Question

3. In a restaurant, the proportion of people who order coffee with their dinner is 0.9. A simple random sample of 144 patrons of the restaurant is taken. a) What are the expected value and standard error? b) What is the probability that the proportion of people who will order coffee with their meal is between 0.85 and 0.875? c) What is the probability that the proportion of people who will order coffee with their meal is at least 0.945?

Answer 1

aill Given, m - 144 Pa 0.9 a) Taking normal approximation to The Binomial, the Sampling distribution of sample proportion follows normal distribution with mean, Mo=p=0.9 i.e expected value = 0.9 standard error, SE= p(1-P) 0.9) (1-0.1) 144 n = 0.025 o Ê = 0.025 Then Ŝ wn (M@= 0.9,6 = 0.025) P-MÊ P - 0.9 follows standard Normal "Б Za 0.025

P (0.85 € 8 20.875) = P(0.85-0.9 LP-MEC 0.875-0.9 0.025 0.025 P(-2 CZ c-i) P(ZL-1) - P( ZL-2) - 0.158655-0.02275 (from slanderd 0.135905 > 0.1359 Normal table 0.1359 0.85 0.88 Z Z=-2.00 Z-1.00 Here, we need to find P($ 3, 0.945) P ($20.945) Plt PP-P 0.945- 0.9 - 0.025 P(zz 1.8) = 1- P( ZL1.8) = 1 - 0.96407 0.03593 0.6359 0.0359 Z-1.8