Question

Based on historical data, your manager believes that 31% of the company's orders come from first-time customers. A random sample of 128 orders will be used to estimate the proportion of first-time-customers. What is the probability that the sample proportion is less than 0.32?

Note: You should carefully round any intermediate 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 = 0.31, n = 128, p^ = 0.32

P(p^ < 0.32) = P((p^-p)/sqrt(pq/n) < (0.32-0.31)/sqrt(0.31*0.69/128))
= P(Z < 0.2446)
= 0.5948

Use the standard normal table to find this area :

  

Z	0.00	0.01	0.02	0.03	0.04	0.05	0.06	0.07	0.08	0.09
0.0	0.5	0.504	0.508	0.512	0.516	0.5199	0.5239	0.5279	0.5319	0.5359
0.1	0.5398	0.5438	0.5478	0.5517	0.5557	0.5596	0.5636	0.5675	0.5714	0.5753
0.2	0.5793	0.5832	0.5871	0.591	0.5948	0.5987	0.6026	0.6064	0.6103	0.6141