Question

Find the normal approximation for the binomial probability of (don't use binomial probability) A) P(x=4) where n=13 and P=.5 B) P (X<3) where n =13 and P=.5

Answer #1

A)

Mean = np = 13 * 0.5 = 6.5

Standard deviation = sqrt(np(1-p)) = sqrt(13 * 0.5 * 0.5) = 1.8028

Using normal approximation,

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

With continuity correction,

P(X = 4) = P( 3.5 < X < 4.5)

= P( X < 4.5) - P( X < 3.5)

= P( X < 4.5 - 6.5 / 1.8028 ) - P (Z < 3.5 - 6.5 / 1.8028)

= P (Z < -1.1094) - P( Z < -1.6641)

= 0.1336 - 0.048

= **0.0856**

b)

P( X < x) = P( Z < x-0.5 - mean / SD) ( With continuity correction)

P( X < 3) = P( X < 2.5 - 6.5 / 1.8028)

= P( Z < -2.2188)

= **0.0133**

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=n=, 13p, 0.5
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. Round
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