5 Marks Questions

Question 15 Marks

The following are the marks of 150 students in an examination. Calculate Karl Pearson's coefficient of skewness.

Answer

First we calculate the mean and median from the given distribution. Mode is not well-defined here

Here, $N =\Sigma f=150, h=10$. Also, $\frac{N}{2}=\frac{150}{2}=75$, therefore median class $=40-50$
$\begin{aligned} \text { Median } & =l+\frac{\left(\frac{N}{2}-c\right)}{f} \times h \\ & =40+\left(\frac{75-70}{10}\right) \times 10 \\ & =45 \\ \text { Mean }(\bar{x}) & =a+\frac{\Sigma f d}{N} \times h\end{aligned}$
$=35+\frac{64}{150} \times 10 \qquad(\because a=35)$
$=39.27$ (Approx.)
Standard Deviation $(\sigma)=h \times \sqrt{\frac{\Sigma f d^2}{N}-\left(\frac{\Sigma f d}{N}\right)^2}$
$=10 \times \sqrt{\frac{808}{150}-\left(\frac{64}{150}\right)^2}=10 \times \sqrt{5.2047}=22.8$
Therefore, coefficient of skewness :
$
\begin{aligned}
S_k & =\frac{3(\text { Mean }- \text { Median })}{\sigma} \\
& =\frac{3(39.27-45)}{22.8}=-0.754
\end{aligned}
$

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Question 25 Marks

Calculate the first four moments about the arbitrary origin and then calculate the first four moments about mean.

x	10-13	13-16	16-19	19-22	22-25	25-28
f	2	4	26	47	15	6

Answer

Moments about origin :
$\begin{array}{l}\mu_1^{\prime}=\frac{\Sigma f d^{\prime}}{N} \times i=\frac{87}{100} \times 3=2.61 \\ \mu_2^{\prime}=\frac{\Sigma f d^{\prime 2}}{N} \times i^2=\frac{173}{100} \times(3)^2=15.57 \\ \mu_3^{\prime}=\frac{\Sigma f d^{\prime 3}}{N} \times i^3=\frac{309}{100} \times(3)^3=83.43 \\ \mu_4^{\prime}=\frac{\Sigma f d^{\prime 4}}{N} \times i^4=\frac{809}{100} \times(3)^4=655.29\end{array}$
Moments about mean :
$\begin{aligned} \mu_1 & =\mu_1^{\prime}-\mu_1^{\prime}=2.61-2.61=0 \\ \mu_2 & =\mu_2^{\prime}-\left(\mu_1^{\prime}\right)^2=15.57-(2.61)^2=8.76 \\ \mu_3 & =\mu_3^{\prime}-3 \mu_2^{\prime} \mu_1^{\prime}+2\left(\mu_1^{\prime}\right)^3=83.43-3(2.61)(15.57)+2(2.61)^3=-2.91 \\ \mu_4 & =\mu_4^{\prime}-4 \mu_3^{\prime} \mu_1^{\prime}+6 \mu_2^{\prime}\left(\mu_1^{\prime}\right)^2-3\left(\mu_1^{\prime}\right)^4 \\ & =665.29-4(83.43)(2.61)+6(15.57)(2.61)^2-3(2.61)^4 \\ & =291.454\end{aligned}$

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Question 35 Marks

For the following distribution, find central moments, $\beta_1$ and $\beta_2$.

Class	1.5 - 2.5	2.5 - 3.5	3.5 - 4.5	4.5 - 5.5	5.5 - 6.5
Frequency	1	3	7	3	1

Answer

Here,
$
\text { mean, } \bar{x}=\frac{\Sigma f x}{\Sigma N}=\frac{60}{15}=4
$
We therefore have,
$\begin{array}{l}\mu_1=\frac{\Sigma f d}{N}=\frac{0}{15}=0 \\ \mu_2=\frac{\Sigma f d^2}{N}=\frac{14}{15}=0.933 \text { (Approx.) } \\ \mu_3=\frac{\Sigma f d^3}{N}=\frac{0}{15}=0 \\ \mu_4=\frac{\Sigma f d^4}{N}=\frac{38}{15}=2.533 \text { (Approx.) }\end{array}$
Thus,
$
\beta_1=\frac{\mu_3^2}{\mu_2^3}=\frac{0}{(0.933)^3}=0
$
Since, $\beta_1=0$ it means the distribution is symmetrical. and
$
\beta_2=\frac{\mu_4}{\mu_2^2}=\frac{2.53}{(0.933)^2}=2.91
$
Since, $\beta_2<3$ it means curve is platykurtic.

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Question 45 Marks

The mean and standard deviation of 20 observations are found to be 10 and 2 , respectively. On rechecking, it was found that an observation 8 was incorrect. Calculate the correct mean and standard deviation in each of the following cases :
(i) If wrong item is omitted.
(ii) If it is replaced by 12 .

Answer

(i)
$\begin{aligned} \text { Number of observations }(n) & =20 \\ \text { Incorrect mean } & =10 \\ \text { Incorrect standard deviation } & =2 \\ \bar{x} & =\frac{1}{n} \sum_{i=1}^{20} x_i \\ 10 & =\frac{1}{20} \sum_{i=1}^{20} x_i\end{aligned}$
$
\Rightarrow \quad \sum_{i=1}^{20} x_i=200
$
That is, incorrect sum of observations $=200$
Correct sum of observations $=200-8=192$
$\therefore \quad$ Correct mean $=\frac{\text { correct sum }}{19}=\frac{192}{19}=10.1$ (Approx.)
Standard deviation $\sigma=\sqrt{\frac{1}{n} \sum_{i=1}^n x_i^2-\frac{1}{n^2}\left(\sum_{i=1}^n x_i\right)^2}=\sqrt{\frac{1}{n} \sum_{i=1}^n x_i^2-(\bar{x})^2}$
$\begin{aligned} \Rightarrow \quad 2 & =\sqrt{\frac{1}{20} \text { Incorrect } \sum_{i=1}^n x_i^2-(10)^2} \\ \Rightarrow \quad 4 & =\frac{1}{20} \text { Incorrect } \sum_{i=1}^n x_i^2-100 \\ \Rightarrow \quad \text { Incorrect } \sum_{i=1}^{15} x_i & =2080 \\ & =2080-64 \\ & =2016\end{aligned}$
$\begin{aligned} \therefore \quad \text { Correct Standard deviation } & =\sqrt{\frac{\text { Correct } \sum x_i^2}{n}-(\text { Correct mean })^2} \\ & =\sqrt{\frac{2016}{19}-(10.1)^2} \\ & =\sqrt{1061.1-102.1} \\ & =\sqrt{4.09} \\ & =2.02\end{aligned}$
(ii) When 8 is replaced by 12
$\begin{aligned} \text { Incorrect sum of observations } & =200 \\ ∴ \text { Correct sum of observation } & =200-8+12=204 \\ ∴\text { Correct mean } & =\frac{\text { Correct sum }}{20}=\frac{204}{20}=10.2 \\ \text { Standard deviation } \sigma & =\sqrt{\frac{1}{n} \sum_{i=1}^n x_i^2-\frac{1}{n^2}\left(\sum_{i=1}^n x_i\right)^2}=\sqrt{\frac{1}{n} \sum_{i=1}^n x_i^2-(\bar{x})^2} \\⇒ 2 & =\sqrt{\frac{1}{20} \text { Incorrect } \sum_{i=1}^n x_i^2-(10)^2} \\ ⇒ 4 & =\frac{1}{20} \text { Incorrect } \sum_{i=1}^n x_i^2-100 \\⇒ \text { Incorrect } \sum_{i=1}^{15} x_i & =2080\end{aligned}$
∴ Correct $\sum x_i^2=$ Incorrect $\sum_{i=1}^{15} x_j-(8)^2+(12)^2$
$\begin{array}{l}=2080-64+144 \\ =2160\end{array}$
∴ Correct standard deviation $=\sqrt{\frac{\text { Correct } \sum x_i^2}{n}-(\text { Correct mean })^2}$
$\begin{array}{l}=\sqrt{\frac{2160}{20}-(10.2)^2} \\ =\sqrt{108-104.04} \\ =\sqrt{3.96} \\ =1.98\end{array}$

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Question 55 Marks

An analysis of monthly wages paid to workers in two firms A and B, belonging to the same industry, gives the following results :

Particular	Firm A	Firm B
No. of wage earners	586	648
Mean of monthly wages	₹ 5,253	₹ 5,253
Variance of the distribution of wages	100	121

(i) Which firm A or B pays larger amounts as monthly wages ?
(ii) Which firm A or B shows greater variability in individual wages ?

Answer

For Firm A :
No. of wages earners $=586$
Mean of monthly wages, $\bar{x}=$₹ $5253$
Amount paid by firm $A=$ ₹ $(586 \times 5253)$=₹ $ 3078258$
Variance of distribution of wages, $\sigma^2=100$
Standard deviation, $\sigma=\sqrt{\sigma^2}$
$=\sqrt{100}$
$=10$
Coefficient of Variation $=\frac{\sigma}{\bar{x}} \times 100$
$=\frac{10}{5,253} \times 100$
$=0.19$
For Firm B :
No. of wage earners $=648$
Mean of monthly wages, $\bar{x}=$₹ $5253$
Amount paid by firm $B=648 \times 5253= $₹ $3403944$
Standard deviation, $\sigma=\sqrt{\sigma^2}=\sqrt{121}=11$
$
\begin{aligned}
\text { Coefficient of variation } & =\frac{\sigma}{\bar{x}} \times 100 \\
& =\frac{11}{5,253} \times 100 \\
& =0.21
\end{aligned}
$
Monthly wages paid by firm $A=$₹ $3078258$
Monthly wages paid by firm $B=$₹ $ 3403944$
$\therefore$ Firm B pays larger amount as monthly wages. Coefficient of variation of wages, of firm $A=0.19$ Coefficient of variation of wages, of firm $B=0.21$
$\therefore$ Firm B shows greater variability in individual wages.

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Question 65 Marks

From the data given below state which group is more variable, A or B ?

Marks	10-20	20-30	30-40	40-50	50-60	60-70	70-80
Group A	9	17	32	33	40	10	9
Group B	10	20	30	25	43	15	7

Answer

For Group A :

Marks	$f_i$	$x_i$	$d_i=\frac{x_i-a}{n}$	$f_i d_i$	$f_i d_i^2$
10-20	9	15	-3	-27	81
20-30	17	25	-2	-34	68
30-40	32	35	-1	-32	32
40-50	33	45	0	0	0
50-60	40	55	1	40	40
60-70	10	65	2	20	40
70-80	9	75	3	27	81
Total	N=150			-6	342

$a=45$
$\begin{aligned} \text { Variance (for Group A) } & =h^2\left[\frac{\sum f_i d_i^2}{N}-\left(\frac{\sum f_i d_i}{N}\right)^2\right] \\ & =(10)^2\left[\frac{342}{150}-\left(\frac{-6}{150}\right)^2\right] \\ & =100[2.28-0.0016] \\ & =227.84\end{aligned}$
For Group B :

Marks	$f_i$	$x_i$	$d_i=\frac{x_i-a}{n}$	$f_i d_i$	$f_i d_i^2$
10-20	10	15	-3	-30	90
20-30	20	25	-2	-40	80
30-40	30	35	-1	-30	30
40-50	25	45	0	0	0
50-60	43	55	1	43	43
60-70	15	65	2	30	60
70-80	7	75	3	21	63
Total	N=150			-6	366

$a=45$
$\begin{aligned} \text { Variance (for Group B) } & =h^2\left[\frac{\sum f_i d_i^2}{N}-\left(\frac{\sum f_i d_i}{N}\right)^2\right] \\ & =(10)^2\left[\frac{366}{150}-\left(\frac{-6}{150}\right)^2\right] \\ & =100[2.4384] \\ & =243.84\end{aligned}$
Since variance for Group $B$ is more than that of $A$, so $B$ is more variable than $A$.

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Question 75 Marks

Following are the marks obtained out of 100 , by two students Reeta and Seeta in 10 tests.

Reeta	25	50	45	30	70	42	36	48	35	60
Seeta	10	70	50	20	95	55	42	60	48	80

Who is more intelligent and who is more consistent ?

Answer

For Reeta :

$x_i$	$x_i^2$
25	625
50	2500
45	2025
30	900
70	4900
42	1764
36	1296
48	2304
35	1225
60	3600
$\sum_{i=1}^{10} x_i=441$	$\sum_{i=1}^{10} x_i^2=21139$

Mean, $\bar{x}=\frac{\sum x_i}{n}=\frac{441}{10}=44.1$
$\begin{aligned} \sigma & =\sqrt{\frac{\sum x_i^2}{n}-\left(\frac{\sum x_i}{n}\right)^2} \\ & =\sqrt{\frac{21139}{10}-\left(\frac{441}{10}\right)^2} \\ & =\sqrt{\frac{211390-194481}{100}}=\sqrt{169.09}=13 \text { (approx.) }\end{aligned}$
For Reeta C.V. $=\frac{13}{44.1} \times 100$
For Seeta :

$x_i$	$x_i^2$
10	100
70	4900
50	2500
20	400
95	9025
55	3025
42	1764
60	3600
48	2304
80	6400
$\sum_{i=1}^{10} x_i=530$	$\sum_{i=1}^{10} x_i^2=34018$

Mean, $\bar{x}=\frac{\sum x_i}{n}=\frac{530}{10}=53$
$and\qquad\sigma=\sqrt{\frac{\sum x_i^2}{n}-\left(\frac{\sum x_i}{n}\right)^2}$
$\sigma=\sqrt{\frac{34018}{10}-(53)^2}=\sqrt{5928}=24.35$ (Approx.)
For Reeta, C.V. $=\frac{\sigma}{\bar{x}} \times 100=\frac{13}{44.1} \times 100=29.48$ (Approx.)
and $\qquad$ for Seeta, C.V. $=\frac{\sigma}{\bar{x}} \times 100=\frac{24.35}{53} \times 100=45.94$
Since coefficient of variation for Seeta is more than that of Reeta, so Seeta is more variable and hence Reeta is more consistent. And Seeta is more intelligent.

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Question 85 Marks

Calculate the mean deviation about the mean of the set first $n$ natural numbers when $n$ is an even number.

Answer

Let $n=2 k \Rightarrow k=\frac{n}{2}$
$
\begin{aligned}
\bar{x} & =\frac{1+2+3+\ldots+n}{n} \\
& =\frac{n(n+1)}{2 n} \\
& =\frac{n+1}{2} \\
& =\frac{2 k+1}{2}
\end{aligned}
$

$x_i$	$x_i-\bar{x}$	$\left\|x_i-\bar{x}\right\|$
1	$1-\left(\frac{2 k+1}{2}\right)=-\left(\frac{2 k-1}{2}\right)$	$\frac{2 k-1}{2}$
2	$-\left(\frac{2 k-3}{2}\right)$	$\frac{2 k-3}{2}$
3	$-\left(\frac{2 k-5}{2}\right)$	$\frac{2 k-5}{2}$
, , ,	, , ,	, , ,
$k-2$	$-\frac{5}{2}$	$\frac{5}{2}$
$k-1$	$-\frac{3}{2}$	$\frac{3}{2}$
$k$	$-\frac{1}{2}$	$\frac{1}{2}$
$k+1$	$\frac{1}{2}$	$\frac{1}{2}$
$k+2$	$\frac{3}{2}$	$\frac{3}{2}$
, , ,	, , ,	, , ,
$2 k+1$	$\frac{2 k-3}{2}$	$\frac{2 k-3}{2}$
$2 k-1$	$\frac{2 k-3}{2}$	$\frac{2 k-3}{2}$

$\begin{array}{l}\therefore \Sigma|x-\bar{x}| \\ =\left\{\frac{2 k-1}{2}+\frac{2 k-3}{2}+\frac{2 k-5}{2}+\ldots+\frac{5}{2}+\frac{3}{2}+1\right\} \\ =1+3+5+\ldots+(2 k-1) \\ =\frac{k}{2}\{1+(2 k-1)\}=k^2=\left(\frac{n}{2}\right)^2=\frac{n^2}{4} .\end{array}$

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Question 95 Marks

Calculate the mean deviation about the mean of the set of first $n$ natural numbers when ' $n$ ' is an odd number.

Answer

Let $n=2 k+1 \Rightarrow k=\frac{n-1}{2}$
$\therefore$ $\bar{x}=\frac{1+2+3+\ldots+n}{n}$
$=\frac{n(n+1)}{2 n}=\frac{n+1}{2}$
$=\frac{2 k+1+1}{2}=k+1$.

$x_i$	$x_i-\bar{x}$	$\left\|x_i-\bar{x}\right\|$
$\begin{array}{l}1 \\ 2 \\ 3 \\ 4\end{array}$	$\begin{array}{c}1-(k+1)=-k \\ -(k-1) \\ -(k-2) \\ -(k-3)\end{array}$	$\begin{array}{c}k \\ k-1 \\ k-2 \\ k-3\end{array}$
, , ,	, , ,	, , ,
$\begin{array}{c}k-2 \\ k-1 \\ k \\ k+1 \\ k+2 \\ k+3\end{array}$	$\begin{array}{c}-3 \\ -2 \\ -1 \\ 0 \\ 1 \\ 2\end{array}$	$\begin{array}{l}3 \\ 2 \\ 1 \\ 0 \\ 1 \\ 2\end{array}$
, , ,	, , ,	, , ,
$\begin{array}{c}2 k \\ 2 k+1\end{array}$	$\begin{array}{c}k-1 \\ k\end{array}$	$\begin{array}{c}k-1 \\ k\end{array}$

$\Sigma\left|x_i-\bar{x}\right|=2(1+2+3 \ldots+k)=2\left\{\frac{k(k+1)}{2}\right\}$
$\begin{array}{l}=k(k+1) \\ =\left(\frac{n-1}{2}\right)\left(\frac{n-1}{2}+1\right)\end{array}$
$\begin{array}{l}=\frac{n-1}{2} \cdot \frac{n+1}{2} \\ =\frac{n^2-1}{4}\end{array}$
$\therefore$ mean deviation about mean i.e.,
$
\begin{aligned}
\text { M.D. }(\bar{x}) & =\frac{1}{n} \sum\left|x_1-\bar{x}\right| \\
& =\frac{1}{n} \cdot \frac{n^2-1}{4} \\
& =\frac{n^2-1}{4 n}
\end{aligned}
$

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Question 105 Marks

Calculate the mean deviation about mean :

Marks	0-10	10-20	20-30	30-40	40-50
No. of Students	5	8	15	16	6

Also, find coefficient of mean deviation.

Answer

Let us calculate Mean deviation.

Here,
$
\begin{aligned}
\operatorname{Mean}(\bar{x}) & =\text { assumed mean }+\frac{\Sigma f_i d_i}{\Sigma f_i} \times h \\
& =25+\frac{10}{50} \times 10 \\
& =25+2 \\
& =27
\end{aligned}
$
Mean deviation about mean,
$
\begin{aligned}
\operatorname{M.D} \cdot(\bar{x}) & =\frac{\Sigma f_i\left|x_i-27\right|}{\Sigma f_i} \\
& =\frac{472}{50} \\
& =9.44
\end{aligned}
$
$
\begin{aligned}
\text { Coefficient of mean deviation } & =\frac{\text { M.D. }(\bar{x})}{\bar{x}} \times 100 \\
& =\frac{9.44}{27} \times 100 \\
& =34.96(\text { Approx. })
\end{aligned}
$

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Question 115 Marks

Calculate the mean deviation about median for the following data :

Class	0-10	10-20	20-30	30-40	40-50	50-60
Frequency	6	7	15	16	4	2

Also, find coefficient of mean deviation.

Answer

Here, The class interval containing $\left(\frac{N}{2}\right)^{\text {th }}$ or $25^{\text {th }}$ terms is $20-30$.
$\therefore 20-30$ is the median class.
$\therefore$ Median, MD $=l+\frac{\frac{N}{2}-C}{f} \times h$ where, $l=20, c=13, f=15, h=10$ and $N =50$
$
\begin{array}{l}
=20+\frac{25-13}{15} \times 10 \\
=20+8 \\
=28
\end{array}
$
Mean deviation about median,
$\begin{aligned} \text { M.D. }\left(M_d\right) & =\frac{1}{N} \Sigma f_i\left|x_i-M\right| \\ & =\frac{1}{50} \times 508 \\ & =10.16 . \\ \text { Coefficient of mean deviation about median } & =\frac{\text { M.D. }\left(M_d\right)}{\text { Median }} \times 100 \\ & =\frac{10.16}{28} \times 100 \\ & =36.28 . \text { (Approx.) }\end{aligned}$

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Question 125 Marks

The following table gives information regarding weekly income of labourers working at a dam site :

Estimate (i) median (ii) 3rd decile (iii) $P_{95}$.

Answer

Here the classes are already continuous and of uniform width. The cumulative frequency table is

Weekly Income (in ₹)	No. of employees	Cumulative frequency
600- 700	40	40
700 -800	68	108
800 - 900	86	194
900 - 1000	120	314
1000 - 1100	90	404
1100 - 1200	40	444
1200 - 1300	26	470

Total number, $n=470$
(i) Median is $\frac{n}{2}$ th i.e., $\frac{470}{2}$ th value i.e., 235 th value,
which lies in class $900-1000$
$\therefore \quad$ Median $\left(Q_2\right)=l+\frac{\frac{n}{2}-c}{f} \times i$
$=900+\frac{235-194}{120} \times 100$
$=900+34.17=934.17$ i.e., ₹ 934.17
(ii) Third decile $D_3$ is $\frac{3}{10}$ of 470 th value i.e., ( 0.3 ) ( 470 th) value i.e., 141 st value, which lies in class $800-900$.
$\therefore \quad D_3=l+\frac{\frac{3}{10} n-c}{f} \times i$
$=800+\frac{141-108}{86} \times 100$
$=800+38.37=838.37$ i.e., ₹ $ 838.37$
(iii) $P_{95}$ is $\frac{95}{100}$ of 470 th value i.e, 446.5 th value, which lies in class $1200-1300$.
$\therefore \quad P_{95}=l+\frac{\frac{95}{100} n-c}{f} \times i$
$=1200+\frac{446.5-444}{26} \times 100$
$=1200+9.5$
$=1209.6$ i.e., ₹ 1209.6.

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Applied Maths 5 Marks Questions

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