Directions are Use the given degree of confidence and sample

Directions are: Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p.

1. When 328 college students are randomly selected and surveyed, it is found that 122 own a car. Find a 99% confidence interval for the true proportion of all college students who own a car.

2. A newspaper article about the results of poll states: \"In theory, the results of such a poll, in 99 cases out of 100 should differ by no more than 5 percentage points in either direction from what would have been obtained by interviewing all voters in the United States.\" Find the sample size suggested by this statement.

3. The lengths of human pregnancies are normally distributed with a mean of 268 days and a standard deviation of 15 days. What is the probability that a pregnancy last at least 300 days.

4. The amount of rainfall in January in a certain city is normally distributed with a mean of 4.5 inches and a standard deviation of 0.3 inches. Find the value of quartile Q1.

5. Find the indicated value of 0.36

6. The weights of fish in a certain lake are normally distributed with a mean of 20lb and a standard deviation of 9. If 9 fish are randomly selected, what is the probability that the mean weight will be between 17.6 and 23.6 lb? and lastly

7. Find the indicated IQ score. The graph depicts IQ scores of adults, and those scores are normally distributed with a mean of 100 and a standard deviation of 15 shaded area is 0.10

If you are able to show all the work including graphs I would be GRATEFUL :) Thanks

Solution

2) 99 case among 100 gives us confidence level
as given in quesion that it Differ by no more than 5% points in either direction indicates the margin of error.

Margin of error = Confidence coefficient*Standard error of p
Margin of error = 5% = 0.05
Confidence coefficient is the critical value of z for 99% confidence level = 2.575
Standard error of p = sqrt [p*(1-p)/n]
= sqrt [0.5*0.5/n]
= 0.5/sqrt n
[When p value is not given , p = 0.5 will be taken as the standard error is maximum at that level]
Therefore,
0.05 = 2.575*0.5/sqrt n
n = [2.575*0.5/0.05]^2 = 664
Sample size is 664