Statistics

We have

$\sum\limits_{i = 1}^{100} {{x_i} = 400} $

$\sum\limits_{i = 1}^{100} {x_i^2 = 2425} $

The variance of the remaining observations is

${\sigma ^2} = {{\sum {x_i^2} } \over N} - {\left( {{{\sum {{x_i}} } \over N}} \right)^2}$

$ \Rightarrow {{2425} \over {97}} - {\left( {{{388} \over {97}}} \right)^2}$

$ \Rightarrow {{2425} \over {97}} - 16$

$ \Rightarrow {{2425 - 1552} \over {97}} = {{873} \over {97}} = 9$

Mean $\left( {\overline x } \right)$ = ${{{x_1} + {x_2}..... + {x_n}} \over n}$ = ${{\sum x } \over n}$

Here, Mean = 40 of 25 teachers

$\therefore$ 40 = ${{\sum x } \over {25}}$

$ \Rightarrow $ $\sum x $ = 40 $ \times $ 25 = 1000

After retireing of a 60 year old teacher, total age of 24 teachers,

x₁ + x₂ + . . . . . .x₂₄ = 1000 $-$ 60 = 940

Now a new teacher of age A year is appointed.

$\therefore$ Now total age of this 25 teachers

x₁ + x₂ + x₃ + . . . . . + x₂₅ = 940 + A

$\therefore$ Mean age = ${{940 + A} \over {25}}$

According to question,

${{940 + A} \over {25}}$ = 39

$ \Rightarrow $ A = 35

Let 5 observations are x₁, x₂, x₃, x₄, x₅

given,   x₁ = 1, x₂ = 2, x₃ = 6

Mean = 5

$ \therefore $   Mean$\left( {\overline x } \right)$ = ${{{x_1} + {x_2} + {x_3} + {x_4} + {x_5}} \over 5}$ = 5

$ \Rightarrow $   1 + 2 + 6 + x₄ + x₅ = 25

$ \therefore $   x₄ + x₅ = 16

$ \Rightarrow $   (x₄ $-$ 5) + (x₅ $-$ 5) + 10 = 16

$ \Rightarrow $   (x₄ $-$ 5) + (x₅ $-$ 5) = 6

$ \therefore $   Mean deviation about mean,

=   ${{\sum {\left| {{x_i} - \overline x } \right|} } \over n}$

=    ${{\left| {1 - 5} \right| + \left| {2 - 5} \right| + \left| {6 - 5} \right| + \left| {{x_4} - 5} \right| + \left| {{x_5} - 5} \right|} \over 5}$

=   ${{4 + 3 + 1 + 6} \over 5}$

=    ${{14} \over 5}$

=   2.8

Given numbers are,

1, 1 + d, 1 + 2d . . . . . 1 + 100d

$ \therefore $   Total 101 number are present.

$ \therefore $   n = 101

$ \therefore $   mean $\left( {\overline x } \right)$ = ${{1 + \left( {1 + d} \right) + ......\left( {1 + 100d} \right)} \over {101}}$

=   ${1 \over {101}} \times {{101} \over 2}$ [1 + (1 + 100d)]

=   1 + 50d

$ \therefore $   Mean deviation from mean

=  ${1 \over {101}}$ $\left[ {\left| {1 - \left( {1 + 50d} \right)} \right| + \left| {\left( {1 + d} \right) - \left( {1 + 50d} \right)} \right|} \right.$

          $\left. { + ...... + \left| {\left( {1 + 100d} \right) - \left( {1 + 50d} \right)} \right|} \right]$

=   ${{2\left| d \right|} \over {101}}$ ( 1 + 2 + 3 + . . . . . + 50)

=   ${{2\left| d \right|} \over {101}} \times {{50 \times 51} \over 2}$

=   ${{2550} \over {101}}\left| d \right|$

From question,

${{2550} \over {101}}\left| d \right|$ = 255

$ \Rightarrow $   $\left| d \right|$ = 10.1

The formula for standard deviation (S.D)

$ = \sqrt {{{\sum {x_i^2} } \over n} - {{\left( {{{\sum {{x_i}} } \over n}} \right)}^2}} $

Where $\sum {x_i^2 = } $ Sum of square of the numbers

$ = {2^2} + {3^2} + {a^2} + {11^2}$

$ = 4 + 9 + {a^2} + 121$

$ = 134 + {a^2}$

$\sum {{x_i}} = $ Sum of numbers

$ = 2 + 3 + a + 11$

$ = 16 + a$

$\therefore\,\,\,$ $SD = \sqrt {{{134 + a{}^2} \over 4} - {{\left( {{{16 + a} \over 4}} \right)}^2}} $

$ \Rightarrow \sqrt {{{134 + {a^2}} \over 4} - {{\left( {{{16 + a} \over 4}} \right)}^2}} = 3.5 = {7 \over 2}$ (given)

$ \Rightarrow {{134 + {a^2}} \over 4} - {\left( {{{16 + a} \over 4}} \right)^2} = {{49} \over 4}$

$ \Rightarrow 4\left( {134 + {a^2}} \right) - \left( {256 + 32a + {a^2}} \right) = 4 \times 49$

$ \Rightarrow 3{a^2} - 32a + 84 = 0$

Initially we have $16$ observations and among them one is $16.$

So, we have $15$ unknowns. Let those are ${a_1},a{}_2,{a_3}.....{a_{15}}$

$\therefore\,\,\,$ Mean of $16$ datal set

$ = {{{a_1} + {a_2} + .....{a_{15}} + 16} \over {16}}$

According to the question,

${{{a_1} + {a_2} + .....{a_{15}} + 16} \over {16}} = 16$

$ \Rightarrow {a_1} + {a_2} + ...... + {a_{15}} = 256 - 16 = 240$

Now we deleted $16$ and replaced by there new numbers $3,4,$ and $5.$

So, new mean

$ = {{{a_1} + {a_2} + .......{a_{15}} + \left( {3 + 4 + 5} \right)} \over {18}}$

$ = {{240 + \left( {3 + 4 + 5} \right)} \over {18}}$

$ = {{240 + 12} \over {18}}$

$=14$

Here is total $50$ numbers, so $N=50$

Variance $ = $ ${{\sum {{x^2}} } \over {50}} - {\left( {{{\sum x } \over {50}}} \right)^2}$

Here $\sum {{x^2}} = $ sum of square of first $50$ even natural number.

$ = {2^2} + {4^2} + ..... + {100^2}$

$ = {2^2}\left[ {{1^2} + {2^2} + ....... + {{50}^2}} \right]$

$ = 4\left[ {{{50 \times 51 \times 101} \over 6}} \right]$

So, ${{\sum {{x^2}} } \over {50}} = {{4 \times 51 \times 101} \over 6} = 3434$

$\sum {x = } $ sum of first $50$ even natural numbers

$ = 2 + 4 + ...... + 100$

$ = 2\left[ {1 + 2 + .... + 50} \right]$

$ = 2\left[ {{{50 \times 51} \over 2}} \right]$

$ = 50 \times 51$

$\therefore\,\,\,$ ${\left( {{{\sum x } \over {50}}} \right)^2} = {\left( {{{50 \times 51} \over {50}}} \right)^2} = 2601$

$\therefore\,\,\,$ Variance $ = 3434 - 2601$ $=833$

As we know variance does not change with the change of origin. So, here even after adding grace marks $10$, the variance will be same.

Let's see with an example,

Assume initial variance $ = {{\sum {{{\left( {{x_i} - \overline x } \right)}^2}} } \over N}$

After adding grace marks $10$ with each student,

the final variance $ = {{{\sum {\left[ {\left( {{x_i} + 10} \right) - \left( {\overline x + 10} \right)} \right]} } \over N}^2}$

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

$ = $ Initial variance.

Given that,

for ${x_1},{x_2},....{x_n},$ $A.M = \overline x $

and variance $ = {\sigma ^2}$

Now A.M of

$2{x_1},2x{}_2.....2{x_n} = {{2\left( {{x_1} + {x_2} + ....{x_n}} \right)} \over n} = 2\overline x $

But given $A.M = 4\overline x $

$\therefore\,\,\,$ Statement ${\rm I}{\rm I}$ is false.

Variance of $2{x_1},2{x_2}......2{x_n}$

$=$ Variance of $\left\{ {2{x_i}} \right\}$

$ = {2^2}$ Variance of $\left\{ {{x_i}} \right\} = 4{\sigma ^2}$

So, statement ${\rm I}$ is correct.

NOTE :

If total no of terms are even then median

$ = {1 \over 2}$ [ ${n \over 2}$th term $ + \left( {{n \over 2} + 1} \right)$ th term]

Here total terms $ = 50,$ which is even

$\therefore$ $\,\,\,$ Median $ = {1 \over 2}$ [ ${{50} \over 2}$ th term $ + \left( {{{50} \over 2} + 1} \right)$ th term]

$ = {1 \over 2}$ [ $25$ th term $+$ $26$ th term ]

$ = {1 \over 2}$ [ $25a$ $+$ $26a$ ]

$=25.5a$

Mean deviation (M.D.) about the median

$ = {{\sum\limits_{i = 1}^{50} {\left| {{x_i} - Median} \right|} } \over N} = 50$ (given)

$\therefore$ ${1 \over {50}}\left[ {2 \times \left| a \right| \times \left( {0.5 + 1.5 + 2.5 + ....24.5} \right)} \right] = 50$

$ \Rightarrow 2\left| a \right| \times {{25} \over 2} \times 25 = 2500$

$ \Rightarrow \left| a \right| = {{2500} \over {25 \times 25}} = 4$

Given that,

${\sigma _1}^2 = 4$

and ${\sigma _2}^2 = 5$

And also given,

$\overline x = 2\,\,$ and $\overline y = 4\,\,$

So, $\,\,\,$ ${{\sum {{x_i}} } \over 5} = 2$

$ \Rightarrow \sum {{x_i}} = 10$ v

and ${{\sum {{y_i}} } \over 5} = 4$

$ \Rightarrow \,\,\,\sum {{y_i}} = 20$

$\therefore\,\,\,$ ${\sigma _1}^2 = {{\sum {x_i^2} } \over 5} - {\left( {\overline x } \right)^2}$

$ \Rightarrow \,\,\,4 = {{\sum {x_i^2} } \over 5} - 4$

$ \Rightarrow \,\,\,\sum {x_i^2} = 40$

and ${\sigma _2}^2 = {{\sum {y_i^2} } \over 5} - {\left( {\overline y } \right)^2}$

$ \Rightarrow \,\,\,\,5 = {{\sum {y_i^2} } \over 5} - 16$

$ \Rightarrow \,\,\,\sum {y_i^2} = 105$

Variance of combined data set

${\sigma ^2} = {1 \over {10}}\left( {\sum {x_i^2 + \sum {y_i^2} } } \right) - {\left( {{{\overline x + \overline y } \over 2}} \right)^2}$

$ = {1 \over {10}}\left( {40 + 105} \right) - 9$

$ = {{145 - 90} \over {10}}$

$ = {{55} \over {10}}$

$ = {{11} \over {2}}$

Mean $\,\,\,\,\left( {\overline x } \right) = {{Sum\,\,of\,\,numbers} \over n}$

$ = {{{n \over 2}\left( {a + l} \right)} \over n}$

$ = {1 \over 2}\left( {1 + l + 100d} \right)$

$ = 1 + 50\,d.$

Mean deviation (M.D) $ = {1 \over n}\sum\limits_{i = 1}^{101} {\left| {{x_i} - \overline x } \right|} $

$ = {1 \over {101}}$[ 50d + 49d + 48d + .......d + 0 + ..... + 50d]

$ = {1 \over {101}}.2d\left( {1 + 2 + .... + 50} \right)$

$ = {1 \over {101}}.2d.{{50 \times 51} \over 2}$

$ = {{50 \times 51\,d} \over {101}}$

Given that M.D = 255

$\therefore\,\,\,$ ${{50 \times 51\,d} \over {101}} = 255$

$ \Rightarrow \,\,\,d = {{101 \times 255} \over {51 \times 50}}$

$ = 10.1$

Let first n even natural numbers = 2,4, 6, 8 ...... 2n

$\therefore$ Sum of those num = 2 + 4 + 6 + ..... 2n

= 2 (1 + 2 + ..... n)

= $2.{{n\left( {n + 1} \right)} \over 2}$

= n (n + 1)

$\therefore\,\,\,$ Mean $\left( {\overline x } \right) = {{n\left( {n + 1} \right)} \over n}$ $ = n + 1$

$\therefore\,\,\,$ Variance $ = {1 \over n}\sum {x_i^2} - {\left( {\overline x } \right)^2}$

$ = {1 \over n}\left[ {{2^2} + {4^2} + ...... + {{\left( {2n} \right)}^2}} \right] - {\left( {n + 1} \right)^2}$

$ = {1 \over n}{2^2}\left[ {{1^2} + {2^2} + .......n{}^2} \right] - \left( {n + 1} \right){}^2$

$ = {4 \over n}\left[ {{{n\left( {n + 1} \right)\left( {2n + 1} \right)} \over 6}} \right] - {\left( {n + 1} \right)^2}$

$ = {{\left( {n + 1} \right)\left[ {2\left( {2n + 1} \right) - 3\left( {n + 1} \right)} \right]} \over 3}$

$ = {{\left( {n + 1} \right)\left( {n - 1} \right)} \over 3}$

$ = {{{n^2} - 1} \over 3}$

$\therefore$ Statement 1 is false.

Statement 2 is true as those are standard formula.

Given that,

Mean of a, b, 8, 5, 10 = 6

$\therefore\,\,\,$ ${{a + b + 8 + 5 + 10} \over 5} = 6$

$ \Rightarrow \,\,\,$ a + b + 23 = 30

$ \Rightarrow \,\,\,$ a + b = 7 ....... (1)

Variance $ = {{\sum {{{\left( {{x_i} - A} \right)}^2}} } \over n} = 6.8$

$ \Rightarrow \,\,\,{{{{\left( {6 - a} \right)}^2} + {{\left( {6 - b} \right)}^2} + {{\left( {6 - 8} \right)}^2} + {{\left( {6 - 5} \right)}^2} + {{\left( {6 - 10} \right)}^2}} \over 5} = 6.5$

$ \Rightarrow \,\,\,{\left( {6 - a} \right)^2} + {\left( {6 - b} \right)^2} + 4 + 1 + 16 = 34$

$ \Rightarrow {\left( {6 - a} \right)^2} + {\left( {6 - b} \right)^2} = 13$

$ \Rightarrow \,\,\,{a^2} + {b^2} = 25$

By using (1) we get,

${a^2} + \left( {7 - a} \right){}^2 = 25$

$ \Rightarrow \,\,\,a{}^2 - 7a + 12 = 0$

$\therefore\,\,\,$ a = 3, 4

then b = 4, 3.

Let x and y are number of boys and girls in a class respectively.

$\therefore$ 52x + 42y = 50 (x + y)

$ \Rightarrow$ 52x + 42y = 50x + 50y

$ \Rightarrow$ 2x = 8y

$ \Rightarrow$ x = 4y

$\therefore$ Total no. of students = x + y = 4y + y = 5y

$\therefore$ Percentage of boys = ${{4y} \over {5y}} \times 100\,\% $ = 80%

Series A = 101, 102 ............ 200

Series B = 151, 152 ............ 250

Here series B can be obtained if we change the origin of A by 50 units.

And we know the variance does not change by changing the origin.

So, $\,\,\,\,$ ${V_A} = {V_B}$

$ \Rightarrow \,\,\,\,\,{{{V_A}} \over {{V_B}}} = 1$

Given that,

Mean = 21 and median = 22

We know,

Mode + 2 Mean = 3 Median

$\therefore$ Mode = 3 $ \times $ 22 $-$ 2 $ \times $ 21

= 66 $-$ 42

= 24

As we know,

${\sigma ^2} \ge 0$

$\therefore\,\,\,$ ${{\sum {x_i^2} } \over n} - {\left( {{{\sum {{x_i}} } \over n}} \right)^2} \ge 0$

$ \Rightarrow \,\,\,{{400} \over n} - {{6400} \over {{n^2}}} \ge 0$

$ \Rightarrow \,\,\,n \ge 16$

$\therefore\,\,\,$ Possible value of n according to the option is = 18

The statements are analyzed as follows :

(a) Mode can be computed from histogram : This is correct. The mode is the value that appears most frequently in a data set. A histogram provides a graphical representation of the frequency of each data value. The data value corresponding to the highest bar in a histogram is the mode.

(b) Median is not independent of change of scale : This is correct. The median is the middle value in a data set when the values are arranged in ascending or descending order. When you change the scale (e.g., by multiplying all data points by a constant), the median changes as well.

(c) Variance is independent of change of origin and scale : This is not correct. Variance, which measures the dispersion of a set of data points, is affected by changes in scale (it is not independent of scale). If you multiply all the values in a dataset by a constant, the variance gets multiplied by the square of that constant. However, it is true that variance is independent of the change of origin (it remains the same if you add or subtract a constant from all data points in the set).

Therefore, the correct answer is Option C : only (a) and (b) are correct.

Mean $\left( A \right) = {{a - a} \over {2n}} = 0$

Given standard deviation (S.D) = 2

$\therefore\,\,\,$ $\sqrt {{{\sum {{{\left( {x - A} \right)}^2}} } \over {2n}}} = 2$

$ \Rightarrow \,\,\,\sqrt {{{{{\left( {a - 0} \right)}^2} + {{\left( {a - 0} \right)}^2} + ..... + {{\left( {0 - a} \right)}^2}} \over {2n}}} = 2$

$ \Rightarrow \,\,\,\sqrt {{{{a^2} + {a^2}........2n\,times} \over {2n}}} = 2$

$ \Rightarrow \,\,\,\sqrt {{{2n\,.\,{a^2}} \over {2n}}} = 2$

$ \Rightarrow \,\,\,\sqrt {{a^2}} = 2$

$ \Rightarrow \,\,\,\left| a \right| = 2$

Here total no of observation is 9 which is a odd number. As we know for odd number 9 the median will be the 5^th term.

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

Given that,

N = 15, $\sum {x{}^2} = 2830,\,\sum x = 170$

As, 20 was replaced by 30 then,

$\sum x = 170 - 20 + 30 = 180$

and $\sum {{x^2}} = 2830 - 400 + 900 = 3330$

So, the corrected variance

$ = {{\sum {{x^2}} } \over N} - {\left( {{{\sum x } \over N}} \right)^2}$

$ = {{3330} \over {15}} - {\left( {{{180} \over {15}}} \right)^2}$

$ = 222 - {12^2}$

$ = 222 - 144$

$ = 78$

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