)$	Mid-values $\left(x_i\right)$	$\quad$$u_i=\frac{x_i-35}{10}$	$\quad$$f_i u_i$
0-10	16	5	-3	-48
10-20	12	15	-2	-24
20-30	18	25	-1	-18
30-40	6	35	0	0
40-50	4	45	1	4
50-60	0	55	2	0
	$N=\Sigma f_i=56$			$N=\Sigma f_i u_i=-86$

$\therefore \quad \bar{X}=a+\left(\frac{1}{N} \Sigma f_i u_i\right) h=35+\frac{1}{56} \times-86 \times 10=19.64$
(iii) (c). The class with maximum frequency is 20-30. So, the modal class is 20-30.
(iv) (b): The modal class is 40 40-50 with $l=40, f=40, f_1=20, f_2=5$ and $h=10$.
$\therefore \quad$ Mode $=1+\frac{f-f_1}{2 f-f_1-f_2}=40+\frac{40-20}{80-20-5} \times 10=40+\frac{200}{55}=43.6$

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Question 104 Marks

The COVID-19 pandeic, also known as coronavirus, is an ongoing pandeic of corona-virus disease by the transission of severe acute respiratory syndroe coronavirus-2 (SARS - CoV-2) aong hu

The following table shows the age distribution of case aditted during a day in two different hospitals.

Age (in years)	5-15	15-25	25-35	35-45	45-55	55-65
Nuber of cases in hospital-1	6	11	21	23	14	5
Nuber of cases in hospital-2	8	16	10	42	24	12

(i) In hospital-1, the average age for which axiu cases occurred is
(a) 32.24$\quad$ (b) 34.36$\quad$ (c) 36.82$\quad$ (d) 42.24
(ii) In hospital-1, the upper liit of odal class is
(a) 15$\quad$ (b) 25$\quad$ (c) 35$\quad$ (d) 45
(iii) In hospital-1, the ean of the given data is
(a) 26.2$\quad$ (b) 32.4$\quad$ (c) 33.5$\quad$ (d) 35.4
(iv) In hospital-2, the ode of the given data is
(a) 41.4$\quad$ (b) 48.2$\quad$ (c) 55.3$\quad$ (d) 64.6

Answer

(i) (c): We have to find the mode of the frequency distribution. The modal class is$35-45$ with $l=35, f=23, f_1=21, f_2=14$ and $h=10$.
$\therefore$ Mode $=35+\frac{23-21}{46-21-14} \times 10=35+\frac{20}{11}=36.82 \quad\left[\because\right.$ Mode $\left.=l+\frac{f-f_1}{2 f-f_1-f_2} \times h\right]$
(ii) (c): Clearly, 35-45 is the modal class and its upper limits is 45.
(iii) (d):

Computation of average age

Age (in years)	Number of cases$\left(f_i\right)$	Mid-values$\left(x_i\right)$	$u_i=\frac{x_i-30}{10}$	$f_1 u_i$
5-15	6	10	-2	-12
15-25	11	20	-1	-11
25-35	21	30	0	0
35-45	23	40	1	23
45-55	14	50	2	28
55-65	5	60	3	15
	$\quad$$\quad$N = 80			$\Sigma f_i u_i=43$

$\therefore \quad \bar{X}=a+h\left(\frac{1}{N} \Sigma f_1 u_i\right) \Rightarrow \bar{X}=30+10\left(\frac{43}{80}\right)=35.37 \simeq 35.4$
(iv) (a): We observe that 35 - 45 is the modal class with $l=35, f=42, f_1=10, f_2=24$ and $h=10$.
$\therefore \quad$ Mode $=35+\frac{42-10}{84-10-24} \times 10=35+\frac{320}{50}=41.4$

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Maths Case study (4 Marks)

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