Question

1. Standard Normal Distribution P(z<c) = 0.7652 (two decimal places)

2. The mean score is 70, standard deviation is 11, P(x>c) = 0.44

3. Out of 100 people sampled, 61 had kids. Construct a 90% confidence interval for the true population proportion of people with kids. __<p<__ (three decimal places)

4. P(-0.88<z<0.4) (four decimal places)

5. Mean of 1500, standard deviation of 300. Estimating the average SAT score, limit the margin of error to 95% confidence interval to 25 points, how many students to sample?

6. Population proportion is 43%, would like to be 95% confident that your estimate is within 4.5% of the true population proportion. How large of a sample is required?

7. P(z<1.34) (four decimal places)

8. Candidate only wants a 2.5% margin error at a 97.5% confidence level, what size of sample is needed?

9. Estimate this proportion to within 4% at the 95% confidence level, how many randomly selected college students must we survey?

10. 420 people were asked if they like dogs, 22% said they did. Find the margin of error for the poll at 95% confidence level. (four decimals

Answer 1

1. Standard Normal Distribution P(z<c) = 0.7652 (two decimal places)

Answer)

From z table, P(z<0.72) = 0.7652

C = 0.72

2. The mean score is 70, standard deviation is 11, P(x>c) = 0.44

Answer)

Z = (x - mean)/s.d

From z table, P(z>0.15) = 0.44

0.15 = (x - 70)/11

X = 71.65

So answer is 71.65

3. Out of 100 people sampled, 61 had kids. Construct a 90% confidence interval for the true population proportion of people with kids. __<p<__ (three decimal places)

Answer)

N = 100

P = 61/100

First we need to check the conditions of normality that is if n*p and n*(1-p) both are greater than 5 or not

N*p = 61

N*(1-p) = 39

Both the conditions are met so we can use standard normal z table to estimate the interval

From z table critical value z for 90% confidence level is 1.645

Margin of error (MOE) = Z*√{P*(1-P)}/√N

MOE = 0.08023486446

Interval is given by

P-MOE < P < P+MOE

0.52976513553 < P < 0.69023486446

4. P(-0.88<z<0.4) (four decimal places)

= p(z<0.4) - p(z<-0.88)

= 0.466