STT 201 Business Statistics Chapter-wise Past Questions — TU BBM Semester 3

Asked on 2022 Exam[5 marks]2022 paper

Consider the following set of sample data:

78	121	143	88	110	107	62	122	130	95	78	139	89	125

Calculate the lower and upper quartiles.

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Calculation of Lower ( $Q_1$ ) and Upper ( $Q_3$ ) Quartiles:

Step 1: Arrange the Data in Ascending Order

Given sample data ( $n = 14$ ): 78, 121, 143, 88, 110, 107, 62, 122, 130, 95, 78, 139, 89, 125

Arranging in ascending order: 62, 78, 78, 88, 89, 95, 107, 110, 121, 122, 125, 130, 139, 143

Step 2: Calculate Lower Quartile ( $Q_1$ )

\text{Position of } Q_1 = \left(\frac{n + 1}{4}\right)^{\text{th}} \text{ item} = \left(\frac{14 + 1}{4}\right)^{\text{th}} = 3.75^{\text{th}} \text{ item}

Q_1 = \text{Value of } 3^{\text{rd}} \text{ item} + 0.75 \times (\text{Value of } 4^{\text{th}} \text{ item} - \text{Value of } 3^{\text{rd}} \text{ item})

Q_1 = 78 + 0.75 \times (88 - 78) = 78 + 0.75 \times 10 = 78 + 7.5 = \mathbf{85.5}

(Note: Under the alternative integer-rank formula $n/4 = 3.5^{\text{th}}$ item, $Q_1 = 88$ ).

Step 3: Calculate Upper Quartile ( $Q_3$ )

\text{Position of } Q_3 = \left(\frac{3(n + 1)}{4}\right)^{\text{th}} \text{ item} = \left(\frac{3 \times 15}{4}\right)^{\text{th}} = 11.25^{\text{th}} \text{ item}

Q_3 = \text{Value of } 11^{\text{th}} \text{ item} + 0.25 \times (\text{Value of } 12^{\text{th}} \text{ item} - \text{Value of } 11^{\text{th}} \text{ item})

Q_3 = 125 + 0.25 \times (130 - 125) = 125 + 0.25 \times 5 = 125 + 1.25 = \mathbf{126.25}

(Note: Under the alternative integer-rank formula $3n/4 = 10.5^{\text{th}}$ item, $Q_3 = 125$ ).

Final Answer:

Lower Quartile ( $Q_1$ ): 85.5
Upper Quartile ( $Q_3$ ): 126.25

Asked on 2022 Exam[5 marks]2022 paper

If the coefficient of skewness is 0.5, first quartile is 8 and third quartile is 16, find the median of the distribution.

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Calculation of Median from Bowley’s Skewness:

Given Data:

First Quartile ( $Q_1$ ) = $8$
Third Quartile ( $Q_3$ ) = $16$
Bowley’s Coefficient of Skewness ( $Sk_B$ ) = $0.5$

Formula:

Bowley’s coefficient of skewness based on quartiles is given by:

Sk_B = \frac{Q_3 + Q_1 - 2M_d}{Q_3 - Q_1}

Step-by-Step Substitution:

0.5 = \frac{16 + 8 - 2M_d}{16 - 8}

0.5 = \frac{24 - 2M_d}{8}

Multiplying both sides by $8$ :

0.5 \times 8 = 24 - 2M_d

4 = 24 - 2M_d

2M_d = 24 - 4

2M_d = 20

M_d = \frac{20}{2} = \mathbf{10}

Final Answer: The median ( $M_d$ ) of the distribution is 10.

Asked on 2022 Exam[5 marks]2022 paper

Compute the five number summaries from the following data and comment on the shape of the distribution:

7	11	25	23	19	34	29	31	9	15	30

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Five-Number Summary and Distribution Shape:

Step 1: Arrange Data in Ascending Order

Given data ( $n = 11$ ): 7, 11, 25, 23, 19, 34, 29, 31, 9, 15, 30

Ordered data: 7, 9, 11, 15, 19, 23, 25, 29, 30, 31, 34

Step 2: Compute the Five-Number Summary

Minimum ( $Min$ ): 7
First Quartile ( $Q_1$ ):
$\text{Position of } Q_1 = \left(\frac{n + 1}{4}\right)^{\text{th}} = \left(\frac{11 + 1}{4}\right)^{\text{th}} = 3^{\text{rd}} \text{ value} = \mathbf{11}$
Median ( $Q_2$ ):
$\text{Position of } Q_2 = \left(\frac{n + 1}{2}\right)^{\text{th}} = \left(\frac{11 + 1}{2}\right)^{\text{th}} = 6^{\text{th}} \text{ value} = \mathbf{23}$
Third Quartile ( $Q_3$ ):
$\text{Position of } Q_3 = \left(\frac{3(n + 1)}{4}\right)^{\text{th}} = 9^{\text{th}} \text{ value} = \mathbf{30}$
Maximum ( $Max$ ): 34

Summary Vector: $[Min, Q_1, Q_2, Q_3, Max] = [7, 11, 23, 30, 34]$

Step 3: Comment on the Shape of the Distribution

Compare quartile distances:

Distance from $Q_1$ to Median: $Q_2 - Q_1 = 23 - 11 = 12$
Distance from Median to $Q_3$ : $Q_3 - Q_2 = 30 - 23 = 7$ $Sk_B = \frac{Q_3 + Q_1 - 2Q_2}{Q_3 - Q_1} = \frac{30 + 11 - 2(23)}{30 - 11} = \frac{-5}{19} \approx -0.263$ $

Conclusion on Shape: Since $(Q_2 - Q_1) > (Q_3 - Q_2)$ ( $12 > 7$ ) and Bowley’s skewness is negative ( $-0.263$ ), the distribution has a longer tail to the left and is negatively skewed (skewed to the left).

Asked on 2022 Exam[5 marks]2022 paper

Find the mean and standard deviation from the following data related to age distribution of a class.

Age (yrs)	15	16	17	18	19	20
No of students	5	7	12	15	7	4

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Mean and Standard Deviation of Age Distribution:

Calculation Table:

Age ( $X$ )	Students ( $f$ )	$fX$	$X^2$	$fX^2$
15	5	75	225	1,125
16	7	112	256	1,792
17	12	204	289	3,468
18	15	270	324	4,860
19	7	133	361	2,527
20	4	80	400	1,600
Total	$N = \sum f = \mathbf{50}$	$\sum fX = \mathbf{874}$	—	$\sum fX^2 = \mathbf{15,372}$

Step 1: Calculation of Mean ( $\bar{X}$ )

\bar{X} = \frac{\sum fX}{N} = \frac{874}{50} = \mathbf{17.48 \text{ years}}

Step 2: Calculation of Standard Deviation ( $\sigma$ )

\sigma = \sqrt{\frac{\sum fX^2}{N} - \left(\frac{\sum fX}{N}\right)^2}

\sigma = \sqrt{\frac{15372}{50} - (17.48)^2} = \sqrt{307.44 - 305.5504} = \sqrt{1.8896} \approx \mathbf{1.3746 \text{ years}}

Final Answer:

Mean Age: 17.48 years
Standard Deviation: 1.37 years

Asked on 2022 Exam[5 marks]2022 paper

The following table shows the monthly expenditure of people living in village A and village B of a country.

Expenditure (000 Rs)	5	10	15	20	25	30
No. of families in village A	4	14	51	20	10	4
No. of families in village B	8	18	40	18	12	9

Which village people have uniform expenditure on the basis of coefficient of variation?

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Comparison of Expenditure Uniformity Between Village A and Village B Using CV:

Calculation Table:

Expenditure $X$ (Rs 000)	$X^2$	$f_A$	$f_A X$	$f_A X^2$	$f_B$	$f_B X$	$f_B X^2$
5	25	4	20	100	8	40	200
10	100	14	140	1,400	18	180	1,800
15	225	51	765	11,475	40	600	9,000
20	400	20	400	8,000	18	360	7,200
25	625	10	250	6,250	12	300	7,500
30	900	4	120	3,600	9	270	8,100
Total	—	$N_A = 103$	$\mathbf{1,695}$	$\mathbf{30,825}$	$N_B = 105$	$\mathbf{1,750}$	$\mathbf{33,800}$

Step 1: Village A Statistics

\bar{X}_A = \frac{\sum f_A X}{N_A} = \frac{1695}{103} = \mathbf{16.46 \text{ thousand Rs}}

\sigma_A = \sqrt{\frac{\sum f_A X^2}{N_A} - \bar{X}_A^2} = \sqrt{\frac{30825}{103} - (16.456)^2} = \sqrt{299.27 - 270.81} = \sqrt{28.46} = \mathbf{5.33}

CV_A = \frac{\sigma_A}{\bar{X}_A} \times 100 = \frac{5.33}{16.46} \times 100 = \mathbf{32.38\%}

Step 2: Village B Statistics

\bar{X}_B = \frac{\sum f_B X}{N_B} = \frac{1750}{105} = \mathbf{16.67 \text{ thousand Rs}}

\sigma_B = \sqrt{\frac{\sum f_B X^2}{N_B} - \bar{X}_B^2} = \sqrt{\frac{33800}{105} - (16.667)^2} = \sqrt{321.90 - 277.78} = \sqrt{44.12} = \mathbf{6.64}

CV_B = \frac{\sigma_B}{\bar{X}_B} \times 100 = \frac{6.64}{16.67} \times 100 = \mathbf{39.83\%}

Conclusion:

Since $CV_A (32.38\%) < CV_B (39.83\%)$ , Village A has a lower coefficient of variation, meaning the people of Village A have more uniform (consistent) monthly expenditure compared to Village B.

Asked on 2023 Exam[2 marks]2023 paper

If the values of upper quartile and lower quartile are 63 and 59 respectively, then calculate the coefficient of quartile deviation.

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Coefficient of Quartile Deviation:

Given:

Upper Quartile ( $Q_3$ ) = $63$
Lower Quartile ( $Q_1$ ) = $59$

Formula:

\text{Coefficient of } QD = \frac{Q_3 - Q_1}{Q_3 + Q_1}

Calculation:

\text{Coefficient of } QD = \frac{63 - 59}{63 + 59} = \frac{4}{122} \approx \mathbf{0.0328} \quad (\text{or } 3.28\%)

Final Answer: The coefficient of quartile deviation is 0.0328.

Asked on 2023 Exam[2 marks]2023 paper

In a moderately asymmetric distribution, the values of mean and mode are 20 and 23 respectively. Compute the value of median.

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Calculation of Median from Empirical Relationship:

Given:

$\text{Mean} = 20$
$\text{Mode} = 23$

Empirical Formula for Moderately Asymmetric Distributions:

\text{Mode} = 3\text{Median} - 2\text{Mean}

Calculation:

23 = 3\text{Median} - 2(20)

23 = 3\text{Median} - 40

3\text{Median} = 23 + 40

3\text{Median} = 63 \implies \text{Median} = \frac{63}{3} = \mathbf{21}

Final Answer: The value of the median is 21.

Asked on 2023 Exam[2 marks]2023 paper

Calculate coefficient of variation, when values of mean and standard deviation are 32 and 17 respectively.

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Calculation of Coefficient of Variation (CV):

Given:

$\text{Mean } (\bar{X}) = 32$
$\text{Standard Deviation } (\sigma) = 17$

Formula:

CV = \frac{\sigma}{\bar{X}} \times 100

Calculation:

CV = \frac{17}{32} \times 100 = \mathbf{53.125\%}

Final Answer: The Coefficient of Variation is 53.125%.

Asked on 2023 Exam[6 marks]2023 paper

The following table shows the marks distribution of students in a subject:

Marks	0-10	10-20	20-30	30-40	40-50	50-60
Frequency	10	25	45	20	18	12

Find the lowest marks of the top 20% of the students.

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Lowest Marks of the Top 20% of Students ( $P_{80}$ ):

Concept:

The lowest marks obtained by the top 20% of students corresponds to the $80^{\text{th}}$ Percentile ( $P_{80}$ ) of the distribution.

Cumulative Frequency Table:

Marks	Frequency ( $f$ )	Cumulative Frequency ( $cf$ )
0 – 10	10	10
10 – 20	25	35
20 – 30	45	80
30 – 40	20	100
40 – 50	18	118
50 – 60	12	130
Total	$N = 130$	—

Step 1: Locate $P_{80}$ Class Position

\text{Position of } P_{80} = \frac{80 \times N}{100} = \frac{80 \times 130}{100} = 104^{\text{th}} \text{ item}

The cumulative frequency just greater than 104 is 118, which corresponds to the class 40 – 50.

Step 2: Apply the Percentile Interpolation Formula

P_{80} = L + \left(\frac{\frac{80N}{100} - cf}{f}\right) \times h

Where:

$L = 40$ (Lower limit of $P_{80}$ class)
$cf = 100$ (Cumulative frequency of the preceding class)
$f = 18$ (Frequency of $P_{80}$ class)
$h = 10$ (Class width)

P_{80} = 40 + \left(\frac{104 - 100}{18}\right) \times 10 = 40 + \left(\frac{4}{18}\right) \times 10 = 40 + 2.22 = \mathbf{42.22}

Final Answer: The lowest marks scored by the top 20% of the students is 42.22 marks.

Asked on 2023 Exam[6 marks]2023 paper

The following table represents the marks of 100 students.

Marks	0-20	20-40	40-60	60-80	80-100
Students	14	-	27	-	15

Find the missing frequencies when modal marks is 48.

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Calculation of Missing Frequencies Given Mode = 48:

Given Data:

Total students ( $N$ ) = $100$
$\text{Mode} = 48$

Let the missing frequency for class $20-40$ be $f_1$ and for $60-80$ be $f_2$ .

Marks	Frequency ( $f$ )
0 – 20	14
20 – 40	$f_1$
40 – 60	27
60 – 80	$f_2$
80 – 100	15
Total	$N = 100$

Step 1: Formulate Total Frequency Equation

\sum f = 14 + f_1 + 27 + f_2 + 15 = 100

f_1 + f_2 + 56 = 100 \implies f_1 + f_2 = 44 \quad \text{--- (1)}

Step 2: Use Mode Formula

Since $\text{Mode} = 48$ , it lies in the modal class 40 – 60.

Lower limit ( $L$ ) = $40$
Frequency of modal class ( $f_m$ ) = $27$
Frequency of pre-modal class ( $f_0$ ) = $f_1$
Frequency of post-modal class ( $f_2'$ ) = $f_2$
Class height ( $h$ ) = $20$ $\text{Mode} = L + \left[\frac{f_m - f_0}{2f_m - f_0 - f_2'}\right] \times h$ $
$48 = 40 + \left[\frac{27 - f_1}{2(27) - f_1 - f_2}\right] \times 20$

$8 = \left[\frac{27 - f_1}{54 - (f_1 + f_2)}\right] \times 20$

Substitute $f_1 + f_2 = 44$ from Equation (1):

8 = \left[\frac{27 - f_1}{54 - 44}\right] \times 20

8 = \left[\frac{27 - f_1}{10}\right] \times 20

8 = 2(27 - f_1)

4 = 27 - f_1 \implies f_1 = 27 - 4 = \mathbf{23}

Now find $f_2$ from Equation (1):

f_2 = 44 - 23 = \mathbf{21}

Final Answer: The missing frequencies are $f_1 = 23$ (for 20–40) and $f_2 = 21$ (for 60–80).

Asked on 2023 Exam[6 marks]2023 paper

From the following income distribution of 500 workers in a locality, find the median value of the income distribution.

Income (000 Rs)	No. of workers
Below 20	50
20-40	90
40-60	150
60 - 80	100
80 - 100	60
100 & above	50

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Calculation of Median Income of 500 Workers:

Cumulative Frequency Table:

Income (Rs '000)	Workers ( $f$ )	Cumulative Frequency ( $cf$ )
0 – 20	50	50
20 – 40	90	140
40 – 60	150	290
60 – 80	100	390
80 – 100	60	450
100 and above	50	500
Total	$N = 500$	—

Step 1: Identify Median Class

\text{Position of Median} = \frac{N}{2} = \frac{500}{2} = 250^{\text{th}} \text{ item}

The cumulative frequency just greater than 250 is 290, so the Median Class is 40 – 60.

Step 2: Apply Median Formula

M_d = L + \left(\frac{\frac{N}{2} - cf}{f}\right) \times h

Where:

$L = 40$
$cf = 140$
$f = 150$
$h = 20$ $M_d = 40 + \left(\frac{250 - 140}{150}\right) \times 20 = 40 + \left(\frac{110}{150}\right) \times 20 = 40 + 14.67 = \mathbf{54.67 \text{ (in '000 Rs)}}$ $

Final Answer: The median income of the workers is Rs 54,667 (or 54.67 thousand Rs).

Asked on 2023 Exam[6 marks]2023 paper

From the following income distribution, calculate the appropriate measures of dispersion.

Daily income (Rs)	Number of workers
Below 50	10
51-100	18
101-150	25
151-200	20
201-250	16
250 and above	11

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Appropriate Measure of Dispersion for Open-Ended Distribution:

Justification:

The given distribution has open-ended classes (“Below 50” and “250 and above”). For open-ended distributions, neither the Range nor the Standard Deviation can be calculated without arbitrary class boundaries. Therefore, the Quartile Deviation (Semi-Interquartile Range) is the uniquely appropriate measure of dispersion.

Cumulative Frequency with True Class Boundaries:

(Correction factor $= (51 - 50)/2 = 0.5$ )

Daily Income (Rs)	$f$	Cumulative Frequency ( $cf$ )
Below 50.5	10	10
50.5 – 100.5	18	28
100.5 – 150.5	25	53
150.5 – 200.5	20	73
200.5 – 250.5	16	89
250.5 & above	11	100
Total	$N = 100$	—

Step 1: Lower Quartile ( $Q_1$ )

\text{Position of } Q_1 = \frac{N}{4} = \frac{100}{4} = 25^{\text{th}} \text{ item} \implies \text{Class: 50.5 – 100.5}

Q_1 = 50.5 + \left(\frac{25 - 10}{18}\right) \times 50 = 50.5 + \left(\frac{15}{18}\right) \times 50 = 50.5 + 41.67 = \mathbf{92.17}

Step 2: Upper Quartile ( $Q_3$ )

\text{Position of } Q_3 = \frac{3N}{4} = \frac{300}{4} = 75^{\text{th}} \text{ item} \implies \text{Class: 200.5 – 250.5}

Q_3 = 200.5 + \left(\frac{75 - 73}{16}\right) \times 50 = 200.5 + \left(\frac{2}{16}\right) \times 50 = 200.5 + 6.25 = \mathbf{206.75}

Step 3: Quartile Deviation ( $QD$ )

QD = \frac{Q_3 - Q_1}{2} = \frac{206.75 - 92.17}{2} = \frac{114.58}{2} = \mathbf{57.29}

\text{Coefficient of } QD = \frac{Q_3 - Q_1}{Q_3 + Q_1} = \frac{114.58}{298.92} = \mathbf{0.3833}

Final Answer:

Quartile Deviation: Rs 57.29
Coefficient of QD: 0.3833

Asked on 2023 Exam[6 marks]2023 paper

The following information was obtained from two investment companies A and B:

	Company A	Company B
Average return (000Rs)	28	37
Variance	25	36
No. of observations	100	100

Calculate combined standard deviation.

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Calculation of Combined Standard Deviation:

Given Data:

Company A: $n_1 = 100$ , $\bar{X}_1 = 28$ , $\sigma_1^2 = 25$
Company B: $n_2 = 100$ , $\bar{X}_2 = 37$ , $\sigma_2^2 = 36$

Step 1: Combined Mean ( $\bar{X}_{12}$ )

\bar{X}_{12} = \frac{n_1 \bar{X}_1 + n_2 \bar{X}_2}{n_1 + n_2} = \frac{100(28) + 100(37)}{100 + 100} = \frac{2800 + 3700}{200} = \frac{6500}{200} = \mathbf{32.5}

Step 2: Deviations of Individual Means from Combined Mean

d_1 = \bar{X}_1 - \bar{X}_{12} = 28 - 32.5 = -4.5 \implies d_1^2 = 20.25

d_2 = \bar{X}_2 - \bar{X}_{12} = 37 - 32.5 = +4.5 \implies d_2^2 = 20.25

Step 3: Combined Variance ( $\sigma_{12}^2$ )

\sigma_{12}^2 = \frac{n_1(\sigma_1^2 + d_1^2) + n_2(\sigma_2^2 + d_2^2)}{n_1 + n_2}

\sigma_{12}^2 = \frac{100(25 + 20.25) + 100(36 + 20.25)}{200}

\sigma_{12}^2 = \frac{100(45.25) + 100(56.25)}{200} = \frac{4525 + 5625}{200} = \frac{10150}{200} = \mathbf{50.75}

Step 4: Combined Standard Deviation ( $\sigma_{12}$ )

\sigma_{12} = \sqrt{50.75} \approx \mathbf{7.1239 \text{ thousand Rs}}

Final Answer: The combined standard deviation is Rs 7,124 (or 7.12 thousand Rs).

Asked on 2023 Exam[10 marks]2023 paper

From the following distribution, find the coefficient of skewness. Also comment the result.

Monthly income (Rs 000)	Number of workers
0-100	15
100-200	50
200-300	75
300-400	40
400-500	30
500- 600	10

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Calculation of Karl Pearson’s Coefficient of Skewness:

Calculation Table:

Monthly Income (Rs 000)	Mid-point ( $m$ )	$f$	$d = \frac{m - 250}{100}$	$fd$	$fd^2$
0 – 100	50	15	-2	-30	60
100 – 200	150	50	-1	-50	50
200 – 300	250	75	0	0	0
300 – 400	350	40	1	40	40
400 – 500	450	30	2	60	120
500 – 600	550	10	3	30	90
Total	—	$N = 220$	—	$\sum fd = 50$	$\sum fd^2 = 360$

Step 1: Mean ( $\bar{X}$ )

\bar{X} = A + \left(\frac{\sum fd}{N}\right) \times h = 250 + \left(\frac{50}{220}\right) \times 100 = 250 + 22.73 = \mathbf{272.73 \text{ thousand Rs}}

Step 2: Mode ( $M_o$ )

Modal class is 200 – 300 (highest frequency $f_1 = 75$ ).

$L = 200$ , $f_0 = 50$ , $f_2 = 40$ , $h = 100$ $M_o = L + \left[\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right] \times h = 200 + \left[\frac{75 - 50}{2(75) - 50 - 40}\right] \times 100 = 200 + \left(\frac{25}{60}\right) \times 100 = \mathbf{241.67}$ $

Step 3: Standard Deviation ( $\sigma$ )

\sigma = h \times \sqrt{\frac{\sum fd^2}{N} - \left(\frac{\sum fd}{N}\right)^2} = 100 \times \sqrt{\frac{360}{220} - \left(\frac{50}{220}\right)^2} = 100 \times \sqrt{1.6364 - 0.05165} = 100 \times 1.2589 = \mathbf{125.89}

Step 4: Karl Pearson’s Coefficient of Skewness ( $Sk_p$ )

Sk_p = \frac{\bar{X} - M_o}{\sigma} = \frac{272.73 - 241.67}{125.89} = \frac{31.06}{125.89} = \mathbf{+0.2467}

Interpretation: The distribution exhibits a positive skewness ( $Sk_p = +0.2467$ ), indicating that the distribution has a longer tail to the right and a concentration of lower to middle-income workers.

Asked on 2023 Exam[10 marks]2023 paper

From the following marks distribution of students of a campus, calculate the coefficient of kurtosis. Also, interpret the result.

Marks	Below 10	10 - 20	20 - 30	30 - 40	40 - 50	50 & above
No. of students	50	100	150	90	60	50

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Calculation of Percentile Coefficient of Kurtosis ( $K$ ):

Concept:

The Percentile Coefficient of Kurtosis is given by:

K = \frac{QD}{P_{90} - P_{10}} = \frac{\frac{Q_3 - Q_1}{2}}{P_{90} - P_{10}}

Cumulative Frequency Table:

Marks	Frequency ( $f$ )	Cumulative Frequency ( $cf$ )
0 – 10	50	50
10 – 20	100	150
20 – 30	150	300
30 – 40	90	390
40 – 50	60	450
50 – 60	50	500
Total	$N = 500$	—

Step 1: Calculate $Q_1$ and $Q_3$

$Q_1$ position: $\frac{N}{4} = 125^{\text{th}} \implies$ Class 10 – 20
$Q_1 = 10 + \left(\frac{125 - 50}{100}\right) \times 10 = 10 + 7.5 = \mathbf{17.5}$
$Q_3$ position: $\frac{3N}{4} = 375^{\text{th}} \implies$ Class 30 – 40
$Q_3 = 30 + \left(\frac{375 - 300}{90}\right) \times 10 = 30 + 8.33 = \mathbf{38.33}$
Semi-Interquartile Range ( $QD$ ):
$QD = \frac{38.33 - 17.5}{2} = \frac{20.83}{2} = \mathbf{10.415}$

Step 2: Calculate $P_{10}$ and $P_{90}$

$P_{10}$ position: $\frac{10N}{100} = 50^{\text{th}} \implies$ Exactly the upper boundary of class 0–10:
$P_{10} = \mathbf{10.0}$
$P_{90}$ position: $\frac{90N}{100} = 450^{\text{th}} \implies$ Exactly the upper boundary of class 40–50:
$P_{90} = \mathbf{50.0}$

Step 3: Compute Kurtosis ( $K$ )

K = \frac{10.415}{50.0 - 10.0} = \frac{10.415}{40} = \mathbf{0.2604}

Interpretation:

For a normal (mesokurtic) distribution, $K = 0.263$ .
Since $K = 0.2604 < 0.263$ , the distribution is slightly platykurtic (flatter-topped than normal).

Asked on 2024 Exam[2 marks]2024 paper

Find the quartile deviation if first and third quartiles are 18 and 30 respectively.

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Quartile Deviation ( $QD$ ):

Given:

First Quartile ( $Q_1$ ) = $18$
Third Quartile ( $Q_3$ ) = $30$

Formula:

QD = \frac{Q_3 - Q_1}{2}

Calculation:

QD = \frac{30 - 18}{2} = \frac{12}{2} = \mathbf{6}

Final Answer: The Quartile Deviation is 6.

Asked on 2024 Exam[2 marks]2024 paper

Calculate the combined mean from the following information.

	Group A	Group B
Mean	65	60
Number of observations	40	60

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Combined Mean Calculation:

Given:

Group A: $n_1 = 40$ , $\bar{X}_1 = 65$
Group B: $n_2 = 60$ , $\bar{X}_2 = 60$

Formula:

\bar{X}_{12} = \frac{n_1 \bar{X}_1 + n_2 \bar{X}_2}{n_1 + n_2}

Calculation:

\bar{X}_{12} = \frac{40(65) + 60(60)}{40 + 60} = \frac{2600 + 3600}{100} = \frac{6200}{100} = \mathbf{62}

Final Answer: The combined mean of the two groups is 62.

Asked on 2024 Exam[2 marks]2024 paper

In a moderately asymmetrical distribution, the mode and mean are 32 and 35 respectively. Calculate the median.

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Calculation of Median from Mode and Mean:

Given:

$\text{Mode} = 32$
$\text{Mean} = 35$

Empirical Formula:

\text{Mode} = 3\text{Median} - 2\text{Mean}

Calculation:

32 = 3\text{Median} - 2(35)

32 = 3\text{Median} - 70

3\text{Median} = 32 + 70 = 102

\text{Median} = \frac{102}{3} = \mathbf{34}

Final Answer: The median of the distribution is 34.

Asked on 2024 Exam[2 marks]2024 paper

Calculate the Pearson’s Coefficient of skewness when mean, mode and standard deviation are 25, 20 and 10 respectively.

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Pearson’s Coefficient of Skewness:

Given:

$\text{Mean} = 25$
$\text{Mode} = 20$
$\text{Standard Deviation } (\sigma) = 10$

Formula:

Sk_p = \frac{\text{Mean} - \text{Mode}}{\sigma}

Calculation:

Sk_p = \frac{25 - 20}{10} = \frac{5}{10} = \mathbf{+0.5}

Final Answer: Pearson’s coefficient of skewness is +0.5 (moderately positively skewed).

Asked on 2024 Exam[2 marks]2024 paper

If first quartile and third quartile of a distribution are 28 and 52 and their 90th and 10th percentiles are 61 and 17 respectively then, find the value of kurtosis.

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Calculation of Percentile Coefficient of Kurtosis:

Given:

$Q_1 = 28$ , $Q_3 = 52$
$P_{90} = 61$ , $P_{10} = 17$

Formula:

K = \frac{QD}{P_{90} - P_{10}} = \frac{\frac{Q_3 - Q_1}{2}}{P_{90} - P_{10}}

Calculation:

QD = \frac{52 - 28}{2} = \frac{24}{2} = 12

P_{90} - P_{10} = 61 - 17 = 44

K = \frac{12}{44} = \frac{3}{11} \approx \mathbf{0.2727}

Final Answer: The coefficient of kurtosis is 0.2727 (since $0.2727 > 0.263$ , it is leptokurtic).

Asked on 2024 Exam[6 marks]2024 paper

The following table represents the weekly expenditure of 100 families.

Expenditure	0-20	20-40	40-60	60-80	80-100
No. of Families	14	?	27	?	15

If the mean value is 48 find the missing frequencies.

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Finding Missing Frequencies When Mean = 48:

Given Data:

Total families ( $N$ ) = $100$
Mean expenditure ( $\bar{X}$ ) = $48$

Let the missing frequency of class 20–40 be $f_1$ and class 60–80 be $f_2$ .

Expenditure	Mid-point ( $m$ )	Frequency ( $f$ )	$fm$
0 – 20	10	14	140
20 – 40	30	$f_1$	$30f_1$
40 – 60	50	27	1,350
60 – 80	70	$f_2$	$70f_2$
80 – 100	90	15	1,350
Total	—	$N = 100$	$\sum fm = 2,840 + 30f_1 + 70f_2$

Step 1: Total Frequency Equation

14 + f_1 + 27 + f_2 + 15 = 100

f_1 + f_2 + 56 = 100 \implies f_1 + f_2 = 44 \quad \text{--- (1)}

Step 2: Mean Equation

\bar{X} = \frac{\sum fm}{N} \implies 48 = \frac{2840 + 30f_1 + 70f_2}{100}

4,800 = 2,840 + 30f_1 + 70f_2

30f_1 + 70f_2 = 1,960

Divide by 10:

3f_1 + 7f_2 = 196 \quad \text{--- (2)}

Step 3: Solve Simultaneously

Multiply Equation (1) by 3:

3f_1 + 3f_2 = 132 \quad \text{--- (3)}

Subtract Equation (3) from Equation (2):

(3f_1 + 7f_2) - (3f_1 + 3f_2) = 196 - 132

4f_2 = 64 \implies f_2 = \frac{64}{4} = \mathbf{16}

Substitute $f_2 = 16$ into Equation (1):

f_1 + 16 = 44 \implies f_1 = 44 - 16 = \mathbf{28}

Final Answer: The missing frequencies are $f_1 = 28$ (for class 20–40) and $f_2 = 16$ (for class 60–80).

Asked on 2024 Exam[6 marks]2024 paper

Calculate coefficient of quartile deviation from the following distribution of expenditure of one thousand households of Pokhara city.

Expenditure (Rs.'00)	No. of households
30-50	54
50-70	100
70-90	140
90-110	300
110-130	230
130-150	125
Above 150	51

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Coefficient of Quartile Deviation of Household Expenditure:

Cumulative Frequency Table:

Expenditure (Rs '00)	Households ( $f$ )	Cumulative Frequency ( $cf$ )
30 – 50	54	54
50 – 70	100	154
70 – 90	140	294
90 – 110	300	594
110 – 130	230	824
130 – 150	125	949
Above 150	51	1000
Total	$N = 1000$	—

Step 1: Lower Quartile ( $Q_1$ )

\text{Position of } Q_1 = \frac{N}{4} = \frac{1000}{4} = 250^{\text{th}} \implies \text{Class: 70 – 90}

Q_1 = 70 + \left(\frac{250 - 154}{140}\right) \times 20 = 70 + \left(\frac{96}{140}\right) \times 20 = 70 + 13.71 = \mathbf{83.71}

Step 2: Upper Quartile ( $Q_3$ )

\text{Position of } Q_3 = \frac{3N}{4} = \frac{3000}{4} = 750^{\text{th}} \implies \text{Class: 110 – 130}

Q_3 = 110 + \left(\frac{750 - 594}{230}\right) \times 20 = 110 + \left(\frac{156}{230}\right) \times 20 = 110 + 13.57 = \mathbf{123.57}

Step 3: Coefficient of Quartile Deviation

\text{Coefficient of } QD = \frac{Q_3 - Q_1}{Q_3 + Q_1} = \frac{123.57 - 83.71}{123.57 + 83.71} = \frac{39.86}{207.28} \approx \mathbf{0.1923}

Final Answer: The coefficient of quartile deviation is 0.1923 (or 19.23%).

Asked on 2024 Exam[6 marks]2024 paper

The mark distribution of 104 students are given below.

Marks	0 - 10	10 - 20	20 - 30	30 - 40	40 - 50	50 - 60	60 - 70
No. of Students	7	8	13	29	35	9	3

Find the lowest marks of the top 10% of the students.

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Lowest Marks of the Top 10% of Students ( $P_{90}$ ):

Concept:

The cutoff score separating the top 10% of students is the $90^{\text{th}}$ Percentile ( $P_{90}$ ).

Cumulative Frequency Table:

Marks	Frequency ( $f$ )	Cumulative Frequency ( $cf$ )
0 – 10	7	7
10 – 20	8	15
20 – 30	13	28
30 – 40	29	57
40 – 50	35	92
50 – 60	9	101
60 – 70	3	104
Total	$N = 104$	—

Step 1: Position of $P_{90}$ $\text{Position} = \frac{90 \times N}{100} = \frac{90 \times 104}{100} = 93.6^{\text{th}} \text{ item}$ $

The cumulative frequency just exceeding 93.6 is 101, corresponding to class 50 – 60.

Step 2: Percentile Interpolation Formula

P_{90} = L + \left(\frac{\frac{90N}{100} - cf}{f}\right) \times h

Where $L = 50$ , $cf = 92$ , $f = 9$ , $h = 10$ :

P_{90} = 50 + \left(\frac{93.6 - 92}{9}\right) \times 10 = 50 + \left(\frac{1.6}{9}\right) \times 10 = 50 + 1.78 = \mathbf{51.78}

Final Answer: The lowest marks of the top 10% of students is 51.78 marks.

Asked on 2024 Exam[10 marks]2024 paper

Calculate the percentile coefficient of kurtosis from the income distribution of sixty-two randomly selected employees of Nepal Bank Limited. Also interpret the result.

Income (Rs '000)	20-30	30-40	40-50	50-60	60-70	Total
No. of employees	7	10	20	18	7	62

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Percentile Coefficient of Kurtosis for Nepal Bank Limited Employees:

Cumulative Frequency Table:

Income (Rs '000)	Employees ( $f$ )	Cumulative Frequency ( $cf$ )
20 – 30	7	7
30 – 40	10	17
40 – 50	20	37
50 – 60	18	55
60 – 70	7	62
Total	$N = 62$	—

Step 1: Calculate $Q_1$ and $Q_3$

$Q_1$ position: $\frac{N}{4} = \frac{62}{4} = 15.5^{\text{th}} \implies$ Class 30 – 40
$Q_1 = 30 + \left(\frac{15.5 - 7}{10}\right) \times 10 = 30 + 8.5 = \mathbf{38.5}$
$Q_3$ position: $\frac{3N}{4} = \frac{186}{4} = 46.5^{\text{th}} \implies$ Class 50 – 60
$Q_3 = 50 + \left(\frac{46.5 - 37}{18}\right) \times 10 = 50 + \left(\frac{9.5}{18}\right) \times 10 = 50 + 5.28 = \mathbf{55.28}$
Semi-Interquartile Range ( $QD$ ):
$QD = \frac{Q_3 - Q_1}{2} = \frac{55.28 - 38.5}{2} = \frac{16.78}{2} = \mathbf{8.39}$

Step 2: Calculate $P_{10}$ and $P_{90}$

$P_{10}$ position: $\frac{10 \times 62}{100} = 6.2^{\text{th}} \implies$ Class 20 – 30
$P_{10} = 20 + \left(\frac{6.2 - 0}{7}\right) \times 10 = 20 + 8.86 = \mathbf{28.86}$
$P_{90}$ position: $\frac{90 \times 62}{100} = 55.8^{\text{th}} \implies$ Class 60 – 70
$P_{90} = 60 + \left(\frac{55.8 - 55}{7}\right) \times 10 = 60 + 1.14 = \mathbf{61.14}$

Step 3: Compute Kurtosis ( $K$ )

K = \frac{QD}{P_{90} - P_{10}} = \frac{8.39}{61.14 - 28.86} = \frac{8.39}{32.28} \approx \mathbf{0.2599}

Interpretation:

The standard normal (mesokurtic) value is $K = 0.263$ .
Since $K = 0.2599 < 0.263$ , the distribution is slightly platykurtic, indicating a slightly flatter peak with thinner tails compared to a normal distribution.

Asked on 2024 Exam[10 marks]2024 paper

The running capacity of two horses is given below, state which is more consistent and why?

Horse A	250	255	280	290	295	300
Horse B	280	282	290	295	298	295

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Consistency Comparison of Two Horses Using Coefficient of Variation (CV):

Data ( $n = 6$ observations each):

Horse A: $[250, 255, 280, 290, 295, 300]$
Horse B: $[280, 282, 290, 295, 298, 295]$

Step 1: Horse A Calculations

\sum X_A = 250 + 255 + 280 + 290 + 295 + 300 = 1,670

\bar{X}_A = \frac{1670}{6} = \mathbf{278.33}

Deviations squared $\sum (X_A - \bar{X}_A)^2$ :

(250 - 278.33)^2 + \dots + (300 - 278.33)^2 = 802.78 + 544.44 + 2.78 + 136.11 + 277.78 + 469.44 = 2,233.33

\sigma_A = \sqrt{\frac{2233.33}{6}} = \sqrt{372.22} = \mathbf{19.29}

CV_A = \frac{\sigma_A}{\bar{X}_A} \times 100 = \frac{19.29}{278.33} \times 100 = \mathbf{6.93\%}

Step 2: Horse B Calculations

\sum X_B = 280 + 282 + 290 + 295 + 298 + 295 = 1,740

\bar{X}_B = \frac{1740}{6} = \mathbf{290.00}

Deviations squared $\sum (X_B - \bar{X}_B)^2$ :

(-10)^2 + (-8)^2 + (0)^2 + (5)^2 + (8)^2 + (5)^2 = 100 + 64 + 0 + 25 + 64 + 25 = 278.00

\sigma_B = \sqrt{\frac{278}{6}} = \sqrt{46.33} = \mathbf{6.81}

CV_B = \frac{\sigma_B}{\bar{X}_B} \times 100 = \frac{6.81}{290} \times 100 = \mathbf{2.35\%}

Conclusion:

A lower Coefficient of Variation signifies greater stability and consistency.
Since $CV_B (2.35\%) < CV_A (6.93\%)$ , Horse B is significantly more consistent in its running performance than Horse A.

Asked on 2026 Exam[2 marks]2026 paper

The values of lower and upper quartiles are 40 and 70 respectively, then calculate coefficient of quartile deviation.

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Coefficient of Quartile Deviation:

Given:

Lower Quartile ( $Q_1$ ) = $40$
Upper Quartile ( $Q_3$ ) = $70$

Formula:

\text{Coefficient of } QD = \frac{Q_3 - Q_1}{Q_3 + Q_1}

Calculation:

\text{Coefficient of } QD = \frac{70 - 40}{70 + 40} = \frac{30}{110} = \frac{3}{11} \approx \mathbf{0.2727} \quad (\text{or } 27.27\%)

Final Answer: The coefficient of quartile deviation is 0.2727.

Asked on 2026 Exam[2 marks]2026 paper

The mean and coefficient of variation of a certain data are 20 and 30% respectively. Calculate standard deviation.

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Calculation of Standard Deviation ( $\sigma$ ):

Given:

$\text{Mean } (\bar{X}) = 20$
$\text{Coefficient of Variation } (CV) = 30\%$

Formula:

CV = \frac{\sigma}{\bar{X}} \times 100

Calculation:

30 = \frac{\sigma}{20} \times 100

30 = 5\sigma \implies \sigma = \frac{30}{5} = \mathbf{6}

Final Answer: The standard deviation is 6.

Asked on 2026 Exam[2 marks]2026 paper

Give that quartile deviation = 20, 10th percentile = 30 and 90th percentile = 90, find the coefficient of kurtosis.

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Calculation of Coefficient of Kurtosis ( $K$ ):

Given:

Quartile Deviation ( $QD$ ) = $20$
$10^{\text{th}}$ Percentile ( $P_{10}$ ) = $30$
$90^{\text{th}}$ Percentile ( $P_{90}$ ) = $90$

Formula:

K = \frac{QD}{P_{90} - P_{10}}

Calculation:

K = \frac{20}{90 - 30} = \frac{20}{60} = \frac{1}{3} \approx \mathbf{0.3333}

Final Answer: The coefficient of kurtosis is 0.3333 (since $0.3333 > 0.263$ , the distribution is leptokurtic).

Asked on 2026 Exam[2 marks]2026 paper

For a moderately asymmetrical distribution, if the values of mean and median are 30 and 28, find the value of mode.

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Calculation of Mode from Empirical Relationship:

Given:

$\text{Mean} = 30$
$\text{Median} = 28$

Empirical Formula:

\text{Mode} = 3\text{Median} - 2\text{Mean}

Calculation:

\text{Mode} = 3(28) - 2(30) = 84 - 60 = \mathbf{24}

Final Answer: The value of the mode is 24.

Asked on 2026 Exam[6 marks]2026 paper

The mean of a set of 100 observations were found to be 40. On checking it was found that two observations, which were wrongly taken as 30 and 72 instead of 3 and 27. Calculate the corrected mean.

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Calculation of Corrected Mean:

Given Data:

Number of observations ( $n$ ) = $100$
Incorrect mean ( $\bar{X}$ ) = $40$
Incorrect values included: $30$ and $72$
Correct values to include: $3$ and $27$

Step 1: Compute Incorrect Sum ( $\sum X$ )

\text{Incorrect } \sum X = n \times \bar{X} = 100 \times 40 = 4,000

Step 2: Compute Corrected Sum

\text{Correct } \sum X = \text{Incorrect } \sum X - (\text{Wrong Values}) + (\text{Correct Values})

\text{Correct } \sum X = 4000 - (30 + 72) + (3 + 27)

\text{Correct } \sum X = 4000 - 102 + 30 = \mathbf{3,928}

Step 3: Calculate Corrected Mean

\text{Correct Mean } (\bar{X}_{\text{corr}}) = \frac{\text{Correct } \sum X}{n} = \frac{3928}{100} = \mathbf{39.28}

Final Answer: The corrected mean of the observations is 39.28.

Asked on 2026 Exam[6 marks]2026 paper

Sample of 50 mobile of two brands Samsung and Apple are taken and average running life (years) of each recorded as shown below.

	Samsung	Apple
Average life in years	11	12
variance	25	36

Which of the two brands show greater uniformity in life?

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Comparison of Uniformity in Running Life: Samsung vs. Apple:

Given Data ( $n = 50$ for both brands):

Samsung:
- Mean life ( $\bar{X}_1$ ) = $11 \text{ years}$
- Variance ( $\sigma_1^2$ ) = $25 \implies \sigma_1 = \sqrt{25} = 5 \text{ years}$
Apple:
- Mean life ( $\bar{X}_2$ ) = $12 \text{ years}$
- Variance ( $\sigma_2^2$ ) = $36 \implies \sigma_2 = \sqrt{36} = 6 \text{ years}$

Step 1: Compute Coefficient of Variation (CV)

CV = \frac{\sigma}{\bar{X}} \times 100

For Samsung:
$CV_1 = \frac{5}{11} \times 100 = \mathbf{45.45\%}$
For Apple:
$CV_2 = \frac{6}{12} \times 100 = \mathbf{50.00\%}$

Conclusion:

A lower Coefficient of Variation indicates greater consistency, stability, and uniformity.
Since $CV_{\text{Samsung}} (45.45\%) < CV_{\text{Apple}} (50.00\%)$ , Samsung shows greater uniformity in running life compared to Apple.

Asked on 2026 Exam[6 marks]2026 paper

Calculate an appropriate measure of central tendency from the following distribution and support for your choice of measure.

Monthly income (in Rs.)	No. of Families
Below 1000	50
1000-1999	500
2000-2999	555
3000-3999	100
4000-4999	30
5000 & above	15

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Appropriate Measure of Central Tendency for Open-Ended Income:

Choice of Measure and Support:

The given income distribution features open-ended terminal classes (“Below 1000” and “5000 & above”).
The arithmetic mean cannot be determined without arbitrary assumptions about class limits. The Median is the most appropriate measure because it is a positional average independent of extreme or undefined boundary values.

Cumulative Frequency Table:

(Using continuous class boundaries with correction factor $0.5$ )

Monthly Income (Rs)	Families ( $f$ )	Cumulative Frequency ( $cf$ )
Below 999.5	50	50
999.5 – 1999.5	500	550
1999.5 – 2999.5	555	1,105
2999.5 – 3999.5	100	1,205
3999.5 – 4999.5	30	1,235
4999.5 & above	15	1,250
Total	$N = 1,250$	—

Step 1: Locate Median Class

\text{Position of Median} = \frac{N}{2} = \frac{1250}{2} = 625^{\text{th}} \text{ item}

The cumulative frequency just greater than 625 is 1,105, which falls in class 1999.5 – 2999.5.

Step 2: Compute Median

M_d = L + \left(\frac{\frac{N}{2} - cf}{f}\right) \times h

Where $L = 1999.5$ , $cf = 550$ , $f = 555$ , $h = 1000$ :

M_d = 1999.5 + \left(\frac{625 - 550}{555}\right) \times 1000 = 1999.5 + \left(\frac{75}{555}\right) \times 1000 = 1999.5 + 135.14 = \mathbf{Rs \; 2,134.64}

Final Answer: The appropriate measure is the Median, and its value is Rs 2,134.64.

Asked on 2026 Exam[6 marks]2026 paper

Calculate the five number summary from the following data and comment on the shape of the distribution. 39, 26, 15, 8, 70, 11, 45, 60, 20, 32, 52

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Five-Number Summary and Distribution Shape:

Step 1: Arrange Data in Ascending Order

Given data ( $n = 11$ ): 39, 26, 15, 8, 70, 11, 45, 60, 20, 32, 52

Sorted order: 8, 11, 15, 20, 26, 32, 39, 45, 52, 60, 70

Step 2: Calculate Five-Number Summary

Minimum ( $Min$ ): 8
First Quartile ( $Q_1$ ):
$\text{Position} = \left(\frac{11 + 1}{4}\right)^{\text{th}} = 3^{\text{rd}} \text{ value} = \mathbf{15}$
Median ( $Q_2$ ):
$\text{Position} = \left(\frac{11 + 1}{2}\right)^{\text{th}} = 6^{\text{th}} \text{ value} = \mathbf{32}$
Third Quartile ( $Q_3$ ):
$\text{Position} = \left(\frac{3 \times 12}{4}\right)^{\text{th}} = 9^{\text{th}} \text{ value} = \mathbf{52}$
Maximum ( $Max$ ): 70

Five-Number Summary: $[8, 15, 32, 52, 70]$

Step 3: Comment on the Shape of the Distribution

Left quartile distance: $Q_2 - Q_1 = 32 - 15 = 17$
Right quartile distance: $Q_3 - Q_2 = 52 - 32 = 20$
Bowley’s Skewness:
$Sk_B = \frac{Q_3 + Q_1 - 2Q_2}{Q_3 - Q_1} = \frac{52 + 15 - 64}{52 - 15} = \frac{3}{37} \approx \mathbf{+0.081}$

Conclusion: Since $(Q_3 - Q_2) > (Q_2 - Q_1)$ ( $20 > 17$ ) and $Sk_B > 0$ , the distribution exhibits a slight positive skewness (skewed to the right).

Asked on 2026 Exam[10 marks]2026 paper

From the following income distribution:

Income (Rs '000)	0 - 4	5 - 9	10 - 14	15 - 19	20 - 24
No of persons	160	200	430	140	70

Compute lowest income of richest 10% of the people.

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Lowest Income of the Richest 10% of People ( $P_{90}$ ):

Concept:

The cutoff separating the lowest income of the richest 10% corresponds to the $90^{\text{th}}$ Percentile ( $P_{90}$ ).

Cumulative Frequency Table:

(Converted to continuous class boundaries)

Income (Rs '000)	True Boundaries	Persons ( $f$ )	Cumulative Frequency ( $cf$ )
0 – 4	0.0 – 4.5	160	160
5 – 9	4.5 – 9.5	200	360
10 – 14	9.5 – 14.5	430	790
15 – 19	14.5 – 19.5	140	930
20 – 24	19.5 – 24.5	70	1000
Total	—	$N = 1000$	—

Step 1: Locate $P_{90}$ Class

\text{Position} = \frac{90 \times N}{100} = \frac{90 \times 1000}{100} = 900^{\text{th}} \text{ item}

The cumulative frequency exceeding 900 is 930, so the $P_{90}$ Class is 14.5 – 19.5.

Step 2: Percentile Interpolation Formula

P_{90} = L + \left(\frac{\frac{90N}{100} - cf}{f}\right) \times h

Where:

$L = 14.5$
$cf = 790$
$f = 140$
$h = 5$ $P_{90} = 14.5 + \left(\frac{900 - 790}{140}\right) \times 5 = 14.5 + \left(\frac{110}{140}\right) \times 5 = 14.5 + 3.9286 = \mathbf{18.43 \text{ (in '000 Rs)}}$ $

Final Answer: The lowest income of the richest 10% of the people is Rs 18,429 (or 18.43 thousand Rs).

Asked on 2026 Exam[10 marks]2026 paper

Calculate Karl Pearson’s coefficient of skewness from the following income distribution of hundred families.

Income (Rs.'00)	0 - 20	20 - 40	40 - 60	60 - 80	80 - 100
No. of families	13	21	27	23	16

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Karl Pearson’s Coefficient of Skewness for 100 Families:

Calculation Table:

Income (Rs '00)	Mid-point ( $m$ )	Families ( $f$ )	$d = \frac{m - 50}{20}$	$fd$	$fd^2$
0 – 20	10	13	-2	-26	52
20 – 40	30	21	-1	-21	21
40 – 60	50	27	0	0	0
60 – 80	70	23	1	23	23
80 – 100	90	16	2	32	64
Total	—	$N = 100$	—	$\sum fd = 8$	$\sum fd^2 = 160$

Step 1: Mean ( $\bar{X}$ )

\bar{X} = A + \left(\frac{\sum fd}{N}\right) \times h = 50 + \left(\frac{8}{100}\right) \times 20 = 50 + 1.6 = \mathbf{51.60 \text{ (in '00 Rs)}}

Step 2: Mode ( $M_o$ )

Modal class is 40 – 60 ( $f_1 = 27, f_0 = 21, f_2 = 23, L = 40, h = 20$ ):

M_o = 40 + \left[\frac{27 - 21}{2(27) - 21 - 23}\right] \times 20 = 40 + \left(\frac{6}{54 - 44}\right) \times 20 = 40 + \left(\frac{6}{10}\right) \times 20 = 40 + 12 = \mathbf{52.00}

Step 3: Standard Deviation ( $\sigma$ )

\sigma = h \times \sqrt{\frac{\sum fd^2}{N} - \left(\frac{\sum fd}{N}\right)^2} = 20 \times \sqrt{\frac{160}{100} - \left(\frac{8}{100}\right)^2} = 20 \times \sqrt{1.60 - 0.0064} = 20 \times \sqrt{1.5936} = 20 \times 1.2624 = \mathbf{25.25}

Step 4: Karl Pearson’s Skewness ( $Sk_p$ )

Sk_p = \frac{\bar{X} - M_o}{\sigma} = \frac{51.60 - 52.00}{25.25} = \frac{-0.40}{25.25} \approx \mathbf{-0.0158}

Interpretation: The distribution has a very slight negative skewness ( $-0.0158$ ), indicating it is nearly symmetrical.

Chapter-wise questions

Unit 1: Unit 2: Describing Data Using Numerical Measures

Step 1: Arrange the Data in Ascending Order

Step 2: Calculate Lower Quartile (Q1Q_1Q1​)

Step 3: Calculate Upper Quartile (Q3Q_3Q3​)

Given Data:

Formula:

Step-by-Step Substitution:

Step 1: Arrange Data in Ascending Order

Step 2: Compute the Five-Number Summary

Step 3: Comment on the Shape of the Distribution

Calculation Table:

Step 1: Calculation of Mean (Xˉ\bar{X}Xˉ)

Step 2: Calculation of Standard Deviation (σ\sigmaσ)

Calculation Table:

Step 1: Village A Statistics

Step 2: Village B Statistics

Conclusion:

Given:

Formula:

Calculation:

Given:

Empirical Formula for Moderately Asymmetric Distributions:

Calculation:

Given:

Formula:

Calculation:

Concept:

Cumulative Frequency Table:

Step 1: Locate P80P_{80}P80​ Class Position

Step 2: Apply the Percentile Interpolation Formula

Given Data:

Step 1: Formulate Total Frequency Equation

Step 2: Use Mode Formula

Cumulative Frequency Table:

Step 1: Identify Median Class

Step 2: Apply Median Formula

Justification:

Cumulative Frequency with True Class Boundaries:

Step 1: Lower Quartile (Q1Q_1Q1​)

Step 2: Upper Quartile (Q3Q_3Q3​)

Step 3: Quartile Deviation (QDQDQD)

Given Data:

Step 1: Combined Mean (Xˉ12\bar{X}_{12}Xˉ12​)

Step 2: Deviations of Individual Means from Combined Mean

Step 3: Combined Variance (σ122\sigma_{12}^2σ122​)

Step 4: Combined Standard Deviation (σ12\sigma_{12}σ12​)

Calculation Table:

Step 1: Mean (Xˉ\bar{X}Xˉ)

Step 2: Mode (MoM_oMo​)

Step 3: Standard Deviation (σ\sigmaσ)

Step 4: Karl Pearson’s Coefficient of Skewness (SkpSk_pSkp​)

Concept:

Cumulative Frequency Table:

Step 1: Calculate Q1Q_1Q1​ and Q3Q_3Q3​

Step 2: Calculate P10P_{10}P10​ and P90P_{90}P90​

Step 3: Compute Kurtosis (KKK)

Given:

Formula:

Calculation:

Given:

Formula:

Calculation:

Given:

Empirical Formula:

Calculation:

Given:

Formula:

Calculation:

Given:

Formula:

Calculation:

Given Data:

Step 1: Total Frequency Equation

Step 2: Mean Equation

Step 3: Solve Simultaneously

Cumulative Frequency Table:

Step 1: Lower Quartile (Q1Q_1Q1​)

Step 2: Upper Quartile (Q3Q_3Q3​)

Step 3: Coefficient of Quartile Deviation

Step 2: Calculate Lower Quartile ( $Q_1$ )

Step 3: Calculate Upper Quartile ( $Q_3$ )

Step 1: Calculation of Mean ( $\bar{X}$ )

Step 2: Calculation of Standard Deviation ( $\sigma$ )

Step 1: Locate $P_{80}$ Class Position

Step 1: Lower Quartile ( $Q_1$ )

Step 2: Upper Quartile ( $Q_3$ )

Step 3: Quartile Deviation ( $QD$ )

Step 1: Combined Mean ( $\bar{X}_{12}$ )

Step 3: Combined Variance ( $\sigma_{12}^2$ )

Step 4: Combined Standard Deviation ( $\sigma_{12}$ )

Step 1: Mean ( $\bar{X}$ )

Step 2: Mode ( $M_o$ )

Step 3: Standard Deviation ( $\sigma$ )

Step 4: Karl Pearson’s Coefficient of Skewness ( $Sk_p$ )

Step 1: Calculate $Q_1$ and $Q_3$

Step 2: Calculate $P_{10}$ and $P_{90}$

Step 3: Compute Kurtosis ( $K$ )

Step 1: Lower Quartile ( $Q_1$ )

Step 2: Upper Quartile ( $Q_3$ )

Step 1: Position of $P_{90}$ $\text{Position} = \frac{90 \times N}{100} = \frac{90 \times 104}{100} = 93.6^{\text{th}} \text{ item}$ $

Step 1: Calculate $Q_1$ and $Q_3$

Step 2: Calculate $P_{10}$ and $P_{90}$

Step 3: Compute Kurtosis ( $K$ )

Data ( $n = 6$ observations each):

Step 1: Compute Incorrect Sum ( $\sum X$ )

Given Data ( $n = 50$ for both brands):

Step 1: Locate $P_{90}$ Class

Step 1: Mean ( $\bar{X}$ )

Step 2: Mode ( $M_o$ )

Step 3: Standard Deviation ( $\sigma$ )

Step 4: Karl Pearson’s Skewness ( $Sk_p$ )