Mathematics Tricks of statistics For NDA

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Tricks of statistics For NDA

Trick For Average Problems

Trick 1 : Problems on Average

1. Arithmatic mean or average of n values `x_1,x_2,x_3,.......,x_n`

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

2. If `barx` is the mean of `x_1,x_2,x_3,.......,x_n` and `bary ` is the mean of `y_1,y_2,y_3,.......,y_m`

Average of hole numbers `x_1,x_2,x_3,.......,x_n, y_1,y_2,y_3,.......,y_m` is `(nbarx + mbary)/(m+n)`

3.if average of `x_1,x_2,x_3,.......,x_n, y_1,y_2,y_3,.......,y_m` is `barz` & If we subtract `y_1,y_2,y_3,.......,y_m` new average

`=((m+n)barz - mbary)/(n)`

Q 2703191948

A sample of 5 observations has mean 32 and median 33. Later it is.found that an observation was recorded incorrectly as 40 instead of 35. If we correct the data, then which one of the . following is correct?
NDA Paper 1 2017

(A)

The mean and median remain the same

(B)

The median remains the same but the mean will decrease

(C)

The mean and median both will decrease

(D)

The mean remains ·the same but median will decrease

Solution:

Let the numbers be

20 , 32 , 33, 35, 40

If we replace 40 by 35 then

median will remain same

and mean will become = ` (20+32+33+35+35) / 5 `

= 31

`=>` mean will decrease

Correct Answer is `=>` (B) The median remains the same but the mean will decrease

Q 2763891745

The mean weight of 150 students in a certain class is 60 kg. The mean weight of boys in the class is 70 kg and that of
girls is 55 kg. What is the number of boys in the class?
NDA Paper 1 2017

(A)

`50`

(B)

`55`

(C)

`60`

(D)

`100`

Solution:

Let no of boy is b & no. of girls is g , `barz = 60, b+g = 150`

`barb = 70, ` `barg = 55`

Total weight

`150*60 = b*70 + (150-b)55`

`b= 50`

Alternatively :

let No. of Boys `=n`

No of girls `= 150-n`

`(sum_(i=1)^n x_i + sum_(n-150)^n x_i)/(150) = 60`.................(i)

`(sum_(i=1)^n x_i)/n = 70`........................(ii)

`(sum_(n-150)^150 x_i)/(150-n) = 55`...............(iii)

by (i), (ii), & (iii)

`70 n+ 55 (150-n) = 60 xx 150`

`n=50`

Correct Answer is `=>` (A) `50`

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 2723891741

The mean of a group of `100` observations was found to be `20`. Later it was found that four observations were ·incorrect, which were recorded as `21, 21, 18` and `20`. What is the mean if the incorrect observations are omitted?
NDA Paper 1 2017

(A)

`18`

(B)

`20`

(C)

`21`

(D)

`22`

Solution:

As the mean of incorrect observations is 20, removing these observations will not effect average.

Correct Answer is `=>` (B) `20`

Q 1761401325

The weighted arithmetic mean of first `10` natural
numbers whose weights are equal to the corresponding numbers is equal to
CDS 2015

(A)

`7`

(B)

`14`

(C)

`35`

(D)

`38.5`

Solution:

Weighted `AM=(1 xx 1+2xx2+3 xx 3+....+10 xx 10)/10`

`=((10 xx 11 xx 21)/6)/10`

`=(21 xx 11)/6=77/2=38.5`

Correct Answer is `=>` (D) `38.5`

Q 1731501422

If each of n nmnbers `x_i = i (i = 1, 2, 3, ... , n)` is replaced by `(i + 1)x_i` then the new mean is
CDS 2015

(A)

`(n+3)/2`

(B)

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

(C)

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

(D)

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

Solution:

We have, `x_i = `i

mean `=(2x_1+3x_2+4x_3+....+(n+1)x_n)/n`

`=(2 xx 1+3 xx 2+4 xx 3+...+(n+1)n)/n`

`=(Sigma n(n+1))/n=(Sigma n^2+ Sigma n)/n`

New mean `=((n(n+1)(2n+1))/6 +(n(n+1))/2)/n`

`=(n(n+1))/(2n)[(2n+1)/3+1]`

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

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

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

Q 2553280144

The harmonic mean between two numbers is `14 2/5` and the geometric mean is `24`. The greater number between them is

UPSEE 2013

(A)

`72`

(B)

`54`

(C)

`36`

(D)

None of these

Solution:

Harmonic mean between two numbers P, Q is

`HM = (2PQ)/(P+Q)` and Geometric mean `sqrt(PQ)`

Let two numbers be P and Q

`therefore` Given `(2PQ)/(P+Q) = 14 2/5 = 72/5` ..........(i)

and `sqrt(PQ) = 24`

`=> PQ = (24)^2 = 576` .........(ii)

`therefore P+Q = (10PQ)/(72)` [from Eq. (i)]

`=> P+Q = (10xx576)/(72) = 80` ..........(iii) [from Eq. (ii)]

`=> (P-Q)^2 = (P+Q)^2-4PQ`

` = 80^2-4xx476`

` = (80+48)(80-48) = 128xx32`

`=> P-Q = 64` .........(iv)
On solving Eqs. (iii) and (iv) we get

`therefore` Greater number =` 72`.

Correct Answer is `=>` (A) `72`

Trick 2 : Change in origin & Scale

Q 1742580433

The variance of numbers `x_1, x_2, x_3 ,........., x_n` is `V`. Consider
the following statements.

I. If every `x_i` is increased by `2`, the variance of the new
set of numbers is `V`.
II. If the numbers `x_i` is squared, the variance of the new
set is `V^2`

Which of the following statements is/are correct?
NDA Paper 1 2014

(A)

Only 1

(B)

Only 2

(C)

Both 1 and 2

(D)

Neither 1 nor 2

Solution:

1. We know that, variance is not dependent on

change of origin. i.e., independent on change of origin.

So, if every `x_i` is increased b `y_2`, the variance of the new set of

numbers is not change i.e., `V`.

2. We know that, variance is dependent on change of scale.

So, if the number `x_i` is squared, the variance of the new set `v^2`.

i.e. If ` V(x_i) = V`

Then, `V_((x_i xx x_i)) = V _((x_i)) V_((x_i ))`

` = V.V`

` = V^2`

Correct Answer is `=>` (C) Both 1 and 2

Q 1770601516

The variance of `20` observations is `5`. If each observation
is multiplied by `2`, then what is the new variance of the
resulting observations?
NDA Paper 1 2014

(A)

`5`

(B)

`10`

(C)

`20`

(D)

`40`

Solution:

New variance on scaling` = a^2 . old \ \ variance `

`a=2`

New variance `= 2^2 * 5 = 20`

Alternatively :

Let `x_1 , x_2 , ..... , x_(20)` be the given observations.

We have,

` 1/(20) sum_( i = 1)^(20) (x_i - bar x)^2 = 5`

To find Variance of `2x_1 , 2x_2 , 2x_3 , .... , 2x_(20)` . Let `bar x` denotes

the mean of new observation.

Clearly, `bar X = ( sum_( i = 1)^(20) 2x_i)/(20) = ( sum_( i = 1)^(20) x_i)/(20) = 2 barx`

Now, variance of new observation

` = 1/(20) sum_( i = 1)^(20) (2x_i - bar X)^2 = 1/(20) sum_( i = 1)^(20) (2x_i -2bar x)^2`

` = 1/(20) sum_( i = 1)^(20) 4(x_i - bar x)^2 = 4 ( 1/(20) sum_( i = 1)^(20) (x_i - bar x )^2 ) = 4 xx 5 = 20`

Correct Answer is `=>` (C) `20`

Q 1732780632

`p, q, r, s` and `t` are five numbers such that the average of
`p, q` and `r` is `5` and that of `s` and `t` is `10`. What is the
average of all the five numbers?
NDA Paper 1 2014

(A)

`7.75`

(B)

`7.5`

(C)

`7`

(D)

`5`

Solution:

Given that, `p, q, r, s` and `t` are five numbers,

`:.` Average of `p, q` and `r = 5`

` => (p+q+r)/3 = 5`

`=> p + q + r =15` ......(1)

and average of `s` and `t = 10`

` => (s +t)/2 =10`

` => s + t = 20` ......(2)

Now, average of all five numbers

` = (p +q + r + s + t)/5`

` = ((p + q + r) + (s + t))/5`

` = (15+ 20)/5` [from Eq. (1) and (2)]

`= (35)/5 = 7`

Correct Answer is `=>` (C) `7`

Q 1730401312

If `barx` and `bary` are the means of two distributions such that
`barx < bary` and `bar z` is the mean of the combined distribution,
then which one of the following statements is correct?
NDA Paper 1 2014

(A)

`bar x < bar y < bar z`

(B)

`bar x > bar y > bar z`

(C)

` bar z = (bar x + bar y ) /2`

(D)

`bar x < bar z < bar y`

Solution:

It is obvious that , `bar x < bar z < bar y`

Correct Answer is `=>` (D) `bar x < bar z < bar y`