Chapter 5 · 10 hours

Statistics

SEE past exam questions

Past questions and answers

32 questions set from this chapter; 12 are most repeated (set, or a close variant set, in 3 or more exams). Most repeated first.

Most repeated · 6 of 10 exams
2081 Grade increment · 3 marks

Find the standard deviation from the given data.

Class Interval 60-65 65-70 70-75 75-80 80-85
Frequency 7 5 6 4 3

Class Interval	60-65	65-70	70-75	75-80	80-85
Frequency	7	5	6	4	3

Similar questions: Standard deviation, 0-100 (2082 Grade increment) · Coefficient of standard deviation (2079 Bagmati) · Standard deviation, 10-60 (2082 Preparation · Bhaktapur · Bhaktapur Municipality)

Answer

Take assumed mean $A=72.5$ and $d=x-A$ .

Class interval	$x$	$f$	$d$	$fd$	$fd^2$
60–65	62.5	7	−10	−70	700
65–70	67.5	5	−5	−25	125
70–75	72.5	6	0	0	0
75–80	77.5	4	5	20	100
80–85	82.5	3	10	30	300
Total		$N=25$		$\Sigma fd=-45$	$\Sigma fd^2=1225$

$\sigma=\sqrt{\frac{\Sigma fd^2}{N}-\left(\frac{\Sigma fd}{N}\right)^2}=\sqrt{\frac{1225}{25}-\left(\frac{-45}{25}\right)^2}=\sqrt{49-3.24}=\sqrt{45.76}\approx 6.76$

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Most repeated · 6 of 10 exams
2082 Grade increment · 3 marks

Find the standard deviation from the given data.

Class interval 0 - 20 20 - 40 40 - 60 60 - 80 80 - 100
Frequency 2 5 6 4 3

Class interval	0 - 20	20 - 40	40 - 60	60 - 80	80 - 100
Frequency	2	5	6	4	3

Similar questions: Find the standard deviation from the given (2081 Grade increment) · Coefficient of standard deviation (2079 Bagmati) · Standard deviation, 10-60 (2082 Preparation · Bhaktapur · Bhaktapur Municipality)

Answer

Mid-values $m=10,30,50,70,90$ . Take assumed mean $50$ , $i=20$ , $d'=\dfrac{m-50}{20}$ .

$m$	$f$	$d'$	$fd'$	$fd'^2$
10	2	$-2$	$-4$	8
30	5	$-1$	$-5$	5
50	6	0	0	0
70	4	1	4	4
90	3	2	6	12
	$N=20$		$\Sigma fd'=1$	$\Sigma fd'^2=29$

$\sigma=i\sqrt{\frac{\Sigma fd'^2}{N}-\left(\frac{\Sigma fd'}{N}\right)^2}=20\sqrt{\frac{29}{20}-\left(\frac1{20}\right)^2}=20\sqrt{1.4475}$

$\sigma\approx20\times1.2031=\boxed{24.06}$

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Most repeated · 6 of 10 exams
2079 Bagmati · 4 marks

Find the coefficient of standard deviation of the given data.

Class interval 30-40 40-50 50-60 60-70 70-80
Frequency 2 3 6 5 4

Class interval	30-40	40-50	50-60	60-70	70-80
Frequency	2	3	6	5	4

Similar questions: Find the standard deviation from the given (2081 Grade increment) · Standard deviation, 0-100 (2082 Grade increment) · Standard deviation, 10-60 (2082 Preparation · Bhaktapur · Bhaktapur Municipality)

Answer

Take assumed mean $A=55$ and class width $h=10$ , with $d=\dfrac{m-55}{10}$ .

Class	$f$	$m$	$d$	$fd$	$fd^2$
30-40	2	35	-2	-4	8
40-50	3	45	-1	-3	3
50-60	6	55	0	0	0
60-70	5	65	1	5	5
70-80	4	75	2	8	16
	$N=20$			$\sum fd=6$	$\sum fd^2=32$

$\bar x=A+\frac{\sum fd}{N}\times h=55+\frac{6}{20}\times10=58$

$\sigma=h\sqrt{\frac{\sum fd^2}{N}-\left(\frac{\sum fd}{N}\right)^2}=10\sqrt{1.6-0.09}=10\sqrt{1.51}\approx12.29$

$\text{Coefficient of S.D.}=\frac{\sigma}{\bar x}=\frac{12.29}{58}\approx 0.212$

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Most repeated · 6 of 10 exams
2082 Preparation · Bhaktapur · Bhaktapur Municipality · 3 marks

Find the standard deviation of the data given below:

Class interval 10-20 20-30 30-40 40-50 50-60
Frequency 4 10 12 8 6

Class interval	10-20	20-30	30-40	40-50	50-60
Frequency	4	10	12	8	6

Similar questions: Find the standard deviation from the given (2081 Grade increment) · Standard deviation, 0-100 (2082 Grade increment) · Coefficient of standard deviation (2079 Bagmati)

Answer

Take assumed mean $A=35$ and class width $i=10$ , with $d'=\dfrac{m-35}{10}$ .

Class	$m$	$f$	$d'$	$fd'$	$fd'^2$
10-20	15	4	$-2$	$-8$	16
20-30	25	10	$-1$	$-10$	10
30-40	35	12	0	0	0
40-50	45	8	1	8	8
50-60	55	6	2	12	24
		40		2	58

$\sigma=i\sqrt{\frac{\sum fd'^2}{N}-\left(\frac{\sum fd'}{N}\right)^2}=10\sqrt{\frac{58}{40}-\left(\frac{2}{40}\right)^2}=10\sqrt{1.45-0.0025}=10\sqrt{1.4475}$

$\boxed{\sigma\approx12.03}$

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Most repeated · 5 of 10 exams
2082 Chaitra · Set B · 3 marks

Calculate the standard deviation from the given data.

Class interval 0 - 4 4 - 8 8 - 12 12 - 16 16 - 20 20 - 24
Frequency 7 7 10 15 7 6

Class interval	0 - 4	4 - 8	8 - 12	12 - 16	16 - 20	20 - 24
Frequency	7	7	10	15	7	6

Similar questions: Calculate the mean deviation from mean of (2081 Grade increment) · Find the standard deviation from the given (2081 Grade increment) · Standard deviation, 0-100 (2082 Grade increment)

Answer

Class interval	$x$	$f$	$fx$	$fx^2$
0–4	2	7	14	28
4–8	6	7	42	252
8–12	10	10	100	1000
12–16	14	15	210	2940
16–20	18	7	126	2268
20–24	22	6	132	2904
Total		$N=52$	$\Sigma fx=624$	$\Sigma fx^2=9392$

$\bar x=\frac{624}{52}=12$

$\sigma=\sqrt{\frac{\Sigma fx^2}{N}-\bar x^2}=\sqrt{\frac{9392}{52}-144}=\sqrt{180.615-144}=\sqrt{36.615}\approx 6.05$

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Most repeated · 5 of 10 exams
2081 Chaitra · Set A · 3 marks

Find the mean deviation from mean of the given data.

Class Interval 0-10 10-20 20-30 30-40 40-50
Frequency 5 8 15 16 6

Class Interval	0-10	10-20	20-30	30-40	40-50
Frequency	5	8	15	16	6

Answer

Class interval	$x$	$f$	$fx$	$\lvert x-\bar x\rvert$	$f\lvert x-\bar x\rvert$
0–10	5	5	25	22	110
10–20	15	8	120	12	96
20–30	25	15	375	2	30
30–40	35	16	560	8	128
40–50	45	6	270	18	108
Total		$N=50$	$\Sigma fx=1350$		$\Sigma f\lvert x-\bar x\rvert=472$

$\bar x=\frac{1350}{50}=27$

$\text{M.D. (from mean)}=\frac{\Sigma f\lvert x-\bar x\rvert}{N}=\frac{472}{50}=9.44$

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Most repeated · 3 of 10 exams
2082 Chaitra · Set A · 2 marks

In a continuous series if $\sum f|m - \bar{x}| = 1134$ , $N = 100$ and mean $(\bar{x}) = 40$ , find the mean deviation and its coefficient.

Similar questions: In a continuous series, median (Md) =40, (2082 Chaitra · Set B) · Coefficient of mean deviation (2082 Grade increment)

Answer

Given: $\sum f|m-\bar{x}|=1134$ , $N=100$ , $\bar{x}=40$

Mean deviation from mean:

MD = \frac{\sum f|m-\bar{x}|}{N} = \frac{1134}{100} = 11.34

Coefficient of mean deviation:

\text{Coefficient} = \frac{MD}{\bar{x}} = \frac{11.34}{40} = 0.2835

Mean deviation = 11.34 and its coefficient = 0.2835.

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Most repeated · 3 of 10 exams
2082 Chaitra · Set B · 2 marks

In a continuous series, median ( $M_d$ ) $=40$ , $\Sigma f=50$ and $\Sigma f|m-M_d|=530$ . Find the coefficient of mean deviation.

Similar questions: Coefficient of mean deviation (2082 Grade increment) · In a continuous series if sum f|m (2082 Chaitra · Set A)

Answer

Mean deviation about median,

$MD = \frac{\Sigma f|m-M_d|}{\Sigma f} = \frac{530}{50}=10.6$

Coefficient of mean deviation about median,

$= \frac{MD}{M_d}=\frac{10.6}{40}=0.265$

Coefficient of mean deviation = 0.265

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Most repeated · 3 of 10 exams
2081 Chaitra · Set B · 2 marks

In a continuous series, third quartile is two times of the first quartile. If the sum of the first quartile and third quartile is 90, find the quartile deviation.

Similar questions: If the third quartile (Q3) of a (2081 Grade increment) · Find Q1 from Q1+Q3 and Q3-Q1 (2082 Pre-board · N-PABSAN)

Answer

Let the first quartile be $Q_1=x$ .
Then the third quartile is $Q_3=2x$ .

Given, $Q_1+Q_3=90$ $x+2x=90$ $3x=90$ $x=30$

So, $Q_1=30,\qquad Q_3=60$

Quartile deviation, $QD=\frac{Q_3-Q_1}{2}=\frac{60-30}{2}=15$

Quartile deviation = 15

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Most repeated · 3 of 10 exams
2081 Grade increment · 3 marks

Calculate the mean deviation from mean of the given data.

Class Interval 0-10 10-20 20-30 30-40 40-50
Frequency 5 8 15 16 6

Class Interval	0-10	10-20	20-30	30-40	40-50
Frequency	5	8	15	16	6

Similar questions: Calculate the standard deviation from the given (2082 Chaitra · Set B) · Find the mean deviation from mean of (2081 Chaitra · Set A)

Answer

Class interval	$x$	$f$	$fx$	$\lvert x-\bar x\rvert$	$f\lvert x-\bar x\rvert$
0–10	5	5	25	22	110
10–20	15	8	120	12	96
20–30	25	15	375	2	30
30–40	35	16	560	8	128
40–50	45	6	270	18	108
Total		$N=50$	$\Sigma fx=1350$		$\Sigma f\lvert x-\bar x\rvert=472$

$\bar x=\frac{1350}{50}=27$

$\text{M.D. (from mean)}=\frac{\Sigma f\lvert x-\bar x\rvert}{N}=\frac{472}{50}=9.44$

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Most repeated · 3 of 10 exams
2082 Grade increment · 2 marks

In a continuous series, if $\Sigma f|m-M_d|=220$ , $N=20$ and median $(M_d)=44$ , find the coefficient of mean deviation.

Similar questions: In a continuous series, median (Md) =40, (2082 Chaitra · Set B) · In a continuous series if sum f|m (2082 Chaitra · Set A)

Answer

Mean deviation about the median:

$\text{MD}=\frac{\Sigma f|m-M_d|}{N}=\frac{220}{20}=11$

Coefficient of mean deviation:

$\frac{\text{MD}}{M_d}=\frac{11}{44}=0.25$

$\boxed{0.25}$

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Most repeated · 3 of 10 exams
CDC Model Question 2080/2081 Optional Mathematics · 4 marks

Find the mean deviation and its coefficient of the given data.

Marks obtained 0-10 10-20 20-30 30-40 40-50
No. of students 2 3 6 5 4

Marks obtained	0-10	10-20	20-30	30-40	40-50
No. of students	2	3	6	5	4

Similar questions: Find the mean deviation from median of (2081 Chaitra · Set B) · Mean deviation from median (2082 Pre-board · N-PABSAN) · Find the coefficient of variation from the (2081 Chaitra · Set A)

Answer

Marks	$f$	$x$	$fx$	$\lvert x-\bar x\rvert$	$f\lvert x-\bar x\rvert$
0–10	2	5	10	23	46
10–20	3	15	45	13	39
20–30	6	25	150	3	18
30–40	5	35	175	7	35
40–50	4	45	180	17	68
Total	$N=20$		$\Sigma fx=560$		$206$

$\bar x=\frac{560}{20}=28$

$\text{M.D.}=\frac{\Sigma f\lvert x-\bar x\rvert}{N}=\frac{206}{20}=10.3$

$\text{Coefficient of M.D.}=\frac{\text{M.D.}}{\bar x}=\frac{10.3}{28}\approx 0.368$

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2081 Chaitra · Set A · 2 marks

If the value of first quartile ( $Q_1$ ) of any data is 43 and the quartile deviation (Q.D.) is 6.5, find the coefficient of quartile deviation.

Similar questions: QD and its coefficient (2082 Preparation · Bhaktapur · Bhaktapur Municipality)

Answer

Given: $Q_1 = 43$ , Q.D. $= 6.5$

We know, $\text{Q.D.} = \frac{Q_3 - Q_1}{2}$ $6.5 = \frac{Q_3 - 43}{2}$ $Q_3 - 43 = 13$ $Q_3 = 56$

Now, $\text{Coefficient of Q.D.} = \frac{Q_3 - Q_1}{Q_3 + Q_1}$ $= \frac{56 - 43}{56 + 43}$ $= \frac{13}{99}$ $= 0.1313$

Coefficient of quartile deviation = 0.1313 (approx.)

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2081 Chaitra · Set B · 3 marks

Find the mean deviation from median of the given data.

Marks obtained 0-10 10-20 20-30 30-40 40-50
No. of students 3 8 5 6 4

Marks obtained	0-10	10-20	20-30	30-40	40-50
No. of students	3	8	5	6	4

Similar questions: Mean deviation from median (2082 Pre-board · N-PABSAN) · Find the mean deviation and its coefficient

Answer

Marks	$x$	$f$	$cf$	$\lvert x-M\rvert$	$f\lvert x-M\rvert$
0–10	5	3	3	19	57
10–20	15	8	11	9	72
20–30	25	5	16	1	5
30–40	35	6	22	11	66
40–50	45	4	26	21	84
Total		$N=26$			$284$

$\dfrac N2=13$ , which lies in the class $20$ – $30$ .

$M=L+\frac{\frac N2-cf}{f}\times h=20+\frac{13-11}{5}\times10=24$

$\text{M.D. (from median)}=\frac{\Sigma f\lvert x-M\rvert}{N}=\frac{284}{26}\approx 10.92$

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2081 Grade increment · 2 marks

If the third quartile ( $Q_3$ ) of a continuous series is 68 and quartile deviation (Q.D.) is 15.43 then find the first quartile ( $Q_1$ ).

Similar questions: In a continuous series, third quartile is (2081 Chaitra · Set B)

Answer

Given:

$Q_3 = 68$
Q.D. $= 15.43$

Formula: $Q.D. = \frac{Q_3 - Q_1}{2}$

Substitute: $15.43 = \frac{68 - Q_1}{2}$ $30.86 = 68 - Q_1$ $Q_1 = 68 - 30.86 = 37.14$

Therefore, the first quartile $(Q_1)$ is 37.14.

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2082 Pre-board · N-PABSAN · 2 marks

In a continuous series, the sum of the first quartile and third quartile is 70 and the difference between the third quartile and first quartile is 30. Find the first quartile.

Similar questions: In a continuous series, third quartile is (2081 Chaitra · Set B)

Answer

Given $Q_3+Q_1=70$ and $Q_3-Q_1=30$ .

Subtracting the second from the first:

$2Q_1=40\Rightarrow Q_1=20$

(Then $Q_3=50$ , which satisfies both conditions.)

Hence the first quartile is $20$ .

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2082 Pre-board · N-PABSAN · 3 marks

Find the mean deviation from median of the given continuous data.

Marks obtained (x) 10-20 20-30 30-40 40-50 50-60 60-70
No. of students (f) 6 8 11 14 8 3

Marks obtained (x)	10-20	20-30	30-40	40-50	50-60	60-70
No. of students (f)	6	8	11	14	8	3

Similar questions: Find the mean deviation from median of (2081 Chaitra · Set B) · Find the mean deviation and its coefficient

Answer

Marks	$f$	c.f.	Mid-value $m$	$\lvert m-\text{Med}\rvert$	$f\lvert m-\text{Med}\rvert$
10-20	6	6	15	25	150
20-30	8	14	25	15	120
30-40	11	25	35	5	55
40-50	14	39	45	5	70
50-60	8	47	55	15	120
60-70	3	50	65	25	75

$N=50$ , so $\dfrac N2=25$ . The c.f. $25$ is first reached in the class $30$ - $40$ , so it is the median class: $L=30$ , c.f. before it $=14$ , $f=11$ , $h=10$ .

$\text{Med}=L+\frac{\frac N2-c.f.}{f}\times h=30+\frac{25-14}{11}\times10=40$

(The deviations in the table are taken from $40$ .)

$\sum f\lvert m-\text{Med}\rvert=590$

$\text{M.D. (median)}=\frac{\sum f\lvert m-\text{Med}\rvert}{N}=\frac{590}{50}=11.8$

Hence the mean deviation from the median is $11.8$ .

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2082 Preparation · Bhaktapur · Bhaktapur Municipality · 2 marks

In a data, $Q_1=20$ and $Q_3=30$ . Find the quartile deviation and its coefficient.

Similar questions: If the value of first quartile (Q1) (2081 Chaitra · Set A)

Answer

Quartile deviation:

$QD=\frac{Q_3-Q_1}{2}=\frac{30-20}{2}=5$

Coefficient of quartile deviation:

$\frac{Q_3-Q_1}{Q_3+Q_1}=\frac{30-20}{30+20}=\frac{10}{50}=0.2$

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2082 Chaitra · Set A · 3 marks

Calculate the quartile deviation of the data given below where the first quartile is 16.25.

Marks obtained 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50
No. of Students 6 8 12 11 7
Figure: Frequency table of marks obtained (classes 0-10 to 40-50) against number of students 6, 8, 12, 11, 7.

Marks obtained	0 - 10	10 - 20	20 - 30	30 - 40	40 - 50
No. of Students	6	8	12	11	7

Answer

Marks	$f$	$cf$
0–10	6	6
10–20	8	14
20–30	12	26
30–40	11	37
40–50	7	44

$N=44$ , so $\dfrac{3N}{4}=33$ , which lies in the class $30$ – $40$ .

$Q_3=L+\frac{\frac{3N}{4}-cf}{f}\times h=30+\frac{33-26}{11}\times10=30+6.36=36.36$

Given $Q_1=16.25$ ,

$\text{Q.D.}=\frac{Q_3-Q_1}{2}=\frac{36.36-16.25}{2}=\frac{20.11}{2}\approx 10.06$

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2082 Chaitra · Set A · 3 marks

Calculate the standard deviation from the given data.

Marks obtained 0 - 10 0 - 20 0 - 30 0 - 40 0 - 50
No. of Students 5 15 35 40 50
Figure: Cumulative frequency table: marks less than 10, 20, 30, 40, 50 with students 5, 15, 35, 40, 50.

Marks obtained	0 - 10	0 - 20	0 - 30	0 - 40	0 - 50
No. of Students	5	15	35	40	50

Answer

Convert the "less than" cumulative frequencies into class frequencies:

Marks	$f$	Mid-value $x$	$fx$	$fx^2$
0–10	5	5	25	125
10–20	15 − 5 = 10	15	150	2250
20–30	35 − 15 = 20	25	500	12500
30–40	40 − 35 = 5	35	175	6125
40–50	50 − 40 = 10	45	450	20250
Total	$N=50$		$\Sigma fx=1300$	$\Sigma fx^2=41250$

$\bar x=\frac{1300}{50}=26$

$\sigma=\sqrt{\frac{\Sigma fx^2}{N}-\bar x^2}=\sqrt{\frac{41250}{50}-26^2}=\sqrt{825-676}=\sqrt{149}\approx 12.21$

Standard deviation $\approx 12.21$ marks.

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2082 Chaitra · Set B · 3 marks

Calculate the quartile deviation of the given data when the third quartile is 38.75.

Class Interval 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60
Frequency 8 12 20 40 12 8

Class Interval	0 - 10	10 - 20	20 - 30	30 - 40	40 - 50	50 - 60
Frequency	8	12	20	40	12	8

Answer

Class interval	$f$	$cf$
0–10	8	8
10–20	12	20
20–30	20	40
30–40	40	80
40–50	12	92
50–60	8	100

$N=100$ , so $\dfrac N4=25$ , which lies in the class $20$ – $30$ .

$Q_1=L+\frac{\frac N4-cf}{f}\times h=20+\frac{25-20}{20}\times10=22.5$

Given $Q_3=38.75$ ,

$\text{Q.D.}=\frac{Q_3-Q_1}{2}=\frac{38.75-22.5}{2}=\frac{16.25}{2}=8.125$

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2081 Chaitra · Set A · 3 marks

Find the coefficient of variation from the given data.

Marks obtained 0-6 6-12 12-18 18-24 24-30
No. of students 4 3 5 6 12

Marks obtained	0-6	6-12	12-18	18-24	24-30
No. of students	4	3	5	6	12

Similar questions: Find the mean deviation and its coefficient

Answer

Marks	$f$	$x$	$fx$	$fx^2$
0–6	4	3	12	36
6–12	3	9	27	243
12–18	5	15	75	1125
18–24	6	21	126	2646
24–30	12	27	324	8748
Total	$N=30$		$\Sigma fx=564$	$\Sigma fx^2=12798$

$\bar x=\frac{564}{30}=18.8$

$\sigma=\sqrt{\frac{12798}{30}-(18.8)^2}=\sqrt{426.6-353.44}=\sqrt{73.16}\approx 8.55$

$\text{C.V.}=\frac{\sigma}{\bar x}\times100=\frac{8.55}{18.8}\times100\approx 45.5\%$

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2081 Chaitra · Set B · 3 marks

Find the standard deviation from the given data:

Age (in years) 0-4 4-8 8-12 12-16 16-20 20-24
No. of students 7 8 10 12 9 6

Age (in years)	0-4	4-8	8-12	12-16	16-20	20-24
No. of students	7	8	10	12	9	6

Answer

Take assumed mean $A=12$ and $d=x-12$ .

Age	$x$	$f$	$d$	$fd$	$fd^2$
0–4	2	7	−10	−70	700
4–8	6	8	−6	−48	288
8–12	10	10	−2	−20	40
12–16	14	12	2	24	48
16–20	18	9	6	54	324
20–24	22	6	10	60	600
Total		$N=52$		$\Sigma fd=0$	$\Sigma fd^2=2000$

$\sigma=\sqrt{\frac{\Sigma fd^2}{N}-\left(\frac{\Sigma fd}{N}\right)^2}=\sqrt{\frac{2000}{52}-0}=\sqrt{38.46}\approx 6.20\text{ years}$

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2082 Grade increment · 3 marks

Find the quartile deviation from the given data, where the first quartile $(Q_1)=50$ .

Marks obtained 30 - 40 40 - 50 50 - 60 60 - 70 70 - 80
No. of students 2 3 6 5 4

Marks obtained	30 - 40	40 - 50	50 - 60	60 - 70	70 - 80
No. of students	2	3	6	5	4

Answer

Here $N=2+3+6+5+4=20$ . Cumulative frequencies: $2,\ 5,\ 11,\ 16,\ 20$ .

$Q_1$ : $\frac N4=5$ th item lies in $40-50$ , giving $Q_1=40+\frac{5-2}{3}\times10=50$ , which agrees with the given value.

$Q_3$ : $\frac{3N}4=15$ th item lies in $60-70$ (cf $16$ ), with $L=60,\ cf=11,\ f=5,\ i=10$ :

$Q_3=60+\frac{15-11}{5}\times10=68$

$\text{Quartile deviation}=\frac{Q_3-Q_1}2=\frac{68-50}2=\boxed{9}$

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2079 Bagmati · 2 marks

In a continuous data, if mean value $(\bar{x})=58$ , sum of the frequencies $(\sum f)=20$ and $\sum f|D|=306$ , find the coefficient of mean deviation.

Answer

$\text{M.D.}=\frac{\sum f|D|}{\sum f}=\frac{306}{20}=15.3$

$\text{Coefficient of M.D.}=\frac{\text{M.D.}}{\bar x}=\frac{15.3}{58}\approx 0.264$

Answer checked by Claude Sonnet · Oct 21, 2026

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2079 Bagmati · 4 marks

The value of the first quartile $(Q_1)$ of the given data is 21.71. Find the value of the quartile deviation.

Class interval 10-20 20-30 30-40 40-50 50-60 60-70
Frequency 12 19 5 10 9 6

Class interval	10-20	20-30	30-40	40-50	50-60	60-70
Frequency	12	19	5	10	9	6

Answer

Class	$f$	c.f.
10-20	12	12
20-30	19	31
30-40	5	36
40-50	10	46
50-60	9	55
60-70	6	61

$N=61$ . Given $Q_1=21.71$ (check: $\frac N4=15.25$ lies in 20-30, so $Q_1=20+\frac{15.25-12}{19}\times10=21.71$ ).

Third quartile. $\frac{3N}{4}=45.75$ , which lies in the class 40-50 (c.f. before it is 36).

$Q_3=40+\frac{45.75-36}{10}\times 10=49.75$

Quartile deviation

$\text{Q.D.}=\frac{Q_3-Q_1}{2}=\frac{49.75-21.71}{2}=\frac{28.04}{2}=14.02$

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2082 Pre-board · N-PABSAN · 3 marks

Calculate the standard deviation from the given information.

Marks obtained (x) 30-40 40-50 50-60 60-70 70-80 80-90
No. of students (f) 4 8 10 16 8 6

Marks obtained (x)	30-40	40-50	50-60	60-70	70-80	80-90
No. of students (f)	4	8	10	16	8	6

Answer

Take the assumed mean $A=65$ and $h=10$ , with $d'=\dfrac{m-65}{10}$ .

Marks	$f$	$m$	$d'$	$fd'$	$fd'^2$
30-40	4	35	-3	-12	36
40-50	8	45	-2	-16	32
50-60	10	55	-1	-10	10
60-70	16	65	0	0	0
70-80	8	75	1	8	8
80-90	6	85	2	12	24

$N=52$ , $\sum fd'=-18$ , $\sum fd'^2=110$ .

$\sigma=h\sqrt{\frac{\sum fd'^2}{N}-\left(\frac{\sum fd'}{N}\right)^2}=10\sqrt{\frac{110}{52}-\left(\frac{-18}{52}\right)^2}$

$=10\sqrt{2.1154-0.1198}=10\sqrt{1.9956}\approx14.13$

Hence the standard deviation is about $14.13$ marks.

Answer checked by Claude Sonnet · Oct 21, 2026

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2082 Preparation · Bhaktapur · Bhaktapur Municipality · 3 marks

Compute the mean deviation from median of the data given below.

Marks 0-10 10-20 20-30 30-40 40-50
No. of students 3 8 5 6 4

Marks	0-10	10-20	20-30	30-40	40-50
No. of students	3	8	5	6	4

Answer

$N=3+8+5+6+4=26$ , so $\dfrac N2=13$ .

Marks	Mid-value $m$	$f$	$cf$	$\lvert m-24\rvert$	$f\lvert m-24\rvert$
0-10	5	3	3	19	57
10-20	15	8	11	9	72
20-30	25	5	16	1	5
30-40	35	6	22	11	66
40-50	45	4	26	21	84
		26			284

The median class is $20-30$ (first $cf\ge13$ ):

$\text{Median}=L+\frac{\frac N2-cf}{f}\times i=20+\frac{13-11}{5}\times10=24.$

$\text{Mean deviation from median}=\frac{\sum f\lvert m-\text{Med}\rvert}{N}=\frac{284}{26}\approx\boxed{10.92}$

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CDC Model Question 2083, Optional Mathematics · 1 mark

The Coefficient of Variation (CV) is used to compare two datasets. What does a higher CV indicate? (a) More consistent data (b) Greater variability (c) Lower mean (d) Higher median

Answer

(b) Greater variability

The coefficient of variation is the ratio of the standard deviation to the mean, $\text{CV}=\dfrac{\sigma}{\bar x}\times 100\%$ . A higher CV means the standard deviation is large relative to the mean, so the data are more scattered — i.e. less consistent, not more.

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CDC Model Question 2083, Optional Mathematics · 1+1+1 marks

The box plots below show the information about the battery life (in hours) of Battery X and Battery Y.
(a) Which battery has higher battery life in terms of median? [1]
(b) The coefficient of quartile deviation of battery Y is 0.098. Calculate the coefficient of quartile deviation of battery X. [1]
(c) A telecommunication company needs to choose between Battery X and Battery Y to power a remote signal tower. Based on the available information, which battery would you suggest to the company and why? [1]
Figure: Two horizontal box plots on a common axis 'Battery Life (Hours)' from 60 to 110. Battery Y: minimum 60, Q1 74, median 80, Q3 90, maximum 110. Battery X: minimum 70, Q1 75, median 79, Q3 84, maximum 90.

Answer

(a) Battery Y has the higher median: its median is 80 hours, while Battery X has 79 hours.

(b) For Battery X, $Q_1=75$ and $Q_3=84$ .

$\text{Coefficient of Q.D.}=\frac{Q_3-Q_1}{Q_3+Q_1}=\frac{84-75}{84+75}=\frac{9}{159}\approx 0.057$

(c) I would suggest Battery X. Its median is only 1 hour lower than Y's, but its coefficient of quartile deviation (0.057) is smaller than Y's (0.098) and its range (90 − 70 = 20 hours) is much smaller than Y's (110 − 60 = 50 hours). So Battery X gives more consistent, predictable life, which matters for a remote signal tower that cannot be serviced easily.

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CDC Model Question 2083, Optional Mathematics · 3+1+1 marks

Department of Road wants to place a bus stop at position $k$ metres along a straight road. The table below shows how many residents live in each section of the road.

Length of section of the road (in metres) Mid value ( $m$ ) Population ( $f$ )
0 – 200 100 20
200 – 400 300 30
400 – 600 500 40
600 – 800 700 10
Department of Road measures how good a position $k$ is by calculating the total squared walking distance $\Sigma f(m - k)^2$ because smaller the value of $\Sigma f(m - k)^2$ , the better the position. Anu found that the total squared walking distance can be represented by a function $s(k) = 10k^2 - 76k + 178$ .
(a) Calculate the coefficient of variation of the residence distribution along the road. [3]
(b) Is the function $s(k)$ a continuous function for all values of $k$ ? [1]
(c) Anu calculated $s(3.8) = 30$ . With this value, will the function $s(k)$ be continuous at 3.8? [1]
Figure: A road from 0 m to 800 m with circles above each 200 m section showing residents (20, 30, 40, 10) and a bus stop marked at position k between 200 m and 400 m... near 400 m.

Length of section of the road (in metres)	Mid value ( $m$ )	Population ( $f$ )
0 – 200	100	20
200 – 400	300	30
400 – 600	500	40
600 – 800	700	10

Answer

(a) $N=\Sigma f=20+30+40+10=100$ .

$m$	$f$	$fm$	$f(m-\bar x)^2$
100	20	2000	$20(100-380)^2=1{,}568{,}000$
300	30	9000	$30(300-380)^2=192{,}000$
500	40	20000	$40(500-380)^2=576{,}000$
700	10	7000	$10(700-380)^2=1{,}024{,}000$
Total	100	38000	3,360,000

$\bar x=\frac{38000}{100}=380\text{ m},\qquad \sigma=\sqrt{\frac{3{,}360{,}000}{100}}=\sqrt{33600}\approx 183.30\text{ m}$

$\text{C.V.}=\frac{\sigma}{\bar x}\times100=\frac{183.30}{380}\times100\approx 48.24\%$

(b) Yes. $s(k)=10k^2-76k+178$ is a polynomial function, and every polynomial is continuous for all real values of $k$ .

(c) Actual value (with $m$ in hundreds of metres and $f$ in tens of residents):

$s(3.8)=10(3.8)^2-76(3.8)+178=144.4-288.8+178=33.6$

Since $\displaystyle\lim_{k\to3.8}s(k)=33.6$ , a value $s(3.8)=30$ would not equal the limit. So with $s(3.8)=30$ the function would not be continuous at $k=3.8$ ; Anu's calculation is wrong, since the correct value $33.6$ makes the function continuous there.

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CDC Model Question 2080/2081 Optional Mathematics · 4 marks

Find the standard deviation from given data.

Age 0-10 10-20 20-30 30-40 40-50 50-60
No. of Persons 4 6 10 20 6 4

Age	0-10	10-20	20-30	30-40	40-50	50-60
No. of Persons	4	6	10	20	6	4

Answer

Take $A=35$ , $h=10$ and $d=\dfrac{x-35}{10}$ .

Age	$x$	$f$	$d$	$fd$	$fd^2$
0–10	5	4	−3	−12	36
10–20	15	6	−2	−12	24
20–30	25	10	−1	−10	10
30–40	35	20	0	0	0
40–50	45	6	1	6	6
50–60	55	4	2	8	16
Total		$N=50$		$\Sigma fd=-20$	$\Sigma fd^2=92$

$\sigma=h\sqrt{\frac{\Sigma fd^2}{N}-\left(\frac{\Sigma fd}{N}\right)^2}=10\sqrt{1.84-0.16}=10\sqrt{1.68}\approx 12.96$

Answer checked by Claude Sonnet · Oct 21, 2026

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Questions from CDC Model Question 2083, Optional Mathematics (new grid) with marking scheme (Official CDC model paper for new SEE 2083 format, with marking scheme (PDF text)), SEE 2082 Optional Mathematics, RE-2021 'A' (Board exam, images), SEE 2082 Optional Mathematics, RE-2021 'B' (Board exam, images), SEE 2081 Optional Mathematics, RE-2021 'A' (Board exam, images + PDF), SEE 2081 Optional Mathematics, second set (Board exam, images), SEE 2081 Grade Increment Optional Mathematics (Grade increment exam, images), CDC Model Question 2080/2081 Optional Mathematics (old grid) (CDC model paper, images), SEE (Grade Increment) 2082 (2026) Optional Mathematics, GI-2021 (Board exam, Grade Increment sitting of SEE 2082 (75 marks, current curriculum); page shows the paper as images), SEE 2079 (2023) Optional Mathematics (Optional I), Bagmati Province, RE-231 'B' (Board exam, regular SEE 2079 on the older 100-mark pattern; only questions that fit the current outline are used), SEE Pre-Board Examination 2082, Optional I Additional Maths, N-PABSAN (Pre-board exam by the National Private and Boarding Schools' Association Nepal, 2082 (75 marks)) and SEE Preparation Examination 2082, Optional Mathematics, Bhaktapur Nagarpalika (Municipality-level preparation exam, 2082 (75 marks)). Answers are written for this site; check them against your class notes.

Chapter titles and hours from the official syllabus ↗

Past questions and answers

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6.Find the mean deviation from mean of the given data. Class Interval0-1010-2020-3030-4040-50Frequency5815166

7.In a continuous series if ∑f∣m−xˉ∣=1134\sum f|m - \bar{x}| = 1134∑f∣m−xˉ∣=1134, N=100N = 100N=100 and mean (xˉ)=40(\bar{x}) = 40(xˉ)=40, find the mean deviation and its coefficient.

8.In a continuous series, median (MdM_dMd​) =40=40=40, Σf=50\Sigma f=50Σf=50 and Σf∣m−Md∣=530\Sigma f|m-M_d|=530Σf∣m−Md​∣=530. Find the coefficient of mean deviation.

9.In a continuous series, third quartile is two times of the first quartile. If the sum of the first quartile and third quartile is 90, find the quartile deviation.

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In a continuous series if $\sum f|m - \bar{x}| = 1134$ , $N = 100$ and mean $(\bar{x}) = 40$ , find the mean deviation and its coefficient.

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No. of Students 6 8 12 11 7
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Calculate the standard deviation from the given data.

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No. of students 7 8 10 12 9 6

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In a continuous data, if mean value $(\bar{x})=58$ , sum of the frequencies $(\sum f)=20$ and $\sum f|D|=306$ , find the coefficient of mean deviation.

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Class interval 10-20 20-30 30-40 40-50 50-60 60-70
Frequency 12 19 5 10 9 6

Calculate the standard deviation from the given information.

Marks obtained (x) 30-40 40-50 50-60 60-70 70-80 80-90
No. of students (f) 4 8 10 16 8 6

Compute the mean deviation from median of the data given below.

Marks 0-10 10-20 20-30 30-40 40-50
No. of students 3 8 5 6 4

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Find the standard deviation from given data.

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No. of Persons 4 6 10 20 6 4