Question

Based on historical data, your manager believes that 37% of the company's orders come from first-time customers. A random sample of 81 orders will be used to estimate the proportion of first-time-customers. What is the probability that the sample proportion is between 0.31 and 0.48? Note: You should carefully round any z-values you calculate to 4 decimal places to match wamap's approach and calculations. Answer = (Enter your answer as a number accurate to 4 decimal places.)

Answer 1

Solution :

Given that ,

p = 0.37

1 - p = 0.63

n = 81

= p = 0.37  

= (p*(1-p))/n =  (0.37*0.63)/ 81= 0.05364

P(0.31 < < 0.48 ) = P((0.31-0.37)/0.05364 ) < ( - ) /  < (0.48-0.37) /0.05364 ) )

= P(-1.1186 < z < 2.0507 )

= P(z < 2.0507) - P(z < -1.1186)

= 0.9799 - 0.1317

= 0.8482

Probability = 0.8482