Question

Use the appropriate normal distribution to approximate the
resulting binomial distribution.

A basketball player has a 75% chance of making a free throw.
(Assume that the throws are independent of each other.) What is the
probability of her making 100 or more free throws in 112 trials?
(Round your answer to four decimal places.)

You may need to use the appropriate table in the Appendix of Tables
to answer this question.

Answer #1

Using Normal Approximation to Binomial

Mean = n * P = ( 112 * 0.75 ) = 84

Variance = n * P * Q = ( 112 * 0.75 * 0.25 ) = 21

Standard deviation = √(variance) = √(21) = 4.5826

P(X < x) = P(Z < ( x - mean) / SD )

P ( X >= 100 ) = ?

Using continuity correction

P ( X > n - 0.5 ) = P ( X > 100 - 0.5 ) =P ( X > 99.5
)

P ( X > 99.5 ) = 1 - P ( X < 99.5 )

Standardizing the value

Z = ( X - µ ) / σ

Z = ( 99.5 - 84 ) / 4.5826

Z = 3.38

P ( ( X - µ ) / σ ) > ( 99.5 - 84 ) / 4.5826 )

P ( Z > 3.38 )

P ( X > 99.5 ) = 1 - P ( Z < 3.38 )

P ( X > 99.5 ) = 1 - 0.9996

P ( X > 99.5 ) = **0.0004**

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