Page 192 - DMTH404_STATISTICS

Statistics

Notes 4 3
5 5 2
C p (1 p ) = 0.4096 and C p (1 p ) = 0.2048
1 2
Using these conditions, we can write

5p (1 p ) 4 0.4096 (1 p ) 1
= = 2 or = 4 . This gives p =
3
10p 2 (1 p ) 0.2048 p 5

1 4
Thus, mean is np = ´ = 1 and npq = ´ = 0.8
1
5
5 5
1
Since (n +1)p, i.e., 6 ´ is not an integer, mode is its integral part, i.e., = 1.
5

Example 9: 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 F 1I 5
The probability mass function is P r a f = C r 2 H K , r = 0, 1, 2, 3, 4, 5.
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
5
Mean = np = ´ = 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 and 3.
2

14.1.3 Fitting of Binomial Distribution

The fitting of a distribution to given data implies the determination of expected (or theoretical)
frequencies for different values of the random variable on the basis of this data.
The purpose of fitting a distribution is to examine whether the observed frequency distribution
can be regarded as a sample from a population with a known probability distribution.
To fit a binomial distribution to the given data, we find its mean. Given the value of n, we can
compute the value of p and, using n and p, the probabilities of various values of the random
variable. These probabilities are multiplied by total frequency to give the required expected
frequencies. In certain cases, the value of p may be determined by the given conditions of the
experiment.

184 LOVELY PROFESSIONAL UNIVERSITY