X | successful throw | unsuccessful throw |
P(X) | 0.3 | 0.7 |
Probsbility of successful throw
=P = successful throw / total Throw = 0.3
Probability of unsuccessful throw
= 1 - Probability of successful throw
= 0.7
2.
For a single trial, shooting guards throw is a bernoulli distribution, but for n trials it is sume of bernoulli distribution which is binomial distribution.
Expected value of binomial distribution = n*p
Expected value of shooting guards throw
= E(X) = n*p = 10 * 0.3 = 3
Variance of binomial distribution = n*p*(1-p)
Var(x) = 10 * 0.3 * 0.7 = 2.1
3.
Now there are two points for each throw
Hence we find expectation of 2*X
Hence,
Expected value of shooting guards throw
= E(2*X) = 2*E(X) = 2*3 = 6
Variance is
Var(2X) = 4*var(x) = 4 * 2.1 = 8.4
{since var(a*X) = a^2*var(x),
Where a is constant
E(a*X) = a*E(X)}
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