Questions and Answers Section

Question 1

The mean of 5 observations is 4.4 and their variance is 8.24. If three of the observations are 1, 2, and 6, find the other two observations.

Solution: Let the two unknown observations be $a$ and $b$. The complete set of observations is {1, 2, 6, $a$, $b$}.

Step 1: Use the Mean Formula to find the sum of the unknown observations.

The mean is given by $\bar{x} = \frac{\sum x_i}{n}$. $$4.4 = \frac{1+2+6+a+b}{5}$$ $$4.4 \times 5 = 9+a+b$$ $$22 = 9+a+b$$ $$a+b = 13 \quad \cdots(1)$$

Step 2: Use the Variance Formula to find the sum of the squares.

The formula for variance is $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$. $$8.24 = \frac{1^2+2^2+6^2+a^2+b^2}{5} - (4.4)^2$$ $$8.24 = \frac{1+4+36+a^2+b^2}{5} - 19.36$$ $$8.24 + 19.36 = \frac{41+a^2+b^2}{5}$$ $$27.6 = \frac{41+a^2+b^2}{5}$$ $$27.6 \times 5 = 41+a^2+b^2$$ $$138 = 41+a^2+b^2$$ $$a^2+b^2 = 97 \quad \cdots(2)$$

Step 3: Solve the system of two equations.

From equation (1), we have $b = 13-a$. Substitute this into equation (2): $$a^2 + (13-a)^2 = 97$$ $$a^2 + (169 - 26a + a^2) = 97$$ $$2a^2 - 26a + 169 - 97 = 0$$ $$2a^2 - 26a + 72 = 0$$ Divide by 2: $a^2 - 13a + 36 = 0$.

Factor the quadratic equation: $(a-4)(a-9) = 0$.

This gives two possible values for $a$: $a=4$ or $a=9$.

• If $a=4$, then from (1), $b = 13-4 = 9$.
• If $a=9$, then from (1), $b = 13-9 = 4$.

Answer: The other two observations are 4 and 9.

Question 2

Find the mean deviation about the median for the frequency distribution:

Marks	0-10	10-20	20-30	30-40	40-50	50-60
No. of Students	6	8	14	16	4	2

Solution: Step 1: Create a table with Mid-points and Cumulative Frequencies.

Marks	Freq ($f_i$)	Mid-point ($x_i$)	C.F.
0-10	6	5	6
10-20	8	15	14
20-30	14	25	28
30-40	16	35	44
40-50	4	45	48
50-60	2	55	50

Step 2: Calculate the Median. Total frequency $N=\sum f_i = 50$. We find the position $N/2 = 25$. The cumulative frequency just greater than 25 is 28, so the median class is 20-30. Using the median formula: $M = l + \frac{N/2 - C}{f} \times h$.

• $l = 20$ (lower limit of median class)
• $C = 14$ (C.F. of class preceding median class)
• $f = 14$ (frequency of median class)
• $h = 10$ (class size) $M = 20 + \frac{25-14}{14} \times 10 = 20 + \frac{11}{14} \times 10 \approx 20 + 7.86 = 27.86$.

Step 3: Calculate the Mean Deviation. We now find the sum of $f_i |x_i - M|$.

$x_i$	$f_i$	$|x_i - 27.86|$	$f_i|x_i - 27.86|$
5	6	22.86	137.16
15	8	12.86	102.88
25	14	2.86	40.04
35	16	7.14	114.24
45	4	17.14	68.56
55	2	27.14	54.28

Sum $\sum f_i |x_i - M| = 137.16+102.88+40.04+114.24+68.56+54.28 = 517.16$.

Finally, M.D.(M) = $\frac{\sum f_i |x_i - M|}{N} = \frac{517.16}{50} = 10.3432$.

Answer: The mean deviation about the median is approximately 10.34.

Question 3

The mean and S.D. of 100 observations were calculated as 40 and 5.1. It was later found that one observation was misread as 50 instead of 40. Find the correct mean and S.D.

Solution: Part A: Correcting the Mean

1. Find the incorrect sum: Incorrect $\sum x_i = n \times \bar{x}_{inc} = 100 \times 40 = 4000$.

2. Calculate the correct sum: We remove the incorrect value and add the correct one.

   Correct $\sum x_i = \text{Incorrect } \sum x_i - (\text{wrong value}) + (\text{correct value}) = 4000 - 50 + 40 = 3990$.

3. Calculate the correct mean: Correct Mean $\bar{x}_{corr} = \frac{3990}{100} = 39.9$.

Part B: Correcting the Standard Deviation

1. Find the incorrect sum of squares: Use the variance formula $$\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$$ So, $\sum x_i^2 = n(\sigma^2 + \bar{x}^2)$.

   Incorrect $\sum x_i^2 = 100(5.1^2 + 40^2) = 100(26.01 + 1600) = 100(1626.01) = 162601$.

2. Calculate the correct sum of squares: We remove the square of the incorrect value and add the square of the correct one.

   Correct $\sum x_i^2 = 162601 - 50^2 + 40^2 = 162601 - 2500 + 1600 = 161701$.

3. Calculate the correct variance: Correct Variance $\sigma^2_{corr} = \frac{\text{Correct } \sum x_i^2}{n} - (\bar{x}_{corr})^2 = \frac{161701}{100} - (39.9)^2$.

   $\sigma^2_{corr} = 1617.01 - 1592.01 = 25$.

4. Calculate the correct S.D.: Correct S.D. $\sigma_{corr} = \sqrt{25} = 5$.

Answer: The correct mean is 39.9 and the correct S.D. is 5.

Question 4

Calculate the variance for the following discrete frequency distribution:

$x_i$	4	8	11	17	20	24	32
$f_i$	3	5	9	5	4	3	1

Solution: Step 1: Calculate the Mean ($\bar{x}$) We need to compute $\sum f_i$ and $\sum f_i x_i$. $N = \sum f_i = 3+5+9+5+4+3+1 = 30$. $\sum f_i x_i = 3(4)+5(8)+9(11)+5(17)+4(20)+3(24)+1(32)$ $= 12+40+99+85+80+72+32=420$. Mean $\bar{x} = \frac{\sum f_i x_i}{N} = \frac{420}{30}=14$.

Step 2: Calculate the sum of squared deviations $\sum f_i(x_i - \bar{x})^2$. We create a table to organize the calculation.

$x_i$	$f_i$	$x_i - \bar{x}$	$(x_i - \bar{x})^2$	$f_i(x_i - \bar{x})^2$
4	3	-10	100	300
8	5	-6	36	180
11	9	-3	9	81
17	5	3	9	45
20	4	6	36	144
24	3	10	100	300
32	1	18	324	324
			Sum:	1374

The sum is $\sum f_i(x_i - \bar{x})^2 = 1374$.

Step 3: Calculate the Variance. The formula for variance is $\sigma^2 = \frac{\sum f_i(x_i - \bar{x})^2}{N}$. $\sigma^2 = \frac{1374}{30} = 45.8$.

Answer: The variance is 45.8.

Question 5

The Coefficient of Variation of two distributions are 60% and 70%, and their standard deviations are 21 and 16 respectively. What are their arithmetic means?

Solution: Recall the Formula: The Coefficient of Variation (C.V.) is defined as: $$\text{C.V.} = \frac{\sigma}{\bar{x}} \times 100$$ where $\sigma$ is the standard deviation and $\bar{x}$ is the mean. We can rearrange this formula to solve for the mean: $\bar{x} = \frac{\sigma}{\text{C.V.}} \times 100$.

Calculation for Distribution 1:

• Given: C.V. = 60 and $\sigma_1 = 21$.
• $\bar{x}_1 = \frac{21}{60} \times 100 = \frac{2100}{60} = 35$.

Calculation for Distribution 2:

• Given: C.V. = 70 and $\sigma_2 = 16$.
• $\bar{x}_2 = \frac{16}{70} \times 100 = \frac{1600}{70} = \frac{160}{7} \approx 22.86$.

Answer: The mean of the first distribution is 35 and the mean of the second distribution is approximately 22.86.

Question 6

Find the mean deviation about the mean for the first $n$ natural numbers.

Solution: Let the set of observations be $x_i = \{1, 2, \dots, n\}$.

Step 1: Find the Mean. The mean of the first $n$ natural numbers (which form an A.P.) is $$\bar{x} = \frac{\text{first term} + \text{last term}}{2} = \frac{n+1}{2}$$

Step 2: Set up the Sum of Deviations.

The mean deviation is $$M.D. = \frac{1}{n} \sum_{i=1}^n |x_i - \bar{x}| = \frac{1}{n} \sum_{i=1}^n |i - \frac{n+1}{2}|$$ The calculation differs slightly depending on whether n is odd or even.

Case 1: n is odd. Let $n=2k+1$. The mean is $\bar{x} = \frac{2k+1+1}{2} = k+1$. The mean is an integer.

The sum of deviations is $2 \times [|1-(k+1)| + ... + |k-(k+1)|] = 2 \times [k + (k-1) + ... + 1]$.

This is twice the sum of the first $k$ natural numbers: $2 \times \frac{k(k+1)}{2} = k(k+1)$.

Since $k=\frac{n-1}{2}$, the sum is $(\frac{n-1}{2})(\frac{n+1}{2}) = \frac{n^2-1}{4}$.

The M.D. is $\frac{\text{Sum}}{n} = \frac{n^2-1}{4n}$.

Case 2: n is even. Let $n=2k$. The mean is $\bar{x} = \frac{2k+1}{2} = k+0.5$.

The sum of deviations is $2 \times [|1-(k+0.5)| + ... + |k-(k+0.5)|] = 2 \times [(k-0.5) + (k-1.5) + ... + 0.5]$.

This sum simplifies to $2 \times \frac{k^2}{2} = k^2$. Since $k = n/2$, the sum is $(n/2)^2 = n^2/4$.

The M.D. is $\frac{\text{Sum}}{n} = \frac{n^2/4}{n} = \frac{n}{4}$.

Answer: The mean deviation is $\frac{n^2-1}{4n}$ if n is odd, and $\frac{n}{4}$ if n is even.

Question 7

The variance of 15 observations is 4. If each observation is increased by 9, what is the new variance?

Solution: 1. Understand the Property: Variance and standard deviation are measures of dispersion, which is the spread of the data around a central point (the mean). They are independent of the actual location of the data on the number line.

2. Change of Origin: Adding or subtracting a constant value to every observation in a dataset is called a change of origin. This action shifts the entire dataset along the number line.

• Let the original data be $x_i$ and the mean be $\bar{x}$.
• Let the new data be $y_i = x_i + 9$.
• The new mean will be $\bar{y} = \bar{x} + 9$.

3. Effect on Deviations: The deviation of a new data point from the new mean is: $y_i - \bar{y} = (x_i + 9) - (\bar{x} + 9) = x_i - \bar{x}$. The deviations remain exactly the same.

4. Conclusion: Since variance is the average of the squared deviations, and the deviations themselves are unchanged, the variance will also be unchanged. New Variance = Old Variance = 4.

Answer: The new variance is 4.

Question 8

The standard deviation of 20 observations is 5. If each observation is multiplied by 2, find the new standard deviation and new variance.

Solution: 1. Understand the Property: Unlike a change of origin, multiplying or dividing each observation by a constant (a change of scale) directly affects the spread of the data.

2. Change of Scale:

• Let the original data be $x_i$ with mean $\bar{x}$ and standard deviation $\sigma_x$.
• Let the new data be $y_i = k \cdot x_i$. In this case, $k=2$.
• The new mean will be $\bar{y} = k \cdot \bar{x}$.

3. Effect on Standard Deviation: The new standard deviation, $\sigma_y$, is $|k|$ times the original standard deviation. $\sigma_y = |k| \cdot \sigma_x$. Given $\sigma_x = 5$ and $k=2$: New S.D. = $|2| \times 5 = \mathbf{10}$.

4. Effect on Variance: The variance is the square of the standard deviation, so $\sigma_y^2 = (|k| \cdot \sigma_x)^2 = k^2 \cdot \sigma_x^2$.

New Variance = $k^2 \times$ Old Variance.

Old Variance = $\sigma_x^2 = 5^2 = 25$.

New Variance = $2^2 \times 25 = 4 \times 25 = \mathbf{100}$.

Alternatively, we can simply square the new standard deviation: New Variance = $(10)^2 = 100$.

Answer: The new standard deviation is 10 and the new variance is 100.

Question 9

Two cricket players, A and B, have the following scores in 10 innings:

• Player A: $\bar{x}_A = 45$, $\sigma_A = 12$
• Player B: $\bar{x}_B = 60$, $\sigma_B = 15$ Who is the more consistent batsman?

Solution: 1. The Need for a Relative Measure: Standard deviation ($"\sigma"$) is an absolute measure of spread. We cannot directly compare the standard deviations (12 vs. 15) to judge consistency because the players have different average scores (means). A variation of 15 runs is less significant for a player averaging 60 than a variation of 12 runs for a player averaging 45.

2. Using the Coefficient of Variation (C.V.): To compare consistency, we must use a relative measure of dispersion, the C.V. It expresses the standard deviation as a percentage of the mean. A lower C.V. indicates higher consistency. $$\text{C.V.} = \frac{\sigma}{\bar{x}} \times 100$$

3. Calculate C.V. for Player A: C.V.(A) = $\frac{\sigma_A}{\bar{x}_A} \times 100 = \frac{12}{45} \times 100 = \frac{4}{15} \times 100 \approx 26.67\%$.

4. Calculate C.V. for Player B: C.V.(B) = $\frac{\sigma_B}{\bar{x}_B} \times 100 = \frac{15}{60} \times 100 = \frac{1}{4} \times 100 = 25\%$.

5. Compare and Conclude: Since C.V.(B) (25%) is less than C.V.(A) (26.67%), Player B has a smaller relative spread in his scores.

Answer: Player B is the more consistent batsman.

Question 10

Find the standard deviation of the first 10 natural numbers.

Solution: The data set is {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}.

Step 1: Calculate the Mean ($\bar{x}$): For the first $n$ natural numbers, the mean is $\frac{n+1}{2}$. $\bar{x} = \frac{10+1}{2} = 5.5$.

Step 2: Calculate the Sum of Squares ($\sum x_i^2$): The sum of the squares of the first $n$ natural numbers is given by the formula $\frac{n(n+1)(2n+1)}{6}$. $\sum x_i^2 = \frac{10(10+1)(2 \cdot 10+1)}{6} = \frac{10(11)(21)}{6} = \frac{2310}{6} = 385$.

Step 3: Calculate the Variance ($\sigma^2$) using the Shortcut Formula: The shortcut formula is $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$. $\sigma^2 = \frac{385}{10} - (5.5)^2 = 38.5 - 30.25 = 8.25$.

Alternatively, for the first n natural numbers, the variance has a direct formula: $\sigma^2 = \frac{n^2-1}{12}$. For n=10, $\sigma^2 = \frac{10^2-1}{12} = \frac{99}{12} = \frac{33}{4} = 8.25$.

Step 4: Calculate the Standard Deviation ($\sigma$): Standard deviation is the square root of the variance. $\sigma = \sqrt{8.25} \approx 2.87$.

Answer: The standard deviation is $\sqrt{8.25}$ (or approximately 2.87).

Question 11

The mean and variance of 7 observations are 8 and 16, respectively. If 5 of the observations are 2, 4, 10, 12, 14, find the remaining two observations.

Solution: Let the two unknown observations be $a$ and $b$. The complete set of observations is {2, 4, 10, 12, 14, $a, b$}.

Step 1: Use the Mean to find $a+b$. $\bar{x} = \frac{\sum x_i}{n} \implies 8 = \frac{2+4+10+12+14+a+b}{7}$.

$8 \times 7 = 42 + a + b$

$56 = 42 + a + b$

$a+b=14 \quad \cdots(1)$.

Step 2: Use the Variance to find $a^2+b^2$. Using the shortcut formula for variance: $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$.

$16 = \frac{2^2+4^2+10^2+12^2+14^2+a^2+b^2}{7} - (8)^2$.

$16 = \frac{4+16+100+144+196+a^2+b^2}{7} - 64$.

$16+64 = \frac{460+a^2+b^2}{7}$.

$80 \times 7 = 460+a^2+b^2$.

$560 = 460+a^2+b^2$.

$a^2+b^2=100 \quad \cdots(2)$.

Step 3: Solve the system of equations. From (1), we have $b=14-a$. Substitute this into (2):

$a^2 + (14-a)^2 = 100$

$a^2 + (196 - 28a + a^2) = 100$

$2a^2 - 28a + 96 = 0$

$a^2 - 14a + 48 = 0$.

Factor the quadratic equation: $(a-6)(a-8) = 0$.

This gives two possible values: $a=6$ or $a=8$.

• If $a=6$, then $b=14-6=8$.
• If $a=8$, then $b=14-8=6$.

Answer: The remaining two observations are 6 and 8.

Question 12

Calculate the mean deviation about the mean for the data:

$x_i$	10	30	50	70	90
$f_i$	4	24	28	16	8

Solution: Step 1: Calculate the Mean ($\bar{x}$) We create a table to help with calculations.

$x_i$	$f_i$	$f_i x_i$
10	4	40
30	24	720
50	28	1400
70	16	1120
90	8	720
Sum	N=80	4000

Mean $\bar{x} = \frac{\sum f_i x_i}{N} = \frac{4000}{80} = 50$.

Step 2: Calculate the sum of absolute deviations, $\sum f_i|x_i - \bar{x}|$ Now we add columns for the deviations.

$x_i$	$f_i$	$|x_i - 50|$	$f_i|x_i - 50|$
10	4	40	160
30	24	20	480
50	28	0	0
70	16	20	320
90	8	40	320
		Sum:	1280

The sum is $\sum f_i|x_i - \bar{x}| = 160+480+0+320+320 = 1280$.

Step 3: Calculate the Mean Deviation. M.D.($\bar{x}$) = $\frac{\sum f_i|x_i - \bar{x}|}{N} = \frac{1280}{80} = 16$.

Answer: The mean deviation about the mean is 16.

Question 13

The sum of 10 items is 12 and the sum of their squares is 18. Find the standard deviation.

Solution: We are given the following information:

• Number of items, $n=10$.
• Sum of items, $\sum x_i = 12$.
• Sum of their squares, $\sum x_i^2 = 18$.

Step 1: Calculate the Mean ($\bar{x}$): $\bar{x} = \frac{\sum x_i}{n} = \frac{12}{10} = 1.2$.

Step 2: Calculate the Variance ($\sigma^2$) using the Shortcut Formula: It is most efficient to use the formula $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$. $\sigma^2 = \frac{18}{10} - (1.2)^2$ $\sigma^2 = 1.8 - 1.44 = 0.36$.

Step 3: Calculate the Standard Deviation ($\sigma$): Standard deviation is the square root of the variance. $\sigma = \sqrt{0.36} = 0.6$.

Answer: The standard deviation is 0.6.

Question 14

If the mean and C.V. of a distribution are 20 and 40% respectively, find the variance.

Solution: This is a two-step problem where we first find the standard deviation from the C.V., and then find the variance.

Step 1: Find the Standard Deviation ($\sigma$) from the C.V.: The formula for the Coefficient of Variation is: $$\text{C.V.} = \frac{\sigma}{\bar{x}} \times 100$$ We are given C.V. = 40 and $\bar{x} = 20$. $40 = \frac{\sigma}{20} \times 100$ To solve for $\sigma$, we can first divide by 100: $0.4 = \frac{\sigma}{20}$ $\sigma = 0.4 \times 20 = 8$. So, the standard deviation is 8.

Step 2: Find the Variance ($\sigma^2$) from the S.D.: Variance is the square of the standard deviation. Variance = $\sigma^2 = 8^2 = 64$.

Answer: The variance is 64.

Question 15

Calculate the standard deviation for the following continuous distribution:

Class	30-40	40-50	50-60	60-70	70-80	80-90	90-100
Freq	3	7	12	15	8	3	2

Solution: For a continuous distribution with equal class sizes, the step-deviation method is the most efficient way to calculate variance and standard deviation.

Step 1: Set up the Calculation Table.

• Class size is $h=10$. Let's choose the assumed mean $a=65$ (the mid-point of the class with the highest frequency).
• Total frequency $N = 3+7+12+15+8+3+2 = 50$.
• We will calculate $y_i = \frac{x_i - a}{h} = \frac{x_i - 65}{10}$.

Class	Freq ($f_i$)	Mid-point ($x_i$)	$y_i$	$f_i y_i$	$y_i^2$	$f_i y_i^2$
30-40	3	35	-3	-9	9	27
40-50	7	45	-2	-14	4	28
50-60	12	55	-1	-12	1	12
60-70	15	65	0	0	0	0
70-80	8	75	1	8	1	8
80-90	3	85	2	6	4	12
90-100	2	95	3	6	9	18
	N=50			$\sum f_i y_i = -15$		$\sum f_i y_i^2 = 105$

Step 2: Calculate the Variance ($\sigma^2$). Using the step-deviation formula for variance: $$\sigma^2 = h^2 \left[ \frac{\sum f_i y_i^2}{N} - \left(\frac{\sum f_i y_i}{N}\right)^2 \right]$$ $$\sigma^2 = 10^2 \left[ \frac{105}{50} - \left(\frac{-15}{50}\right)^2 \right]$$ $$\sigma^2 = 100 \left[ 2.1 - (-0.3)^2 \right] = 100 [2.1 - 0.09] = 100 [2.01] = 201$$.

Step 3: Calculate the Standard Deviation ($\sigma$). $$\sigma = \sqrt{\text{Variance}} = \sqrt{201} \approx 14.18$$

Answer: The standard deviation is approximately 14.18.

Question 16

The mean of a set of 20 observations is 10 and its standard deviation is 2. If each observation is multiplied by 3 and then increased by 4, find the new mean and new standard deviation.

Solution: This question tests the effect of change of origin and change of scale on measures of central tendency and dispersion. Let the original observations be $x_i$ and the new observations be $y_i$. The transformation is $y_i = 3x_i + 4$.

1. Effect on the Mean: The mean is affected by both change of scale (multiplication) and change of origin (addition). New Mean $\bar{y} = 3\bar{x} + 4$. Given $\bar{x} = 10$. $\bar{y} = 3(10) + 4 = 30 + 4 = \mathbf{34}$.

2. Effect on the Standard Deviation: Standard deviation is a measure of spread.

• Change of Origin (+4): Adding a constant to every value shifts the entire dataset but does not change its spread. So, this part has no effect on the S.D.
• Change of Scale (x3): Multiplying every value by a constant stretches or shrinks the data, so it does affect the spread. The new S.D. is multiplied by the absolute value of the constant. New S.D. $\sigma_y = |3| \times \sigma_x$. Given $\sigma_x = 2$. $\sigma_y = 3 \times 2 = \mathbf{6}$.

Answer: The new mean is 34 and the new standard deviation is 6.

Question 17

Find the range of the data: 148, 154, 158, 160, 162, 166, 170, 175, 180, 185.

Solution: 1. Understand Range: The range is the simplest measure of dispersion. It is the difference between the highest (maximum) and lowest (minimum) values in a dataset. $$\text{Range} = \text{Maximum Value} - \text{Minimum Value}$$

2. Identify Maximum and Minimum Values: The data is already sorted in ascending order, which makes it easy to identify the extremes.

• Minimum Value = 148
• Maximum Value = 185

3. Calculate the Range: Range = $185 - 148 = 37$.

Answer: The range of the data is 37.

Question 18

For a group of 50 male workers, the mean age is 40 years with a S.D. of 5 years. For a group of 40 female workers, the mean age is 35 years with a S.D. of 4 years. Find the combined standard deviation of the 90 workers.

Solution: Step 1: Find the Combined Mean ($\bar{x}_{12}$)

$n_1=50, \bar{x}_1=40, \sigma_1=5$ $n_2=40, \bar{x}_2=35, \sigma_2=4$. $$\bar{x}_{12} = \frac{n_1\bar{x}_1+n_2\bar{x}_2}{n_1+n_2} = \frac{50(40)+40(35)}{50+40} = \frac{2000+1400}{90} = \frac{3400}{90} \approx 37.78$$

Step 2: Calculate $d_1$ and $d_2$

These values represent the difference between each group's mean and the combined mean.

$d_1 = \bar{x}_1 - \bar{x}_{12} = 40 - 37.78 = 2.22$.

$d_2 = \bar{x}_2 - \bar{x}_{12} = 35 - 37.78 = -2.78$.

Step 3: Calculate the Combined Variance ($\sigma^2_{12}$) The formula for combined variance is: $$\sigma^2_{12} = \frac{n_1(\sigma_1^2 + d_1^2) + n_2(\sigma_2^2 + d_2^2)}{n_1+n_2}$$ First, find the components:

• $$\sigma_1^2 = 5^2 = 25$$
• $d_1^2 = (2.22)^2 \approx 4.93$
• $\sigma_2^2 = 4^2 = 16$
• $d_2^2 = (-2.78)^2 \approx 7.73$

Now, substitute into the formula:

$\sigma^2_{12} = \frac{50(25 + 4.93) + 40(16 + 7.73)}{90} = \frac{50(29.93)+40(23.73)}{90}$.

$\sigma^2_{12} = \frac{1496.5+949.2}{90} = \frac{2445.7}{90} \approx 27.17$.

Step 4: Calculate the Combined Standard Deviation. $\sigma_{12} = \sqrt{\text{Combined Variance}} = \sqrt{27.17} \approx 5.21$.

Answer: The combined standard deviation is approximately 5.21 years.

Question 19

Find the mean deviation from the mode for the data: 3, 5, 6, 6, 7, 10, 10, 10, 12, 12.

Solution: Step 1: Find the Mode. The mode is the value that appears most frequently in the dataset.

• 6 appears 2 times.
• 10 appears 3 times.
• 12 appears 2 times. The observation 10 occurs most often, so the Mode = 10.

Step 2: Calculate the Absolute Deviations from the Mode. We find the absolute difference between each observation and the mode (10).

• $|3-10|=7$
• $|5-10|=5$
• $|6-10|=4$
• $|6-10|=4$
• $|7-10|=3$
• $|10-10|=0$
• $|10-10|=0$
• $|10-10|=0$
• $|12-10|=2$
• $|12-10|=2$

Step 3: Sum the Absolute Deviations. Sum = $7+5+4+4+3+0+0+0+2+2 = 27$.

Step 4: Calculate the Mean Deviation. There are $n=10$ observations. M.D.(Mode) = $\frac{\text{Sum of Deviations}}{n} = \frac{27}{10} = 2.7$.

Answer: The mean deviation from the mode is 2.7.

Question 20

If the sum of 10 observations is 100 and the sum of their squares is 1090, find the Coefficient of Variation.

Solution: To find the C.V., we first need to calculate the mean and the standard deviation.

Step 1: Calculate the Mean ($\bar{x}$): Given $n=10$ and $\sum x_i = 100$. $\bar{x} = \frac{\sum x_i}{n} = \frac{100}{10} = 10$.

Step 2: Calculate the Variance ($\sigma^2$): Using the shortcut formula $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$, with $\sum x_i^2 = 1090$. $\sigma^2 = \frac{1090}{10} - (10)^2 = 109 - 100 = 9$.

Step 3: Calculate the Standard Deviation ($\sigma$): $\sigma = \sqrt{\text{Variance}} = \sqrt{9} = 3$.

Step 4: Calculate the Coefficient of Variation (C.V.): C.V. = $\frac{\sigma}{\bar{x}} \times 100 = \frac{3}{10} \times 100 = 30\%$.

Answer: The Coefficient of Variation is 30%.

Question 21

The first of two samples has 100 items with mean 15 and S.D. 3. If the whole group has 250 items with mean 15.6 and S.D. $\sqrt{13.44}$, find the S.D. of the second group.

Solution: This is a reverse problem using the combined variance formula.

Given:

• Group 1: $n_1=100, \bar{x}_1=15, \sigma_1=3$
• Combined Group: $N=250, \bar{x}_{12}=15.6, \sigma_{12}^2=13.44$
• Group 2: $n_2=N-n_1 = 150$. We need to find $\sigma_2$.

Step 1: Find the mean of the second group ($\bar{x}_2$).

Using the combined mean formula: $$\bar{x}_{12} = \frac{n_1\bar{x}_1+n_2\bar{x}_2}{n_1+n_2}$$ $$15.6 = \frac{100(15)+150(\bar{x}_2)}{250}$$ $$15.6 \times 250 = 1500 + 150\bar{x}_2$$ $$3900 = 1500 + 150\bar{x}_2 \implies 150\bar{x}_2 = 2400 \implies \bar{x}_2=16$$

Step 2: Find $d_1$ and $d_2$.

$d_1 = \bar{x}_1 - \bar{x}_{12} = 15 - 15.6 = -0.6$.

$d_2 = \bar{x}_2 - \bar{x}_{12} = 16 - 15.6 = 0.4$.

Step 3: Use the combined variance formula to solve for $\sigma_2^2$. $$\sigma^2_{12} = \frac{n_1(\sigma_1^2 + d_1^2) + n_2(\sigma_2^2 + d_2^2)}{n_1+n_2}$$ $$13.44 = \frac{100(3^2 + (-0.6)^2) + 150(\sigma_2^2 + 0.4^2)}{250}$$ $$13.44 \times 250 = 100(9 + 0.36) + 150(\sigma_2^2 + 0.16)$$ $$3360 = 100(9.36) + 150\sigma_2^2 + 150(0.16)$$ $$3360 = 936 + 150\sigma_2^2 + 24$$ $$3360 = 960 + 150\sigma_2^2$$ $$2400 = 150\sigma_2^2 \implies \sigma_2^2 = \frac{2400}{150} = 16$$.

Step 4: Find the S.D. of the second group. $\sigma_2 = \sqrt{16} = 4$.

Answer: The S.D. of the second group is 4.

Question 22

Find the variance of the first 50 even natural numbers.

Solution: The data set is {2, 4, 6, …, 100}.

Method 1: Using Properties of Variance

1. Relate to Natural Numbers: Each observation $x_i$ in our set is twice the corresponding natural number $i$. So, $x_i = 2i$ for $i=1, 2, ..., 50$.

2. Apply Change of Scale Property: The variance of a dataset $kx_i$ is related to the variance of $x_i$ by the formula $\text{Var}(kx) = k^2 \text{Var}(x)$.

   Here, $k=2$. So, $\text{Var}(2,4,...,100) = 2^2 \text{Var}(1,2,...,50) = 4 \times \text{Var}(1,2,...,50)$.

3. Use Formula for Variance of First n Natural Numbers: The variance of the first $n$ natural numbers is $\sigma_n^2 = \frac{n^2-1}{12}$.

   For $n=50$, the variance of {1, 2, …, 50} is $\frac{50^2-1}{12} = \frac{2499}{12} = \frac{833}{4}$.

4. Calculate Final Variance:

   Variance = $4 \times \frac{833}{4} = 833$.

Method 2: Direct Calculation

1. Mean: The mean of an A.P. is the average of the first and last term: $\bar{x} = \frac{2+100}{2} = 51$.

2. Sum of Squares: $\sum x_i^2 = \sum_{i=1}^{50} (2i)^2 = 4 \sum_{i=1}^{50} i^2$.

   The sum of squares formula is $\frac{n(n+1)(2n+1)}{6}$.

   $\sum x_i^2 = 4 \times \frac{50(51)(101)}{6} = 4 \times 42925 = 171700$.

3. Variance: $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2 = \frac{171700}{50} - (51)^2 = 3434 - 2601 = 833$.

Answer: The variance is 833.

Question 23

Calculate the mean deviation from the mean for the observations: 1, 2, 3, 4, 5.

Solution: Step 1: Calculate the Mean ($\bar{x}$): For the numbers {1, 2, 3, 4, 5}, the sum is 15 and the number of observations is $n=5$. $\bar{x} = \frac{1+2+3+4+5}{5} = \frac{15}{5} = 3$.

Step 2: Calculate the Absolute Deviations from the Mean: We find the absolute difference between each observation and the mean (3).

• $|1-3| = 2$
• $|2-3| = 1$
• $|3-3| = 0$
• $|4-3| = 1$
• $|5-3| = 2$

Step 3: Sum the Absolute Deviations: Sum = $2+1+0+1+2 = 6$.

Step 4: Calculate the Mean Deviation: The mean deviation is the average of these deviations. M.D.($\bar{x}$) = $\frac{\text{Sum of Deviations}}{n} = \frac{6}{5} = 1.2$.

Answer: The mean deviation is 1.2.

Question 24

If the C.V. of a stock is 8% and its price has a standard deviation of $2, what is the mean price?

Solution: 1. Recall the C.V. Formula:

The Coefficient of Variation (C.V.) relates the mean, standard deviation, and C.V. as follows: $$\text{C.V.} = \frac{\sigma}{\bar{x}} \times 100$$

2. Rearrange the Formula to Solve for the Mean : $$\bar{x} = \frac{\sigma}{\text{C.V.}} \times 100$$

3. Substitute the Given Values:

• C.V. = 8
• Standard Deviation, $${\sigma} = 2$$

$${\bar{x}} = \frac{2}{8} \times 100 = \frac{1}{4} \times 100 = 25$$

Answer:The mean price of the stock is 25.

Question 25

The mean and S.D. of a group of 100 observations are 20 and 3, respectively. Later it was found that three observations were recorded incorrectly as 21, 21, and 18, whereas they were 20, 20, and 20. Find the correct mean and variance.

Solution: Part A: Correcting the Mean

1. Find the incorrect sum: Incorrect $\sum x_i = n \times \bar{x}_{inc} = 100 \times 20 = 2000$.

2. Calculate the correct sum:

   Sum of wrong items = $21+21+18 = 60$.

   Sum of correct items = $20+20+20 = 60$.

   Correct $\sum x_i = 2000 - 60 + 60 = 2000$.

3. Calculate the correct mean: Since the sum is unchanged, the correct mean is still $\frac{2000}{100} = 20$.

Part B: Correcting the Variance

1. Find the incorrect sum of squares: Use $\sum x_i^2 = n(\sigma^2 + \bar{x}^2)$.

   Incorrect $\sum x_i^2 = 100(3^2 + 20^2) = 100(9 + 400) = 40900$.

2. Calculate the correct sum of squares:

   Sum of squares of wrong items = $21^2+21^2+18^2 = 441+441+324 = 1206$.

   Sum of squares of correct items = $20^2+20^2+20^2 = 400+400+400 = 1200$.

   Correct $\sum x_i^2 = 40900 - 1206 + 1200 = 40894$.

3. Calculate the correct variance:

   Correct Variance $\sigma^2_{corr} = \frac{\text{Correct } \sum x_i^2}{n} - (\bar{x}_{corr})^2 = \frac{40894}{100} - (20)^2$.

   $\sigma^2_{corr} = 408.94 - 400 = 8.94$.

Answer: The correct Mean is 20, and the correct Variance is 8.94.

Question 26

Find the variance of the numbers 1, 3, 5, 7, 9.

Solution: The data set {1, 3, 5, 7, 9} is a small set of ungrouped data.

Step 1: Calculate the Mean ($\bar{x}$): The numbers form an Arithmetic Progression, so the mean is the middle value, which is 5. Alternatively, $\bar{x} = \frac{1+3+5+7+9}{5} = \frac{25}{5} = 5$.

Step 2: Calculate the Squared Deviations from the Mean: We find the square of the difference between each observation and the mean (5).

• $(1-5)^2 = (-4)^2 = 16$
• $(3-5)^2 = (-2)^2 = 4$
• $(5-5)^2 = 0^2 = 0$
• $(7-5)^2 = 2^2 = 4$
• $(9-5)^2 = 4^2 = 16$

Step 3: Sum the Squared Deviations: Sum = $16+4+0+4+16 = 40$.

Step 4: Calculate the Variance ($\sigma^2$): Variance is the average of the squared deviations.

$\sigma^2 = \frac{\text{Sum of Squared Deviations}}{n} = \frac{40}{5} = 8$.

Answer: The variance is 8.

Question 27

The variance of the data 2, 4, 6, 8, 10 is 8. Find the variance of 4, 8, 12, 16, 20.

Solution: 1. Analyze the Relationship Between the Datasets:

• First dataset ($X$): {2, 4, 6, 8, 10}
• Second dataset ($Y$): {4, 8, 12, 16, 20} We can see that each observation in dataset Y is obtained by multiplying the corresponding observation in dataset X by 2. So, $y_i = 2x_i$.

2. Apply the Change of Scale Property for Variance: If each observation in a dataset is multiplied by a constant $k$, the new variance is $k^2$ times the old variance. New Variance = $k^2 \times$ Old Variance.

3. Calculate the New Variance:

• Old Variance = 8
• Constant, $k = 2$

New Variance = $2^2 \times 8 = 4 \times 8 = 32$.

Answer: The variance of the new dataset is 32.

Question 28

For a frequency distribution, the mean is 45. If the assumed mean $a=30$, the class size is 10, and $\sum f_i y_i = 150$, find the total number of observations N.

Solution: This problem involves working backwards with the step-deviation formula for the mean.

1. Recall the Step-Deviation Formula for Mean: $$\bar{x} = a + \left( \frac{\sum f_i y_i}{N} \right) \times h$$ where:

• $\bar{x}$ = Actual Mean
• $a$ = Assumed Mean
• $h$ = Class size
• $N$ = Total number of observations ($\sum f_i$)
• $y_i = \frac{x_i - a}{h}$

2. Substitute the Given Values:

• $\bar{x} = 45$
• $a = 30$
• $h = 10$
• $\sum f_i y_i = 150$

$45 = 30 + \left( \frac{150}{N} \right) \times 10$

3. Solve for N: First, subtract 30 from both sides: $45 - 30 = \frac{1500}{N}$ $15 = \frac{1500}{N}$ Now, solve for N: $N = \frac{1500}{15} = 100$.

Answer: The total number of observations (N) is 100.

Question 29

Which is more consistent? Team A which scored a mean of 200 runs with S.D. 10, or Team B which scored a mean of 150 runs with S.D. 8?

Solution: 1. The Need for a Relative Measure: To compare the consistency of two datasets with different means, we cannot simply compare their standard deviations. We must use a relative measure of dispersion, the Coefficient of Variation (C.V.).

2. The Rule of Consistency: A lower C.V. indicates higher consistency (less variability relative to the mean).

3. Calculate C.V. for Team A: $$\text{C.V.}(A) = \frac{\sigma_A}{\bar{x}_A} \times 100$$ C.V.(A) = $\frac{10}{200} \times 100 = 0.05 \times 100 = 5\%$.

4. Calculate C.V. for Team B: $$\text{C.V.}(B) = \frac{\sigma_B}{\bar{x}_B} \times 100$$ C.V.(B) = $\frac{8}{150} \times 100 \approx 0.0533 \times 100 = 5.33\%$.

5. Compare and Conclude: Since C.V.(A) (5%) is less than C.V.(B) (5.33%), Team A has a smaller relative spread in its scores.

Answer: Team A is more consistent.

Question 30

Find the mean deviation from the median for the data: 15, 25, 18, 12, 13, 28, 30, 14, 20, 22.

Solution: Step 1: Sort the Data. To find the median, we must first arrange the data in ascending order. Sorted Data: 12, 13, 14, 15, 18, 20, 22, 25, 28, 30.

Step 2: Find the Median (M). There are $n=10$ observations (an even number). The median is the average of the two middle terms, which are the $(\frac{n}{2})^{th}$ and $(\frac{n}{2}+1)^{th}$ terms.

• $5^{th}$ term = 18
• $6^{th}$ term = 20 Median (M) = $\frac{18+20}{2} = 19$.

Step 3: Calculate the Absolute Deviations from the Median (19).

• $|12-19|=7$
• $|13-19|=6$
• $|14-19|=5$
• $|15-19|=4$
• $|18-19|=1$
• $|20-19|=1$
• $|22-19|=3$
• $|25-19|=6$
• $|28-19|=9$
• $|30-19|=11$

Step 4: Sum the Absolute Deviations. Sum = $7+6+5+4+1+1+3+6+9+11 = 53$.

Step 5: Calculate the Mean Deviation. M.D.(M) = $\frac{\text{Sum of Deviations}}{n} = \frac{53}{10} = 5.3$.

Answer: The mean deviation from the median is 5.3.