Question

A baseball player has a lifetime batting average of 0.318. If, in a season, this player has 330 "at bats", what is the probability he gets 124 or more hits?

Probability of 124 or more hits =

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Answer 1

X ~ Bin ( n , p)  

n = 330 , p = 0.318

Mean = np = 330 * 0.318 = 104.94

Standard deviation = sqrt ( n p ( 1 - p) )

= sqrt ( 330 * 0.318 ( 1 - 0.318) )

= 8.4599

Using normal approximation with continuity correction,

P(X >= x) = P(Z > ( x-0.5 - Mean ) / SD )

So,

P(X >= 124) = P(Z > ( 123.5 - 104.94) / 8.4599)

= P(Z > 2.19)

= 0.0143 (From Z table)