Mathematics Must Do Problems Of Statistics For NDA

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Must Do Problems Of Statistics For NDA

Must Do Problems Of Statistics

Q 2886134077

An incomplete frequency distribution is given below.
Total of the frequency is `229`.
The value of missing frequency x is

(A)

`35`

(B)

`24`

(C)

`34`

(D)

`44`

Solution:

` sum_(i = 1)^n f_i = 229`

`=> 12 + 30 + x + 65 + 45 + 25 + 18 = 229`

`=> x + 195 = 229 => x = 34`

Correct Answer is `=>` (C) `34`

Q 2826823771

If ` sum_(i=1)^n (x_1 - 2) = 110, sum_(i = 1)^n (x_1 - 5) = 20`, then what is the mean?

(A)

`(11)/2`

(B)

`2/(11)`

(C)

`(17)/3`

(D)

`(17)/9`

Solution:

`∵ sum_(i = 1)^n (x_i - 2) = 110`

`:. x_1 +x_2 + ... + x_n - 2 n = 110`

`=> x_1 + x_2 + ... + x_n = 2n + 110` .....(i)

and ` sum_(i = 1)^n ( x_i - 5) = 20 => x_1 + x_2 + ... + x_n = 20`

`=> x_1 + x_2 + ... + x_n = 5n + 20` .......(ii)

From Eqs. (i) and (ii), we get ` 5n + 20 = 2n + 110`

` => 3n = 90 => n = 30`

Now, mean ` = ( x_1 + x_2 + ... + x_n)/n = ( 5 xx 30 + 20)/(30) = (170)/(30) = (17)/3`

Correct Answer is `=>` (C) `(17)/3`

Q 1777101086

If `A` is the arithmetic mean and `G_1, G_2` be two geometric mean between

any two numbers, then prove that `2A = G_1^2/G_2 + G_2^2/G_1` .

NCERT Exemplar

Solution:

Let the numbers be `a` and `b`.

Then, `A = (a + b)/2 `

` => 2A = a + b` ........(1)

and `G_1, G_2` be geometric mean between `a` and `b`, then `a, G_1, G_2 , b` are in `GP`.

Let `r` be the common ratio.

Then, `b = ar^(4 - 1)`

`=> b = ar^3 => b/a = r^3`

`:. r = (b/a)^(1//3)`

Now. ` G_1 = ar = a(b/a)^(1//3) quad [∵ r = (b/a)^(1//3)]`

and ` G_2 = ar^2 = a(b/a)^(2//3)`

` RHS = G_1^2/G_2 + G_2^2/G_1 = ( [ a(b/a)^(1//3)]^2)/(a(b/a)^(2//3)) + ([a(b/a)^(2//3)]^2)/(a(b/a)^(1//3))`

` = ( a^2 (b/a)^(2//3))/( a(b/a)^(2//3)) + ( a^2 (b/a)^(4//3))/(a(b/a)^(1//3))`

` = a + a(b/a) = a + b = 2A quad` [using Eq. (1)]

` = LHS`

Q 1710478319

If a variable takes discrete values `a+ 4, a- 35,
a- 2.5, a- 3, a- 2, a+ 0.5, a+ 5` and `a- 0.5`,
where `a > 0`, then the median of the data set is
CDS 2015

(A)

`a-2.5`

(B)

`a-1.25`

(C)

`a-1.5`

(D)

`a-0.75`

Solution:

We have, discrete value `a + 4, a - 3.5, a - 2.5, a - 3, a- 2,
a 0.5,a +5` and `a -0.5`

Now, ascending order is `a - 3.5, a - 3, a - 2.5, a -2,`

` a - 0.5,a+0.5,a+4,a+5`

Now ' median `=(a-2+a-0.5)/2`

`=(2a-2.5)/2=a-1.25`

Correct Answer is `=>` (B) `a-1.25`

Q 1669378215

Consider the following (fequency distribution).

What is the median of the distribution?
CDS 2015

(A)

`37`

(B)

`38`

(C)

`39`

(D)

`40`

Solution:

From the table,

`I_1=30,I_2=40,f=10` and `C=16`

`:.` Mean `=I_1+(I_2-I_1)/f(N/2-C)`

`=30+(40-30)/10(25-16)`

`=30+10/10 xx 9=30+9=39`

Correct Answer is `=>` (C) `39`

Q 2836623572

The standard deviation in a variable `x` is `sigma`. The standard deviation of the variable `(ax + b)/c` ; where a, b and c are constants, is

(A)

`(a/c) sigma`

(B)

`|a/c| sigma`

(C)

`(a^2/c^2) sigma`

(D)

None of these

Solution:

Let ` y = (ax + b)/c`

`=> y = a/c x + b/c`

` => y = Ax + B`,

where `A= a/c , B= b/c`

So, `y = A bar x + B`

`:. y - bar y =(Ax + B)- (A bar x + B)`

` = A (x - bar x)`

`=> (y - bar y)^2 = A^2 (x - bar x)^2`

`=> sum (y - bar y)^2 = A^2 sum (x - bar x)^2`

`=> n sigma _y^2 = A^2 (n sigma_x^2)`

` => sigma _y = A | A | sigma_x`

`:. sigma _y = |a/c| sigma`

Correct Answer is `=>` (B) `|a/c| sigma`

Q 2816823770

In a study on the relationship between investment (x) and profit (y), the two regression equations `3x + y - 12 = 0` and `x + 2y - 14 = 0` were obtained based on the data on x and y. What is the mean `bar x` ?

(A)

`6`

(B)

`5`

(C)

`4`

(D)

`2`

Solution:

Since, lines of regression passes through `(bar x , bar y )`.

`:. 3 bar x + bar y - 12 = 0` .......(i)

and `bar x + 2 bar y - 14 = 0` ........(ii)

On solving Eqs. (i) and (ii), we get `bar x = 2` and `bar y = 6`

Correct Answer is `=>` (D) `2`

Q 2856123974

The mean of `7` observations is `10` and that of `3` observations is `5`. What is the mean of all the `10` observations?

(A)

`15`

(B)

`10`

(C)

`8.5`

(D)

`7.5`

Solution:

Given, mean of `7` observations `= 10`

` => ( sum_(i = 1)^7 X_i)/7 = 10 => sum_(i = 1)^7 X_i = 70` ........(i)

and mean of `3` observations ` = 5 => ( sum_(i = 1)^3 X_i)/3 = 5`

`=> sum_(i = 1)^3 X_i = 15` .......(ii)

On adding Eqs. (i) and (ii), we get

` = sum_(i = 1)^7 X_i + sum_(i = 1)^3 X_i = 70 + 15 => sum_(i = 1)^(10) X_i = 85`

`:.` Mean of `10` observations ` = ( sum_(i = 1)^(10) X_i)/(10) = (85)/(10) = 8.5`

Correct Answer is `=>` (C) `8.5`

Q 2876823776

What is the value of `n` for which the numbers `1, 2, 3, ... , n` have variance `2`?

(A)

`4`

(B)

`5`

(C)

`6`

(D)

`8`

Solution:

Mean of the numbers ` = ((n(n+1))/2)/n = (n + 1)/2`

`:.` Variance ` = ((1 - (n +1)/2)^2 + ( 2 - (n + 1)/2)^2 + ... + ( n - (n + 1)/2)^2)/n`

`=> 2 = ((1^2 + 2^2 + 3^2 + ... + n^2) + n( (n + 1)/2)^2 - 2 ( (n + 1)/2) ( 1 + 2 + 3 + ...) )/n`

`=> 2n = 1/6 n (n + 1)(2n + 1) + (n (n + 1)^2)/4 - 2 ( (n+1)/2) { (n(n+1))/2 }`

`=> 2n = n(n +1) [ (2n+1)/6 + (n +1 )/4 - (n =1)/2 ]`

`=> 2 = (n + 1) [ ( 4n + 2 - 3n - 3)/(12) ]`

`=> n^2 - 1 = 24 => n^2 = 25 => n = ± 5`

`:. n = 5` [since, n cannot be negative]

Correct Answer is `=>` (B) `5`

Q 2314134050

If each observation of a raw data whose

variance is `sigma^2`, is multiplied by `h`, then

the variance of new set is
BITSAT Mock

(A)

`sigma^2`

(B)

`h^2 sigma^2`

(C)

`h sigma^2`

(D)

`h + sigma^2`

Solution:

`sigma^2 = 1/n sum x_(i)^2 - (1/n sum x_i)^2`

Now `x_i` is changed by `hx_i`, let

variance be `(sigma ')^2`.

`(sigma ')^2 =1/n sum (hx_i)^2 - [1/n sum (hx_i) ]^2`

`= h^2 [1/n sum x_(i)^2 - (1/n sum x_i)^2 ]`

`= h^2 sigma^2`

Correct Answer is `=>` (B) `h^2 sigma^2`

Q 1956101074

The following is the record of goals scored by'
team `A` in a football session.
For the team `B`, mean number of goals scored
per match was `2` with a standard deviation `1.25`
goals. Find which team may be considered more
consistent?
Class 11 Exercise 15.3 Q.No. 4

Solution:

For Team `A`:

` bar x = (50)/(25) = 2`

` S.D. = sqrt ((sum f_i x_i^2)/n - ( (sum f_i x_i^2)/n)^2 ) = sqrt ((130)/(25) - ( (50)/(25))^2 )`

`= sqrt (5.2 - 4) = sqrt ( 1.2) = 1.09`

For team `B, bar x = 2, S.D.= 1.25`

Since their means are same, `:. sigma_A` is less than

that `sigma_b` , therefore team `A` is more consistent than

team `B`.

Q 1114223159

The mean of four observations is `3`. If the sum of the squares of these observations is `48` then their standard deviation is
EAMCET 2014

(A)

`sqrt7`

(B)

`sqrt2`

(C)

`sqrt3`

(D)

`sqrt5`

Solution:

The variance for a given set of observations is given by :

`sigma^2=(sum(x_i)^2)/n-((sumx_i)/n)^2`

`sigma^2=(x_1^2+x_2^2+x_3^2+x_4^2)/4-((x_1+x_2+x_3+x_4)/4)^2`

`=48/4-(3)^2=12-9=3`

Hence, the deviation is given by :

`sigma=sqrt(3)`

Correct Answer is `=>` (C) `sqrt3`

Q 2530467312

The variance of first `20` natural numbers is
WBJEE 2015

(A)

`133/4`

(B)

`279/12`

(C)

`133/2`

(D)

`399/4`

Solution:

Variance of `n` natwal numbers

`= (n^2 -1)/12 = ((20)^2 -1)/12` `[ :. n =20 ]`

`= (400 -1)/12 = 399/12 = 133/4`

Correct Answer is `=>` (A) `133/4`

Q 2549691513

The variance of the series ` a, a+ d, a+ 2d , ···, a + (2n - 1) d , a + 2nd` is

BCECE Mains 2015

(A)

`(n (n +1))/2 d^2`

(B)

`(n (n -1))/6 d^2`

(C)

`(n (n +1))/6 d^2`

(D)

`(n (n +1))/3 d^2`

Solution:

The mean of the given series is `a + nd`.

`:. ` Variance ` = 1/(2n + 1) sum_(r = 0)^(2n) {(a + rd) - (a + nd)}^2`

` =>` Variance ` = d^2/(2n + 1) sum_(r = 0)^(2n) | r - n |^2`

`=>` Variance ` = d^2/(2n + 1) sum_(r = 0)^(2n) ( n - r)^2`

`=>` Variance ` = (2d^2)/(2n + 1) sum_(r = 1)^n r^2`

`=>` Variance ` = (2d^2)/(2n + 1) xx (n (n + 1) (2n + 1))/6`

` = (n (n + 1))/3 d^2`

Correct Answer is `=>` (D) `(n (n +1))/3 d^2`

Q 2488556407

If the mean deviation of number `1, 1 + d,
1 + 2d, ... , 1 + 100d` from their mean is `255`,
then `d` is equal to
BCECE Stage 1 2016

(A)

`10.0`

(B)

`20.0`

(C)

`10.1`

(D)

`20.2`

Solution:

Since, we know `bar x = (text(sum of quantities))/n = (n//2(a+l))/n`

`=1/2 [1+1+ 100d]= 1+500 d`

`MD=1/n sum | x_i- bar x|`

`=> 255=1/101[50 d+49 d +.......+d+0+d+......+ 50 d]`

`=> 255=(2d)/101[(50 xx 51)/2]`

`:. d=(255 xx 101)/(50 xx 51)=10.1`

Correct Answer is `=>` (C) `10.1`

Q 2478256106

The mean of the numbers `a, b, 8, 5, 10` is `6`
and the variance is `6.80`. Then, which one
of the following gives possible values of a
and b?
BCECE Stage 1 2016

(A)

`a=3, b=4`

(B)

`a=0,b=7`

(C)

`a=5, b=2`

(D)

`a=1, b=6`

Solution:

According to the question

`6.80=[(6-a)^2+(6-b)^2+(6-8)^2+(6-5)^2+(6-10)^2]//5`

`34 = (6- a)^2 + (6- b)^2 + 4 + 1 + 16`

`(6 -a)^2 + (6 - b)^2 = 3^2 + 2^2`

`a=3,b=4`

Correct Answer is `=>` (A) `a=3, b=4`

Q 2816723670

If `bar x = bar y = 0 , sum x_i y_i = 12, sigma_x = 2 , sigma_y = 3` and `n = 10`, then the coefficient of correlation is

(A)

`0.4`

(B)

`0.3`

(C)

`0.2`

(D)

`0.1`

Solution:

`∵ r = (sum (x_i - bar x) (y_i - bar y) )/( n sigma_x sigma_y)`

` = (sum (x_i - 0) (y_i -0))/( (10) (2) (3)) = ( sum (x_i) (y_i) )/( (10) (2) (3))`

` ( sum x_i y_i)/( (10) (2) (3)) = (12)/((10) (2) (3)) = 0.2`

Correct Answer is `=>` (C) `0.2`

Q 2886723677

The two lines of regression are `8x - 10y = 66` and `40x - 18y = 214` and variance of x series is `9`. What is the standard deviation of `y` series ?

(A)

`3`

(B)

`4`

(C)

`6`

(D)

`9`

Solution:

The regression coefficient `y` on `x` on

the line `8x - 10y = 66`,

`b_(yx) = 4//5`

The regression cocfficient x on y on the line

` 40x - 18y = 214, b_(yx) = 9/(20)`

` => r^2 = (36)/(100) => r = 0.6`

` => b_(yx) = ( r sigma_y)/sigma_x`

` :. sigma_y = ( 4/5 xx 3)/(0.6) = (12)/3 = 4`

Correct Answer is `=>` (B) `4`

Q 2876123976

The coefficient of regressions `b_(y x)` and `b_(yx)` from the set of observations `{(x, y)} = {( 4, 2), (2, 3), (3, 2), ( 4, 4), (2, 4)}` will be

(A)

`1/4 , 1/4`

(B)

`(-1)/4 , 1/4`

(C)

`(-1)/4 , (-1)/4`

(D)

`1/4 , (-1)/4`

Solution:

For calculation of `b_(yx)` and `b_(xy)` we have

to calculate ` sum x , sum y, sum x y, sum x^2` and `sum y^2` .

We have

` :. b_(yx) = ( sum xy - 1/n sum x sum y)/( sum x^2 - 1/n ( sum x)^2)`

` = ( 44 - 1/5 (15)(15) )/( 49 - 1/5 (15)^2) = ( 44 - 45)/(49 - 45) = (-1)/4`

` :. b_(yx) = ( sum xy - 1/n sum x sum y)/( sum y^2 - 1/n ( sum y)^2) = (44 - 45)/( 49 - 45) = (-1)/4`

Correct Answer is `=>` (C) `(-1)/4 , (-1)/4`

Q 2886723677

The two lines of regression are `8x - 10y = 66` and `40x - 18y = 214` and variance of x series is `9`. What is the standard deviation of `y` series ?

(A)

`3`

(B)

`4`

(C)

`6`

(D)

`9`

Correct Answer is `=>` (B) `4`

Q 2816723679

If the monthly expenditure pattern of a person who earns a monthly salary of Rs. `15000` is represented in a pie diagram, then the sector angle of an item on transport expenses measures `15°`. What is his monthly expenditure on transport ?

(A)

Rs. `450`

(B)

Rs. `625`

(C)

Rs. `675`

(D)

insufficient data

Solution:

Since, monthly salary = Rs. `15000`

and sector angle of expenses `= 15°`

`:.` Amount `= (15^o)/(360^o) xx 15000 =` Rs. `625`

Correct Answer is `=>` (B) Rs. `625`

Q 1710756610

The election result in which six parties contested
was depicted by a pie chart. Party A had an angle
`135^0` on this pie chart. If it secured `21 960` votes,
then how many valid votes in total were cast?
CDS 2016

(A)

`51240`

(B)

`58560`

(C)

`78320`

(D)

`87840`

Solution:

Let the total votes be `x`.
Then, central angie of party `A` `= (360^0/x)xx21960`

`135^0 = (360^0/x)xx21960`

`x = (360^0xx21960)/135^0 = 58560`

Correct Answer is `=>` (B) `58560`