Question

In a past presidential election, the actual voter turnout was 63%. In a survey, 1126 subjects were asked if they voted in the presidential election.

Find the mean and standard deviation for the numbers of actual voters in groups of 1126.
(Round answer to one decimal place.)
μ=

(Round answer to two decimal places.)
σ=

Give the interval of usual values for the number of voters in groups of 1126.
(Enter answer as an interval using square-brackets only with whole numbers.)
usual values =

In the survey of 1126 people, 782 said that they voted in the last presidential election. Is this result consistent with the actual voter turnout, or is this result unlikely to occur with an actual voter turnout of 63%?

( )this result is consistent with the actual voter turnout

( )this result is unlikely to occur with the actual voter turnout

Answer 1

(a)

n = 1126

p = 0.63

q= 1- p =0.37

= np = 1126 X 0.63 = 709.4

(b)

=

(c)

Usual values:

[709.4 - (2 X 16.2), 709.4 + (2 X 16.2)]

= [ 709.4 - 32.4, 709.4 + 32.4]

= [677, 742]

(d)

Correct option:

this result is unlikely to occur with the actual voter turnout,