Question

A)

Given a random sample size n=1600 from a binomial probability distribution with P=0.40 do the following... Find the probability that the percentage of successes is between 0.37 and 0.44. The probability is O A. 0.0095 O B. 0.9701 O C. 0.9021 O D. None of these answers are correct.

B)

Answer 1

Problem A)

Given p=0.40 and n=1600 Mean of the sampling distribution p is Mo =p=0.40 Standard deviation of the sampling distribution p is p*(1-P) V 0.40*(1-0.40) –0,012247 V 1600 PCO 37c AcoM=P0.37–40P-M, 0.44–40 P(0.37<p <0.44)=P(0.012247 35 0.012247 ) =P(-2.45<z<3.27) =P(z<3.27)-P(Z<-2.45) =0.9995-0.0071 =0.9923 =99.23% Option Dis correct

Problem B)

Problem 2 Given p=0.23 and n=952 Mean =np=952*0.23= 218.96 Standard deviation - Vnpq = 952 * 0.23 * (1 – 0.23) =12.9846 We have P(x>210) =P(> 210.384.36) =P(27-0.69) =1-P(z<-0.69) =1-0.2451 =0.7549 (using z value table) 1st option is correct