Class 10 Statistics Mock Test (CBSE)

Topics in Statistics

• Mean of grouped data
• Mode of grouped data
• Median of grouped data
• Cumulative frequency

Statistics practice questions with answers

1. Q1Multiple choice·1 mark
   For a given grouped frequency distribution, if the modal class is $20-30$, the lower limit of the modal class is:

   1. (a)$10$
   2. (b)$20$
   3. (c)$30$
   4. (d)$25$
   Show answer and step marks

   1. Correct option: (b) $20$. The lower limit of the modal class $20-30$ is $20$.+1

   Answer: (b)
2. Q2Assertion and reason·1 mark
   Assertion (A): In a less than type cumulative frequency distribution, the cumulative frequency of a class equals the sum of the frequencies of that class and all classes before it. Reason (R): The cumulative frequency of a class is found by subtracting the frequency of the previous class from its own frequency.

   1. (a)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
   2. (b)Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
   3. (c)Assertion (A) is true but Reason (R) is false.
   4. (d)Assertion (A) is false but Reason (R) is true.
   Show answer and step marks

   1. Correct option: (c) Assertion (A) is true but Reason (R) is false.+1

   Answer: (c)
3. Q3Multiple choice·1 mark
   While computing the median of grouped data, if the sum of frequencies is $\sum f_i = 50$, then the value of $\frac{N}{2}$ used to locate the median class is:

   1. (a)$25$
   2. (b)$50$
   3. (c)$25.5$
   4. (d)$24.5$
   Show answer and step marks

   1. Correct option: (a) $25$. Here $N = \sum f_i = 50$, so $\frac{N}{2} = \frac{50}{2} = 25$.+1

   Answer: (a)
4. Q4Assertion and reason·1 mark
   Assertion (A): The arithmetic mean of grouped data can be calculated by direct method, assumed mean method, or step-deviation method. Reason (R): All three methods yield the same value of the arithmetic mean for a given grouped data.

   1. (a)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
   2. (b)Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
   3. (c)Assertion (A) is true but Reason (R) is false.
   4. (d)Assertion (A) is false but Reason (R) is true.
   Show answer and step marks

   1. Correct option: (a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A). Both statements are correct, and Reason correctly explains why different computational methods are used for the mean.+1

   Answer: (a)
5. Q5Short answer·2 marks
   Write the formula for the mean of grouped data by the direct method. State what $x_i$ and $f_i$ represent in this formula.
   Show answer and step marks

   1. Formula: $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$+1
   2. $x_i$ is the class mark of the i-th class interval+0.5
   3. $f_i$ is the frequency of the i-th class interval+0.5

   Answer: $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$, where $x_i$ is the class mark and $f_i$ is the frequency of the i-th class.
6. Q6Short answer·2 marks
   The number of text messages sent in a day by 52 students is grouped as follows: $0-10$: 4, $10-20$: 8, $20-30$: 20, $30-40$: 12, $40-50$: 8. Explain why the class $20-30$ is the modal class, and then find the mode of the data.
   Show answer and step marks

   1. The modal class is the class with the highest frequency; $20-30$ has the highest frequency 20, so it is the modal class, with $l = 20$, $h = 10$, $f_1 = 20$, $f_0 = 8$, $f_2 = 12$+1
   2. Mode $= l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h = 20 + \frac{20-8}{40-8-12} \times 10 = 20 + \frac{12}{20} \times 10 = 26$+1

   Answer: Modal class is $20-30$; mode $= 26$ messages.

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