Quiz Review for Weeks 5 and 6

1. Find the area under the standard normal curve between z = 1.6 and z = 2.6. First we look for the area of both by doing “2nd ,Vars, NORMALCDF” and inputting “-1000, “Z,” 0, 1 then find the difference between both.

2. A business wants to estimate the true mean annual income of its customers. It randomly samples 220 of its customers. The mean annual income was $61,400 with a standard deviation of $2,200. Find a 95% confidence interval for the true mean annual income of the business’ customers.

First we find E by doing Zc(standard deviation/square root of number of trials.) Now we add and subtract that number from the mean income to find both endpoints. The Zc of 95% is 1.96 so we would do 1.96((2200)/square root of 220) giving us 290.71. Now if we subtract that from the mean income, we get 61109.29 as the left endpoint and 61690.71 as the right endpoint.

3. IQ test scores are normally distributed with a mean of 100 and a standard deviation of 15. An individual's IQ score is found to be 120. Find the z-score corresponding to this value. “X-mean/standard deviation. (120-100)/15

4. Two high school students took equivalent language tests, one in German and one in French. The student taking the German test, for which the mean was 66 and the standard deviation was 8, scored an 82, while the student taking the French test, for which the mean was 27 and the standard deviation was 5, scored a 35. Compare the scores. Same equation as the question above to find the corresponding z-score. 2 would be better.

5. A business wants to estimate the true mean annual income of its customers. The business needs to be within $250 of the true mean. The business estimates the true population standard deviation is around $2,400. If the confidence level is 90%, find the required sample size in order to meet the desired accuracy. first we find the Z score for the percentage which ends up being 1.645. Then we multiply that times the standard