JEE Main Maths Statistics Previous Year Questions With Solutions

Statistics for IIT JEE covers important concepts such as measures of Dispersion, calculation of mean, median, mode of grouped and ungrouped data, calculation of standard deviation, variance and mean deviation for grouped and ungrouped data. In this article, aspirants can practice previous year’s questions on Statistics asked in IIT JEE exams. We have included all the important questions and a detailed solution for each question.

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JEE Main Statistics Previous Year Questions With Solutions

Question 1: The mean of n items is x\overline{x}. If the first term is increased by 1, second by 2 and so on, then what is the new mean?

Solution:

Let x1, x2,……. xn be n items.

Then, x\overline{x} = [1 / n] Σ xi

Let y1 = x1 + 1, y2 = x2 + 2, y3 = x3 + 3,…………,yn = xn + n

Then the mean of the new series is [1 / n] Σ yi = [1 / n] ∑i=1n (xi + i)

= [1 / n] ∑i=1n xi + 1/n (1 + 2 + 3 + ….. + n)

=x\overline{x} + [1 / n] * [n (n + 1) / 2]

=x\overline{x} + (n+1)/2.

Question 2: The average of n numbers x1, x2,……. xn is M. If xn is replaced by x′, then what is the new average?

Solution:

M = x1,x2,.xnn\frac{x_1, x_2,……. x_n}{n} i.e.,

nM = x1, x2,……. xn−1 + xn

nM – xn = x1, x2,……. xn−1

[nM – xn + x′] / n = [x1, x2,……. xn−1 + x′] / n

New average = [nM − xn + x′] / n

Question 3: Mean of 100 observations is 45. It was later found that two observations 19 and 31 were incorrectly recorded as 91 and 13. The correct mean is _________.

Solution:

Sum of 100 items = 45 × 100 = 4500

Sum of items added = 19 + 31 = 50

Sum of items replaced = 91 + 13 = 104

New sum = 4500 − 104 + 50 = 4446

New mean= 4446 / 100 = 44.46

Question 4: The following data gives the distribution of the height of students.

Height (in cm) 160 150 152 152 161 154 155
Name of students 12 8 4 4 3 3 7

What is the median of the distribution?

Solution:

Arranging the data in ascending order of magnitude, we obtain

Height (in cm) 150 152 154 155 156 160 161
Number of students 8 4 3 7 3 12 4
Cumulative frequency 8 12 15 22 25 37 41

Here, the total number of items is 41 i.e., an odd number.

Hence, the median is [(41 + 1) / 2]th i.e., 21st item.

From the cumulative frequency table, we find that median i.e., 21st item is 155.

(All items from 16 to 22nd are equal, each 155).

Question 5: The mean and S.D. of the marks of 200 candidates were found to be 40 and 15 respectively. Later, it was discovered that a score of 40 was wrongly read as 50. The correct mean and S.D. respectively are ______________.

Solution:

Corrected Σx = 40 × 200 − 50 + 40 = 7990

Corrected x bar = 7990 / 200 = 39.95

Incorrect Σx2 = n [σ2 + (xˉ2\bar{x}^2) = 200 [152 + 402] = 365000

Correct Σx2 = 365000 − 2500 + 1600 = 364100

Corrected σ = 364100200\sqrt{\frac{364100 }{ 200}} − (39.95)2

= √(1820.5 − 1596)

= √224.5

= 14.98

Question 6: Let r be the range and S2 = 1/(n-1) be the S.D. of a set of observations x1, x2,….. xn, then what is the condition for S?

Solution:

r = max |xi − xj|; i ≠ j and S2 = 1n1i=1n(x1xˉ)2\frac{1}{n-1} \sum_{i=1}^n (x_1- \bar x)^2

Now, (xixˉ)2(x_i – \bar x)^2 = (xi − [x1 + x2+….. +xn] / n)2

= [1 / n2] [(xi − x1) + (xi − x2) +…..+(xi − xi − 1) + (xi − xi + 1)+…..+(xi − xn)]

≤ [1 / n2] [(n − 1)r]2,

[∵|xi − xj| ≤ r]

≤ r2

i=1n(xixˉ)2\sum_{i=1}^n(x_i – \bar x)^2 ≤ nr2

=> 1n1i=1n(xixˉ)2 \frac{1}{n-1} \sum_{i=1}^n(x_i – \bar x)^2 ≤ [nr2 / (n − 1)]

or S2nr2n1\frac{nr^2}{n-1}

S ≤ rnn1\frac{\sqrt{r n}}{n-1}

Question 7: If a variable takes the discrete values α − 4, α − 7 / 2, α − 5 / 2, α − 3, α − 2, α + 1 / 2, α − 1 / 2, α + 5 (α > 0), then the median is ____________.

Solution:

Arrange the data as α − 7/2, α − 3, α − 5/2, α − 2, α − 1/2, α + 1/2, α – 4, α + 5

Median = [1 / 2] [value of 4th item + value of 5th item ]

Median = [(α − 2) + (α − 1 / 2)] / 2

= [2α − 5 / 2] / 2

= α − 5 / 4

Question 8: The median of a set of 9 distinct observations is 20.5. If each of the largest 4 observations of the set is increased by 2, then the median of the new set is _______.

Solution:

Since n = 9, then median term = ([9 + 1] / 2)th = 5th term.

Now, the last four observations have increased by 2.

The median is 5th observation, which is remaining unchanged.

There will be no change in the median.

Question 9: Runs scored by a batsman in 10 innings are: 38, 70, 48, 34, 42, 55, 63, 46, 54, 44 the mean deviation is ___________.

Solution:

Arranging the given data in ascending order, we have 34, 38, 42, 44, 46, 48, 54, 55,63, 70, Median, M = [46 + 48] / [2] = 47

(Here, n = 10, median is the mean of 5th and 6th items)

Mean deviation = xiMn=xi4710\frac{\sum|x_i – M|}{n} = \frac{\sum|x_i – 47|}{10}

= 13+9+5+3+1+1+7+8+16+2310\frac{13 + 9 + 5 + 3 + 1 + 1 + 7 + 8 + 16 + 23}{10}

= 8.6

Question 10: Computer the median from the following table.

Marks obtained No. of students
0-10 2
10-20 18
20-30 30
30-40 45
40-50 35
50-60 20
60-70 6
70-80 3

Solution:

Marks obtained No. of students Cumulative frequency
0-10 2 2
10-20 18 20
20-30 30 50
30-40 45 95
40-50 35 130
50-60 20 150
60-70 6 156
70-80 3 159

N = Σf = 159 (Odd number)

Median is [1 / 2] (n + 1) = [1 / 2] [(159 + 1)] = 80th value, which lies in the class [30 – 40] (see the row of cumulative frequency 95, which contains 80).

Hence the median class is [30 – 40].

We have l = Lower limit of median class = 30

f = frequency of median class = 45

C = Total of all frequencies preceding median class = 50

i = width of class interval of median class=10

Required median = l + ([N / 2 − C] / f) * i

= 30 + ([159 / 2 − 50] / 45) × 10

= 3 + 295 / 45

= 36.55

Question 11: The number of observations in a group is 40. If the average of the first 10 is 4.5 and that of the remaining 30 is 3.5, then the average of the whole group is _________.

Solution:

[x1 + x2 + ….. + x10] / [10] = 4.5

[x1 + x2 + ….. + x10] = 45 and [x11 + x12 +….. +x40] / [30] = 3.5

x11 + x12 +….. + x40 = 105

x1 + x2 + ….. + x40 = 150

[x1 + x2 + ….. + x40] / [40] = 150 / 40 = 15 / 4.

Question 12: A school has four sections of chemistry in class XII having 40, 35, 45 and 42 students. The mean marks obtained in chemistry tests are 50, 60, 55 and 45 respectively for the four sections, the overall average of marks per student is _____________.

Solution:

Total number of students = 40 + 35 + 45 + 42 = 162

Total marks obtained = (40 × 50) + (35 × 60) + (45 × 55) + (42 × 45) = 8465

Overall average of marks per student = 8465 / 162 = 52.25.

Question 13: The mean weight per student in a group of seven students is 55 kg If the individual weights of 6 students are 52, 58, 55, 53, 56 and 54; then weights of the seventh student is __________.

Solution:

Total weight of 7 students is = 55× 7 = 385 kg

Sum of weight of 6 students = 52 + 58 + 55 + 53 + 56 + 54 = 328 kg

Weight of seventh student = 385 – 328 = 57kg.

Question 14: In a class of 100 students there are 70 boys whose average marks in a subject are 75. If the average marks of the complete class are 72, then what are the average marks of the girls?

Solution:

Let the average marks of the girls students be x, then

72 = [70 × 75 + 30 x] / 100

or [7200 − 5250] / [30] = x

x = 65.

Question 15: The average weight of students in a class of 35 students is 40 kg. If the weight of the teacher is included, the average rises by 12kg; the weight of the teacher is __________.

Solution:

Let the weight of the teacher is w kg , then

40 + [1 / 2] = [35 × 40 + w] / [35 + 1]

36 × 40 + 36 × 12 = 35 × 40 + w

w=58

Weight of the teacher = 58kg.

Example 16: The mean of the values 0, 1, 2,……,n having corresponding weight nC0,nC1,nC2,..,nCn^{n}{{C}_{0}},{{\,}^{n}}{{C}_{1}},{{\,}^{n}}{{C}_{2}},……..\,,{{\,}^{n}}{{C}_{n}} respectively is

A)2nn+1B)2n+1n(n+1)C)n+12D)n2A) \frac{{{2}^{n}}}{n+1}\\ B) \frac{{{2}^{n+1}}}{n(n+1)}\\ C) \frac{n+1}{2}\\ D) \frac{n}{2}

Solution:

 The required mean is 

xˉ=0.1+1.nC1+2.nC2+3.nC3++n.nCn1+nC1+nC2+.+nCn=r=0nr.nCrr=0nnCr=r=1nr.nrn1Cr1r=0nnCr=nr=1nn1Cr1r=0nnCr=n.2n12n=n2.\bar{x}=\frac{0.1+{{1.}^{n}}{{C}_{1}}+{{2.}^{n}}{{C}_{2}}+{{3.}^{n}}{{C}_{3}}+……+n{{.}^{n}}{{C}_{n}}}{1{{+}^{n}}{{C}_{1}}{{+}^{n}}{{C}_{2}}+….{{+}^{n}}{{C}_{n}}}\\ =\frac{\sum\limits_{r=0}^{n}{r.\,{{\,}^{n}}{{C}_{r}}}}{\sum\limits_{r=0}^{n}{^{n}{{C}_{r}}}}\\=\frac{\sum\limits_{r=1}^{n}{r.\frac{n}{r}\,{{\,}^{n-1}}{{C}_{r-1}}}}{\sum\limits_{r=0}^{n}{^{n}{{C}_{r}}}}\\= \frac{n\sum\limits_{r=1}^{n}{^{n-1}{{C}_{r-1}}}}{\sum\limits_{r=0}^{n}{^{n}{{C}_{r}}}}\\ =\frac{n{{.2}^{n-1}}}{{{2}^{n}}}\\ =\frac{n}{2}.  

Example 17: In an experiment with 15 observations on x, the following results were available x2=2830,x=170.\sum {{x}^{2}}=2830, \sum x=170. On observation that was 20 was found to be wrong and was replaced by the correct value 30. Then the corrected variance is      

A) 78.00      

B) 188.66

C) 177.33    

D) 8.33

Solution:

x=170,x2=2830\sum x=170, \sum {{x}^{2}}=2830

Increase in x=10,\sum x=10,, then x=170+10=180\sum {x}’=170+10=180 

Increase in x2=900400=500,\sum {{x}^{2}}=900-400=500, then x2=2830+500=3330\sum {{{x}’}^{2}}=2830+500=3330

Variance =1nx2(xn)2=333015(18015)2=222144=78.=\frac{1}{n}\sum {{{x}’}^{2}}-{{\left( \frac{\sum {x}’}{n} \right)}^{2}}\\ =\frac{3330}{15}-{{\left( \frac{180}{15} \right)}^{2}}=222-144=78.

Example 18: The means of five observations is 4 and their variance is 5.2. If three of these observations are 1, 2 and 6, then the other two are                                       

A) 2 and 9   

B) 3 and 8   

C) 4 and 7   

D) 5 and 6

Solution:

Let the two unknown items be x and y, then                    

Mean =41+2+6+x+y5=4x+y=11(i)=4\Rightarrow \frac{1+2+6+x+y}{5}=4 \\x+y=11 \rightarrow (i)

Variance = 5. 2   

12+22+62+x2+y25(mean)2=5.241+x2+y2=5[5.2+(4)2]41+x2+y2=106x2+y2=65(ii)\frac{{{1}^{2}}+{{2}^{2}}+{{6}^{2}}+{{x}^{2}}+{{y}^{2}}}{5}-{{(\text{mean})}^{2}}=5.2 41+{{x}^{2}}+{{y}^{2}}=5[5.2+{{(4)}^{2}}]\\ 41+{{x}^{2}}+{{y}^{2}}=106\\ {{x}^{2}}+{{y}^{2}}=65 \rightarrow (ii)

Solving (i) and (ii) for x and y, we get    

x=4,y=7 or x=7,y=4.x=4,y=7 \text \ or \ x=7,y=4.

Example 19: If xˉ1{{\bar{x}}_{1}} and xˉ2{{\bar{x}}_{2}} are the means of two distributions such that xˉ1<xˉ2{{\bar{x}}_{1}}<{{\bar{x}}_{2}} and xˉ\bar{x} is the mean of the combined distribution, then

A)xˉ<xˉ1B)xˉ>xˉ2C)Xˉ=Xˉ1+Xˉ22D)xˉ1<xˉ<xˉ2A)\bar{x}<{{\bar{x}}_{1}}\\ B)\bar{x}>{{\bar{x}}_{2}}\\ C)\bar{X}=\frac{{{{\bar{X}}}_{1}}+{{{\bar{X}}}_{2}}}{2}\\ D){{\bar{x}}_{1}}<\bar{x}<{{\bar{x}}_{2}}

Solution: 

Let n1 and n2{{n}_{1}} \text \ and \ {{n}_{2}} be the number of observations in two groups having means xˉ1{{\bar{x}}_{1}} and xˉ2{{\bar{x}}_{2}} respectively. Then, 

xˉ=n1xˉ1+n2xˉ2n1+n2\bar{x}=\frac{{{n}_{1}}{{{\bar{x}}}_{1}}+{{n}_{2}}{{{\bar{x}}}_{2}}}{{{n}_{1}}+{{n}_{2}}}

Now,

xˉxˉ1=n1xˉ1+n2xˉ2n1+n2xˉ1=n2(xˉ2xˉ1)n1+n2>0,[ because xˉ2>xˉ1]xˉ>xˉ1(i)xˉxˉ2=n(xˉ1xˉ2)n1+n2<0,[ because xˉ2>xˉ1]xˉ<xˉ2(ii)\bar{x}-{{\bar{x}}_{1}}=\frac{{{n}_{1}}{{{\bar{x}}}_{1}}+{{n}_{2}}{{{\bar{x}}}_{2}}}{{{n}_{1}}+{{n}_{2}}}-{{\bar{x}}_{1}}\\ =\frac{{{n}_{2}}({{{\bar{x}}}_{2}}-{{{\bar{x}}}_{1}})}{{{n}_{1}}+{{n}_{2}}}>0,\,\,\,\,\,[\text \ because \ {{\bar{x}}_{2}}>{{\bar{x}}_{1}}] \\ \bar{x}\,\,\,>\,\,\,{{\bar{x}}_{1}} \rightarrow (i) \\ \bar{x}-{{\bar{x}}_{2}}=\frac{n({{{\bar{x}}}_{1}}-{{{\bar{x}}}_{2}})}{{{n}_{1}}+{{n}_{2}}}<0, \left[\text \ because \ {{{\bar{x}}}_{2}}\,\,>\,{{{\bar{x}}}_{1}} \right] \\ \bar{x}\,\,<\,\,\,{{\bar{x}}_{2}} \rightarrow (ii)

From (i) and (ii),

xˉ1<xˉ<xˉ2.{{\bar{x}}_{1}}<\bar{x}\,<\,\,{{\bar{x}}_{2}}.

How IIT JEE Statistics Question Papers Help Students?

JEE Main Statistics questions are important and very useful for JEE preparation. This set of questions will help students by how one question is asked in different ways. Here, we are providing a wide collection of Statistics questions asked in previous years question papers so that candidates can access all of them in one place. Students can download JEE Main previous year Statistics questions PDFs along with their solutions to practice offline.