Based on historical data, your manager believes that 30% of the company's orders come from first-time customers. A random sample of 64 orders will be used to estimate the proportion of first-time-customers. What is the probability that the sample proportion is greater than than 0.2?

0.9594

Step-by-step explanation:

Given that:

Sample size (n) = 64

p = 30% = 0.3 ; 1 - p = 0.7

Probability that sample proportion is greater than 0.2

Mean (m) = n*p = 64 * 0.3 = 19.2

Standard deviation (s) = √n*p*(1-p) = √(19.2*0.7)= 3.67

0.2 of 64 = 12.8

Z > [(x - m) / s]

Z > (12.8 - 19.2) / 3.67

Z > (-6.4 /3.67)

Z > - 1.7438

= 0.9594