Question

A population proportion is 0.5. A sample of size 150 will be taken and the sample proportion p will be used to estimate the population proportion. Use z-table Round your answers to four decimal places. a. what is the probability that the sample proportion will be within ±0.03 of the population proportion? b, what is the probability that the sample proportion will be within ±0.08 ofthe population proportion?

Answer 1

1. Given that a population proportion is P=0.5. A sample of size n=150 will be taken and the sample proportion p will be used to estimate the population proportion. a. The probability that the sample proportion will be within +0.03 of the population proportion is obtained as follows: P(P-P $10.03)=pl_P-P P(1-P) 0.03 P(1-P) 0.03 = PZS 0.5(1-0.5) V 150 = P(Z <+0.74) = P(-0.74 52 50.74) = P(2 50.74)-P(25-0.74) = 0.7704 - 0.2296 = 0.5408 Thus, the probability that the sample proportion will be within +0.03 of the population proportion is 0.5408.

b. The probability that the sample proportion will be within +0.08 of the population proportion is obtained as follows: POD=A5 2009 r. Po por p-P P(\p-P|S=0.08)=P - P(1-P) 0.08 P(1-P) 0.08 = PZS 10.5(1-0.5) V 150 = P(Z <+1.95) = P(-1.95<251.95) = P(z 51.95)- P(Z S-1.95) = 0.9744 -0.0256 = 0.9488 Thus, the probability that the sample proportion will be within +0.08 of the population proportion is 0.9488.

2. The president of Doerm Distributors, Inc., believes that 32% of the firm's orders come from first-time customers. Sample random of 100 orders will be used to estimate the proportion of first-time customers. a. Assume that the president is correct and p= 0.32. The sampling distribution of P for this study will be a normal distribution because np and n(1-p) are both greater than 5. b. The probability that the sample proportion will be between 0.24 and 0.40 is derived as follows: P P(0.24 <p<0.40) = P 0.24- P – P(1-P) -P P(1-P) 0.40-P P(1-P) 0.24-0.32 0.40 -0.32 = P =<<- 0.32(1-0.32) 10.32(1-0.32) V 100 100 = P(-1.71<z<1.71) = P(Z <1.71) - P(z<-1.71) = 0.9564 -0.0436 = 0.9128 Thus, the probability that the sample proportion will be between 0.24 and 0.40 is 0.9128.

The probability that the sample proportion will be between 0.29 and 0.35 is derived as follows: P 0.29- P P(0.29 <p<0.35)= P - P(1-P) -P P(1-P) 0.35-P P(1-P) 0.29 -0.32 0.35 -0.32 = P < <- 10.32(1-0.32) 10.32 (1-0.32) (V 100 V 100 = P(-0.64 <z<0.64) = P(z<0.64)-P(z<-0.64) = 0.7389 -0.2611 = 0.4778 Thus, the probability that the sample proportion will be between 0.29 and 0.35 is 0.4778.