Sol. Mean = Y x. - P(x ) = - [1 + 2 + 3 + 4 + 5 + 6] = 3.5
= 1 [(1 - 3.5)2 + (2 - 3.5)2 + (3 - 3.5)2 + (4 - 3.5)2 + (5 - 3.5)2 + (6 - 3.5)2] 6
Var(X) = 2.92. Standard deviation, o = ^Var(X) = ^2.92
If X and Y are independent random variable, the mean and variance of sum or difference is given by
Bernouilli or binomial distribution. The distribution of x repetitions of an event (say head of toss) with two possible outcomes is called a binomial distribution and the numbers above are called binomial coefficients. Consider the series of trials (n) satisfies the following conditions :
(a) Each trial can have two possible outcomes. e.g. success or failure with probabilities p and 1 - p .
(b) The outcome of each trial is an independent random event.
(c) Statistical equilibrium (i.e. the probabilities do not change). The number of ways of choosing an x things out of n trial is given as
C(n, x) is called binomial coefficient.
The probability of one particular combination of x success and n - x failures is px (1 - p)n-x.
Given two disjoint events with probabilities p and (1 -p), the probability of first occurring x times and second occurring (n - x) times in n trial, the most general form of binomial distribution is
P(n, x, p) = C(n, x) px (1 - p)n-x or simply P(x) = ( n px (1 - p)n-x ...(8.28)
Example 8.5. Suppose that the prob of a person being on the phone 0.1 and there are 10 people in the office. Dicide number of lines required in order to ensure that everyone one get reasonable chance.
Sol. For n = 10,p = 0.1, 1-p = 0.9 and for various x (no of persons), using equation (8.28), binomial distribution is tabulated in table 8.1.
P(x) = Fnj px (1 - p)n-x
P > x
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