In a random experiment of throwing 5 coins,th | vedclass

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(D) In a random experiment of throwing $n = 5$ coins,the number of heads $X$ follows a binomial distribution $B(n, p)$,where $n = 5$ and $p = \frac{1}

Vedclass · Accepted Answer

$\frac{5}{2}$

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(D) In a random experiment of throwing $n = 5$ coins,the number of heads $X$ follows a binomial distribution $B(n, p)$,where $n = 5$ and $p = \frac{1}{2}$ (probability of getting a head).The mean of a binomial distribution is given by the formula $E(X) = np$.Substituting the values,we get $E(X) = 5 	imes \frac{1}{2} = \frac{5}{2}$.Alternatively,using the probability distribution table:$X$$0$$1$$2$$3$$4$$5$$P(X)$$\frac{1}{32}$$\frac{5}{32}$$\frac{10}{32}$$\frac{10}{32}$$\frac{5}{32}$$\frac{1}{32}$The mean is calculated as $\sum X P(X) = (0 	imes \frac{1}{32}) + (1 	imes \frac{5}{32}) + (2 	imes \frac{10}{32}) + (3 	imes \frac{10}{32}) + (4 	imes \frac{5}{32}) + (5 	imes \frac{1}{32})$$= 0 + \frac{5}{32} + \frac{20}{32} + \frac{30}{32} + \frac{20}{32} + \frac{5}{32} = \frac{80}{32} = \frac{5}{2}$.

In a random experiment of throwing $5$ coins,the number of heads is defined as a random variable. The mean of the random variable is

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