Mathematics Problem Solving Techniques of statistics For NDA
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Problem Solving Techniques of statistics For NDA

Tip 1 : How to solve Problems related to regression line

(i) The line of regression of `y` on `x` or regression line of `y` on `x` is given by

`y- bar y = r (sigma_y)/(sigma_x) (X- bar x)`

(ii) The line of regression of `x` on `y` or regression line of `x` on `y` is given by `x-bar x= r (sigma_x)/(sigma_y) (y- bar y)`

Here r is correlation coefficient

`r(x,y) =(cov (x,y))/(sigma_x sigma_y)`

(iii) Regression coefficient of `y` on `x`, is denoted by `yx`,

`b_(yx)= r (sigma_y)/(sigma_x) =(cov (x,y))/(sigma_x^2)`

(iv) Regression coefficient of `x` on `y`, is denoted by `xy`,

`b_(xy) = r (sigma_x)/(sigma_y) =(cov (x,y))/(sigma_y^2)`

(v) If `theta` is the angle between the two regression lines, then

`tan theta =((1-r^2))/(|r|) * (sigma_x sigma_y)/(sigma_x^2 + sigma_y^2)`

where , `tan theta=(M_2-M_1)/(1+ M_1 M_2)`

(a) If `r = 0, theta = pi/2`, then the two regression lines are perpendicular to each other.

(b) If `r =1` or `- 1, theta = 0, pi` then the regression lines coincide.


Coefficient of Correlation :

1. Karl Pearson's Coefficient of Correlation :

Covariance `(x, y) cov (x, y)`

`=1/n sum_(i=1)^n (x_i - bar x) (y_i - bar y) = 1/n sum_(i=1)^n x_i y_i - bar x bar y`

Let `sigma_x` and `sigma_r` the the SD of variables `x` and `y`, respectively. Then coefficient of correlation

`r(x,y) =(cov (x,y))/(sigma_x sigma_y) = (sum_(i=1)^n (x_i - bar x)(y_i - bar y))/(sqrt(sum_(i=1)^n (x_i - bar x)^2 sum_(i=1)^n (y_i - bar y)^2))`

`= (n sum x_i y_i -(sum x_i)(sum y_i))/(sqrt((n sum x_i^2 -(sum x_i)^2)(n sum y_i^2 -(sum y_i)^2))`

In general

` color{red} {b_(yx) = ( sum xy - 1/n sum x sum y)/( sum x^2 - 1/n ( sum x)^2)}`

2. Rank Correlation (Spearman's) Let `d` be the difference between paired ranks and `n` be the number of items ranked. Then, `rho` the coefficient of rank correlation is given by `rho = 1- (6 sum_(i=1)^n d^2)/(n(n^2-1))`
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 2347656583

Consider the following data



What is the regression equation of y on `x` ?
NDA Paper 1 2010
(A)

`y = 0.6 + 0.4x`

(B)

`y = 0.7 + 0.3x`

(C)

`y = 6 + 5x`

(D)

`y = 4 + 9x`

Solution:

`barx = 30/5 = 6 ` and `bary = 15/5 = 3`


`therefore b_(yx) = (nSigmaxy -(Sigmax)(Sigmay))/(nSigmax^2- (Sigmax)^2)`


` = (5xx94-30xx15)/(5xx190-(30)^2)`


` = (470-450)/(950-900)`


` = 20/50 = 2/5 = 0.4`


`y -3 = 0.4(x-6)`

` => y = 0.4x+3-2.4`

`y = 0.4x+0.6`
Correct Answer is `=>` (A) `y = 0.6 + 0.4x`
Q 2387034887

If two lines of regression are perpendicular, then
the correlation coefficient r is
NDA Paper 1 2013
(A)

`2`

(B)

`1/2`

(C)

`0`

(D)

None of these

Solution:

We know that, `tan theta = (1-r^2)/(|r|) *(sigma_x * sigma_y)/(sigma_x^2 + sigma_y^2)` Where `theta` is the angle between the two regression lines.
Given that, two lines of regression at perpendicular.
i.e.,`theta = 90^0`

`therefore tan90^0 = (1- r^2)/(|r|) * (sigma_x * sigma_y)/(sigma_x^2+sigma_y^2) = 1/0`


`=> |r| *( sigma_x^2+sigma_y^2) = 0`

`=> |r| = 0` (`because sigmax^2+sigmay^2 = 0)`

`therefore r = 0`

i.e., Correlation coefficient (r)= 0
Correct Answer is `=>` (C) `0`
Q 2753191944

If the regression coefficient of x on y and y on x are `-1/2` and `-1/8` respectively, then what is the correlation coefficient
between x and y?
NDA Paper 1 2017
(A)

`-1/4`

(B)

`-1/16`

(C)

`1/16`

(D)

`1/4`

Solution:

`b_(yx) =(-1)/2`

`b_(xy) =(-1)/8`

`r= sqrt(b_(yx) * b_(xy)`

`= sqrt((-1/2) (-1/8))`

`=1/4`
Correct Answer is `=>` (A) `-1/4`
Q 2387467387

In a study on the relationship between investment
(x) and profit (y), the following two regression
equations were obtained based on the data on x
and y.

`3x + y -12 = 0`
`x+2y-14=0`
What is the mean `barx`' ?
NDA Paper 1 2009
(A)

`6`

(B)

`5`

(C)

`4`

(D)

`2`

Solution:

Since, lines of regression passes through `(barx, bary)`


`therefore 3barx+bary-12 = 0` .........(i)

and `barx+2 bary-14 = 0` .................(ii)

on solving Eqs (i) and(ii) we get

`barx = 2 ` and `bar y = 6`
Correct Answer is `=>` (D) `2`
Q 1648601503

The regression coefficients of a bivariate distribution are
`-0.64` and `-0.36` Then, the correlation coefficient of the
distribution is
DSSB Paper 1 2015
(A)

`0.48`

(B)

`-0.48`

(C)

`0.50`

(D)

`- 0.50`

Solution:

We have, `b_(xy) =- 0.64, b_(yx) = - 0.36`

`:. ` Correlation coefficient `(sigma) = sqrt (b_(xy) xx b_(yx) )`

` = pm sqrt ( ( 0.64) (- 0.36) = pm 0.48)`

`=> sigma =- 0.48`

Because `b_(xy)` and `b_(yx)` both are negative.
Correct Answer is `=>` (B) `-0.48`
Q 2511845729

The two regression lines are `2x - 7 y + 6 = 0` and `7 x - 2y + 1 = 0`. The correlation coefficient between x and y is :
BCECE Stage 1 2015
(A)

`-2/3`

(B)

`2/7`

(C)

`4/9`

(D)

None of these

Solution:

Let us assume that line of regression of x on y is

`2x-7y+6=0`
or `2x=7y - 6` ...................(i)

and line of regression of y on x is
`7x- 2y + 1 = 0`

or `2y = 7x +1` ..............(ii)

From (i) and (ii)

`b_(xy) = 7/2` and `b_(yx) =7/2`

Now `sqrt(b_(xy) * b_(yx) ) = sqrt(7/2 * 7/2)`

`=> r = 7/2 > 1`

`:.` Our assumption is wrong.

`=>` Eq. (i) is of yon x and Eq. (ii) is of x on y

`:. r = sqrt(2/7 * 2/7) =2/7`
Correct Answer is `=>` (B) `2/7`
Q 2886123977

Consider the following statements with regard to correlation coefficient r between random variables x and y.
I. `r = + 1` or `- 1` means there is a linear relation between x and y.
II. `- 1 <= r <= 1` and `r^2` is a measure of the linear
relationship between the variables.
Which of the above statement(s) is/are correct?

(A)

Only I

(B)

Only II

(C)

Both I and II

(D)

Neither I nor II

Solution:

We know that, when two lines are

coincident linearly, then `r = ± 1`

Also, if ` r in [- 1, 1]`, then there is a

relation measure of the linear

relationship between the variables.
Correct Answer is `=>` (C) Both I and II
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 2856723674

If `b_(yx)` and `b_(xy)` are regression coefficients of `y` on `x` and x on y respectively, then which of the following statements is true?

(A)

`b_(xy) = 1.5` and `b_(yx) = 1.4`

(B)

`b_(xy) = 1.5` and `b_(yx) = 0.9`

(C)

`b_(xy) = 1.5` and `b_(yx) = 0.8`

(D)

`b_(xy) = 1.5` and `b_(yx) = 0.6`

Solution:

In option (d),

`r = sqrt( 1.5 xx 0.6) = sqrt(0.9) < 1`
Correct Answer is `=>` (D) `b_(xy) = 1.5` and `b_(yx) = 0.6`

Tip 2 : Calculation of Mean, Median, mode, SD and Variance

Formulae :

1. `=> text( Arithmetic Mean) = text ( sum of all values)/ text(Total number of values)`

`AM = (x_1+x_2+.....+x_n)/n`

` bar(X) = (sum x_(i))/n =(80)/5 = 16 => MD = (sum |x_(i) - bar(X)|)/n`

2. `=> text( Geometric Mean) = text (( Product of all values))^(text(Total number of values))`

`GM = (x_1.x_2......x_n)^(1/n)`

3. Mode = 3 Median - 2 Mean

4. Deviation form mean `d_i = |x_i - barx|`

5. Variance `= 1/n sum_(i=1)^n (x_i - bar x)^2 = (sum_(i=1)^n x_i^2)/n - bar x^2`

`= [(sum_(i=1)^n d_i^2)/n - (( sum_(i=1)^n d_i)/n)^2]`

6. standard deviation `= sqrt(var (x))`

`SD = sigma= sqrt([ 1/n sum_(i=1)^n (x_i - bar x)^2])` or `sigma= sqrt([(sum_(i=1)^n x_i^2)/n - bar x^2]`

Q 2137180082

What is the mean deviation from the mean of the
numbers `10, 9, 21, 16, 24?`
NDA Paper 1 2016
(A)

`5.2`

(B)

`5.0`

(C)

`4.5`

(D)

`4.0`

Solution:

Given, `x_(i) = 10, 9, 21, 16, 24`

`:. sum x_(i) = 10 + 9 + 21 + 16 + 24 = 80`

Now ` bar(X) = (sum x_(i))/n =(80)/5 = 16 => MD = (sum |x_(i) - bar(X)|)/n`


` (|10 - 16 | + |9-16| + | 21 -16| + | 16 -16| + | 24 -16|)/5`

`= (6+7+5+0+8)/5 = (26)/5 = 5.2`
Correct Answer is `=>` (A) `5.2`
Q 2117780680

The mean of the series `x_(1) , x_( 2) , ... , x_(n),`. is `bar(X)`. If `x_( 2)` is

replaced by `lamda` , then what is the new mean?
NDA Paper 1 2016
(A)

` bar (X) - x_(2) + lamda`

(B)

` (bar (X) - x_(2) - lamda)/n `

(C)

` (bar (X) - x_(2) + lamda)/n `

(D)

` (nbar (X) - x_(2) + lamda)/n `

Solution:

We know,` bar(X) = (x_(1) + x_(2) + ... + x_(n))/n`

` => x_(1) + x_(2) + ... + x_(n) = n bar(X)`

` => x_(1) + x_(3) + ... + x_(n) = n bar(X) - x_(2)`

` => x_(1) + x_(3) + ... + x_(n) + lamda`

` = n bar(X) - x_(2) + lamda`

`=> text( Mean) = text ( sum of all values)/ text(Total number of values)`

`= ( x_(1) + x_(3) + ... + x_(n) + lamda)/n`

` = (n bar(X) - x_(2) + lamda)/n`
Correct Answer is `=>` (D) ` (nbar (X) - x_(2) + lamda)/n `
Q 2167780685

For the data
`3, 5, 1, 6, 5, 9, 5, 2, 8, 6`
the mean, median and mode are `x, y` and `z`, respectively.
Which one of the following is correct ?
NDA Paper 1 2016
(A)

`x = y != z`

(B)

`x != y = z`

(C)

`x != y != z`

(D)

`x = y = z`

Solution:

Mean= ` (sum x_(i))/n`

`= (3+5+1+6+5+9+5+2 + 8+6)/(10)`

` =(50)/(10) = 5`

Now, the data in 'Ascending' order is;

`1,2,3,5,5,5,6,6,8,9`

Clearly, median (mid value) is `5` and mode (most

appeared value) is also `5`.

`x = y = z`
Correct Answer is `=>` (D) `x = y = z`
Q 1730501412

What is the mean deviation about the mean for the data
`4, 7, 8, 9, 10, 12, 13` and `17?`
NDA Paper 1 2014
(A)

`2.5`

(B)

`3`

(C)

`3.5`

(D)

`4`

Solution:

Mean deviation about the mean

` = (sum |x_i - bar x|)/N`

Here, `bar x = (4 + 7 + 8 + 9 + 10 + 12 + 13 + 17)/8 = 10`

`:.` Mean deviation about mean

` ( | 4 - 10 | + |7 - 10 | + |8-10 | + |9- 10 | + |10 -10| +| 12-10| + |13-10| +| 17-10 | )/8`

`= (6+3+2+1+0+2+3+7)/8 = (24)/8 = 3`
Correct Answer is `=>` (B) `3`
Q 2723191941

For given statistical data, the graphs for less than ogive and more' than ogive are "drawn. If the point at which the two curves intersect is P, then abscissa of point . P gives the value of which one of the following measures of central tendency?
NDA Paper 1 2017
(A)

Median

(B)

Mean

(C)

Mode

(D)

Geometric mean

Solution:

A curve that represents the cumulative frequency distribution of grouped data is called an ogive or cumulative frequency curve.

The two types of Ogives are more than type ogive and less than type ogive. An ogive representing a cumulative frequency distribution of ‘more than’ type is called a more than ogive. An ogive representing a cumulative frequency distribution of ‘less than’ type is called a less than ogive.

Ogives can be used to find the median of a grouped data. The median of grouped data can be obtained graphically by plotting the Ogives of the less than type and more than type and locate the point of intersection of both the Ogives. The x-coordinate of the point of intersection of two Ogives gives the median of the grouped data
Correct Answer is `=>` (A) Median
Q 2167780685

For the data
`3, 5, 1, 6, 5, 9, 5, 2, 8, 6`
the mean, median and mode are `x, y` and `z`, respectively.
Which one of the following is correct ?
NDA Paper 1 2016
(A)

`x = y != z`

(B)

`x != y = z`

(C)

`x != y != z`

(D)

`x = y = z`

Solution:

Mean= ` (sum x_(i))/n`

`= (3+5+1+6+5+9+5+2 + 8+6)/(10)`

` =(50)/(10) = 5`

Now, the data in 'Ascending' order is;

`1,2,3,5,5,5,6,6,8,9`

Clearly, median (mid value) is `5` and mode (most

appeared value) is also `5`.

`x = y = z`
Correct Answer is `=>` (D) `x = y = z`
Q 2703091848

If the data are moderately non-symmetrical, then which one of the following empirical relationships is
correct?
NDA Paper 1 2017
(A)

2 x Standard deviation= 5 x Mean deviation

(B)

5 x Standard deviation = 2 x Mean deviation

(C)

4 x Standard deviation = 5 x Mean deviation

(D)

5 x Standard deviation = 4 x Mean deviation

Solution:

we know that, If the data are moderately non-symmetrical, following relationships hold :

4 S.D = 5 M.D = 6 Q.D

here Q.D. = Quartile Deviation
Correct Answer is `=>` (C) 4 x Standard deviation = 5 x Mean deviation
Q 2357345284

If a variate `X` takes values `2 ,9, 3, 7, 5, 4, 3, 2` ·and
`10`, then what is the median ?
NDA Paper 1 2012
(A)

`2`

(B)

`4`

(C)

`7`

(D)

`9`

Solution:

The given observations are arranged in ascending
order
`2,2,3,3, 4,5, 7,9, 10`
Here, total term = `9`

`therefore` Median = `((9+1)/2) th ` term = `(10/2) th` term


` = 5th` term = `4`
Correct Answer is `=>` (B) `4`
Q 2630580412

Three numbers are selected at random (without replacement) from first six positive integers. Let X denote the largest of the three numbers obtained. Find the probability distribution of X.Also, find the mean and variance of the distribution.
CBSE-12th 2016
Solution:

`X =` larger of of three numbers

`X = 3, 4, 5, 6`

` P ( x = 3) = 6 xx 1/6 xx 1/5 xx 1/4 = 1/(20)`

` P ( x = 4) = 18 xx 1/6 xx 1/5 xx 1/4 = 3/(20)`

`P ( x = 5) = 36 xx 1/6 xx 1/5 xx 1/4 = 6/(20)`

`P ( x = 6) = 60 xx 1/6 xx 1/5 xx 1/4 = (10)/(20)`

Mean `sum P_i X_i^2 = ((105)/(20))^2 = 5.25`

` sum P_i X_i^2 = ( 567)/(20)`

var`(X) = sum, P_i X_i^2 = ( sum P_i X_i )^2 `

` = (576)/(20) - ((105)/(20))^2 = 0.787`
Q 2137180082

What is the mean deviation from the mean of the
numbers `10, 9, 21, 16, 24?`
NDA Paper 1 2016
(A)

`5.2`

(B)

`5.0`

(C)

`4.5`

(D)

`4.0`

Solution:

Given, `x_(i) = 10, 9, 21, 16, 24`

`:. sum x_(i) = 10 + 9 + 21 + 16 + 24 = 80`

Now ` bar(X) = (sum x_(i))/n =(80)/5 = 16 => MD = (sum |x_(i) - bar(X)|)/n`


` (|10 - 16 | + |9-16| + | 21 -16| + | 16 -16| + | 24 -16|)/5`

`= (6+7+5+0+8)/5 = (26)/5 = 5.2`
Correct Answer is `=>` (A) `5.2`
Q 2448180903

The standard deviation for the scores 1, 2,
3, 4, 5, 6 and 7 is 2. Then, the standard
deviation of 12, 23, 34, 45, 56,67 and 78 is
BCECE Stage 1 2016
(A)

2

(B)

4

(C)

22

(D)

11

Solution:

Here, `n = 7` sum `= 315`

`:.` Mean `=315/7=45`

`:.` standard deviation

`=sqrt((12-45)^2(23-45)^2+(34-45)^2+(45-45)^2+(56-45)^2+(67-45)^2+(78-45)^2)/7)`

`=sqrt((2(1089+484+121))/7)`

`=sqrt 484`

`=22`
Correct Answer is `=>` (C) 22
Q 1629478311

Consider the following frequency distribution.

What is the mode of the distribution'?

CDS 2015
(A)

`38.33`

(B)

`40.66`

(C)

`42.66`

(D)

`42.33`

Solution:

Modal class of the given data is `40-50`, because it has largest
frequency among the given classes of the data i.e. `12`.

Here, `I = 40, f_1 =12, f_0 = 10, f_2 = 8` and `h = 10`

`:. ` Mode`=I+{(f_1-f_0)/(2f_1-f_0-f_2)} xx h`

`=40 +{(12-10)/(2 xx 12 xx 10-8)} xx 10`

`=40+(2 xx 10)/(24-18)=40+20/6`

`=40+3.33=43.33`

Hence, the mode of given data is `43 33`.
Correct Answer is `=>` (D) `42.33`
Q 1750056814

The mean and median of `5` observations are `9` and
`8`, respectively If `1` is subtracted from each
observation, then the new mean and the new
median will respectively be
CDS 2016
(A)

`text(8 and 7)`

(B)

`text(9 and 7)`

(C)

`text(8 and 9)`

(D)

`text(Cannot be determined due to insufficient data)`

Solution:

`(text(sum of 5 observations))/(text(Number of observations)) = 9`


=> Sum of 5 observations `= 9 xx 5 = 45`

If 1 is subtmcted from each observation, then

New mean of 5 observations `= text(sum of 5 observations-5)/5`

` = (45-5)/5 = 8`

median of `5` observations `= (5+1)/2` th term `= 3rd` term `= 8`


If 1 is subtracted from each observation, then
New median `= 8 - 1 = 7`
Hence, the new mean anci median are `8` and `7`, respectively
Correct Answer is `=>` (A) `text(8 and 7)`
Q 2651523424

Two the numbers are selected at random (without replacement) from first six positive integers. Let X denote the larger of the two numbers obtained. Find the probability distribution of X. Find the mean and variance of this distribution.
CBSE-12th 2015
Solution:

First six positive integers are `{1, 2, 3, 4, 5, 6}`

No. of ways of selecting `2` numbers from `6` numbers without replacement `= text()^6C_2 = 15`

X denotes the larger of the two numbers, so X can take the values `2, 3, 4, 5, 6`.

Probability distribution of X :

mean ` = sum p_i x_i = (70)/(15) = 4.67`

Variance ` = sum p_ix_i^2 - ( sum p_i x_i )^2 = (70)/3 = (196)/9 = (210 - 196)/9 = (14)/9`
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 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`

Pie Chart related problems

Central angle = [Frequency x 360° / Total frequency]
Q 2367745685

Study the pie chart given below
and answer the questions that follow.

The following pie chart gives the distribution of funds in a
five yea.r plan under the major heads of development
expenditures
Agriculture (A), Industry (B), Education (C),
Employment (D) and Miscellaneous (E).
Which head is allocated the maximum funds'?
NDA Paper 1 2011
(A)

Agriculture

(B)

Industry

(C)

Employment

(D)

Miscellaneous

Solution:

Employment is allocated maximum funds.
Correct Answer is `=>` (C) Employment
Q 2367745685

Study the pie chart given below
and answer the questions that follow.

The following pie chart gives the distribution of funds in a
five yea.r plan under the major heads of development
expenditures
Agriculture (A), Industry (B), Education (C),
Employment (D) and Miscellaneous (E).
How much money (in crore) is allocated to Education'?
NDA Paper 1 2011
(A)

3000

(B)

6000

(C)

9000

(D)

10800

Solution:

Money allocated to Education ` = (30^0)/(360^0) xx36000 = 3000`
Correct Answer is `=>` (A) 3000
Q 2367745685

Study the pie chart given below
and answer the questions that follow.

The following pie chart gives the distribution of funds in a
five yea.r plan under the major heads of development
expenditures
Agriculture (A), Industry (B), Education (C),
Employment (D) and Miscellaneous (E).
How much money (in crore) is allocated to both
Agriculture and Employment?
NDA Paper 1 2011
(A)

20000

(B)

21000

(C)

24000

(D)

27000

Solution:

Money allocated to both Agriculture and Employment

` = (90^0+120^0)/(360^0)xx36000`


` = (210)/(360) xx 36000 = 21000`
Correct Answer is `=>` (B) 21000
Q 2367745685

Study the pie chart given below
and answer the questions that follow.

The following pie chart gives the distribution of funds in a
five yea.r plan under the major heads of development
expenditures
Agriculture (A), Industry (B), Education (C),
Employment (D) and Miscellaneous (E).
How much excess money (in crore) is allocated to
Miscellaneous over Education?
NDA Paper 1 2011
(A)

3600

(B)

4200

(C)

4500

(D)

4800

Solution:

Required value of money ` = (75-30)/(360^0) xx36000`

` = (45)/(360)xx36000 = 4500`
Correct Answer is `=>` (C) 4500
Q 2816334270

Study the pie chart given below and answer the questions that follow. The following pie chart gives the distribution of funds in a
five year plan under the major heads of development expenditures Agriculture (A), Industry (8), Education (C), Employment (D) and Miscellaneous (E). The total allocation is 36000 (in crore of rupees).
Which head is allocated maximum funds?

(A)

Agriculture

(B)

Industry

(C)

Employment

(D)

Miscellaneous

Solution:

Employment is allocated maximum

funds.
Correct Answer is `=>` (C) Employment
Q 2733091842

In an examination, `40 %` of candidates
got second class. When the data are
represented by a pie chart, what is the
angle· corresponding to second class?
NDA Paper 1 2017
(A)

`40^o`

(B)

`90^o`

(C)

`144 ^o`

(D)

`320 ^o`

Solution:

`40 %` of candidates got second class.

then angel will be `40 %` of `360^o`

`360^o xx 40/100`

`=144^o`
Correct Answer is `=>` (C) `144 ^o`
 
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