Question

In a past​ election, the voter turnout was 75​%.In a​ survey, 1147 subjects were asked if they voted in the election. a. Find the mean and standard deviation for the numbers of voters in groups of 1147. b. In the survey of 1147 people, 829 said that they voted in the election. Is this result consistent with the​ turnout, or is this result unlikely to occur with a turnout of 75​%?Why or why​ not? c. Based on these​ results, does it appear that accurate voting results can be obtained by asking voters how they​ acted?

a. μ=____(Round to one decimal place as​ needed.)

  σ=____​(Round to one decimal place as​ needed.)

b. Is the result of 829 voting in the election usual or​ unusual?

A.This result is unusual because 829 is greater than the maximum usual value.

B.This result is unusual because 829 is below the minimum usual value.

C.This result is unusual because 829 is within the range of usual values.

D. This result is usual because 829 is within the range of usual values.

c. Does it appear that accurate voting results can be obtained by asking voters how they​ acted?

A.No comma because it appears that substantially more people say that they voted than the proportion of people who actually did vote.

B.Yes, because the results indicate that 75 % is a possible turnout.

C. No, because it appears that substantially fewer people say that they voted than the proportion of people who actually did vote.

Answer 1

a) a. μ= np= 1147*0.75= 860.2
σ= sqrt(npq)= sqrt(1147*0.75*0.25)= sqrt(215.0625)= 14.7

b)  The range rule of thumb suggests that values are unusual
if they lie outside of these limits:

Maximum usual values = µ + 2 σ = 860.2+2*14.7= 860.2+29.4= 889.6

Minimum usual values  = µ – 2 σ = 860.2-2*14.7 = 860.2-29.4 = 830.8

Therefore 829 value lie within it therefore it is not unusual.

This result is usual because 829 is within the range of usual values. OPTION D

c) Yes, because the results indicate that 75 % is a possible turnout. OPTION B