I NEED HELP WITH THESE PROBLEMS ASAP 1500 POINTS 1 In a past

I NEED HELP WITH THESE PROBLEMS ASAP. 1500 POINTS

1. In a past presidential election, the actual voter turnout was 55%. In a survey, 1142 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 1142.
(Round answer to one decimal place.)
?=  
(Round answer to two decimal places.)
?=

Give the interval of usual values for the number of voters in groups of 1142.
(Enter answer as an interval using square-brackets only with whole numbers.)
usual values =

In the survey of 1142 people, 702 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 55%?

this result is consistent with the actual voter turnout

this result is unlikely to occur with the actual voter turnout

2. Several economics students are unprepared for a surprise true/false test with 11 questions, and all of their answers are guesses.

Find the mean for the number of correct answers for such students.
(Round answer to one decimal place.)
?=

Find the standard deviation for the number of correct answers for such students.
(Round answer to two decimal places.)
?=

Give the range for the usual number of correct answers.
(Enter answer as an interval using square-brackets only with whole numbers.)
usual values =

Would it be unusual for a student to pass by guessing and getting at least 7 correct answers?

yes, it would be unusual

no, it would not be unusual

3. Assume that a procedure yields a binomial distribution with n=400 trials and the probability of success for one trial is p=1415.

Find the mean for this binomial distribution.
(Round answer to one decimal place.)
? =

Find the standard deviation for this distribution.
(Round answer to two decimal places.)
? =

Use the range rule of thumb to find the minimum usual value ?

Solution

Mean of sample = 55% of 1142 = 78.1

Std dev = rt of (0.55)(0.45)/1142 = 0.0147

= 0.01

95% conf interval for no of voters in percent

52.11% to 57.89%

no of voters 95% = (606.56, 661.10)

H0: p = 55%

Ha:: p not equals 55%

p hat = sample proportion = 702/1142 = 0.615

Difference = 6.5%

z statistic = 6.5/rt 0.55(0.45) = 2.81

The P-Value is 0.004954.

The result is significant at p < 0.05.