Question

A baseball player has a lifetime batting average of 0.208. If, in a season, this player has 300 "at bats", what is the probability he gets 55 or more hits?

Probability of 55 or more hits

Answer 1

X ~ Binomial (n,p)

Where n = 300 , p = 0.208

Mean = n * p = 300 * 0.208 = 62.4

Standard deviation = Sqrt(np(1-p))

= Sqrt( 300 * 0.208 * 0.792)

= 7.03

Using normal approximation to binomial distribution with continuity correction,

P( X >= x) = P( X > x - 0.5) ( Using continuity correction )

P( X >= x) = P( Z > x - Mean / Standard deviation) ( Using normal approximation)

P( X >= 55) = P( X > 54.5)

= P( Z > 54.5 - 62.4 / 7.03)

= P( Z > -1.1238)

= P( Z < 1.1238)

= 0.8694