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CAIIB - ABM- CASE STUDIES / NUMERICAL QUESTIONS


Find Correlation coefficient for X and Y values given below :
X= (1,2,3,4,5)
Y= {11,22,34,43,56}

a. 0.8899
b. 0.9989
c. 1.0899
d. 1.0989

Ans - b

Explanation :

Step 1: Find Mean for X and Y
X=15/5=3
Y=166/5=33.2

Step 2: Calculate Standard Deviation for Y inputs:
σx=
√(1/(N-1)*((x1-xm)2+(x2-xm)2+..+(xn-xm)2))

=√(1/(5-1)((11-33.2)2+(22-33.2)2+(34-33.2)2+(43-33.2)2+(56-33.2)2))

=√(1/4((-22.2)2+(-11.2)2+(0.8)2+(9.8)2+(22.8)2))

=√(1/4((492.84)+(125.44)+(0.64)+(96.04)+(519.84)))

=√(308.7)

=17.5699

Step 3: Standard Deviation for X Inputs:
σx=
√(1/(N-1)*((x1-xm)2+(x2-xm)2+..+(xn-xm)2))

=√(1/(5-1)((1-3)2+(2-3)2+(3-3)2+(4-3)2+(5-3)2))

=√(1/4((-2)2+(-1)2+(0)2+(1)2+(2)2))

=√(1/4((4)+(1)+(0)+(1)+(4)))

=√(2.5)

=1.5811

Σ((X - μx) (Y - μy))
=(1-3)(11-33.2)+(2-3)(22-33.2)+(3-3)(34-33.2)+(4-3)(43-33.2)+(5-3)(56-33.2)
=(-2*-22.2) + (-1*-11.2) + (0* 0.8) + (1 *9.8) + (2* 22.8)
=44.4 + 11.2 + 0 + 9.8 + 45.6
=111

Correlation Coefficient = 111/((5-1)*1.5811*17.5699)
Correlation Coefficient (r) = 0.9989
Hence the correlation coefficient between the two given data set is 0.9989
.............................................

Calculate Standard Error from the given data : X = 10, 20,30,40,50

a. 6.1071
b. 6.0711
c. 7.1071
d. 7.0711

Ans - d

Explanation :

Total Inputs (N) = (10,20,30,40,50)
Total Inputs (N) =5

First find Mean:
Mean (xm) = (x1+x2+x3...xn)/N
Mean (xm) = 150/5
Mean (xm) = 30

Then find SD:
SD = √(1/(N-1)*((x1-xm)2+(x2-xm)2+..+(xn-xm)2))
= √(1/(5-1)((10-30)2+(20-30)2+(30-30)2+(40-30)2+(50-30)2))
= √(1/4((-20)2+(-10)2+(0)2+(10)2+(20)2))
= √(1/4((400)+(100)+(0)+(100)+(400)))
= √(250)
= 15.811

Then Find Standard Error:
Standard Error=SD / √(N)
= 15.8114/√(5)
= 15.8114/2.2361
= 7.0711
.............................................

A sack contains 4 black balls 5 red balls. What is probability to draw 1 black ball and 2 red balls in one draw ?

a. 12/21
b. 9/20
c. 10/21
d. 11/20

Ans – c

Solution :

Out of 9, 3 (1 black & 2 red. are expected to be drawn)

Hence sample space

n(S) = 9c3
= 9!/(6!×3!)
= 362880/4320
= 84

Now out of 4 black ball 1 is expected to be drawn hence
nb. = 4c1
= 4
Same way out of 5 red balls 2 are expected be drawn hence
n(R) = 5c2
= 5!/(3!×2!)
= 120/12
= 10
Then P(B U R) = n(B)×n(R)/n(S)
i.e 4×10/84 = 10/21

……………………………………………………………………………………………………………………………………………

 


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