Question Bank — Statistics

Section A — MCQs (10 questions, 1 mark each)

1. The formula for mean using the direct method for grouped data is:
   (a) \(\bar{x} = \frac{\sum f_i x_i}{\sum f_i}\)
   (b) \(\bar{x} = a + \frac{\sum f_i d_i}{\sum f_i}\)
   (c) \(\bar{x} = a + h \left( \frac{\sum f_i u_i}{\sum f_i} \right)\)
   (d) None of these

2. In the assumed mean method, ‘a’ is chosen as:
   (a) The lowest class mark
   (b) The highest class mark
   (c) A class mark preferably in the middle of the distribution
   (d) Any arbitrary number

3. The modal class in a grouped frequency distribution is the class with:
   (a) Highest frequency
   (b) Lowest frequency
   (c) Cumulative frequency equal to \(n/2\)
   (d) Class size equal to h

4. The formula for mode of grouped data is:
   (a) Mode = \(l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) h\)
   (b) Mode = \(l + \left( \frac{f_1 - f_0}{f_1 - f_0 - f_2} \right) h\)
   (c) Mode = \(l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times 2h\)
   (d) None of these

5. For finding the median of grouped data, the median class is the class whose cumulative frequency is:
   (a) Exactly equal to \(n/2\)
   (b) Just greater than and nearest to \(n/2\)
   (c) Equal to the total frequency
   (d) Less than \(n/2\)

6. The step-deviation method is preferred when:
   (a) Class sizes are unequal
   (b) \(x_i\) and \(f_i\) are small
   (c) All \(d_i\) have a common factor and class sizes are equal
   (d) Data is ungrouped

7. If the mean obtained by direct, assumed mean and step-deviation methods on the same data is different, then:
   (a) Direct method is wrong
   (b) Data has error
   (c) The three values must be the same
   (d) Step-deviation method cannot be used

Assertion-Reason Questions

8. Assertion (A): The assumed mean method and step-deviation method are simplified forms of the direct method.
   Reason (R): Both methods ultimately use the relation \(\bar{x} = a + \frac{\sum f_i d_i}{\sum f_i}\) or its equivalent form.
   (a) Both A and R are true and R is the correct explanation of A.
   (b) Both A and R are true but R is not the correct explanation of A.
   (c) A is true but R is false.
   (d) A is false but R is true.

9. Assertion (A): The mode of grouped data always lies in the modal class.
   Reason (R): The modal class is the class with the maximum frequency and the mode is calculated using the lower limit and frequencies of this class and its adjacent classes.
   (a) Both A and R are true and R is the correct explanation of A.
   (b) Both A and R are true but R is not the correct explanation of A.
   (c) A is true but R is false.
   (d) A is false but R is true.

10. Which of the following is not true for the three methods of finding mean?
    (a) All three give the same value when applied correctly.
    (b) Step-deviation method requires equal class sizes.
    (c) Direct method is always the most accurate.
    (d) Choice of method depends on the numerical values of \(x_i\) and \(f_i\).

Section B — Very Short Answer (6 questions, 2 marks each)

1. State the formula for finding the mean by the step-deviation method. What does \(h\) represent?

2. Write the formula for the mode of grouped data. Identify each symbol used.

3. What is meant by the median class in grouped data? How is it located?

4. In the assumed mean method, if \(a = 47.5\) and \(\frac{\sum f_i d_i}{\sum f_i} = 14.5\), find the mean.

5. Why is the class mark used as the representative value (\(x_i\)) for each class interval?

6. If the cumulative frequency just greater than \(n/2\) belongs to the class 60–70, state the values of \(l\) and \(cf\) to be used in the median formula (assume standard symbols).

Section C — Short Answer (5 questions, 3 marks each)

1. The following table gives the distribution of students of two sections according to the marks obtained. Find the mean marks of section A using the direct method.

2. Using the assumed mean method with \(a = 25\), find the mean of the following data:

3. Find the mode of the following distribution:

4. The following data gives the distribution of total monthly household expenditure of 200 families. Find the modal monthly expenditure (use the mode formula).

5. For the data given below, find the median:

Section D — Long Answer (3 questions, 5 marks each)

1. The marks obtained by 30 students of Class X in a Mathematics test (out of 50) are given below. Form a grouped frequency distribution table with class intervals of width 10 and find the mean marks using (i) direct method, (ii) assumed mean method (take \(a = 25\)) and (iii) step-deviation method. Verify that all three methods give the same result.

Marks: 12, 18, 22, 25, 27, 31, 32, 35, 38, 40, 42, 45, 48, 20, 15, 33, 36, 39, 41, 44, 47, 19, 23, 28, 30, 34, 37, 43, 46, 29.

2. The following table shows the daily wages of 50 workers in a factory. Find the mean daily wage using a suitable method. Also find the modal daily wage.

3. The median of the following distribution is 28.5. Find the values of \(x\) and \(y\) if the total frequency is 60.

Section E — Case/Source-Based (2 questions, 4 marks each)

Case 1: A survey was conducted among 40 households in a locality to find the number of plants they have. The data is summarised below:

Sub-questions:
(a) Identify the modal class and write the values of \(l, f_1, f_0, f_2, h\).
(b) Calculate the mode number of plants per house.
(c) Using the assumed mean method (\(a = 5\)), find the mean number of plants.
(d) Compare the mode and mean values and interpret which measure is more appropriate here.

Case 2: The heights (in cm) of 51 girls of Class X were recorded and presented as cumulative frequency distribution of “less than” type:

Sub-questions:
(a) Convert the data into a grouped frequency distribution table with class intervals of width 5 cm.
(b) Locate the median class and write the values of \(l, cf, f, h\).
(c) Calculate the median height.
(d) Interpret what the median value tells about the heights of the girls.