SOLUTION: A fair coin is tossed 6times. The probability distribution of X, the number of heads is recorded in the table below; x 0 1 2 3 4 5 6 P(X=x) 0.16 0.093 0.23 0.46 0.23

Question 1087230: A fair coin is tossed 6times. The probability distribution of X, the number of heads is recorded in the table below;
x 0 1 2 3 4 5 6
P(X=x) 0.16 0.093 0.23 0.46 0.23 0.093 0.16

Find the expected number of heads.

Answer by mathmate(429) (Show Source): You can put this solution on YOUR website!
Question:
A fair coin is tossed 6times. The probability distribution of X, the number of heads is recorded in the table below;
x 0 1 2 3 4 5 6
P(X=x) 0.16 0.093 0.23 0.46 0.23 0.093 0.16
Find the expected number of heads.

Solution:
The expected value is the sum ∑x*p(x)
where x is the value of the outcome, and p(x) is the corresponding probability.

For the given distribution, first verify that it is in fact is a distribution by adding up the probabilities. If they do not add up to exactly one, something is amiss.

We find that the sum adds up to 1.426, so there is a problem with the given distribution, and work cannot progress unless this is corrected.

The correction can be made by recalculating the probabilities using the binomial distribution, the results of which are as follows:

x P(X=x)
0 0.015625
1 0.09375
2 0.234375
3 0.3125
4 0.234375
5 0.09375
6 0.015625

From the symmetry, we can tell that the expected value (=mean) is 3.
Using the above formula, we find
∑ x*P(X=x)
=(0.0+0.09375+0.46875+0.9375+0.9375+0.46875+0.09375)
=3 as expected.

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