Since the given coin is a fair coin, so we have
P(Xi = 0) = 1/2 = 0.5
P(Xi = 1) = 1/2 = 0.5
So E(Xi) = 0*0.5 + 1*0.5 = 0.5
Let X = X1 + X2 + ... + X10
The covariance terms is vanishes because the coins tossed independently.
Answer:
E(X) = 5
Var(X) = 2.5
b) Since X is the summation of 10 independent Bernoulli's trials with probability of successes 0.5, so the distribution of X is binomial with parameters n = 10 and p = 0.5
Answer: X follows binomial distribution with parameters n = 10 and p = 0.5
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