what is the mean of probability distribution

Question

there is a 0.1141 probability that a best of seven contest will last four games, a 0.157 probability that it will last five games, a 0.1789 probability that it will last 6games, and a0.5563 probability that it will last 7 games. verify this is a probability distribution. find it mean and standard deviation. it is unusual for a team to sweep by winning in four games.

Answer 1

I take it 0.157 should read 0.1507. Check the sum of the probabilities sums to 1. Assuming the correction is right the sum is 1. So the x values are: 4, 5, 6, 7 with corresponding probabilities: 0.1141, 0.1507, 0.1789, 0.5563. The expectation, or mean is 4*0.1141+5*0.1507+6*0.1789+7*0.5563=6.1774. 

variance=SD^2=(16*0.1141+25*0.1507+36*0.1789+49*0.5563)-mean^2=39.2922-38.1602=39.2922-38.1602=1.132 approx. so SD (standard deviation)=1.064 approx.

Therefore mean and SD are 6.177+1.064, making it unusual for a team to sweep in 4 games. The z score for 4 is (4-6.18)/1.06=-2.06 (corresponding to a probability of 1.88%).