expected value --please explain

Mary has a free token to play a game. The probability that Mary will win the game is 0.01, so the probability that she will not win is 0.99. If Mary wins, she will be given $100, while if she loses, she must pay $5. Let X = the amount of money Mary wins (or loses).

what is the expected value of x?

is it (n)(p)? ...(.01)(100) + (.99)(-5) ?


what is the variance of x?

You're on the right track. \(\displaystyle 100\left(\frac{1}{100}\right)-5\left(\frac{99}{100}\right)=-3.95\)

She can expect to lose $3.95 just for playing the game.

so then the variance of x would be...

(100)(.1)(1-.1)= 9

+

(-5)(.99)(1-.99)= -.0495

= 8.95?