ML Aggarwal Solution Class 10 Chapter 21 Measures of Central Tendency Exercise 21.2
Exercise 21.2
Question 1
A student scored the following marks in 11 questions of a question paper : 3, 4, 7, 2, 5, 6, 1, 8, 2, 5, 7 Find the median marks.
Sol :
Arranging in the ascending order, 1, 2, 2, 3, 4, 5, 5, 6, 7, 7, 8
Here, n = 11 i.e. odd,
The middle term
$=\frac{n+1}{2}=\frac{11+1}{2}=\frac{12}{2}$
=6 th terms
Media=5
Question 2
(a) Find the median of the following set of numbers : 9, 0, 2, 8, 5, 3, 5, 4, 1, 5, 2, 7 (1990)
(b)For the following set of numbers, find the median: 10, 75, 3, 81, 17, 27, 4, 48, 12, 47, 9, 15.
Sol :
(a) Arranging in ascending order :
0, 1, 2, 2, 3, 4, 5, 5, 5, 7, 8, 9
Here, n = 12 which is even
$=\frac{1}{2}\left[\frac{12}{2} \mathrm{th}+\left(\frac{12}{2}+1\right) \mathrm{th}\right]$
$=\frac{1}{2}(6 \mathrm{th}+7 \mathrm{th})$
$=\frac{1}{2}(4+5)=\frac{9}{2}=4 \cdot 5$
(b) Arranging the given numbers in ascending order:
3,4,9,10,12,15,17,27,47,48,75,81
Here n=12
∵No. of numbers is even
∴Median =Mean of $\left[\frac{n}{2}\right.$ th $\left.-\left(\frac{n}{2}+1\right) \mathrm{th}\right]$ terms
=Mean of 6th and 7th terms
$=\frac{15+17}{2}=\frac{32}{2}$
=16
Question 3
Calculate the mean and the median of the numbers : 2, 1, 0, 3, 1, 2, 3, 4, 3, 5
Sol :
Writing in ascending order 0, 1, 1, 2, 2, 3, 3, 3, 4, 5
Here, n = 10 which is even
$=\frac{1}{10}(24)$
=2.4
Median$=\frac{1}{2}\left[\frac{n}{2}\right.$ th term $+\left(\frac{n}{2}+1\right)$ th term $]$
$=\frac{1}{2}\left[\frac{10}{2}\right.$ th term $+\left(\frac{10}{2}+1\right)$ th term $]$
$=\frac{1}{2}($ 5th term $+$ 6th term $)=\frac{1}{2}(2+3)$
$=\frac{5}{2}=2 \cdot 5$
Question 4
The median of the observations 11, 12, 14, (x – 2), (x + 4), (x + 9), 32, 38, 47 arranged in ascending order is 24. Find the value of x and hence find the mean.
Sol :
Observation are :
11, 12, 14, (x – 2), (x + 4), (x + 9), 32, 38, 47
n = 9
i.e, 5th term = x + 4
Mean$=\frac{11+12+14+18+24+29+32+38+47}{9}=\frac{225}{9}$
=25
Question 5
The mean of the numbers 1, 7, 5, 3, 4, 4, is m. The numbers 3, 2, 4, 2, 3, 3, p have mean m – 1 and median q. Find
(i) p
(ii) q
(iii) the mean of p and q.
Sol :
(i) Mean of 1, 7, 5, 3, 4, 4 is m.
Here n = 6
Mean of 3,2,4,2,3,3, p is m-1
$\therefore m-1=\frac{3+2+4+2+3+3+p}{7}$
$\Rightarrow 4-1=\frac{17+p}{7} \Rightarrow \frac{17+p}{7}=3$
$\Rightarrow 17+p=21 \Rightarrow p=21-17=4$
(ii) Now median of 3,2,4,2,3,3,4
Writing them in ascending order 2,2,3,3,3,4,4
Here, n=7 which is odd
$\therefore$ Median $=\frac{n+1}{2}$ th term
$=\frac{7+1}{2}=4$ th term $=3$
∴q=3
(iii) Mean of p and q
$=\frac{1}{2}(4+3)=\frac{1}{2} \times 7=\frac{7}{2}$
=3.5
Question 6
Find the median for the following distribution:
| Wages per day (in rupees) | 38 | 45 | 48 | 55 | 62 | 65 |
| No. of workers | 14 | 8 | 7 | 10 | 6 | 2 |
Sol :
Writing the distribution in cumulative frequency table:
| Wages per day (in Rs) |
No. of workers (f) |
c.f. |
| 38 | 14 | 14 |
| 45 | 8 | 22 |
| 48 | 7 | 29 |
| 55 | 10 | 39 |
| 62 | 6 | 45 |
| 65 | 2 | 47 |
Question 7
Find the median for the following distribution.
| Marks | 35 | 45 | 50 | 64 | 70 | 72 |
| No. of students | 3 | 5 | 8 | 10 | 5 | 5 |
Sol :
Writing the distribution in cumulative frequency table :
| Marks | No. of students | c.f |
|---|---|---|
| 35 | 3 | 3 |
| 45 | 5 | 8 |
| 50 | 8 | 16 |
| 64 | 10 | 26 |
| 70 | 5 | 31 |
| 72 | 5 | 36 |
Here n=36 which is even
∴Median$=\frac{1}{2}\left[\frac{n}{2}\right.$ th tenm $+\left(\frac{n}{2}+1\right)$ th term $]$
$=\frac{1}{2}\left[\frac{36}{2}\right.$ th term $+\left(\frac{36}{2}+1\right)$ th tem $]$
$=\frac{1}{2}(18$ th term +19 th term $)$
$=\frac{1}{2}(64+64)=\frac{1}{2} \times 128=64$
(∵Here all the observation from 17 to 26 all are equal to 64)
∴Median=64
Question 8
Marks obtained by 70 students are given below :
| Marks | 20 | 70 | 50 | 60 | 75 | 90 | 40 |
| No. of students | 8 | 12 | 18 | 6 | 9 | 5 | 12 |
Calculate the median marks.
Sol :
Arranging the variates in ascending order and in c.f. table.
| Marks | No. of students | c.f |
|---|---|---|
| 20 | 8 | 8 |
| 40 | 12 | 20 |
| 50 | 18 | 38 |
| 60 | 6 | 44 |
| 70 | 12 | 56 |
| 75 | 9 | 65 |
| 90 | 5 | 70 |
∴Median $=\frac{1}{2}\left[\frac{n}{2}\right.\text{ th term }+\left(\frac{n}{2}+1\right) \text{ th term ]}$
$=\frac{1}{2}\left[\frac{70}{2}\right.\text{ th term }+\left(\frac{70}{2}+1\right)\text{ th term ]}$
$=\frac{1}{2}(35\text{ th term +36 th term )}$
$=\frac{1}{2}(50+50)=\frac{1}{2} \times 100=50$
(∵Here all the observations from 21 to 38 all are equal to 50)
Question 9
Calculate the mean and the median for the following distribution :
| Number | 5 | 10 | 15 | 20 | 25 | 30 | 35 |
| Frequency | 1 | 2 | 5 | 6 | 3 | 2 | 1 |
Writing the distribution in c.f. table :
| Number (x) | Frequency (f) | c.f | f x |
|---|---|---|---|
| 5 | 1 | 1 | 5 |
| 10 | 2 | 3 | 20 |
| 15 | 5 | 8 | 75 |
| 20 | 6 | 14 | 120 |
| 25 | 3 | 17 | 75 |
| 30 | 2 | 19 | 60 |
| 35 | 1 | 20 | 35 |
| Total | 20 | 390 |
(i) Mean $=\frac{\sum f x}{\sum f}=\frac{390}{20}=19 \cdot 5$
(ii) Here n=20, which is even
∴Median $=\frac{1}{2}\left[\frac{n}{2}\right.\text{ th term }+\left(\frac{n}{2}+1\right)\text{ th term ]}$
$=\frac{1}{2}\left[\frac{20}{2}\right.\text{ th tem }+\left(\frac{20}{2}+1\right)\text{ th term ]}$
$=\frac{1}{2}$(10 th term +11 th term )
$=\frac{1}{2}(20+20)=\frac{1}{2} \times 40=20$
(∵Here observation from 9 to 14 all are equal to 20)
Question 10
The daily wages (in rupees) of 19 workers are
41, 21, 38, 27, 31, 45, 23, 26, 29, 30, 28, 25, 35, 42, 47, 53, 29, 31, 35.
Find
(i) the median
(ii) lower quartile
(iii) upper quartile range,
(iv) interquartile range.
Sol :
Arranging the observations in ascending order
21, 23, 25, 26, 27, 28, 29, 29, 30, 31, 31, 35, 35, 38, 41, 42, 45, 47, 53
Here n = 19 which is odd.
$=\frac{19+1}{2}=\frac{20}{2}=10$ th term =31
(ii) Lower quartile
$\left(Q_{1}\right)=\frac{n+1}{4}=\frac{19+1}{4}=\frac{20}{4}=5$ th term=27
(iii) Upper quartile
$\left(\mathrm{Q}_{3}\right)=3\left(\frac{n+1}{4}\right)=3\left(\frac{19+1}{4}\right)=3 \times 5$
=15 th term=41
(iv) Interquartile range
$=Q_{3}-Q_{1}=41-27=14$
(v) Sem- interquartile range
Question 11
From the following frequency distribution, find :
(i) the median
(ii) lower quartile
(iii) upper quartile
(iv) inter quartile range
| Variate | 15 | 18 | 20 | 22 | 25 | 27 | 30 |
| Frequency | 4 | 6 | 8 | 9 | 7 | 8 | 6 |
Writing frequency distribution in c.f. table :
| Marks | No. of students | c.f |
|---|---|---|
| 15 | 4 | 4 |
| 18 | 6 | 10 |
| 20 | 8 | 18 |
| 22 | 9 | 27 |
| 25 | 7 | 34 |
| 27 | 8 | 42 |
| 30 | 6 | 48 |
∴Median $=\frac{1}{2}\left[\frac{n}{2}\right.\text{ th term }+\left(\frac{n}{2}+1\right)\text{ th term ]}$
$=\frac{1}{2}\left[\frac{48}{2}\right.\text{ th term }+\left(\frac{48}{2}+1\right)\text{ th term ]}$
$=\frac{1}{2}$(24th term+25th term)
$=\frac{1}{2}(22+22)=\frac{1}{2} \times 44=22$
(∵Observation from 19 to 27 are all equal to 22)
(ii) Lower quartile
$\left(\mathrm{Q}_{1}\right)=\frac{n}{4}$ th term $=\frac{48}{4}$
=12th term=20
(iii) Upper quartile
$\left(Q_{3}\right)=\frac{3 n}{4}$ th term $=\frac{3 \times 48}{4}$
=36th term=27
(iv) Inter quartile range=upper quartile-lower quartile
=27-20=7
Question 12
For the following frequency distribution, find :
(i) the median
(ii) lower quartile
(iii) upper quartile
| Variate | 25 | 31 | 34 | 40 | 45 | 48 | 50 | 60 |
| Frequency | 3 | 8 | 10 | 15 | 10 | 9 | 6 | 2 |
Sol :
Writing the distribution in cumulative frequency (c.f.) table :
| Variate | Frequency | c.f |
|---|---|---|
| 25 | 3 | 3 |
| 31 | 8 | 11 |
| 34 | 10 | 21 |
| 40 | 15 | 36 |
| 45 | 10 | 46 |
| 48 | 9 | 55 |
| 50 | 6 | 61 |
| 60 | 2 | 63 |
$\left(\mathrm{Q}_{1}\right)=\frac{n+1}{4}$ term
$=\frac{63+1}{4}=16$ th term=34
(iii) Upper quartile
$\left(\mathrm{Q}_{3}\right)=\frac{3 n+1}{4}$
$=\frac{3 \times(63+1)}{4}$ th $=\frac{192}{4}$ th
=48 th term
=48
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