Question 1

In a survey, the number of family members in households is recorded as 1-3, 3-5, 5-7, 7-9, 9-11 with frequencies 7, 8, 2, 2, 1 respectively. What is the mode of this data?

Accepted Answer

Correct Option: A. Solution: The modal class is the one with the highest frequency. Here, the class 3-5 has the highest frequency of 8, so it is the modal class.

Question 2

What is the formula for finding the mode of grouped data?

Accepted Answer

Correct Option: B. Solution: The formula for finding the mode of grouped data is Mode = $l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} ight) 	imes h$, where $l$ is the lower limit of the modal class, $h$ is the class size, $f_1$ is the frequency of the modal class, $f_0$ is the frequency of the class preceding the modal class, and $f_2$ is the frequency of the class succeeding the modal class.

Question 3

Given a grouped frequency distribution with class intervals of equal width, the median is calculated using the formula: $$	ext{Median} = l + \left( \frac{\frac{n}{2} - cf}{f} ight) 	imes h$$. If the lower boundary of the median class is $l = 40$, the cumulative frequency of the class preceding the median class is $cf = 20$, the frequency of the median class is $f = 10$, and the class width is $h = 5$, what is the median?

Accepted Answer

Correct Option: A. Solution: Using the formula, $$	ext{Median} = 40 + \left( \frac{25 - 20}{10} ight) 	imes 5 = 40 + 2.5 = 42.5$$.

Question 4

What is the primary purpose of an ogive in statistics?

Accepted Answer

Correct Option: B. Solution: An ogive is a graphical representation of cumulative frequency used to estimate the median visually and understand distribution trends.

Question 5

If a grouped frequency distribution has the following class intervals: 10-20, 20-30, 30-40, with frequencies 5, 15, and 10 respectively, and the median class is 20-30, what is the cumulative frequency of the class preceding the median class?

Accepted Answer

Correct Option: A. Solution: The cumulative frequency of the class preceding the median class (10-20) is 5.

Question 6

Which method is most suitable for calculating the mean when the class sizes are unequal and $x_i$ values are large?

Accepted Answer

Correct Option: C. Solution: The step-deviation method is suitable when class sizes are unequal and $x_i$ values are large because it simplifies calculations.

Question 7

For a grouped data, the class intervals are 50-60, 60-70, 70-80 with frequencies 5, 8, and 12 respectively. If the median class is determined to be 60-70, what is the class width?

Accepted Answer

Correct Option: B. Solution: The class width is the difference between the upper and lower boundaries of any class interval, which is $70 - 60 = 10$.

Question 8

Given a frequency distribution with class intervals of equal width, if the modal class is 20-30 with a frequency of 15, the preceding class has a frequency of 10, and the succeeding class has a frequency of 5, what is the mode?

Accepted Answer

Correct Option: A. Solution: Using the mode formula: Mode = $l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} ight) 	imes h$. Here, $l = 20$, $h = 10$, $f_1 = 15$, $f_0 = 10$, $f_2 = 5$. Mode = $20 + \left( \frac{15 - 10}{2 	imes 15 - 10 - 5} ight) 	imes 10 = 21.67$.

Question 9

What is the cumulative frequency of the class interval 40-50 in a less than type cumulative frequency distribution if the frequencies of the classes 30-40 and 40-50 are 3 and 3 respectively?

Accepted Answer

Correct Option: C. Solution: The cumulative frequency of the class interval 40-50 is the sum of the frequencies of all preceding classes including itself, which is 3 (for 30-40) + 3 (for 40-50) = 6.

Question 10

If the cumulative frequency of a class is 45 and the cumulative frequency of the preceding class is 30, what is the frequency of the class?

Accepted Answer

Correct Option: A. Solution: The frequency of the class is the difference between the cumulative frequency of the class and the cumulative frequency of the preceding class, i.e., $45 - 30 = 15$.

Question 11

A grouped frequency distribution of student scores is given with class intervals of 50-60, 60-70, 70-80, 80-90, and 90-100. If the frequencies are 5, 10, 15, 10, and 5 respectively, what is the mean score using the assumed mean method with an assumed mean of 75?

Accepted Answer

Correct Option: B. Solution: Using the assumed mean method, deviations $d_i$ are calculated from 75. The mean is $$75 + \frac{5(-25) + 10(-15) + 15(0) + 10(15) + 5(25)}{5 + 10 + 15 + 10 + 5} = 75$$.

Question 12

For a dataset of exam scores with class intervals 30-40, 40-50, 50-60, 60-70, and 70-80, the frequencies are 6, 9, 15, 10, and 5 respectively. Calculate the mean score using the direct method.

Accepted Answer

Correct Option: C. Solution: The class marks are 35, 45, 55, 65, and 75. The mean is calculated as $$\frac{6 	imes 35 + 9 	imes 45 + 15 	imes 55 + 10 	imes 65 + 5 	imes 75}{6 + 9 + 15 + 10 + 5} = 50$$.

Question 13

In a grouped frequency distribution, the class intervals are 10-20, 20-30, 30-40, 40-50, and 50-60. The frequencies are 5, 10, 15, 10, and 5 respectively. What is the median of this distribution?

Accepted Answer

Correct Option: A. Solution: The median class is determined by finding the cumulative frequency that exceeds half of the total frequency. Here, the cumulative frequency just exceeding 22.5 (half of 45) is 30, which lies in the 30-40 class interval. Thus, the median is calculated as $$30 + \left(\frac{22.5 - 15}{15}ight) 	imes 10 = 35$$.

Question 14

Which method is most suitable for calculating the mean when the numerical values of $x_i$ and $f_i$ are large?

Accepted Answer

Correct Option: B. Solution: The Assumed Mean Method is preferred when $x_i$ and $f_i$ are large because it simplifies calculations by reducing the size of numbers involved.

Question 15

Given a cumulative frequency distribution of less than type for a dataset, how would you determine the median class?

Accepted Answer

Correct Option: B. Solution: The median class is determined by finding the class interval where the cumulative frequency just exceeds half of the total number of observations.

Question 16

Why might the mean calculated using the direct method differ from that calculated using the assumed mean method?

Accepted Answer

Correct Option: B. Solution: The difference arises because the assumed mean method uses mid-point assumptions, which can lead to approximate results.

Question 17

A group of students were surveyed about the number of hours they spent on extracurricular activities per week. The results are grouped into intervals: 0-5, 5-10, 10-15, 15-20, and 20-25 with frequencies 8, 12, 20, 15, and 5 respectively. Calculate the mean number of hours using the assumed mean method with an assumed mean of 12.5.

Accepted Answer

Correct Option: B. Solution: With class marks 2.5, 7.5, 12.5, 17.5, and 22.5, the deviations $d_i$ from 12.5 are calculated. The mean is $$12.5 + \frac{8(-10) + 12(-5) + 20(0) + 15(5) + 5(10)}{8 + 12 + 20 + 15 + 5} = 12.5$$ hours.

Question 18

How is the median of grouped data affected by the class size?

Accepted Answer

Correct Option: D. Solution: The class size ($h$) is a component in the formula for calculating the median of grouped data, affecting the calculation of the median value.

Question 19

What is the formula to calculate the mean of grouped data using the direct method?

Accepted Answer

Correct Option: A. Solution: The direct method for calculating the mean of grouped data is given by $$x = \frac{\Sigma f_i x_i}{\Sigma f_i}$$.

Question 20

What is the empirical relationship between the mean, median, and mode?

Accepted Answer

Correct Option: B. Solution: The empirical relationship is given by the formula: $3 	imes 	ext{Median} = 	ext{Mode} + 2 	imes 	ext{Mean}$.

Question 21

If the mode of a grouped data set is found to be less than the mean, which of the following is a possible interpretation?

Accepted Answer

Correct Option: A. Solution: If the mode is less than the mean, it indicates a positively skewed distribution, where the tail on the right side of the distribution is longer or fatter.

Question 22

Given the following class intervals and frequencies for the number of hours students study per week: 10-20 (5 students), 20-30 (8 students), 30-40 (12 students), 40-50 (10 students), and 50-60 (5 students). Calculate the mean study hours per week using the direct method.

Accepted Answer

Correct Option: A. Solution: Using the direct method, the class marks are 15, 25, 35, 45, and 55. The mean is calculated as $$\frac{5 	imes 15 + 8 	imes 25 + 12 	imes 35 + 10 	imes 45 + 5 	imes 55}{5 + 8 + 12 + 10 + 5} = 35$$ hours.

Question 23

In a grouped frequency distribution, which class is considered the modal class?

Accepted Answer

Correct Option: B. Solution: The modal class in a grouped frequency distribution is the class with the largest frequency.

Question 24

Given a dataset where the mean is 50 and the mode is 40, what is the median according to the empirical relationship between mean, median, and mode?

Accepted Answer

Correct Option: A. Solution: Using the empirical relationship $3 	imes 	ext{Median} = 	ext{Mode} + 2 	imes 	ext{Mean}$, we substitute the given values: $3 	imes 	ext{Median} = 40 + 2 	imes 50$. Solving for the median gives $3 	imes 	ext{Median} = 140$, so the median is $\frac{140}{3} = 46.67$. The closest option is 45.

Question 25

In a dataset, the class intervals are 0-10, 10-20, 20-30 with frequencies 8, 12, and 10 respectively. If the median class is 10-20, what is the lower boundary of the median class?

Accepted Answer

Correct Option: B. Solution: The lower boundary of the median class 10-20 is 10.

Question 26

In a dataset with large values and equal class intervals, which method would be most efficient for calculating the mean?

Accepted Answer

Correct Option: C. Solution: The Step-Deviation Method is efficient for large values and equal class intervals because it involves dividing deviations by a common factor, simplifying calculations.

Question 27

In a survey, the number of books read by students in a month is recorded as follows: 0-5 (10 students), 5-10 (20 students), 10-15 (30 students), 15-20 (25 students), and 20-25 (15 students). Calculate the mean number of books read using the step-deviation method with a class width of 5.

Accepted Answer

Correct Option: A. Solution: The class marks are 2.5, 7.5, 12.5, 17.5, and 22.5. Using the step-deviation method with $a = 12.5$ and $h = 5$, the mean is calculated as $$12.5 + 5 	imes \frac{-2 	imes 10 + -1 	imes 20 + 0 	imes 30 + 1 	imes 25 + 2 	imes 15}{10 + 20 + 30 + 25 + 15} = 12.5$$ books.

Question 28

If the median of a data set is 525 and it lies in the class 500-600, with a class size of 100, what is the lower limit of the median class?

Accepted Answer

Correct Option: A. Solution: The lower limit of the median class is the starting point of the class interval where the median lies, which is 500 in this case.

Question 29

What is the formula for calculating the median of grouped data?

Accepted Answer

Correct Option: A. Solution: The formula for the median of grouped data is $$	ext{Median} = l + \left( \frac{\frac{n}{2} - cf}{f} ight) 	imes h$$, where $l$ is the lower limit of the median class, $cf$ is the cumulative frequency of the class preceding the median class, $f$ is the frequency of the median class, and $h$ is the class size.

Question 30

In a dataset where the mean is 50 and the mode is 40, what is the median according to the empirical relationship?

Accepted Answer

Correct Option: A. Solution: Using the empirical relationship $3 	imes 	ext{Median} = 	ext{Mode} + 2 	imes 	ext{Mean}$, we substitute the values: $3 	imes 	ext{Median} = 40 + 2 	imes 50 = 140$. Therefore, Median = $\frac{140}{3} = 46.67$. However, the closest option is 45.

Question 31

In the assumed mean method, what is the role of the assumed mean 'a'?

Accepted Answer

Correct Option: B. Solution: In the assumed mean method, 'a' is chosen to simplify calculations by reducing the size of numbers involved.

Question 32

In a frequency distribution, if the cumulative frequency just before the median class is 30, the frequency of the median class is 12, the class width is 10, and the total number of observations is 100, what is the position of the median?

Accepted Answer

Correct Option: B. Solution: The median position is given by $$\frac{n}{2} = \frac{100}{2} = 50$$. Since the cumulative frequency just before the median class is 30, the 50th observation lies in the median class.

Question 33

In a grouped frequency distribution, what does the cumulative frequency represent?

Accepted Answer

Correct Option: D. Solution: Cumulative frequency is the frequency obtained by adding the frequencies of all the classes preceding the given class.

Question 34

When dealing with a large dataset with equal class intervals, which method is preferred for calculating the mean to reduce calculation complexity?

Accepted Answer

Correct Option: C. Solution: The Step-Deviation Method is preferred for large datasets with equal class intervals as it simplifies calculations by using a common factor.

Question 35

In a cumulative frequency distribution, the cumulative frequency of the class interval 40-50 is 18, and the cumulative frequency of the class interval 50-60 is 22. If the class size is 10, what is the median class if the total number of observations is 53?

Accepted Answer

Correct Option: C. Solution: The median class is the class where the cumulative frequency is greater than or equal to $\frac{n}{2}$. Here, $n = 53$, so $\frac{n}{2} = 26.5$. The cumulative frequency of the class 50-60 (22) is less than 26.5, but the next class 60-70 has a cumulative frequency of 29, which is greater than 26.5. Therefore, the median class is 50-60.

Question 36

Consider a cumulative frequency distribution of marks obtained by students in a test. If the cumulative frequency just below the median class is 22, the frequency of the median class is 7, and the class interval is 10, what is the median if the lower limit of the median class is 60?

Accepted Answer

Correct Option: B. Solution: Using the formula for the median: $$	ext{Median} = l + \left( \frac{\frac{n}{2} - cf}{f} ight) 	imes h$$, where $l = 60$, $cf = 22$, $f = 7$, $h = 10$, and $n = 53$. Substituting these values, we get $$	ext{Median} = 60 + \left( \frac{26.5 - 22}{7} ight) 	imes 10 = 66.4$$.

Question 37

What is the mean of the marks obtained by students using the direct method in Example 1?

Accepted Answer

Correct Option: A. Solution: The mean of the marks obtained by students using the direct method is 59.3.

Question 38

A class of students has their Mathematics scores grouped into intervals as follows: 10-25, 25-40, 40-55, 55-70, 70-85, and 85-100. If the number of students in each interval is 2, 3, 7, 6, 6, and 6 respectively, what is the mean score using the class marks?

Accepted Answer

Correct Option: B. Solution: The mean is calculated using class marks as representative values. The class marks are 17.5, 32.5, 47.5, 62.5, 77.5, and 92.5. The mean is $$\frac{1860}{30} = 62$$.

Question 39

Given a grouped frequency distribution where the class intervals are 10-20, 20-30, 30-40, 40-50, and the corresponding frequencies are 5, 8, 12, and 10, find the mode using the formula for grouped data.

Accepted Answer

Correct Option: B. Solution: The modal class is 30-40 as it has the highest frequency of 12. Using the formula for mode: $$	ext{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} ight) 	imes h$$ where $l = 30$, $f_1 = 12$, $f_0 = 8$, $f_2 = 10$, $h = 10$. Thus, Mode = 30 + ((12 - 8)/(2*12 - 8 - 10)) * 10 = 35.

Question 40

In a cumulative frequency distribution of the 'more than' type, which of the following is true about the cumulative frequency for the first class interval?

Accepted Answer

Correct Option: A. Solution: In a 'more than' type cumulative frequency distribution, the cumulative frequency for the first class interval is equal to the total number of observations.

Question 41

A grouped data has class intervals 60-70, 70-80, 80-90, 90-100 with frequencies 7, 9, 7, 8 respectively. If the mode is calculated to be 75, what is the class size?

Accepted Answer

Correct Option: A. Solution: The mode is calculated using the formula $$	ext{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} ight) 	imes h$$. Given that the mode is 75 and the modal class is 70-80, the class size $h$ is 10.

Question 42

If the cumulative frequency of the class interval 70-80 is 38 and the cumulative frequency of the class interval 80-90 is 45, what does the difference in these cumulative frequencies represent?

Accepted Answer

Correct Option: B. Solution: The difference in cumulative frequencies between two consecutive class intervals represents the frequency of the latter class interval. Therefore, 45 - 38 = 7, which is the frequency of the class interval 80-90.

Question 43

In the context of grouped data, how is the class mark of a class interval calculated?

Accepted Answer

Correct Option: A. Solution: The class mark of a class interval is calculated by taking the average of its upper and lower limits.

Question 44

Which class interval is the median class if the cumulative frequency just before it is 22, the frequency of the class is 7, and the total number of observations is 53?

Accepted Answer

Correct Option: C. Solution: The median class is the class interval where the cumulative frequency just exceeds half of the total number of observations. Here, 53/2 = 26.5, so the class interval 60-70 is the median class as its cumulative frequency is 29.

Question 45

Which of the following statements is true regarding cumulative frequency curves (ogives)?

Accepted Answer

Correct Option: C. Solution: Ogives are primarily used to estimate the median and analyze distribution trends, not for calculating the mean, mode, or standard deviation.

Question 46

In a grouped frequency distribution, the mean is 60, and the median is 65. What is the mode of the dataset?

Accepted Answer

Correct Option: C. Solution: Using the empirical relationship $3 	imes 	ext{Median} = 	ext{Mode} + 2 	imes 	ext{Mean}$, we substitute the given values: $3 	imes 65 = 	ext{Mode} + 2 	imes 60$. Solving for the mode gives $195 = 	ext{Mode} + 120$, so the mode is $195 - 120 = 75$.

Question 47

In a grouped frequency distribution, the cumulative frequency of the class interval 60-70 is 29. If the cumulative frequency of the preceding class interval 50-60 is 22, what is the frequency of the class interval 60-70?

Accepted Answer

Correct Option: A. Solution: The frequency of the class interval 60-70 is the difference between its cumulative frequency and that of the preceding class interval, which is 29 - 22 = 7.

Question 48

For a dataset with class intervals 100-200, 200-300, 300-400, 400-500, and frequencies 5, 12, 18, 10, find the mode if the class size is 100.

Accepted Answer

Correct Option: B. Solution: The modal class is 300-400 with the highest frequency of 18. Using the mode formula: $$	ext{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} ight) 	imes h$$, where $l = 300$, $f_1 = 18$, $f_0 = 12$, $f_2 = 10$, $h = 100$, the mode is 350.

Question 49

In a dataset where the numerical values of $x_i$ and $f_i$ are large, which method would be most efficient to calculate the mean?

Accepted Answer

Correct Option: D. Solution: Both the Assumed Mean Method and the Step-Deviation Method are efficient for large numerical values of $x_i$ and $f_i$ as they simplify the calculations compared to the Direct Method.

Question 50

The mode of grouped data can be determined by identifying the class with the maximum frequency and using the formula: $$	ext{Mode} = l + \frac{(f_1 - f_0)}{2f_1 - f_0 - f_2} 	imes h$$ where $l$ is the lower limit of the modal class, $h$ is the class size, $f_1$ is the frequency of the modal class, $f_0$ is the frequency of the class preceding the modal class, and $f_2$ is the frequency of the class succeeding the modal class.

Accepted Answer

Answer: True. Solution: The formula provided is correct for calculating the mode of grouped data. It considers the frequencies of the modal class and its neighboring classes to determine the mode.

Question 51

The empirical relationship between mean, median, and mode is given by the formula $3 	imes 	ext{Median} = 	ext{Mode} + 2 	imes 	ext{Mean}$.

Accepted Answer

Answer: True. Solution: The empirical relationship $3 	imes 	ext{Median} = 	ext{Mode} + 2 	imes 	ext{Mean}$ connects the three measures of central tendency.

Question 52

The mean obtained from grouped data using the direct method is always more accurate than the mean obtained using the assumed mean method.

Accepted Answer

Answer: False. Solution: The mean obtained using the direct method is not necessarily more accurate than the assumed mean method. The assumed mean method can provide the same mean value as the direct method when calculations are done correctly.

Question 53

The cumulative frequency distribution of the 'more than' type is used to find the median.

Accepted Answer

Answer: False. Solution: The cumulative frequency distribution of the 'less than' type is typically used to find the median.

Question 54

The median of grouped data can be calculated using the formula: $$	ext{Median} = l + \left( \frac{\frac{n}{2} - cf}{f} ight) 	imes h$$ where $l$ is the lower limit of the median class, $cf$ is the cumulative frequency of the class preceding the median class, $f$ is the frequency of the median class, and $h$ is the class size.

Accepted Answer

Answer: True. Solution: This formula is used to find the median of grouped data as it accounts for the distribution of data within the median class.

Question 55

A cumulative frequency distribution can be used to determine the median of a dataset.

Accepted Answer

Answer: True. Solution: Cumulative frequency distributions are used to find the median by identifying the median class and applying the median formula.

Question 56

The cumulative frequency of a class is the frequency obtained by adding the frequencies of all the classes succeeding the given class.

Accepted Answer

Answer: False. Solution: The cumulative frequency of a class is obtained by adding the frequencies of all the classes preceding the given class, not succeeding.

Question 57

The step-deviation method is preferred for calculating the mean when the class intervals are unequal and the numerical values of $x_i$ are large.

Accepted Answer

Answer: True. Solution: The step-deviation method simplifies calculations by reducing the numerical values, making it suitable for large values and unequal class intervals.

Question 58

Ogives are used to visually estimate the median and understand distribution trends.

Accepted Answer

Answer: True. Solution: Ogives, or cumulative frequency curves, are graphical representations that help in estimating the median and understanding distribution trends.

Question 59

The formula for calculating the mean using the assumed mean method is given by $x = a + \frac{\Sigma f_i d_i}{\Sigma f_i}$, where $a$ is the assumed mean.

Accepted Answer

Answer: True. Solution: The formula for calculating the mean using the assumed mean method is indeed $x = a + \frac{\Sigma f_i d_i}{\Sigma f_i}$, where $a$ is the assumed mean, $f_i$ are the frequencies, and $d_i$ are the deviations from the assumed mean.

Question 60

The mode for grouped data is always equal to the mean.

Accepted Answer

Answer: False. Solution: The mode for grouped data is not necessarily equal to the mean. They are different measures of central tendency and can have different values depending on the data distribution.

Question 61

In a grouped frequency distribution, the median is always the middle value of the data set.

Accepted Answer

Answer: False. Solution: In a grouped frequency distribution, the median is not necessarily the middle value but is calculated using the median class and the formula for median.

Question 62

In the assumed mean method, the mean is calculated by adding the assumed mean to the mean of deviations.

Accepted Answer

Answer: True. Solution: In the assumed mean method, the mean is calculated as $x = a + \frac{\Sigma f_i d_i}{\Sigma f_i}$, where $a$ is the assumed mean.

Question 63

The mean of grouped data can be accurately calculated using the mid-point assumption for each class interval.

Accepted Answer

Answer: False. Solution: The mid-point assumption for each class interval provides an approximate mean, not an exact one, as it assumes that all data points in a class are concentrated at the mid-point.

Question 64

The median of grouped data can be found using the formula: $$	ext{Median} = l + \left( \frac{\frac{n}{2} - cf}{f} ight) 	imes h$$, where $l$ is the lower limit of the median class, $n$ is the total number of observations, $cf$ is the cumulative frequency of the class preceding the median class, $f$ is the frequency of the median class, and $h$ is the class size.

Accepted Answer

Answer: True. Solution: This formula is used to calculate the median of grouped data by identifying the median class and applying the given parameters.

Statistics

Learning Objectives

Learning Objectives

Revision Notes & Summary

Chapter Notes

Mean of Grouped Data

Mode of Grouped Data

Median of Grouped Data

Cumulative Frequency Distribution

Empirical Relationship between Mean, Median, and Mode

Exam Tips & Common Mistakes

Common Mistakes & Exam Tips

Mean of Grouped Data

Mode of Grouped Data

Median of Grouped Data

Cumulative Frequency Distribution

Empirical Relationship between Mean, Median, and Mode

General Tips

AI Learning Assistant

Practice Test – MCQs, True/False

Experience the StudyTunnel Method

Step 1: Chapter Practice

Step 2: Term Boss Exam

Step 3: Redemption Arena

Practice, Analyze & Improve 🚀

Multiple Choice Questions

Q1.In a survey, the number of family members in households is recorded as 1-3, 3-5, 5-7, 7-9, 9-11 with frequencies 7, 8, 2, 2, 1 respectively. What is the mode of this data?

Correct Answer: A

Q2.What is the formula for finding the mode of grouped data?

Correct Answer: B

Correct Answer: A

Q4.What is the primary purpose of an ogive in statistics?

Correct Answer: B

Q5.If a grouped frequency distribution has the following class intervals: 10-20, 20-30, 30-40, with frequencies 5, 15, and 10 respectively, and the median class is 20-30, what is the cumulative frequency of the class preceding the median class?

Correct Answer: A

Q6.Which method is most suitable for calculating the mean when the class sizes are unequal and xix_ixi​ values are large?

Correct Answer: C

Q7.For a grouped data, the class intervals are 50-60, 60-70, 70-80 with frequencies 5, 8, and 12 respectively. If the median class is determined to be 60-70, what is the class width?

Correct Answer: B

Q8.Given a frequency distribution with class intervals of equal width, if the modal class is 20-30 with a frequency of 15, the preceding class has a frequency of 10, and the succeeding class has a frequency of 5, what is the mode?

Correct Answer: A

Q9.What is the cumulative frequency of the class interval 40-50 in a less than type cumulative frequency distribution if the frequencies of the classes 30-40 and 40-50 are 3 and 3 respectively?

Correct Answer: C

Q10.If the cumulative frequency of a class is 45 and the cumulative frequency of the preceding class is 30, what is the frequency of the class?

Correct Answer: A

Q11.A grouped frequency distribution of student scores is given with class intervals of 50-60, 60-70, 70-80, 80-90, and 90-100. If the frequencies are 5, 10, 15, 10, and 5 respectively, what is the mean score using the assumed mean method with an assumed mean of 75?

Correct Answer: B

Q12.For a dataset of exam scores with class intervals 30-40, 40-50, 50-60, 60-70, and 70-80, the frequencies are 6, 9, 15, 10, and 5 respectively. Calculate the mean score using the direct method.

Correct Answer: C

Q13.In a grouped frequency distribution, the class intervals are 10-20, 20-30, 30-40, 40-50, and 50-60. The frequencies are 5, 10, 15, 10, and 5 respectively. What is the median of this distribution?

Correct Answer: A

Q14.Which method is most suitable for calculating the mean when the numerical values of xix_ixi​ and fif_ifi​ are large?

Correct Answer: B

Q15.Given a cumulative frequency distribution of less than type for a dataset, how would you determine the median class?

Correct Answer: B

Q16.Why might the mean calculated using the direct method differ from that calculated using the assumed mean method?

Correct Answer: B

Correct Answer: B

Q18.How is the median of grouped data affected by the class size?

Correct Answer: D

Q19.What is the formula to calculate the mean of grouped data using the direct method?

Correct Answer: A

Q20.What is the empirical relationship between the mean, median, and mode?

Correct Answer: B

Q21.If the mode of a grouped data set is found to be less than the mean, which of the following is a possible interpretation?

Correct Answer: A

Q22.Given the following class intervals and frequencies for the number of hours students study per week: 10-20 (5 students), 20-30 (8 students), 30-40 (12 students), 40-50 (10 students), and 50-60 (5 students). Calculate the mean study hours per week using the direct method.

Correct Answer: A

Q23.In a grouped frequency distribution, which class is considered the modal class?

Correct Answer: B

Q24.Given a dataset where the mean is 50 and the mode is 40, what is the median according to the empirical relationship between mean, median, and mode?

Correct Answer: A

Q25.In a dataset, the class intervals are 0-10, 10-20, 20-30 with frequencies 8, 12, and 10 respectively. If the median class is 10-20, what is the lower boundary of the median class?

Correct Answer: B

Q26.In a dataset with large values and equal class intervals, which method would be most efficient for calculating the mean?

Correct Answer: C

Q27.In a survey, the number of books read by students in a month is recorded as follows: 0-5 (10 students), 5-10 (20 students), 10-15 (30 students), 15-20 (25 students), and 20-25 (15 students). Calculate the mean number of books read using the step-deviation method with a class width of 5.

Correct Answer: A

Q28.If the median of a data set is 525 and it lies in the class 500-600, with a class size of 100, what is the lower limit of the median class?

Correct Answer: A