CBSE Class 11 Mathematics Statistics Worksheet Set 01

Multiple Choice Questions

Question. Following are the marks obtained by 9 students in a mathematics test 50, 69, 20, 33, 53, 39, 40, 65, 59
The mean deviation from the median is
(a) 9
(b) 10.5
(c)12.67
(d) 14.76
Answer : C

Question. The mean deviation from the mean of the set of observations −1, 0 and 4 is
(a) 3
(b) 1
(c) −2
(d) 2
Answer : D

Question. Following are the marks obtained by 9 students in a mathematics test 50, 69, 20, 33, 53, 39, 40, 65, 59
The mean deviation from the median is
(a) 9 (b)
10.5
(c) 12.67
(d) 14.76
Answer : C

Question. Consider the following data
36, 72, 46, 42, 60, 45, 53, 46, 51, 49
Then, the mean deviation about the median for the data is
(a) 6
(b) 8
(c) 7
(d) None of the above
Answer : C

Question. The mean deviation about the median for the data
32, 34, 36, 38, 40, 42, 44, 46, 48, 50 is
(a) 4.5
(b) 4
(c) 5.5
(d) 5
Answer : D

Question. The standard deviation of data 6, 5, 9, 13, 12, 8 and 10 is
(a) √52/7
(b) √52/7
(c) √6 (d) 6
Answer : A

Question. The mean deviation of the data 3, 10, 10, 4, 7, 10, 5 from the mean is
(a) 2
(b) 2.57
(c) 3
(d) 3.75
Answer : B

Question. Mean deviation about the median for the data 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17 is
(a) 2.44
(b) 2.33
(c) 1.44
(d) 1.33
Answer : B

Question. The mean deviation about the median for the data 34, 66, 30, 38, 44, 50, 40, 60, 42, 51 is
(a) 8.7
(b) 7.7
(c) 87
(d) 77
Answer : A

Question. The mean and the median of the following ten numbers in increasing order 10, 22, 26, 29, 34, x, 42, 67, 70, y are 42 and 35 respectively, then y/x is equal to:
(a) 9/4
(b) 7/2
(c) 8/3
(d) 7/3
Answer : D

Question. The mean of the data set comprising of 16 observations is 16. If one of the observation valued 16 is deleted and three new observations valued 3, 4 and 5 are added to the data, then the mean of the resultant data, is:
(a) 15.8
(b) 14.0
(c) 16.8
(d) 16.0
Answer : B

Question. If the median and the range of four numbers {x, y, 2x + y, x – y}, where 0 < y < x < 2y, are 10 and 28 respectively, then the mean of the numbers is :
(a) 18
(b) 10
(c) 5
(d) 14
Answer : D

Question. A factory is operating in two shifts, day and night, with 70 and 30 workers respectively. If per day mean wage of the day shift workers is ₹ 54 and per day mean wage of all the workers is ₹ 60, then per day mean wage of the night shift workers (in ₹) is : 
(a) 69
(b) 66
(c) 74
(d) 75
Answer : C

Question. The mean age of 25 teachers in a school is 40 years. A teacher retires at the age of 60 years and a new teacher is appointed in his place. If now the mean age of the teachers in this school is 39 years, then the age (in years) of the newly appointed teacher is :
(a) 25
(b) 30
(c) 35
(d) 40
Answer : C

Question. The average marks of boys in class is 52 and that of girls is 42. The average marks of boys and girls combined is 50.
The percentage of boys in the class is
(a) 80
(b) 60
(c) 40
(d) 20
Answer : D

Question. In a set of 2n distinct observations, each of the observations below the median of all the observations is increased by 5 and each of the remaining observations is decreased by 3. Then the mean of the new set of observations:
(a) increases by 1
(b) decreases by 1
(c) decreases by 2
(d) increases by 2
Answer : A

Question. The median of a set of 9 distinct observations is 20.5. If each of the largest 4 observations of the set is increased by 2, then the median of the new set 
(a) remains the same as that of the original set
(b) is increased by 2
(c) is decreased by 2
(d) is two times the original median.
Answer : A

Question. The mean of a data set consisting of 20 observations is 40. If one observation 53 was wrongly recorded as 33, then the correct mean will be : 
(a) 41
(b) 49
(c) 40.5
(d) 42.5
Answer : A

Question. The median of 100 observations grouped in classes of equal width is 25. If the median class interval is 20 - 30 and the number of observations less than 20 is 45, then the frequency of median class is
(a) 10
(b) 20
(c) 15
(d) 12
Answer : A

Question. In a class of 100 students there are 70 boys whose average marks in a sub ect are 75. If the average marks of the complete class is 72, then what is the average of the girls?
(a) 73
(b) 65
(c) 68
(d) 74
Answer : B

Question. Let x₁, x₂ , .............. xn be n observations such that ∑x_i² = 400 and ∑x_i = 80. Then the possible value of n among the following is 
(a) 15
(b) 18
(c) 9
(d) 12
Answer : B

Question. The mean and the standard deviation (s.d.) of 10 observations are 20 and 2 respectively. Each of these 10 observations is multiplied by p and then reduced by q, where p ≠ 0 and q ≠ 0. If the new mean and new s.d. become half of their original values, then q is equal to:
(a) –5
(b) 10
(c) –20
(d) –10
Answer : C

Question. If the standard deviation of the numbers –1, 0, 1, k is √5 where k > 0, then k is equal to:
(a) 2√6
(b) 2√10/2
(c) 4√5/3
(d) √6
Answer : A

Question. If in a frequency distribution, the mean and median are 21 and 22 respectively, then its mode is approximately 
(a) 22.0
(b) 20.5
(c) 25.5
(d) 24.0
Answer : D

Question. The mean and variance of 7 observations are 8 and 16, respectively. If five observations are 2, 4, 10, 12, 14, then the absolute difference of the remaining two observations is :
(a) 1
(b) 4
(c) 2
(d) 3
Answer : C

Assertion-Reasoning MCQs

Directions Each of these questions contains two statements Assertion (A) and Reason (R). Each of the questions has four alternative choices, any one of the which is the correct answer. You have to select one of the codes (a), (b), (c) and (d) given below.
(a) A is true, R is true; R is a correct explanation of A.
(b) A is true, R is true; R is not a correct explanation of A.
(c) A is true; R is false
(d) A is false; R is true.

Question. Assertion (A) In order to find the dispersion of values of x from mean x , we take absolute measure of dispersion.
Reason (R) Sum of the deviations from mean (x ) is zero.
Answer : A

Question. Assertion (A) The mean deviation about the mean for the data 4, 7, 8, 9, 10, 12, 13, 17 is 3.
Reason (R) The mean deviation about the mean for the data 38, 70, 48, 40, 42, 55, 63, 46, 54, 44 is 8.5.
Answer : C

Question. Assertion (A) The mean deviation about median calculated for series, where variability is very high, cannot be fully relied.
Reason (R) The median is not a representative of central tendency for the series where degree of variability is very high.
Answer : A

Question. Assertion (A) The mean deviation about the mean to find measure of dispersion has certain limitations.
Reason (R) The sum of deviations from the mean is more than the sum of deviations from median. Therefore, the mean deviation about the mean is not very scientific, where degree of variability is very high.
Answer : A

Question. Assertion (A) The average marks of boys in a class is 52 and that of girls is 42. The average marks of boys and girls combined is 50. The percentage of boys in the class is 80%.
Reason (R) Mean marks scored by the students of a class is 53. The mean marks of the girls is 55 and the mean marks of the boys is 50. The percentage of girls in the class is 64%.
Answer : C

Case Based MCQs

You are given the following grouped data.

x_i  2  5   6  8  10  12
f_i  2  8  10  7   8    5

Based on these data, answer the following questions.

Question. Mean of the grouped data is
(a) 7.0
(b) 7.5
(c) 8.0
(d) 8.5
Answer : B

Question. Mean deviation about the mean is
(a) 2.1
(b) 2.2
(c) 2.3
(d) 2.4
Answer : C

Question. The value of median is
(a) 5
(b) 6
(c) 7
(d) 8
Answer : C

Question. Mean deviation about the median is
(a) 1.9
(b) 2.0
(c) 2.2
(d) 2.3
Answer : D

Question. The difference between mean and median is
(a) 0.9
(b) 0.7
(c) 0.5
(d) 0.3
Answer : C

Question 1. Coefficients of variation of 2 distributions are 70% & 75%, & S.D. are 28 & 27 respectively. Find their arithmetic means?
Answer: Let the coefficient of variation, standard deviation, and arithmetic mean of the first distribution be \(C_1\), \(\sigma_1\), and \(\bar{x}_1\) respectively.
Let the coefficient of variation, standard deviation, and arithmetic mean of the second distribution be \(C_2\), \(\sigma_2\), and \(\bar{x}_2\) respectively.
Formula for Coefficient of Variation (C.V.) is given by:
\(\text{C.V.} = \frac{\text{Standard Deviation } (\sigma)}{\text{Arithmetic Mean } (\bar{x})} \times 100\)
For the first distribution:
\(70 = \frac{28}{\bar{x}_1} \times 100\)
\(\bar{x}_1 = \frac{28 \times 100}{70} = \frac{2800}{70} = 40\)
For the second distribution:
\(75 = \frac{27}{\bar{x}_2} \times 100\)
\(\bar{x}_2 = \frac{27 \times 100}{75} = \frac{2700}{75} = 36\)
Thus, the arithmetic means of the two distributions are 40 and 36 respectively.
In simple words: Using the coefficient of variation formula, we divided standard deviation by the percentage C.V. and multiplied by 100 to find both means.
Exam Tip: Remember that C.V. is expressed in percentage, so always divide by 100 or adjust the formula accordingly as \(\text{C.V.} = \frac{\sigma}{\bar{x}} \times 100\).

Question 2. For a group of 200 candidate the mean and S.D. were found to be 40 and 15 respectively. Later on it was found that the score 43 was misread as 34. Find the correct mean and correct S.D.
Answer: Total number of candidates \(n = 200\)
Incorrect mean \(\bar{x} = 40\)
Incorrect standard deviation \(\sigma = 15\)
Incorrect sum of observations \(\sum x = n \times \bar{x} = 200 \times 40 = 8000\)
Correct sum of observations \(\sum x_{\text{correct}} = \sum x - \text{incorrect value} + \text{correct value}\)
\(\sum x_{\text{correct}} = 8000 - 34 + 43 = 8009\)
Correct mean \(\bar{x}_{\text{correct}} = \frac{8009}{200} = 40.045\)
Incorrect variance \(\sigma^2 = 15^2 = 225\)
We know variance formula: \(\sigma^2 = \frac{\sum x^2}{n} - (\bar{x})^2\)
Incorrect sum of squares \(\sum x^2 = n \times (\sigma^2 + (\bar{x})^2) = 200 \times (225 + 40^2) = 200 \times (225 + 1600) = 200 \times 1825 = 365000\)
Correct sum of squares \(\sum x_{\text{correct}}^2 = \sum x^2 - (\text{incorrect value})^2 + (\text{correct value})^2\)
\(\sum x_{\text{correct}}^2 = 365000 - (34)^2 + (43)^2 = 365000 - 1156 + 1849 = 365693\)
Correct variance \(\sigma_{\text{correct}}^2 = \frac{\sum x_{\text{correct}}^2}{n} - (\bar{x}_{\text{correct}})^2 = \frac{365693}{200} - (40.045)^2\)
\(\frac{365693}{200} = 1828.465\)
\((40.045)^2 = 1603.602025\)
Correct variance \(\sigma_{\text{correct}}^2 = 1828.465 - 1603.602025 = 224.862975\)
Correct standard deviation \(\sigma_{\text{correct}} = \sqrt{224.862975} \approx 14.995\)
In simple words: We subtracted the wrong value and added the correct value to the total sum and sum of squares, then recalculated the mean and standard deviation.
Exam Tip: Be very careful with arithmetic when squaring numbers like 34 and 43 for the sum of squares formula.

Question 3. Find the mean deviation about the median of the following frequency distribution:-
Class 0-6 6-12 12-18 18-24 24-30
Frequency 8 10 12 9 5
Answer:

Here \(N = 44\), so \(\frac{N}{2} = 22\).
The cumulative frequency just greater than 22 is 30, corresponding to the class interval 12-18.
Median class = 12-18, \(l = 12\), \(F = 18\), \(f = 12\), \(h = 6\)
Median \(M = l + \frac{\frac{N}{2} - F}{f} \times h = 12 + \frac{22 - 18}{12} \times 6 = 12 + \frac{4}{12} \times 6 = 12 + 2 = 14\)
Mean Deviation about Median \((\text{M.D.}_M) = \frac{\sum f_i |x_i - M|}{N} = \frac{276}{44} = 6.27\)
In simple words: We found the median class, calculated the exact median value, found absolute differences from the median, and applied the mean deviation formula.
Exam Tip: Always find the median class first using \(N/2\) before applying the continuous median formula.

Question 4. The following table gives the distribution of income of 100 families in a village. Calculate mean and variance.
Income (in Rs.) 0-1000 1000-2000 2000-3000 3000-4000 4000-5000 5000-6000
No. of families 18 26 30 12 10 4
Answer: Let \(A = 2500\) and \(h = 1000\). Let \(u_i = \frac{x_i - A}{h}\).

Mean \(\bar{x} = A + \left(\frac{\sum f_i u_i}{N}\right) \times h = 2500 + \left(\frac{-18}{100}\right) \times 1000 = 2500 - 180 = \text{Rs. } 2320\)
Variance \(\sigma^2 = h^2 \left[ \frac{\sum f_i u_i^2}{N} - \left(\frac{\sum f_i u_i}{N}\right)^2 \right]\)
\(\sigma^2 = (1000)^2 \left[ \frac{186}{100} - \left(\frac{-18}{100}\right)^2 \right] = 1000000 \left[ 1.86 - 0.0324 \right]\)
\(\sigma^2 = 1000000 \times 1.8276 = 1827600\)
In simple words: We used step-deviation method with scaled variable \(u_i\) to simplify calculations for mean and variance.
Exam Tip: Step-deviation method \(u_i = \frac{x_i - A}{h}\) avoids large numbers and calculation errors.

Question 5. Find the mean deviation from the mean and from median of the following distribution:-
Marks 0-10 10-20 20-30 30-40 40-50
No. of students 5 8 15 16 6
Answer: Let \(A = 25\) and \(h = 10\).

Here \(N = 50\), \(\sum f_i u_i = 10\).
Mean \(\bar{x} = 25 + \left(\frac{10}{50}\right) \times 10 = 25 + 2 = 27\)
Mean Deviation from Mean \(= \frac{\sum f_i |x_i - \bar{x}|}{N} = \frac{472}{50} = 9.44\)
For Median: \(\frac{N}{2} = 25\), which lies in class 20-30.
Median \(M = 20 + \frac{25 - 13}{15} \times 10 = 20 + \frac{12 \times 10}{15} = 20 + 8 = 28\)
Mean Deviation from Median \(= \frac{\sum f_i |x_i - M|}{N} = \frac{479.98}{50} \approx 9.6\)
In simple words: We calculated both the mean and the median, found the absolute deviations of midpoints from each, and divided the sum of products by total frequency.
Exam Tip: Construct separate columns for mean deviation about mean and median to prevent confusion.

Question 6. Calculate the mean, median and standard deviation of the following distribution:-
C.I. 31-35 36-40 41-45 46-50 51-55 56-60 61-65 66-70
F 2 3 8 12 16 5 2 3
Answer: First, convert the inclusive classes to exclusive classes by subtracting 0.5 from lower limits and adding 0.5 to upper limits:
\(30.5-35.5, 35.5-40.5, 40.5-45.5, 45.5-50.5, 50.5-55.5, 55.5-60.5, 60.5-65.5, 65.5-70.5\).
Let \(A = 53\), \(h = 5\), and \(u_i = \frac{x_i - 53}{5}\).

1. Mean \(\bar{x} = A + \left(\frac{\sum f_i u_i}{N}\right) \times h = 53 + \left(\frac{-27}{51}\right) \times 5 = 53 - 2.65 = 50.35\)
2. Median: \(\frac{N}{2} = \frac{51}{2} = 25.5\), which lies in exclusive class 45.5-50.5.
Median \(M = 45.5 + \frac{25.5 - 25}{12} \times 5 = 45.5 + \frac{0.5 \times 5}{12} = 45.5 + 0.208 = 45.71\)
3. Standard Deviation \(\sigma = h \sqrt{\frac{\sum f_i u_i^2}{N} - \left(\frac{\sum f_i u_i}{N}\right)^2}\)
\(\sigma = 5 \sqrt{\frac{143}{51} - \left(\frac{-27}{51}\right)^2} = 5 \sqrt{2.8039 - 0.2801} = 5 \sqrt{2.5238} = 5 \times 1.5886 = 7.94\)
In simple words: We converted inclusive class intervals to exclusive ones, then calculated mean, median, and standard deviation using step-deviation.
Exam Tip: Always check if class intervals are inclusive (like 31-35) and convert them to continuous exclusive intervals (30.5-35.5) before finding the median.

Question 7. An analysis of monthly wages paid to workers of two firms A & B belonging to the same industry gives following result:
Firm A: No. of workers = 1000, Average monthly wages = Rs. 2800, Variance = 100
Firm B: No. of workers = 1200, Average monthly wages = Rs. 2800, Variance = 169
In which firm, A or B is there greater variability in individual wages?
Answer: For Firm A:
Mean \(\bar{x}_A = 2800\)
Variance \(\sigma_A^2 = 100 \implies \text{S.D. } (\sigma_A) = \sqrt{100} = 10\)
Coefficient of Variation \((\text{C.V.})_A = \frac{\sigma_A}{\bar{x}_A} \times 100 = \frac{10}{2800} \times 100 = \frac{10}{28} = 0.357\%\)
For Firm B:
Mean \(\bar{x}_B = 2800\)
Variance \(\sigma_B^2 = 169 \implies \text{S.D. } (\sigma_B) = \sqrt{169} = 13\)
Coefficient of Variation \((\text{C.V.})_B = \frac{\sigma_B}{\bar{x}_B} \times 100 = \frac{13}{2800} \times 100 = \frac{13}{28} = 0.464\%\)
Since \((\text{C.V.})_B > (\text{C.V.})_A\), Firm B has greater variability in individual wages.
In simple words: Greater coefficient of variation means greater variability. Firm B has a higher C.V., so its wages vary more.
Exam Tip: To compare variability of two distributions, always calculate and compare their Coefficients of Variation (C.V.).

Question 8. The mean of 5 observations is 4.4 & their variance is 8.24. If three of observations are 1, 2 & 6, find other two observations.
Answer: Let the five observations be \(1, 2, 6, x, y\).
Given mean \(\bar{x} = 4.4\) and \(n = 5\).
Sum of 5 observations \(= 5 \times 4.4 = 22\)
\(1 + 2 + 6 + x + y = 22\)
\(9 + x + y = 22 \implies x + y = 13 \implies y = 13 - x\) --- (1)
Given variance \(\sigma^2 = 8.24\)
We know \(\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2\)
\(8.24 = \frac{1^2 + 2^2 + 6^2 + x^2 + y^2}{5} - (4.4)^2\)
\(8.24 = \frac{1 + 4 + 36 + x^2 + y^2}{5} - 19.36\)
\(8.24 + 19.36 = \frac{41 + x^2 + y^2}{5}\)
\(27.6 = \frac{41 + x^2 + y^2}{5}\)
\(41 + x^2 + y^2 = 27.6 \times 5 = 138\)
\(x^2 + y^2 = 138 - 41 = 97\) --- (2)
Substitute \(y = 13 - x\) into equation (2):
\(x^2 + (13 - x)^2 = 97\)
\(x^2 + 169 - 26x + x^2 = 97\)
\(2x^2 - 26x + 169 - 97 = 0\)
\(2x^2 - 26x + 72 = 0\)
Divide by 2:
\(x^2 - 13x + 36 = 0\)
\((x - 4)(x - 9) = 0\)
So, \(x = 4\) or \(x = 9\).
If \(x = 4\), then \(y = 13 - 4 = 9\).
Thus, the other two observations are 4 and 9.
In simple words: We used the given mean and variance to form two simultaneous equations in terms of the two unknown numbers, then solved a quadratic equation.
Exam Tip: Utilize both the mean equation (for sum) and the variance equation (for sum of squares) to set up a system of equations for two missing observations.

Question 9. From the data given below state which group is more variable G1 or G2?
Marks 10-20 20-30 30-40 40-50 50-60 60-70 70-80
Group G1 9 17 32 33 40 10 9
Group G2 10 20 30 25 43 15 7
Answer: To determine which group is more variable, we need to calculate the Coefficient of Variation (C.V.) for both groups G1 and G2. Whichever group has a higher C.V. is more variable.
Let \(A = 45\), \(h = 10\), and \(u_i = \frac{x_i - 45}{10}\).

For Group G1:
Mean \(\bar{x}_1 = 45 + \left(\frac{-6}{150}\right) \times 10 = 45 - 0.4 = 44.6\)
Variance \(\sigma_1^2 = (10)^2 \left[ \frac{342}{150} - \left(\frac{-6}{150}\right)^2 \right] = 100 \left[ 2.28 - 0.0016 \right] = 100 \times 2.2784 = 227.84\)
S.D. \(\sigma_1 = \sqrt{227.84} = 15.09\)
\((\text{C.V.})_1 = \frac{15.09}{44.6} \times 100 = 33.83\%\)
For Group G2:
Mean \(\bar{x}_2 = 45 + \left(\frac{-6}{150}\right) \times 10 = 45 - 0.4 = 44.6\)
Variance \(\sigma_2^2 = (10)^2 \left[ \frac{366}{150} - \left(\frac{-6}{150}\right)^2 \right] = 100 \left[ 2.44 - 0.0016 \right] = 100 \times 2.4384 = 243.84\)
S.D. \(\sigma_2 = \sqrt{243.84} = 15.61\)
\((\text{C.V.})_2 = \frac{15.61}{44.6} \times 100 = 35.0\%\)
Since \((\text{C.V.})_2 > (\text{C.V.})_1\), Group G2 is more variable.
In simple words: We computed the coefficient of variation for both groups, and found that Group G2 has a higher C.V.
Exam Tip: Group with greater C.V. is always more variable, consistent, or unstable depending on the context.

Question 10. Find the mean deviation about the median for the following data:-
\(x_i\): 10 15 20 25 30 35 40 45
\(f_i\): 7 3 8 5 6 8 4 9
Answer:

Here \(N = 50\), which is even.
Median \(M = \frac{\frac{N}{2}\text{-th observation} + \left(\frac{N}{2} + 1\right)\text{-th observation}}{2}\)
\(\frac{N}{2} = 25\text{-th observation}\), and \(\left(\frac{N}{2} + 1\right) = 26\text{-th observation}\).
From cumulative frequency table, both 25th and 26th observations correspond to \(x_i = 30\) and \(x_i = 25\) respectively? Wait, let's look at cumulative frequencies:
7, 10, 18, 23, 29 (24th to 29th observations are 30).
Thus, 25th observation is 30 and 26th observation is 30.
Median \(M = \frac{25 + 30}{2} = 27.5\) (or using discrete series CF, median lies between 25 and 30, exact median = 27.5).
Mean Deviation about Median \(= \frac{\sum f_i |x_i - M|}{N} = \frac{515}{50} = 10.3\)
In simple words: For discrete frequency distribution, we found the median from cumulative frequency, calculated absolute deviations, and divided by total frequency.
Exam Tip: For discrete data, when \(N\) is even, the median is the mean of the \((N/2)\)-th and \((N/2 + 1)\)-th observations.

Question 11. Find the mean deviation about the median for the data given:-
22, 24, 30, 27, 29, 31, 25, 28, 41, 42
Answer: First, arrange the given raw data in ascending order:
22, 24, 25, 27, 28, 29, 30, 31, 41, 42
Number of observations \(n = 10\) (even).
Median \(M = \frac{\left(\frac{n}{2}\right)\text{-th term} + \left(\frac{n}{2} + 1\right)\text{-th term}}{2} = \frac{5\text{-th term} + 6\text{-th term}}{2}\)
5th term = 28, 6th term = 29
Median \(M = \frac{28 + 29}{2} = 28.5\)
Now find absolute deviations \(|x_i - M|\):
\(|22 - 28.5| = 6.5\)
\(|24 - 28.5| = 4.5\)
\(|25 - 28.5| = 3.5\)
\(|27 - 28.5| = 1.5\)
\(|28 - 28.5| = 0.5\)
\(|29 - 28.5| = 0.5\)
\(|30 - 28.5| = 1.5\)
\(|31 - 28.5| = 2.5\)
\(|41 - 28.5| = 12.5\)
\(|42 - 28.5| = 13.5\)
Sum of absolute deviations \(\sum |x_i - M| = 6.5 + 4.5 + 3.5 + 1.5 + 0.5 + 0.5 + 1.5 + 2.5 + 12.5 + 13.5 = 47.0\)
Mean Deviation about Median \(= \frac{\sum |x_i - M|}{n} = \frac{47}{10} = 4.7\)
In simple words: We sorted the numbers, found the median of the 10 items, calculated deviations from the median, and averaged them.
Exam Tip: Always sort raw data in ascending order first before attempting to find the median.

Question 12. Calculate the mean, variance and standard deviation for the following distribution :-
Marks above 30 & below 40 below 50 below 60 below 70 below 80 below 90 below 100
No. of Students 3 10 22 37 45 48 50
Answer: Let us convert the cumulative frequency distribution into a standard frequency distribution with class intervals and frequencies \(f_i\):

Let \(A = 65\) and \(h = 10\).
1. Mean \(\bar{x} = A + \left(\frac{\sum f_i u_i}{N}\right) \times h = 65 + \left(\frac{-15}{50}\right) \times 10 = 65 - 3 = 62\)
2. Variance \(\sigma^2 = h^2 \left[ \frac{\sum f_i u_i^2}{N} - \left(\frac{\sum f_i u_i}{N}\right)^2 \right]\)
\(\sigma^2 = (10)^2 \left[ \frac{105}{50} - \left(\frac{-15}{50}\right)^2 \right] = 100 \left[ 2.1 - (-0.3)^2 \right] = 100 \left[ 2.1 - 0.09 \right]\)
\(\sigma^2 = 100 \times 2.01 = 201\)
3. Standard Deviation \(\sigma = \sqrt{201} \approx 14.18\)
In simple words: We subtracted consecutive cumulative frequencies to obtain class frequencies, then applied the step-deviation method to find mean, variance, and standard deviation.
Exam Tip: For 'below' cumulative frequency data, subtract each cumulative frequency from the next to obtain individual class frequencies \(f_i\).