Page 281 - DCOM203_DMGT204_QUANTITATIVE_TECHNIQUES_I

Quantitative Techniques – I

Notes 2. In a binomial distribution consisting of 5 independent trials, the probability of 1 and 2
successes are 0.4096 and 0.2048 respectively. Calculate the mean, variance and mode of the
distribution.
Solution:
1. It is given that np = 6 and npq = 2
npq 2 1 1 2 3
q so that p 1 and n 6 9
np 6 3 3 3 2
Now P(5 X 7) = P(X = 5) + P(X = 6) +P(X = 7)

5 4 6 3 7 2
9 2 1 9 2 1 9 2 1
C 5 C 6 C 7
3 3 3 3 3 3
5 5
2 9 9 9 2
9 C 5 C 6 2 C 7 4 9 438
3 3
2. Let p be the probability of a success. It is given that
5 4 5 2 3
C p 1 p 0.4096 and C p 1 p 0.2048
1
2
Using these conditions, we can write
4
p
5 1 p 0.4096 1 p 1
2 or 4 . This gives p
2 3
10p 1 p 0.2048 p 5
1 4
Thus, mean is np 5 1 and npq 1 0.8
5 5
1
Since (n +1)p, i.e 6 is not an integer, mode is its integral part, i.e., = 1.
5

Example: 5 unbiased coins are tossed simultaneously and the occurrence of a head is
termed as a success. Write down various probabilities for the occurrence of 0, 1, 2, 3, 4, 5 successes.
Find mean, variance and mode of the distribution.
Solution.
1
Here n = 5 and p q .
2
5
5 1
The probability mass function is P r C r , r = 0, 1, 2, 3, 4, 5.
2
The probabilities of various values of r are tabulated below:

r 0 1 2 3 4 5 Total
1 5 10 10 5 1
P r 1
32 32 32 32 32 32
1 1
Mean = np 5 2.5 and variance 2.5 1.25
2 2
1
Since (n +1)p = 6 3 is an integer, the distribution is bimodal and the two modes are
2
2 and 3.

276 LOVELY PROFESSIONAL UNIVERSITY