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