Question

A tourism magazine surveyed 5000 people for their expected travel plans the following year. The response rate of the survey is 75%. 45% of the respondents of the survey agreed that they already have plans for their visits the following year, while the remaining respondents said they like to plan spontaneously. Assume that while preparing a research report for you hospitality class you conduct a similar survey and select a sample of 40 people.

1. Calc. the probability that the proportion that plans their visit in advance lies between 43% and 47%
2. Calc. the probability that the proportion that plans their visit in advance is less than 60%

Answer 1

for normal distribution z score =(p̂-p)/σp
here population proportion=     p=	0.7500
sample size       =n=	40
std error of proportion=σp=√(p*(1-p)/n)=	0.06847

a) probability that the proportion that plans their visit in advance lies between 43% and 47%

probability =

P(0.43<X<0.47)

=

P(-4.67<Z<-4.09)=

0-0=

0.0000

b)

probability that the proportion that plans their visit in advance is less than 60% :

probability =

P(X<0.6)

=

P(Z<-2.19)=

0.0143