$\begin{aligned} & \text { Mean } \Rightarrow \bar{x}=\frac{\sum\limits_{i=1}^7 x_i}{7}=\frac{\frac{7}{2}[2 a+6 d]}{7}=a+3 d=x_4 \\\\ & \text { Variance }=\frac{\sum\limits_{i=1}^7\left(x_i-\bar{x}\right)^2}{7}=(4)^2 \Rightarrow \frac{\sum\limits_{i=1}^7\left(x_i-x_4\right)^2}{7}=16 \\\\ & \Rightarrow \frac{(3 d)^2+(2 d)^2+d^2+0+d^2+(2 d)^2+(3 d)^2}{7}=16 \\\\ & =4 d^2=16 \Rightarrow d=2 \\\\ & \Rightarrow \bar{x}=9+3(2)=15 \\\\ & x_6=a+5 d=9+5(2)=19 \Rightarrow \bar{x}+x_6=34\end{aligned}$

Statistics Questions — JEE Mathematics

Q1

If the mean of the data : 7, 8, 9, 7, 8, 7,

\lambda

, 8 is 8, then the variance of this data is :

A

{7 \over 8}

B 1

C

{9 \over 8}

D 2

Correct Answer

Option B

Solution

\overline x

=

{{7 + 8 + 9 + 7 + 8 + 7 + \lambda + 8} \over 8}

= 8 $\Rightarrow$

\,\,\,

{{54 + \lambda } \over 8}

= 8 $\Rightarrow$ $\lambda$ = 10 Now variance = $\sigma$ 2 =

{{{{\left( {7 - 8} \right)}^2} + {{\left( {8 - 8} \right)}^2} + {{\left( {9 - 8} \right)}^2} + {{\left( {7 - 8} \right)}^2} + {{\left( {8 - 8} \right)}^2} + {{\left( {7 - 8} \right)}^2} + {{\left( {10 - 8} \right)}^2} + {{\left( {8 - 8} \right)}^2}} \over 8}

$\Rightarrow$ $\sigma$ 2 =

{{1 + 0 + 1 + 1 + 0 + 1 + 4 + 0} \over 8}

=

{8 \over 8}

= 1 Hence, the variance is 1.

Q2

The mean and the variance of five observations are 4 and 5.20, respectively. If three of the observations are 3, 4 and 4 ; then the absolute value of the difference of the other two observations, is :

A 1

B 7

C 3

D 5

Correct Answer

Option B

Solution

mean

\overline x

= 4, $\sigma$ 2 = 5.2, n = 5, . x1 = 3 x2 = 4 = x3

\sum {{x_i}} = 20

x4 + x5 = 9 . . . . . . (i)

{{\sum {x_i^2} } \over x} - {\left( {\overline x } \right)^2} = \sigma \Rightarrow \sum {x_i^2} = 106

x_4^2 + x_5^2 = 65

. . . . . .(ii) Using (i) and (ii) (x4 $-$ x5)2 = 49

\left| {{x_4} - {x_5}} \right| = 7

Q3

If mean and standard deviation of 5 observations x1, x2, x3, x4, x5 are 10 and 3, respectively, then the variance of 6 observations x1, x2, ….., x5 and –50 is equal to

A 582.5

B 507.5

C 586.5

D 509.5

Correct Answer

Option B

Solution

\overline x = 10 \Rightarrow \sum\limits_{i = 1}^5 {{x_i} = 50}

S.D.

= \sqrt {{{\sum\limits_{i = 1}^5 {x_i^2} } \over 5} - {{\left( {\overline x } \right)}^2}} = 8

\Rightarrow \,\sum\limits_{i = 1}^5 {{{\left( {{x_i}} \right)}^2}} = 109

variance

= \,\,{{\sum\limits_{i = 1}^5 {{{\left( {{x_i}} \right)}^2}} + {{\left( { - 50} \right)}^2}} \over 6} - \left( {\sum\limits_{i = 1}^5 {{{{x_i} - 50} \over 6}} } \right)

= \,\,507.5

Q4

Let $$9=x_{1}

A

2\left(9+\dfrac{8}{\sqrt{7}}\right)

B 25

C

18\left(1+\dfrac{1}{\sqrt{3}}\right)

D 34

Correct Answer

Option D

Solution

$\begin{aligned} & \text{ Mean } \Rightarrow \bar{x}=\dfrac{\sum\limits_{i=1}^7 x_i}{7}=\dfrac{\dfrac{7}{2}[2 a+6 d]}{7}=a+3 d=x_4 \\\\ & \text{ Variance }=\dfrac{\sum\limits_{i=1}^7\left(x_i-\bar{x}\right)^2}{7}=(4)^2 \Rightarrow \dfrac{\sum\limits_{i=1}^7\left(x_i-x_4\right)^2}{7}=16 \\\\ & \Rightarrow \dfrac{(3 d)^2+(2 d)^2+d^2+0+d^2+(2 d)^2+(3 d)^2}{7}=16 \\\\ & =4 d^2=16 \Rightarrow d=2 \\\\ & \Rightarrow \bar{x}=9+3(2)=15 \\\\ & x_6=a+5 d=9+5(2)=19 \Rightarrow \bar{x}+x_6=34\end{aligned}$

Q5

The mean of five observations is 5 and their variance is 9.20. If three of the given five observations are 1, 3 and 8, then a ratio of other two observations is -

A 6 : 7

B 10 : 3

C 4 : 9

D 5 : 8

Correct Answer

Option C

Solution

Let two observations are x1 & x2 mean =

{{\sum {{x_i}} } \over 5} = 5

\Rightarrow 1 + 3 + 8 + {x_1} + {x_2} = 25

\Rightarrow {x_1} + {x_2} = 13

. . . . (1) variance

\left( {{\sigma ^2}} \right)

=

{{\sum {x_i^2} } \over 5} - 25 = 9.20

$\Rightarrow$

{\sum {x_i^2 = 171} }

$\Rightarrow$

x_1^2 + x_2^2 = 97

. . . . . (2) $\Rightarrow$ (x1 + x2)2 $-$ 2x1x2 = 97 $\Rightarrow$ 169 - 2x1x2 = 97 or x1x2 = 36 $\therefore$ x1 : x2 = 4 : 9

Q6

5 students of a class have an average height 150 cm and variance 18 cm2. A new student, whose height is 156 cm, joined them. The variance (in cm2) of the height of these six students is :

A 16

B 22

C 20

D 18

Correct Answer

Option C

Solution

Average height of 5 students,

\overline x = {{{x_1} + {x_2} + {x_3} + {x_4} + {x_5}} \over 5} = 150

\Rightarrow \,\,\,\sum\limits_{i = 1}^5 {{x_i}} = 750

We know, Variance

\left( \sigma \right) = {{\sum {x_i^2} } \over 5} - {\left( {\overline x } \right)^2}

given that,

{{\sum {x_i^2} } \over 5} - {\left( {150} \right)^2} = 18

\Rightarrow \,\,\,\sum {x_i^2} = 112590

Height of new student, x6 $=$ 156 cm New average height

\left( {{{\overline x }_{new}}} \right) = {{750 + 156} \over 6} = 151

New variance

= {{\,\sum\limits_{i = 1}^6 {x_i^2} } \over 6} - {\left( {{{\overline x }_{new}}} \right)^2}

= {{112590 + {{\left( {156} \right)}^2}} \over 6} - {\left( {151} \right)^2}

= 22821 - 22801

= 20

Q7

Let sets A and B have 5 elements each. Let the mean of the elements in sets A and B be 5 and 8 respectively and the variance of the elements in sets A and B be 12 and 20 respectively. A new set C of 10 elements is formed by subtracting 3 from each element of

\mathrm{A}

and adding 2 to each element of

\mathrm{B}

. Then the sum of the mean and variance of the elements of

\mathrm{C}

is ___________.

A 36

B 40

C 38

D 32

Correct Answer

Option C

Solution

To solve this problem, let's break it down step by step : Step 1 : Determine the mean of set C The mean of set A = $5$ The mean of set B = $8$ After subtracting 3 from each element in set A, the new mean becomes $5 - 3 = 2$ .

After adding 2 to each element in set B, the new mean becomes $8 + 2 = 10$ .

When we combine both modified sets to form set C, the mean of set C is the weighted average of the means of these modified sets: Mean of C = $\dfrac{5 \times 2 + 5 \times 10}{10} = \dfrac{10 + 50}{10} = 6$ Step 2 : Determine the variance of set C Variance is defined as the expectation of the squared deviation of a random variable from its mean.

One property of variance is that, if you add (or subtract) a constant from each data point in a set, the variance of the set does not change.

Thus, the variance of the elements in set A remains $12$ even after subtracting 3 from each element, and the variance of the elements in set B remains $20$ even after adding 2 to each element.

Now, when combining variances from two datasets into one : Variance of C = $\dfrac{n_1 \times \text{Variance of A} + n_1 \times (\text{Mean of modified A} - \text{Mean of C})^2 + n_2 \times \text{Variance of B} + n_2 \times (\text{Mean of modified B} - \text{Mean of C})^2}{n_1 + n_2}$ Given : $n_1 = n_2 = 5$ Variance of A = $12$ , Variance of B = $20$ Mean of modified A = $2$ , Mean of modified B = $10$ , Mean of C = $6$ Plugging in the values, we get : Variance of C = $\dfrac{5 \times 12 + 5 \times (2 - 6)^2 + 5 \times 20 + 5 \times (10 - 6)^2}{10}$ Variance of C = $\dfrac{60 + 80 + 100 + 80}{10} = \dfrac{320}{10} = 32$ Step 3 : Sum of the mean and variance of set C Sum = Mean of C + Variance of C = $6 + 32 = 38$ So, the correct answer is : Option C : 38.

Q8

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 :

A is increased by 2

B is decreased by 2

C is two times the original median

D remains the same as that of the original set

Correct Answer

Option D

Solution

Here total no of observation is 9 which is a odd number.

As we know for odd number 9 the median will be the 5th term.

Now question says, you increase largest 4 number by 2 which does not affect the 5th term so the new median will be the same.

Q9

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 is 72, then what is the average marks of the girls?

A 73

B 65

C 68

D 74

Correct Answer

Option B

Solution

Given that, total students = 100 and number of boys = 70 $\therefore$ No. of girls = 100 - 70 = 30 Average of 100 students = 72 $\therefore$ Total marks of 100 students = 100 $\times$ 72 = 7200 Average of 70 boys = 75 $\therefore$ Total marks of 70 boys = 70 $\times$ 75 = 5250 $\therefore$ Total marks of 30 girls = 7200 $-$ 5250 = 1950 $\therefore$ Average marks of 30 girls =

{{1950} \over {30}}

= 65

Q10

Consider three observations a, b, and c such that b = a + c. If the standard deviation of a + 2, b + 2, c + 2 is d, then which of the following is true?

A b2 = 3(a2 + c2) + 9d2

B b2 = 3(a2 + c2)

-

9d2

C b2 = 3(a2 + c2 + d2)

D b2 = a2 + c2 + 3d2

Correct Answer

Option B

Solution

For a, b, c mean =

\overline x = {{a + b + c} \over 3}

\overline x = {{2b} \over 3}

We know, S.D. of a + 2, b + 2, c + 2 = S.D. of a, b, c = d

{d^2} = {{{a^2} + {b^2} + {c^2}} \over 3} - {{4{b^2}} \over 9}

{b^2} = 3{a^2} + 3{c^2} - 9{d^2}