Question

The National Coffee Association reported that 64 % of U.S. adults drink coffee daily. A random sample of 225 U.S. adults is selected. Round your answers to at least four decimal places as needed.

probability of adults who drink coffee daily between .60 and .72

Answer 1

Sampling distribution of p̂ is approximately normal if np >=10 and n (1-p) >= 10
n * p = 225 * 0.64 = 144
n * (1 - p ) = 225 * (1 - 0.64) = 81

Mean = = p = 0.64
Standard deviation = = 0.032

X ~ N ( µ = 0.64 , σ = 0.032 )
P ( 0.6 < X < 0.72 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 0.6 - 0.64 ) / 0.032
Z = -1.25
Z = ( 0.72 - 0.64 ) / 0.032
Z = 2.5
P ( -1.25 < Z < 2.5 )
P ( 0.6 < X < 0.72 ) = P ( Z < 2.5 ) - P ( Z < -1.25 )
P ( 0.6 < X < 0.72 ) = 0.9938 - 0.1056
P ( 0.6 < X < 0.72 ) = 0.8881