Question

Based on historical data, your manager believes that 25% of the
company's orders come from first-time customers. A random sample of
165 orders will be used to estimate the proportion of
first-time-customers. What is the probability that the sample
proportion is greater than than 0.32?

Answer #1

Sampling distribution of p̂ is approximately normal if np
>=10 and n (1-p) >= 10

n * p = 165 * 0.25 = 41.25

n * (1 - p ) = 165 * (1 - 0.25) = 123.75

Mean = =
p = 0.25

Standard deviation =
= 0.03371

X ~ N ( µ = 0.25 , σ = 0.03371 )

P ( X > 0.32 ) = 1 - P ( X < 0.32 )

Standardizing the value

Z = ( X - µ ) / σ

Z = ( 0.32 - 0.25 ) / 0.03371

Z = 2.0765

P ( ( X - µ ) / σ ) > ( 0.32 - 0.25 ) / 0.03371 )

P ( Z > 2.0765 )

P ( X > 0.32 ) = 1 - P ( Z < 2.0765 )

P ( X > 0.32 ) = 1 - 0.9811

P ( X > 0.32 ) = 0.0189

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