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CAIIB - ABM- CASE STUDIES / NUMERICAL QUESTIONS


Find Coefficient of Variance for the values given : {13,35,56,35,77}

a. 0.4156
b. 0.5164
c. 0.5614
d. 0.6514

Ans - c

Explanation :

Number of terms (N) = 5

Mean:
Xbar = (13+35+56+35+77)/5
= 216/5
= 43.2

Standard Deviation (SD):
Formula to find SD is

σx= √(1/(N - 1)*((x1-xm)2+(x2-xm)2+..+(xn-xm)2))
=√(1/(5-1)((13-43.2)2+(35-43.2)2+(56-43.2)2+(35-43.2)2+(77-43.2)2))
=√(1/4((-30.2)2+(-8.2)2+(12.9)2+(-8.2)2+(33.8)2))
=√(1/4((912.04)+(67.24)+(163.84)+(67.24)+(1142.44)))
=√(588.2)
=24.2528

Coefficient of variation (CV)
CV = Standard Deviation / Mean
= 24.2528/43.2
= 0.5614

Hence the required Coefficient of Variation is 0.5614
.............................................

A bank calculates that its individual savings accounts are normally distributed with a mean of Rupees 2,000 and a standard deviation of Rupeess600. If the bank takes a random sample of 100 accounts, what is the probability that the sample mean will lie between Rupees 1,900 and Rupees 2,050?

a. 0.4792
b. 0.7492
c. 0.7942
d. 0.9742

Ans - b

Explanation :

Standard Error = SD / √(N)
= 600 / √100
= 600 / 10
= 60

Using the equation
z = (x bar minus Mu)/SE

we get 2 z values

for x bar = Rs. 1900,
z = (1900 - 200) / 60
= (-100) / 60
= -1.67

for x bar = Rs. 2050,
z = (2050 - 200) / 60
= 50 / 60
= 0.83
Probability table gives us probability of 0.4525 corresponding to a z value of –1.67, and it gives probability of 0.2967 for a z value of 0.83. If we add these two together, we get 0.7492 as the total probability that the sample mean will lie between Rs. 1900 and Rs. 2,050.
.............................................

Suppose that a population with N is equal to 144 has p is equal to 24. What is the mean of the sampling distribution of the mean for samples of size 25?

a. 24
b. 12
c. 4.8
d. 2

Ans – a
.............................................

We have six students say A, B, C, D, E, F participating in a quiz contest. Out of six students only two can reach to the final. What is the probability of reaching to the final of each student?

a. 2/5
b. 1/2
c. 1/3
d. 1/4

Ans - c
........................................................

Suppose a population with N = 144 has u(Mean)=24. What is the mean of sampling distribution of the mean for samples of size of Rs 25 ?

a. 24
b. 2
c. 4.8
d. 3.2

Ans - a

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