The mean age is 47.5 years old. The standard deviation is 10.74832 years. http://www.calculator.net/standard-deviation-calculator.html Sample Standard Deviation, s:
Sample Standard Variance, s2
Total Numbers, N
Population Standard Deviation, σ
Population Standard Variance, σ2
If it follows the normal distribution
The 68.3% measure confidence range, σ
36.751683118298 - 58.248316881702
The 90% measure confidence range, 1.645σ
29.819018729601 - 65.180981270399
The 95% measure confidence range, 1.960σ
26.433298911865 - 68.566701088135
The 99% measure confidence …show more content…
On which exam did the student do better?
The student did better on their English test. I found this answer by calculating the z scores on both tests (by subtracting their score from the mean and then dividing by the standard deviation) then looking in the textbook’s appendix for the percentile. The student was in the 62.1 percentile for Math and 68.4 percentile for English.
4. Suppose you administered an anxiety test to a large sample of people and obtained normally distributed scores with a mean of 45 and standard deviation of 4. Do not use web-calculator to answer the following questions. Instead, you need to use the Z distribution table in Appendix A in Jackson’s book.
A. If Andrew scored 45 on this test. What is his Z score?
Andrew’s Z score would be zero.
B. If Anna scored 30 on this test. What is her Z score?
Anna’s Z score would be -.375.
C. If Bill’s Z score was 1.5, what is his real score on this test?
Bill’s test score was 51
D. There are 200 students in a sample. How many of these students will have scores that fall under the score of 41? Z=41-45 Z=