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Statistics questions are a common type of questions regularly asked in competitive exams. These questions carry a weightage of 1-2 questions (2-4 marks) in SSC exams and. To get a good rank in competitive exams, you have a good grasp on the concepts of Statistics as they are asked in scoring questions.

Here are some tips for solving Statistics questions: focus on mastering fundamental concepts, including mean, median, mode, and standard deviation, Use the appropriate formula as there are many different formulas, Practice solving a variety of problems to build problem-solving skills

So, we have attached 10 questions of Statistics for you to practice with. You should aim to solve these questions in less than half a minute for each.

 

<h2 id="content-id-1699107832845-0">Practice Questions on Statistics</h2>

You can also download the Statistics questions and answers pdf. Just click on the Download PDF button. So let's start with the very first question.

Q:1 What is the mode of the following set of data?

2, 4, 5, 6, 2, 4, 8, 6, 2, 6, 8, 5, 8, 4, 9, 11, 8

1. 2

2. 4

3. 6

4. 8

(Difficulty: 3, Estimated Time: 20 Seconds) A calculative one to start. Did get it right?

Q:2 While checking the value of 10 observations, it was noted 75 was wrongly noted as 25 while calculating the mean. The present mean is 50. Find the correct mean.

1. 45

2. 50

3. 55

4. 60

(Difficulty: 3, Estimated Time: 20 Seconds) This was a simple one, don't get stuck in unnecessary calculations! 

Q:3 31st to 40th-rank students are standing around the table. The marks obtained in mathematics in the final semester are 95 by 8 out of 10 and the remaining got 90. Find the standard deviation of their marks.

1. 4

2. 9.5

3. 36.4

4. 12.8

(Difficulty: 3, Estimated Time: 20 Seconds) This was a bit calculative one...

Q:4 Calculate the median of the following data -

<table border="1" cellpadding="1" cellspacing="1">
	<tbody>
		<tr>
			<td>Class</td>
			<td>0 - 5 </td>
			<td>5 - 10 </td>
			<td>10 - 15</td>
			<td>15 - 20</td>
		</tr>
		<tr>
			<td>Frequency</td>
			<td>4</td>
			<td>6</td>
			<td>9</td>
			<td>5</td>
		</tr>
	</tbody>
</table>

1. 14.9

2. 11.1

3. 12.3

4. 15.4

(Difficulty: 3, Estimated Time: 20 Seconds) It is not an easy one but I think now you're prepared for it.

Q:5 If the mode of the following data is 10, then the value of k in the data set 2, 8, 5, 10, 1, 6, 10, 6, 6, 2k + 7, 9, 10, and 12 is:

1. 1.5

2. 2.5

3. 3

4. 4

(Difficulty: 3, Estimated Time: 20 Seconds) We're halfway through. Have you got all your questions correct so far?

Q:6 The data of marks scored by 11 students of Pratham Tuition classes are as follows: 223, 217, 237, 255, 243, 218, 265, 279, 263, 233, 227. Find the median of the marks scored by the students.

1. 227

2. 233

3. 237

4. 243

(Difficulty: 2, Estimated Time: 15 Seconds) Should we raise the level of questions?

Q:7 In the given frequency distribution, the mean is 39.4, which of the following is the value of 'k'.

<table border="1" cellpadding="1" cellspacing="1">
	<tbody>
		<tr>
			<td>Variable</td>
			<td>40</td>
			<td>42</td>
			<td>37</td>
			<td>35</td>
			<td>40</td>
		</tr>
		<tr>
			<td>Frequency</td>
			<td>5</td>
			<td>7</td>
			<td>k</td>
			<td>4</td>
			<td>6</td>
		</tr>
	</tbody>
</table>

1. 3

2. 5

3. 8

4. Cannot be determined

(Difficulty: 3, Estimated Time: 20 Seconds) Try using short tricks to save time..

Q:8 What is the mode of the distribution where the mean and median are 28.33, and 28 respectively?

1. 28

2. 28.5

3. 28.66

4. 27.33

(Difficulty: 2, Estimated Time: 15 Seconds) A question of seconds.....

Q:9 The mean of 5 numbers is 17. If the difference of two consecutive numbers is 3 and the third number is the same as the mean value of all 5 numbers, then find the difference between the sum of the last two numbers and the first two numbers.

1. 15

2. 18

3. 21

4. 12

(Difficulty: 3, Estimated Time: 20 Seconds) Try decreasing time in your calculations

Q:10 Find the sum of the mean values of the three series given below:

(i) 25, 25, 25, 25, 25, and 25

(ii) 5, 20, 35, 15, 55, 20, 40 and 50

(iii) 0.81, 0.098, 12.6, 0.19, 4.902, 4, 40 and 1.4

1. 25

2. 52

3. 63

4. 102.81

(Difficulty: 2, Estimated Time: 15 Seconds) Did you guess them all correctly?

<h2 id="content-id-1699107832893-1">Answer Key</h2>

Let's check out your score in this test.

<table border="0" cellpadding="1" cellspacing="1">
	<tbody>
		<tr>
			<td>1. (4)</td>
			<td>2. (3)</td>
			<td>3. (1)</td>
			<td>4. (2)</td>
			<td>5. (1)</td>
		</tr>
		<tr>
			<td>6. (3)</td>
			<td>7. (1)</td>
			<td>8. (4)</td>
			<td>9. (2)</td>
			<td>10. (3)</td>
		</tr>
	</tbody>
</table>

Comment below your score, considering each question has 1 mark only. If you scored 8 to 10, congratulations! You are one step closer to selection. If you have scored 5 to 8 marks, then you are doing well, keep it up. If you have scored less than 5 marks then you need to work a little harder on this subject. But don't worry, we are here to help you master the subject.

Let's check the answers and solutions and try to find out what went wrong. 
 

<h2 id="content-id-1699107832893-2">Answers and Solutions</h2>

Q:1 The correct answer is Option 4 i.e. 8.

Mode = Number with maximum occurrence

Mode of the given series = 8 (4 times)

 

Q:2 The correct answer is option 3 i.e. 55

Given :

Present mean = 50

number of observations = 10

wrongly noted number = 25 

correct number = 75

Calculations :

Total sum = mean × total observations

⇒ Total sum = 50 × 10

⇒ 500

Let the correct mean be y

⇒ y × observations = Mean - wrongly noted + correct number

⇒ 10y = 500 - 25 + 75

⇒ 10y = 550

⇒ y = 55

 

Q:3 The correct answer is option 1 i.e. 4.

According to the question-

Total marks = (95 × 8) + (90 × 2) = 760 + 180 = 940

Mean of marks = 940/10 = 94

Varience = 8 (95 - 94) + 2 (90 - 94) = 8(1) + 2(4) = (8 + 8) = 16

Formula:

Standard deviation = √(variance) 

Hence,

SD = √(16) = 4

 

Q:4 The correct answer is Option 2 i.e. 11.1.

Median = L + [{(N/2 - C)/f} × h]

N = sum of frequencies, L = lower limit of the median class, f = frequency of median class, h = width of the class, and C = cumulative frequency of the class just preceding the median class.

The median class is the class interval whose

cumulative frequency is ≥ N/2.

<table border="1" cellpadding="1" cellspacing="1">
	<tbody>
		<tr>
			<td>Class</td>
			<td>Frequency</td>
			<td>Cumulative Frequency</td>
		</tr>
		<tr>
			<td>0 - 5</td>
			<td>4</td>
			<td>4</td>
		</tr>
		<tr>
			<td>5 - 10</td>
			<td>6</td>
			<td>10</td>
		</tr>
		<tr>
			<td>10 - 15</td>
			<td>9</td>
			<td>19</td>
		</tr>
		<tr>
			<td>15 - 20</td>
			<td>5</td>
			<td>24</td>
		</tr>
	</tbody>
</table>

Sum of frequency = N/2 = 24/2 = 12

Hence, 10 - 15 is the median class, where -

L = 10, f = 9, h = 5, and C = 10

Median = 10 + [{12 - 10)/9} × 5] = 10 + 10/9 = 100/9 = 11.1

 

Q:5 The correct answer is Option 1 i.e. 1.5

Mode is the value that occurs most often in the data set of values.

Data = 2, 8, 5, 10, 1, 6, 10, 6, 6, 2k + 7, 9, 10, and 12

In the above data set, 10 and 6 has occurred more times i.e., 3 times, but given that mode is 10.

10 should occur more times than 6.

Hence the variable 2k + 7 must be 10

⇒ 2k + 7 = 10

⇒ 2k = 3

⇒ k = 3/2 = 1.5

 

Q:6 The correct answer is option 3 i.e. 237

The median is calculated by arranging the data in increasing or decreasing order; 

217, 218, 223, 227, 233, 237, 243, 255, 263, 265, 279

Median = (n + 1)/2 (Where n = number of terms and number of terms are odd)

Median = (11 + 1)/2 = 6th term = 237

Here, the middle term after arranging in the increasing or decreasing order is 237. Thus, 237 is the median of the given data.

 

Q:7 The correct answer is option 1 i.e. 3.

The sum of observations = Frequency × variable

Mean = Sum of observations/total frequency

Sum of observations= 40 × 5 + 42 × 7 + 37 × k + 35 × 4 + 40 × 6

⇒ 200 + 294 + 37k + 140 + 240

⇒ 874 + 37k

Total frequency = 5 + 7 + k + 4 + 6 = 22 + k

⇒ 39.4(22 + k) = 874 + 37k

⇒ 866.28 + 39.4k = 874 + 37k

⇒ 2.4k = 7.2

⇒ k = 3

Q:8 The correct answer is option 4 i.e. 27.33

As we know,

⇒ Mean – Mode = 3 (Mean – Median)

⇒ 28.33 - mode = 3(28.33 - 28)

⇒ 28.33 - mode = 1

⇒ Mode = 27.33

 

Q:9 The correct answer is option 2 i.e. 18. 

Third number = Mean = 17

Difference = 3

4th number = 17 + 3 = 20

5th number = 20 + 3 = 23

2nd number = 17 - 3 = 14

1st number = 14 - 3 = 11

(11 + 14 + 17 + 20 + 23)/5 = 85/5 = 17

Difference between last two and first two numbers = (23 + 20) - (11 + 14) = (43) - (25) = 18 

 

Q:10 The correct answer is option 3 i.e. 63

Given,

(i) 25, 25, 25, 25, 25, and 25

(ii) 5, 20, 35, 15, 55, 20, 40 and 50

(iii) 0.81, 0.098, 12.6, 0.19, 4.902, 4 and 1.4

Mean = Sum of number/Total number of numbers

 

(i) 25, 25, 25, 25, 25, and 25

Sum of numbers = 6(25) = 150

Total number of numbers = 6

Mean = 150/6 = 25 (Note: When all data is same like 'x' then mean value of data is also be a x)

Now, (ii) 5, 20, 35, 15, 55, 20, 40 and 50

Sum of numbers = 5 + 20 + 35 + 15 + 55 + 20 + 40 + 50 = 240

Total number of numbers = 8

Mean = 240/8 = 30

Now, (iii) 0.81, 0.098, 12.6, 0.19, 4.902, 4, 40 and 1.4

Sum of numbers

 ⇒ (0.81 + 0.19) + (0.098 + 4.902) + (12.6 + 1.4) + 4 + 40

 ⇒ (1) + (5) + (14) + 4 + 40 = 64

Total number of numbers = 8

Mean = 64/8 = 8

The sum of means of (i), (ii), and (iii) = 25 + 30 + 8 = 63

So, this is it for today. We will meet again with another new topic. Till then, you can practice the questions again by downloading the PDF of Statistics.

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