A survey found that 38% of consumers from a country A are more likely to buy sto

Question

A survey found that 38% of consumers from a country A are more likely to buy stock in a company based in Country A, or shop at it stores, if it is making an effort to publicly talk about how it is becoming more sustainable. Suppose you select a sample of 100 respondents from Country A. Complete parts (a) through (d) below. a. What is the probability that in the sample tower than 38% are more likely to buy stock in a company based in Country A, or shop at its stores, if it is making an effort to publicly talk about how it becoming more sustainable? The probability is % b. What is the probability that in the sample, between 34% and 42% are more likely to buy stock in a company based in Country A. or shop at its stores, if it is making an effort to publicly talk about how it is becoming more sustainable? c. What is the probability that in the sample, more than 34% are more likely to buy stock in a company based in Country A, or shop at its stores, if it is making an effort to publicly talk about how it is becoming more sustainable? The probability is %. d. If a sample of 400 is taken, how does this change your answers to (a) through (c)? If a sample of 400 is taken, what is the probability that in the sample fewer than 38% are more likely to buy stock in a company based in Country A, or shop at its stores, if it is making an effort to publicly talk about how it is becoming more sustainable? The probability is %. If a sample of 400 is taken, what is the probability that in the sample between 34% and 42% are more likely to buy stock in a company based in Country A, or shop

Explanation / Answer

let x=number of person like to buy stock

here we use binomial distribution with parameter n=100 and p=0.38

mean=E(x)=np=100*0.38=38

var(x)=n*p*(1-p)=100*0.38*(1-0.38)=23.56

SD(x)=sqrt(var(x))=sqrt(23.56)=4.85

for (a) to (c) we use use standard normal variate z=(x-mean)/SD(x)

(a) for percentate 38%, x=0.38*100=38 and z=(38-38)/4.85=0

P(x<38)=P(z<0)=0.5 =50%( using ms-excel =normsdist(0))

(b) for percentage 34%, x=34 and z=(34-38)/4.85=-0.8247

for percentage 42%, x=42 and z=(42-38)/4.85=0.8247

P(34<x<42)=P(-0.8247<z<0.8247)=P(z<0.8247)-P(z<-0.8247)=0.7952-0.2048=0.5904=59.04%

P(z<0.8247)=0.7952( using ms-excel command=nomrsdist(0.7952)

(c) for percentage 34%, x=34 and z=(34-38)/4.85=-0.8247

P(x>34)=P(z>-0.08247)=1-P(z<-0.8247)=1-0.2047=0.7953=79.53%

part (d)

when n=400 and p=38%=0.38

E(x)=np=400*0.38=152

SD(x)=sqrt(np(1-p))=sqrt(400*0.38*(1-0.38))=9.71

(a) for percentate 38%, x=0.38*400=152 and z=(152-152)/9.71=0

P(x<152)=0.5

(b) for percentage 34%, x=136 and z=(136-152)/9.71=-1.6478

for percentage 42%, x=168 and z=(168-152)/9.71=1.6478

P(34<x<42)=P(-1.6478<z<1.6478)=P(z<1.6478)-P(z<-1.6478)=0.9503-0.0497=0.9006=90.06%

(c)P(x>136)=P(z>-1.6478)=1-P(z<-1.6478)=1-0.0497=0.9503=95.03%

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