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Maths 5 Marks Questions

Statistics · CBSE STD 11 Science (CBSE - English Medium). 15 questions.

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Question 15 Marks
Following are the marks obtained$,$ out of $100, $by two students Ravi and Hashina in $10$ tests.
Ravi
$25$
$50$
$45$
$30$
$72$
$42$
$36$
$48$
$35$
$60$
Hashima
$10$
$70$
$50$
$20$
$95$
$55$
$42$
$60$
$48$
$80$
Who is more intelligent and who is more consistent?
Answer
For Ravi$,$
$x_i$
$d_i = x_i - 45$
$d_i^2$
$25$
$-20$
$400$
$50$
$5$
$25$
$45$
$0$
$0$
$30$
$-15$
$225$
$70$
$25$
$625$
$42$
$-3$
$9$
$36$
$-9$
$81$
$48$
$3$
$9$
$35$
$-10$
$100$
$60$
$15$
$225$
$Total$
$\sum\text{d}_\text{i}=-9$
$\sum\text{d}^2_\text{i}=1699$
$\sigma=\sqrt{\frac{{\sum\text{d}_\text{i}}}{\text{n}}-\Big(\frac{\sum\text{d}_\text{i}}{\text{n}}\Big)^2}=\sqrt{\frac{1699}{10}-\Big(\frac{-9}{10}\Big)^2}$
$=\sqrt{169.9-0.81}=\sqrt{169.09}=13.003$
Now, $\bar{\text{x}}=\text{A}+\frac{\sum\text{d}_\text{i}}{\sum\text{f}_\text{i}}=45-\frac{14}{10}=43.6$
For Hashina$,$
$x_i$
$d_i = x_i - 55$
$d_i^2$
$10$ $-45$ $2025$
$70$ $15$ $225$
$50$ $-5$ $25$
$20$ $-35$ $1225$
$95$ $40$ $1600$
$55$ $0$ $0$
$42$ $-13$ $169$
$60$ $5$ $25$
$48$ $-7$ $49$
$80$ $25$ $625$
$Total$
$\sum\text{d}_\text{i}=-20$
$\sum\text{d}^2_\text{i}=5968$
$\therefore\ \sigma=\sqrt{\frac{5968}{10}-\Big(\frac{-20}{10}\Big)^2}$
$=\sqrt{596.8-4}=\sqrt{592.8}=24.46$
For Ravi, $\text{CV}=\frac{\sigma}{\bar{\text{x}}}\times100=\frac{13.003}{43.6}\times100=29.82$
For Hashina$, \text{CV}=\frac{\sigma}{\bar{\text{x}}}\times100=\frac{24.46}{55}\times100=44.47$
Hence$,$ Hashina is more consistent and intelligent.
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Question 25 Marks
Mean and standard deviation of 100 observations were found to be 40 and 10, respectively. If at the time of calculation two observations were wrongly taken as 30 and 70 in place of 3 and 27 respectively, find the correct standard deviation.
Answer
Given that, $\text{n}=100,\ \bar{\text{x}}=40,\ \sigma=10$
$\therefore\ \bar{\text{x}}=\frac{\sum\text{x}_\text{i}}{\text{N}}$
$\Rightarrow40=\frac{\sum\text{x}_\text{i}}{100}$
$\Rightarrow{\sum\text{x}_\text{i}}=4000$
Corrected ${\sum\text{x}_\text{i}}=4000-30-70+3+27=3930$
and Corrected mean $=\frac{3930}{100}=39.3$
Now, $\sigma^2=\frac{\sum\text{x}_\text{i}^2}{\text{n}}-(40)^2$
$\Rightarrow100=\frac{\sum\text{x}_\text{i}^2}{100}-1600$
$\Rightarrow{\sum\text{x}_\text{i}^2}=1700\times100$
$\Rightarrow{\sum\text{x}_\text{i}^2}=170000$
$\therefore\text{ Corrected }{\sum\text{x}_\text{i}^2}=170000-(30)^2-(70)^2+(3)^2+(27)^2$
 $=170000-900-4900+9+729=164938$
$\therefore\ \text{Correct SD}=\sqrt{\frac{164938}{100}-(39.3)^2}$
$=\sqrt{1649.38-1544.49}$
$=\sqrt{104.89}=10.24$
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Question 35 Marks
The frequency distribution:
$x$ $A$ $2A$ $3A$ $4A$ $5A$ $6A$
$f$ $2$ $1$ $1$ $1$ $1$ $1$
where $A$ is a positive integer$,$ has a variance of $160.$ Determine the value of $A.$
Answer
$x$
$f_i$
$f_ix_i$
$f_ix_i^2$
$A$ $2$ $2A$ $2A^2$
$2A$ $1$ $2A$ $4A^2$
$3A$ $1$ $3A$ $9A^2$
$4A$ $1$ $4A$ $16A^2$
$5A$ $1$ $5A$ $25A^2$
$6A$ $1$ $6A$ $36A^{2 }$
$Total$
$n = 7$
 $\sum\text{f}_\text{i}\text{x}_\text{i}=22\text{A}$
$\sum\text{f}_\text{i}\text{x}_\text{i}^2=92\text{A}^2$
$\therefore\ \sigma^2=\frac{\sum\text{f}_\text{i}\text{x}_\text{i}^2}{\text{n}}-\Big(\frac{\sum\text{f}_\text{i}\text{x}_\text{i}}{\text{n}}\Big)^2$$\Rightarrow160=\frac{92\text{A}^2}{7}-\Big(\frac{224}{7}\Big)^2$
$\Rightarrow160=\frac{92\text{A}^2}{7}-\frac{484\text{A}^2}{49}$
$\Rightarrow160=(644-484)\frac{\text{A}^2}{49}$
$\Rightarrow160=\frac{160\text{A}^2}{49}$
$\Rightarrow\text{A}^2=49\ \therefore\ \text{A}=7$
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Question 45 Marks
Calculate the mean deviation about the mean of the set of first n natural numbers when n is an odd number.
Answer
First n natural numbers are 1, 2, 3, ........, n. Here, n is odd.
$\therefore\text{ Mean }\bar{\text{x}}=\frac{1+2+3+\ ......\ +\text{n}}{\text{n}}=\frac{\frac{\text{n}(\text{n}+1)}{2}}{2}=\frac{\text{n}+1}{2}$
The deviations of numbers from mean $\Big(\frac{\text{n}+1}{2}\Big)$ are
$1-\frac{\text{n}+1}{2},2-\frac{\text{n}+1}{2},3-\frac{\text{n}+1}{2},\ .......\ ,\text{n}-\frac{\text{n}+1}{2}$
$\text{i.e.},-\frac{\text{n}-1}{2},\frac{\text{n}-3}{2},\ ......\ ,-2,-1,0,1,2,\ .....\ ,\frac{\text{n}-1}{2}$
The absolute values of deviation from the mean i.e., $|\text{x}_\text{i}-\bar{\text{x}}|$
$\frac{\text{n}-1}{2},\frac{\text{n}-3}{2},\ ......\ ,2,1,0,1,2,\ .....\ ,\frac{\text{n}-1}{2}$
The sum of absolute values of deviations from the mean i.e. $|\text{x}_\text{i}-\bar{\text{x}}|$
$=2\Big(1+2+3+\ ....\text{ to }\frac{\text{n}-1}{2}\text{terms}\Big)$
$=2\cdot\frac{\frac{\text{n}-1}{2}\big(\frac{\text{n}-1}{2}+1\big)}{2}=\frac{\text{n}-1}{2}\cdot\frac{\text{n}+1}{2}=\frac{\text{n}^2-1}{4}$
$\therefore$ Mean deviation about the mean.
$=\frac{\sum|\text{x}_\text{i}-\bar{\text{x}}|}{\text{n}}=\frac{\frac{\text{n}^2-1}{4}}{\text{n}}=\frac{\text{n}^2-1}{4\text{n}}$
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Question 55 Marks
Calculate the mean deviation about the mean for the following frequency distribution:
Class interval
 $0-4$ $4-8$ $8-12$ $12-16$ $16-20$
Frequency
 $4$ $6$ $8$ $5$ $2$
Answer
$\text{Class interval}$
$f_i$
$X_i$
$f_ix_i$
$\text{d}_\text{i}=|\text{x}_\text{i}-\bar{\text{x}}|$
$f_id_i$
$0-4$ $4$ $2$ $8$ $7.2$ $28.8$
$4-8$ $6$ $6$ $36$ $3.2$ $19.2$
$8-12$ $8$ $10$ $80$ $0.8$ $6.4$
$12-16$ $5$ $14$ $70$ $4.8$ $24.0$
$16-20$ $2$ $18$ $36$ $8.8$ $17.6$
 
$N = 16$
 
 $\sum\text{f}_\text{i}\text{x}_\text{i}=230$
 
$\sum\text{f}_\text{i}\text{d}_\text{i}=96.0$
Mean$=\frac{\sum\text{f}_\text{i}\text{x}_\text{i}}{\text{N}}=\frac{230}{25}=9.2$
And Mean deviation $ =\frac{\sum\text{f}_\text{i}\text{d}_\text{i}}{\text{N}}=\frac{96}{25}=3.84$
Hence$,$ the required $\text{MD} = 3.84$
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Question 65 Marks
The mean life of a sample of 60 bulbs was 650 hours and the standard deviation was 8 hours. A second sample of 80 bulbs has a mean life of 660 hours and standard deviation 7 hours. Find the overall standard deviation.
Answer
Given that $\text{n}_1=60,\ \bar{\text{x}}_1=650,\ \text{s}_1=8$
And $\text{n}_2=80,\ \bar{\text{x}}_2=660,\ \text{s}=7$
We know that for a combined series.
$\sigma=\sqrt{\frac{\text{n}_1\text{s}^2_1+\text{n}_2\text{s}^2_2}{\text{n}_1+\text{n}_2}+\frac{\text{n}_1\text{n}_2(\bar{\text{x}}_1-\bar{\text{x}}_2)^2}{(\text{n}_1+\text{n}_2)^2}}$
$=\sqrt{\frac{60\times(8)^2+80\times(7)^2}{60+80}+\frac{60\times80(650-660)^2}{(60+80)^2}}$
$=\sqrt{\frac{6\times64+8+49}{14}+\frac{60\times80\times100}{140\times140}}$
$=\sqrt{\frac{192+196}{7}+\frac{1200}{49}}=\sqrt{\frac{338}{7}+\frac{1200}{49}}$
$=\sqrt{\frac{2716+1200}{49}}=\sqrt{\frac{3916}{49}}$
$=\frac{62.58}{7}=8.9$
Hence, the required SD = 8.9
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Question 75 Marks
While calculating the mean and variance of 10 readings, a student wrongly used the reading 52 for the correct reading 25. He obtained the mean and variance as 45 and 16 respectively. Find the correct mean and the variance.
Answer
Given, $\text{n}=10,\ \bar{\text{x}}=45,\ \sigma^2=16$
$\bar{\text{x}}=45$
$\Rightarrow\frac{\sum\text{x}_\text{i}}{\text{n}}=45$
$\Rightarrow{\sum\text{x}_\text{i}}=450$
Corrected ${\sum\text{x}_\text{i}}=450-52+25=423$
$\therefore$ Corrected mean, $\bar{\text{x}}=\frac{432}{10}=42.3$ 
 $\Rightarrow\ \sigma^2=\frac{\sum\text{x}_\text{i}^2}{\text{n}}-\Big(\frac{\sum\text{x}_\text{i}}{\text{n}}\Big)^2$
$\Rightarrow16=\frac{\sum\text{x}_\text{i}^2}{10}-(45)^2$
$\Rightarrow{\sum\text{x}_\text{i}^2}=20410$
$\therefore\text{ Corrected }{\sum\text{x}_\text{i}^2}=20410-(53)^2+(25)^2=18331$
And $\text{Corrected }\sigma^2=\frac{18331}{10}-(42.3)^2=43.81$
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Question 85 Marks
There are $60$ students in a class. The following is the frequency distribution of the marks obtained by the students in a test:
Marks
 $0$ $1$ $2$ $3$ $4$ $5$
Frequency
 $x - 2$ $x$ $x^2$ $(x + 1)^2$ $2x$ $x + 1$
where $x$ is a positive integer. Determine the mean and standard deviation of the marks.
Answer
Sum of frequencies$,$
$x - 2 + x + x^2 + (x + 1)^2 + 2x + x + 1 = 60 ($given$)$
$\Rightarrow 2x^2 + 7x - 60 = 0$
$\Rightarrow (2x + 15)(x - 4) = 0$
$\Rightarrow x = 4$
$x_i$
$f_i$
$d_i= x_i - 3$
$f_id_i$
$f_id_i^2$
$0$ $2$ $-3$ $-6$ $18$
$1$ $4$ $-2$ $-8$ $16$
$2$ $16$ $-1$ $-16$ $16$
$A = 3$ $25$ $0$ $0$ $0$
$4$ $8$ $1$ $8$ $8$
$5$ $5$ $2$ $10$ $20$
$Total$
$\sum\text{f}_\text{i}=60$
 
$\sum\text{f}_\text{i}\text{d}_\text{i}=-12$
 $\sum\text{f}_\text{i}\text{d}_\text{i}^2=78$
Mean $=\text{A}+\frac{\sum\text{f}_\text{i}\text{d}_\text{i}}{\sum\text{f}_\text{i}}=3+\Big(\frac{-12}{60}\Big)=2.8$
Standard Deviation$,$
$\sigma=\sqrt{\frac{\sum\text{f}_\text{i}\text{d}^2_\text{i}}{\sum\text{f}_\text{i}}-\Big(\frac{\sum\text{f}_\text{i}\text{d}_\text{i}}{\sum\text{f}_\text{i}}\Big)^2}$
$=\sqrt{\frac{78}{60}-\Big(\frac{-12}{60}\Big)^2}$
$=\sqrt{1.3-0.04}$
$=\sqrt{1.26}=1.12$
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Question 95 Marks
Find the standard deviation of the first $n$ natural numbers.
Answer
$x_i$
 $1$ $2$ $3$ $4$ $5$ $-$ $-$ $n$
$x_i^2$
 $1$ $4$ $9$ $16$ $25$ $-$ $-$ $n^2$
$\sum\text{x}_\text{i}=1+2+3+4+5+\ ....\ +\text{n}=\frac{\text{n}(\text{n}+1)}{2}$
$\sum\text{x}_\text{i}^2=1^2+2^2+3^2+\ ....\ +\text{n}^2=\frac{\text{n}(\text{n}+1)(2\text{n}+1)}{6}$
$\therefore\ \text{S.D.}(\sigma)=\sqrt{\frac{\sum\text{x}_\text{i}^2}{\text{n}}-\Big(\frac{\sum\text{x}_\text{i}}{\text{n}}\Big)^2}$
$=\sqrt{\frac{\text{n}(\text{n}+1)(2\text{n}+1)}{6\text{n}}-\frac{\text{n}^2(\text{n}+1)}{4\text{n}^2}}$
$=\sqrt{\frac{(\text{n}+1)(2\text{n}+1)}{6}-\frac{(\text{n+1}^2)}{4}}$
$=\sqrt{\frac{2\text{n}^2+3\text{n}+1}{6}-\frac{\text{n}^2+2\text{n}+1}{4}}$
$=\sqrt{\frac{4\text{n}^2+6\text{n}+2-3\text{n}^2-6\text{n}-3}{12}}$
$=\sqrt{\frac{\text{n}^2-1}{12}}$
Hence$,$ the required $\text{SD} =\sqrt{\frac{\text{n}^2-1}{12}}$
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Question 105 Marks
The weights of coffee in $70$ jars is shown in the following table:
Weight $($in grams$)$
Frequency
$200-201$ $13$
$201-202$ $27$
$202-203$ $18$
$203-204$ $10$
$204-205$ $1$
$205-206$ $1$
Determine variance and standard deviation of the above distribution.
Answer
$\text{Weight (in grams)}$
$\text{Frequency}$
$x_i$
$d_i = x_i - A$
$f_id_i$
$f_id_i^2$
$200-201$ $13$ $200.5$ $-2$ $-26$ $52$
$201-202$ $27$ $201.5$ $-1$ $-27$ $27$
$202-203$ $18$ $202.5(A)$ $0$ $0$ $0$
$203-204$ $10$ $203.5$ $1$ $10$ $10$
$204-205$ $1$ $204.5$ $2$ $2$ $4$
$205-206$ $1$ $205.5$ $3$ $3$ $9$
 
$N = 70$
 
 
$\sum\text{f}_\text{i}\text{d}_\text{i}=-38$
$\sum\text{f}_\text{i}\text{d}_\text{i}^2=102$
$\therefore\ \text{Variance}=\sigma^2=\frac{\sum\text{f}_\text{i}\text{d}^2_\text{i}}{\text{N}}-\Big(\frac{\sum\text{f}_\text{i}\text{d}_\text{i}}{\text{N}}\Big)^2$
$=\frac{102}{70}-\Big(\frac{-38}{70}\Big)^2=1.457-0.292=1.165$
$\therefore\ \text{SD}=\sigma=\sqrt{1.165}=1.08\text{g}$
Hence$,$ the required variance $= 1.165$ and $\text{SD} = 1.08\ g$
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Question 115 Marks
Determine mean and standard deviation of first n terms of an$ A.P$. whose first term is a and common difference is $d.$
Answer
$x_i$
$x_i - a$
$(x_i - a)^2$
$a$ $0$ $0$
$a + d$ $d$ $d^2$
$a + 2d$ $2d$ $4d^2$
$....$ $....$ $9d^2$
$....$ $....$ $....$
$....$ $....$ $....$
$a + (n - 1)d$ $(n - 1)d$ $(n - 1)2d^2$
$\sum\text{x}_\text{i}=\frac{\text{n}}{2}\big[2\text{a}+(\text{n}-1)\text{d}\big]$
 
  
$\because$ Mean  $ =\frac{\sum\text{x}_\text{i}}{\text{n}}=\frac{1}{\text{n}}\Big[\frac{\text{n}}{2}\big(2\text{a}+(\text{n}-1)\big)\text{d}\Big]=\text{a}+\frac{(\text{n}-1)}{2}\text{d}$
$\therefore\ \sum(\text{x}_\text{i}-\text{a})=\text{d}\big[1+2+3+\ .....\ +(\text{n}-1)\text{d}\big]=\text{d}+\frac{(\text{n}-1)\text{n}}{2}$
and $\sum(\text{x}_\text{i}-\text{a})^2=\text{d}^2\big[1^2+2^2+3^2+\ .....\ +(\text{n}-1)^2\text{d}\big]=\frac{\text{d}^2\text{n}(\text{n}-1)(2\text{n}-1)}{6}$
$\sigma=\sqrt{\frac{(\text{x}_\text{i}-\text{a})^2}{\text{n}}-\Big(\frac{\text{x}_\text{i}-\text{a}}{\text{n}}\Big)^2}$
$=\sqrt{\frac{\text{d}^2(\text{n})(\text{n}-1)(2\text{n}-1)}{6\text{n}}-\Big[\frac{\text{d}(\text{n}-1)\text{n}}{2\text{n}}\Big]^2}$
$=\sqrt{\frac{\text{d}^2(\text{n}-1)(2\text{n}-1)}{6}-\frac{\text{d}^2(\text{n}-1)^2}{4}}$
$=\text{d}\sqrt{\frac{(\text{n}-1)(\text{n}+1)}{12}}$
$=\text{d}\sqrt{\frac{\text{n}^2-1}{12}}$
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Question 125 Marks
The mean and standard deviation of some data for the time taken to complete a test are calculated with the following results: Number of observations $= 25,$ mean $= 18.2$ seconds, standard deviation $= 3.25$ seconds. Further$,$ another set of $15$ observations $x_1, x_2, ..., x_{15},$ also in seconds$,$ is now available and we have $\sum\limits^{15}_{\text{i}=1}\text{x}_\text{i}=279$ and $\sum\limits^{15}_{\text{i}=1}\text{x}_\text{i}^2=5524.$ Calculate the standard derivation based on all $40$ observations.
Answer
Given $\text{n}_1=25,\ \bar{\text{x}}_\text{i}=18.2,\ \sigma_1=3.25$
$\text{n}_2=15,\ \sum\limits^{15}_{\text{i}=1}\text{x}_\text{i}=279$ and $\sum\limits^{15}_{\text{i}=1}\text{x}_\text{i}^2=5524$
For first set $\sum\text{x}_\text{i}=25\times18.2=455$
$\therefore\ \sigma^2_1=\frac{\sum\text{x}^2_\text{i}}{25}-(18.2)^2$
$\Rightarrow(3.25)^2=\frac{\sum\text{x}^2_\text{i}}{25}-(18.2)^32$
$\Rightarrow10.5625+331.24=\frac{\sum\text{x}^2_\text{i}}{25}$
$\Rightarrow\sum\text{x}^2_\text{i}=25\times(10.5625+331.24)$
$\Rightarrow\sum\text{x}^2_\text{i}=25\times341.8025=8545.0625$
For combined $\text{SD}$ of the $40$ observations, $n = 40$
Now$, \sum\limits^{40}_{\text{i}=1}\text{x}^2_\text{i}=5524+8545.0625=14069.0625$
and $\sum\limits^{40}_{\text{i}=1}\text{x}_\text{i}=455+279=734$
$\therefore\ \text{SD}=\sqrt{\frac{14069.0625}{40}-\Big(\frac{734}{40}\Big)^2}$
$=\sqrt{351.1726-(18.35)^2}$
$=\sqrt{351.726-336.7225}$
$=\sqrt{15.0035}$
$=3.87$
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Question 135 Marks
Determine the mean and standard deviation for the following distribution:
Marks
 $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $11$ $12$ $13$ $14$ $15$ $16$
Frequency
 $1$ $6$ $6$ $8$ $8$ $2$ $2$ $3$ $0$ $2$ $1$ $0$ $0$ $0$ $1$
Answer
$\text{Marks}$
$f_i$
$f_ix_i$
$\text{d}_\text{i}=\text{x}_\text{i}-\bar{\text{x}}$
$f_id_i$
$f_id_i^2$
$2$ $1$ $2$ $-4$ $-4$ $16$
$3$ $6$ $18$ $-3$ $-18$ $54$
$4$ $6$ $24$ $-2$ $-12$ $24$
$5$ $8$ $40$ $-1$ $-8$ $8$
$6$ $8$ $40$ $-1$ $-8$ $8$
$7$ $2$ $14$ $1$ $2$ $2$
$8$ $2$ $16$ $2$ $4$ $8$
$9$ $3$ $27$ $3$ $9$ $27$
$10$ $0$ $0$ $4$ $0$ $0$
$11$ $2$ $22$ $5$ $10$ $50$
$12$ $1$ $12$ $6$ $6$ $36$
$13$ $0$ $0$ $7$ $0$ $0$
$14$ $0$ $0$ $8$ $0$ $0$
$15$ $0$ $0$ $9$ $0$ $0$
$16$ $1$ $16$ $10$ $10$ $100$
$Total$
$\sum\text{f}_\text{i}=40$
$\sum\text{f}_\text{i}\text{x}_\text{i}=239$
 
$\sum\text{f}_\text{i}\text{d}_\text{i}=-1$
$\sum\text{f}_\text{i}\text{x}^2_\text{i}=325$
$\therefore$ Mean $\bar{\text{x}}=\frac{\sum\text{f}_\text{i}\text{x}_\text{i}}{\sum\text{f}_\text{i}}=\frac{239}{40}=5.975\approx6$
and $\sigma=\sqrt{\frac{\sum\text{f}_\text{i}\text{x}_\text{i}}{\sum\text{f}_\text{i}}-\Big(\frac{\sum\text{f}_\text{i}\text{d}_\text{i}}{\sum\text{f}_\text{i}}\Big)^2}$
$=\sqrt{\frac{325}{40}-\Big(\frac{-1}{40}\Big)^2}$
$=\sqrt{8.125-0.000625}$
$=\sqrt{8.124375}$
$=2.85$
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Question 145 Marks
Calculate the mean deviation from the median of the following data:
Class interval
 $0-6$ $6-12$ $12-18$ $18-24$ $24-30$
Frequency
 $4$ $5$ $3$ $6$ $2$
Answer
$\text{Class interval}$
$f_i$
$X_i$
$c.f.$
$\text{d}_\text{i}=|\text{x}_\text{i}-{\text{Med}}|$
$f_id_i$
$0-6$ $4$ $3$ $4$ $11$ $44$
$6-12$ $5$ $9$ $9$ $5$ $25$
$12-18$ $3$ $15$ $12$ $1$ $3$
$18-24$ $6$ $21$ $18$ $7$ $42$
$24-30$ $2$ $27$ $20$ $13$ $26$
 
$N = 20$
 
 
 
$\sum\text{f}_\text{i}\text{d}_\text{i}=140$
Median class $=\Big(\frac{\text{N}}{2}\Big)^{\text{th}}\text{ term}=\frac{20}{2}^{\text{th}}\text{ term}$
$=10^{\text{th}}\text{ term i.e. }12-18$
$\therefore$ Median $ =\text{l}+\frac{\frac{\text{N}}{2}-\text{cf}}{\text{f}}\times\text{h}$
$=12+\frac{10-9}{3}\times6=12+\frac{1}{3}\times6=12+2=14$
and $\text{MD}=\frac{\sum\text{f}_\text{i}\text{d}_\text{i}}{\text{N}}=\frac{140}{20}=7$
Hence$,$ the required $\text{MD} = 7$
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Question 155 Marks
The mean and standard deviation of a set of $n_1$ observations are $\bar{\text{x}}_1$ and $S_1,$ respectively while the mean and standard deviation of another set of $n_2$ observations are $\bar{\text{x}}_2$ and $S_2,$ respectively. Show that the standard deviation of the combined set of $(n_1 + n_2 )$ observations is given by $\text{S.D.}=\sqrt{\frac{\text{n}_1(\text{s}_1)^2+\text{n}_2(\text{s}_2)^2}{\text{n}_1+\text{n}_2}+\frac{\text{n}_1\text{n}_2(\bar{\text{x}}_1-\bar{\text{x}}_2)^2}{(\text{n}_1+\text{n}_2)^2}}$
Answer
We have two sets of observations,
$\text{x}_\text{i},\text{ i}=1,2,3,\ .....\ ,\text{n}_1$ and $\text{y}_\text{j},\text{ j}=1,2,3,\ ....,\ \text{n}_2$
$\therefore\ \bar{\text{x}}_1=\frac{1}{\text{n}_\text{1}}\sum\limits^{\text{n}_1}_{\text{i}=1}\text{x}_\text{i}$ and $\bar{\text{x}}_2=\frac{1}{\text{n}_\text{2}}\sum\limits^{\text{n}_2}_{\text{j}=1}\text{y}_\text{j}$
$\Rightarrow\ \sigma^2_1=\frac{1}{\text{n}_1}\sum\limits^{\text{n}_1}_{\text{i}=1}(\text{x}_\text{i}-\bar{\text{x}}_1)^2$ and $\sigma^2_2=\frac{1}{\text{n}_2}\sum\limits^{\text{n}_2}_{\text{j}=1}(\text{y}_\text{j}-\bar{\text{x}}_2)^2$
Now, mean $\bar{\text{x}}$ of the given series is given by
$\bar{\text{x}}=\frac{1}{\text{n}_1+\text{n}_2}\bigg[\sum\limits^{\text{n}_1}_{\text{i}=1}\text{x}_\text{i}+\sum\limits^{\text{n}_2}_{\text{j}=1}\text{y}_\text{j}\bigg]=\frac{\text{n}_1\bar{\text{x}}_1+\text{n}_2\bar{\text{x}}_2}{\text{n}_1+\text{n}_2}$
The variance $\sigma^2$ of the combined series is given by,
$\sigma^2=\frac{1}{\text{n}_1+\text{n}_2}\bigg[\sum\limits^{\text{n}_1}_{\text{i}=1}(\text{x}_{\text{i}}-\bar{\text{x}})^2+\sum\limits^{\text{n}_2}_{\text{j}=1}(\text{y}_\text{j}-\bar{\text{x}})^2\bigg]$
Now, $\sum\limits^{\text{n}_1}_{\text{i}=1}(\text{x}_\text{i}-\bar{\text{x}})^2=\sum\limits^{\text{n}_1}_{\text{i}=1}(\text{x}_\text{i}-\bar{\text{x}}_\text{j}+\bar{\text{x}}_\text{j}-\bar{\text{x}})^2$
But $\sum\limits^{\text{n}_1}_{\text{i}=1}(\text{x}_\text{i}-\bar{\text{x}})=0$ [Algebraic sum of the deviation of values of first series from mean is zero]
Also, $\sum\limits^{\text{n}_1}_{\text{i}=1}(\text{x}_\text{i}-\bar{\text{x}})^2=\text{n}_1\text{s}^2_1=\text{n}_1(\bar{\text{x}}_1-\bar{\text{x}})^2=\text{n}_1\text{s}^2_1+\text{n}_1\text{d}^2_1$
Where, $\text{d}_1=(\bar{\text{x}}_1-\bar{\text{x}})$
Similarly, $\sum\limits^{\text{n}_2}_{\text{j}=1}(\text{y}_\text{j}-\bar{\text{x}})^2=\sum\limits^{\text{n}_2}_{\text{j}=1}(\text{y}_\text{j}-\bar{\text{x}}_\text{i}+\bar{\text{x}}_\text{i}-\bar{\text{x}})^2=\text{n}_2\text{s}^2_2+\text{n}_2\text{d}^2_2$
Where, $\text{d}_2=\bar{\text{x}}_2-\bar{\text{x}}$
Combined $\text{SD}$, $\sigma=\sqrt{\frac{\big[\text{n}_1\big(\text{s}^2_1+\text{d}^2_1\big)+\text{n}_2\big(\text{s}^2_2+\text{d}^2_2\big)\big]}{\text{n}_1+\text{n}_2}}$
Where, $\text{d}_1=\bar{\text{x}}_1-\bar{\text{x}}=\bar{\text{x}}_1-\Big(\frac{\text{n}_1\bar{\text{x}}_1+\text{n}_2\bar{\text{x}_2}}{\text{n}_1+\text{n}_2}\Big)=\frac{\text{n}_2(\bar{\text{x}}_1-\bar{\text{x}}_2)}{\text{n}_1+\text{n}_2}$
and $\text{d}_2=\bar{\text{x}}_2-\bar{\text{x}}=\bar{\text{x}}_2-\frac{\text{n}_1\bar{\text{x}}_1+\text{n}_2\bar{\text{x}_2}}{\text{n}_1+\text{n}_2}=\frac{\text{n}_1(\bar{\text{x}}_2-\bar{\text{x}}_1)}{\text{n}_1+\text{n}_2}$
$\therefore\ \sigma^2=\frac{1}{\text{n}_1+\text{n}_2}\bigg[\text{n}_1\text{s}^2_1+\text{n}_2\text{s}^2_2+\frac{\text{n}_1\text{n}_2(\bar{\text{x}}_1+\bar{\text{x}}_2)^2}{(\text{n}_1+\text{n}_2)^2}+\frac{\text{n}_2\text{n}_1(\bar{\text{x}}_2+\bar{\text{x}}_1)^2}{(\text{n}_1+\text{n}_2)^2}\bigg]$
Also, $\sigma=\sqrt{\frac{\text{n}_1\text{s}^2_1+\text{n}_2\text{s}^2_2}{\text{n}_1+\text{n}_2}+\frac{\text{n}_1\text{n}_2(\bar{\text{x}}_1-\bar{\text{x}}_2)^2}{(\text{n}_1+\text{n}_2)^2}}$
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