Question

A random sample of size 200 is to be taken from a population that has a proportion equal to 0.75. The sample proportion will be used to estimate the population proportion. a. Calculate the probability that the sample proportion will be within 1 0.02 of the population proportion. b. Calculate the probability that the sample proportion will be within +2 standard errors of the population proportion. c. Calculate the probability that the sample proportion will be within +0.04 of the population proportion. a. The probability is (Round to four decimal places as needed.) b. The probability is (Round to four decimal places as needed.) c. The probability is (Round to four decimal places as needed.)

Answer 1

an nd ce P Let denote respectively the population proportion sample proportion and that if the We know size n is sample sufficiently large so a and that np = 10

nlip) > 10, then ê ânle, P(1-P) Here na 200, p=0.75 i np np = 200 x 0.75= 150=10. nll-p) = 2008 11-0.75) 50710 P a N (0.75, 0.0009375)

Ke 0.75 [ Let Z 0.0009 375 then za N10,1).] = P(Z <0.6532) Plze-0.6532)

10.6532) - Ø 1-0.6532) • 10-4532) - 1- 10.1532) know [ $(2) is the cdf cdf of N10,1). we that 01-0) = 1- Ile), for any w 0 74 3186 - {1 -0.743186 0 4864. b) that Note SELê 0009 375

Required probability -el-2x SE LÅS & ê - P + 2X SELÊ 11 - 2 L 2 ê -P SE lê ê 0.75 L2 =p 2 L 000 9375 = P(-22252) Plz=2) - Plze-2) $(2) - $ 1-2) 012) - {1- 512)] 2. 097 7250 - z 0977250 503 N. o 9545 .

e> Required probability =pl-o 0.045 Popco.04 4 ) D P - 0.04 00009375 ê -0.75 Jo 0.0009 375 0.04 - Jo.000 9375 pl-1.3064 22 e 1.3064) Plz=1.3064) -P(2=-1.3064) I 11.3064) 61-3664) )

호(1, 3069) - |회(11·30649] = 0 904292 {1- 0.904292 0.8086 ,