The opera theater manager believes that 23#% of the opera tickets for tonight's show have been sold. If the manager is accurate, what is the probability that the proportion of tickets sold in a sample of 869869 tickets would differ from the population proportion by less than 3%3%

0.9642

Step-by-step explanation:

Sample proportion p follows a Normal distribution with mean = p0 = 0.23 and Standard deviation = √P0(1 - P0) / n = √0.23(1-0.23) / 869 = 0.0143

Then Z = p - mean / Standard deviation = X - 0.23 / 0.01343

P(0.2<p<0.26) = P(p<0.26) - P(p<0.2)

= P(p-023/0.0143 < 0.26-0.23/0.0143) - P(p-0.23/0.0143 < 0.2-0.23/0.0143)

= P(Z < 2.1) - P(Z < -2.1)

= 0.9821 - 0.0179

= 0.9642

answer: i'm not sure

step-by-step explanation: