Kit Library / Mathematics / Statistics & Probability

⚡ Topic Learning Kit

Statistics Mean

English 14 leveled MCQs Free

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1

CBSE Class 10 Statistics: Finding the Mean

~8 min · full explanation, examples & memory tricks in the app

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All 14 questions, laddered from warm-up to topper-level, each with its answer and the reasoning behind it.

Warm-up · 5 questions

Q1. Which formula is used to calculate the mean of grouped data using the direct method?

Beginner
A $\bar{x} = \sum f_i x_i$ B $\bar{x} = \frac{\sum f_i}{\sum f_i x_i}$ C $\bar{x} = \frac{\sum x_i}{\sum f_i}$ D $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$
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$\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$

The direct method formula for mean is $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$, where $f_i$ are frequencies and $x_i$ are class marks.

Q2. For a grouped frequency distribution, the class mark of a class interval is:

Beginner
A Difference of upper and lower limits divided by 2 B Upper limit + lower limit C Upper limit - lower limit D Average of upper and lower limits
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Average of upper and lower limits

Class mark $x_i = \frac{\text{lower limit} + \text{upper limit}}{2}$.

Q3. The mean of a frequency distribution can be found using:

Beginner
A Only the assumed mean method B Only the direct method C Only the step-deviation method D Direct, assumed mean, or step-deviation methods
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Direct, assumed mean, or step-deviation methods

All three methods—direct, assumed mean, and step-deviation—can be used to find the mean, and they give the same result.

Q4. For a grouped frequency distribution, the mean is calculated using:

Beginner
A Only class limits B Class marks and frequencies C Only cumulative frequencies D Only frequencies
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Class marks and frequencies

In grouped data, we use class marks ($x_i$) and their corresponding frequencies ($f_i$) to compute the mean.

Q5. The mean is always one of the observations in the data set.

Beginner
A True B False
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False

The mean is not necessarily an observation in the data set; it is a calculated average and may not coincide with any data point.

Building up · 6 questions

Q6. In the step-deviation method, the formula for mean is $\bar{x} = a + h \times \frac{\sum f_i u_i}{\sum f_i}$. What does $u_i$ represent?

Intermediate
A $\frac{a - x_i}{h}$ B $x_i - a$ C $\frac{x_i - h}{a}$ D $\frac{x_i - a}{h}$
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$\frac{x_i - a}{h}$

In step-deviation method, $u_i = \frac{x_i - a}{h}$, where $a$ is assumed mean and $h$ is class size.

Q7. If the mean of a distribution is 25 and the sum of frequencies is 20, what is the sum of $f_i x_i$?

Intermediate
A 5 B 45 C 500 D 0.8
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500

Using $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$, we get $\sum f_i x_i = \bar{x} \times \sum f_i = 25 \times 20 = 500$.

Q8. Which method of finding the mean is most suitable when the class marks are large numbers?

Intermediate
A Median method B Direct method C Step-deviation method D Assumed mean method
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Step-deviation method

Step-deviation method simplifies calculations when class marks are large by using deviations from an assumed mean.

Q9. In the assumed mean method, the mean is calculated as $\bar{x} = a + \frac{\sum f_i d_i}{\sum f_i}$. What is $d_i$?

Intermediate
A $x_i + a$ B $a - x_i$ C $\frac{x_i}{a}$ D $x_i - a$
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$x_i - a$

In assumed mean method, $d_i = x_i - a$, where $a$ is the assumed mean.

Q10. If the class intervals are not continuous, what should be done before calculating the mean?

Intermediate
A Use only the lower limits B Make the classes continuous by adjusting the limits C Ignore the gaps D Use only the upper limits
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Make the classes continuous by adjusting the limits

For grouped data with discontinuous classes, we first make the classes continuous by subtracting and adding half the gap to lower and upper limits respectively.

Q11. The mean of a grouped frequency distribution can be found using the step-deviation method only if the class sizes are equal.

Intermediate
A True B False
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True

The step-deviation method is most convenient when class sizes are equal, as it simplifies calculations using a common factor $h$.

Topper level · 3 questions

Q12. The algebraic sum of deviations of individual observations from their mean is:

Advanced
A Equal to the mean B Always negative C Always positive D Zero
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Zero

A key property of mean is that the sum of deviations of all observations from the mean is zero: $\sum (x_i - \bar{x}) = 0$.

Q13. If each observation in a data set is increased by 5, the mean of the new data set:

Advanced
A Decreases by 5 B Becomes 5 times the original mean C Remains the same D Increases by 5
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Increases by 5

If each observation is increased by a constant $k$, the mean also increases by $k$. So the new mean is original mean + 5.

Q14. If the mean of 10 observations is 20 and one observation 30 is removed, what is the mean of the remaining 9 observations?

Advanced
A 18.89 B 17 C 19 D 20
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18.89

Sum of 10 observations = $10 \times 20 = 200$. After removing 30, sum = 170. New mean = $\frac{170}{9} \approx 18.89$.

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