Question

Just before a referendum on a school​ budget, a local newspaper polls

398398

voters to predict whether the budget will pass. Suppose the budget has the support of

5555​%

of the voters. What is the probability that the​ newspaper's sample will lead it to predict​ defeat?

The probability is

nothing.

​(Round to three decimal places as​ needed.)

Can you please help me solve this by using the ti-84? I do not have excel

Answer 1

here

std error of proportion=σp=√(p*(1-p)/n)=

0.0249

for ti-83: press 2nd -vars -choose normalcdf : lower: 0 , upper : 0.5 , mean : 0.55 , std deviaiton =0.0249 , paste and you will get 0.022

below is detailed calculation

for normal distribution z score =(p̂-p)/σp
here population proportion=     p=	0.550
sample size       =n=	398
std error of proportion=σp=√(p*(1-p)/n)=	0.0249

probability that the​ newspaper's sample will lead it to predict​ defeat:

probability =

P(X<0.5)

=

P(Z<-2.01)=

0.022

( please try 0.023 if this comes wrong)