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CAIIB-ABM-RECOLLECTED QUESTIONS FROM FEB 2017-2



Friends, Updating here the recollected questions from Feb 2017 Exams. Wish you all the very best for your exam.

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Questions on Propability / Sampling / Standard Deviation, Standard Error, Co-variance

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?

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.
.............................................

A jar contains 3 red marbels, 7 green marbels and 10 white marbles. If a marble is drawn at random, What is the probability that marble drawn is white?

Here Red = 3
Green = 7
White = 10
Hence total sample space is (3+7+10)= 20
Out of 20 one ball is drawn n(S) = {c(20,a.} = 20

To find the probability of occurrence of one White marble out of 10 white ball
n(R)={c(10,a.} = 10

Hence P(R) = n(R)/n(S)
= 10/20 = 1/2
........................................................

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 ?

Since out of 6, 2 can reach the final. Hence sample space is
n(S) = 6 c2 = 6!/(6-b.!×2! = 15
Here event of occurrence of probability of each student out of six (A B C D E F) = (AB AC AD AE AF) = n(E) = 5
Now P(E) = 5/15 = 1/3
........................................................

An bag contains 10 black balls and 5 white balls. 2 balls are drawn from the bag one after other without replacement. What is the probability that both drawn are black ?

Let E and F denote respective events that first and second ball drawn are black.
We have to find here P(E), P(E/F) and P(E n F )

Now P(E) = P(Black in first drawn) = 10/15

Also given that the first ball is drawn i.e events E has occurred. Now there are 9 black balls and 5 white balls left in the urn. Therefore the probability that the second ball drawn is black, given that the ball first drawn is black nothing but conditional probability of F given that E has occurred already.

Hence P(E/F) = 9/14

Now by the multiplication rule of probability
P(E n F) = P(E) × P(E/F)
= 10/15 × 9/14 = 3/7
........................................................

A sum of Rs. 32800 is borrowed to be paid back in 2 years by two equal annual installments allowing 5% compound interest. Find the annual payment.

Here,

PV =?
P = Rs. 32800
T = 2 years
R = 5% = 0.05

PV = P / R * [(1+R)^T - 1]/(1+R)^T

32800 = P × (1.052 – 1) ÷ (0.05 × 1.052)
P = 32800 ÷ 1.8594
P = 17640
.............................................

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

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?

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.
.............................................


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