Statistics — Page 3 | JEE Mathematics

Q21

The sum of 100 observations and the sum of their squares are 400 and 2475, respectively. Later on, three observations, 3, 4 and 5, were found to be incorrect. If the incorrect observations are omitted, then the variance of the remaining observations is :

A 8.25

B 8.50

C 8.00

D 9.00

Correct Answer

Option D

Solution

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

Q22

For two data sets, each of size 5, the variances are given to be 4 and 5 and the corresponding means are given to be 2 and 4, respectively. The variance of the combined data set is

A

{5 \over 2}

B

{11 \over 2}

C 6

D

{13 \over 2}

Correct Answer

Option B

Solution

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}}

Q23

If the mean deviation about the median of the numbers a, 2a,........., 50a is 50, then |a| equals

A 4

B 5

C 2

D 3

Correct Answer

Option A

Solution

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

Q24

Let x1, x2,........., xn be n observations, and let

\overline x

be their arithematic mean and

{\sigma ^2}

be their variance. Statement 1 : Variance of 2x1, 2x2,......., 2xn is 4

{\sigma ^2}

. Statement 2 : : Arithmetic mean of 2x1, 2x2,......, 2xn is 4

\overline x

.

A Statement 1 is false, statement 2 is true

B Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1

C Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1

D Statement 1 is true, statement 2 is false

Correct Answer

Option D

Solution

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.

Q25

All the students of a class performed poorly in Mathematics. The teacher decided to give grace marks of 10 to each of the students. Which of the following statistical measures will not change even after the grace marks were given?

A median

B mode

C variance

D mean

Correct Answer

Option C

Solution

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.

Q26

The variance of first 50 even natural numbers is

A 833

B 437

C

{{437} \over 4}

D

{{833} \over 4}

Correct Answer

Option A

Solution

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

Q27

The mean of the data set comprising of 16 observations is 16. If one of the observation valued 16 is deleted and three new observations valued 3, 4 and 5 are added to the data, then the mean of the resultant data, is :

A 15.8

B 14.0

C 16.8

D 16.0

Correct Answer

Option B

Solution

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

Q28

If the standard deviation of the numbers 2, 3, a and 11 is 3.5, then which of the following is true?

A 3

a

2 - 26

a

+ 55 = 0

B 3

a

2 - 32

a

+ 84 = 0

C 3

a

2 - 34

a

+ 91 = 0

D 3

a

2 - 23

a

+ 44 = 0

Correct Answer

Option B

Solution

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

Q29

The mean age of 25 teachers in a school is 40 years. A teacher retires at the age of 60 years and a new teacher is appointed in his place. If now the mean age of the teachers in this school is 39 years, then the age (in years) of the newly appointed teacher is :

A 25

B 30

C 35

D 40

Correct Answer

Option C

Solution

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

Q30

If

\sum\limits_{i = 1}^9 {\left( {{x_i} - 5} \right)} = 9

and

\sum\limits_{i = 1}^9 {{{\left( {{x_i} - 5} \right)}^2}} = 45

, then the standard deviation of the 9 items

{x_1},{x_2},.......,{x_9}

is

A 3

B 9

C 4

D 2

Correct Answer

Option D

Solution

IMPORTANT POINT :- When every number is added or subtracted by a fixed number then the standard Deviation remain unchanged. so let

{x_i} - 5 = {y_i}

So, new equation is

\sum\limits_{i = 1}^9 {{y_i}} = 9

and

\sum\limits_{i = 1}^9 {y_i^2} = 45

As, we know. Standard Deviation (S.D)

= \sqrt {{{\sum\limits_{i = 1}^9 {y_i^2} } \over 9} - {{\left( {{{\sum\limits_{i = 1}^9 {yi} } \over 9}} \right)}^2}}

= \sqrt {{{45} \over 9} - {{\left( {{9 \over 9}} \right)}^2}}

= \sqrt {5 - 1}

$=$

\sqrt 4

=2