6.(CLASS EXAMPLE) Assuming the test grades in our statistical class with100students are normally distributed. In this distribution, the mean =75and the standard deviation =10.

Note:We do not require you touse the Standard Normal Table to find theareas under the normal curvecorresponding to a given Z score. The easier way is touse the following website to get the proportions instead:

http://davidmlane.com/hyperstat/z_table.html

1)Find the percentage of students whose scores range between the mean and 85.

Myfirstsolution was from using theStandard Normal Table in the textbook.FYI.Feel free to disregard this approach.

N= 100; X= 85; mean = 75; standard deviation = 10

a) Convert 85 to a Z score: Z = (85-75)/10 = +1

b) Look up"1"in the standard normal table (Column A) and find the corresponding area in Column B. The value(in the percentage format)is 34.13.

c)Thus,the percentage of students whose scoresrange between the mean and 85 is34.13%.

Alternatively, using thegivenwebsite is much easier.

http://davidmlane.com/hyperstat/z_table.html

Please pay close attention to where I put the values of mean (=75), standard deviation (=10), and the range of scores required in this question (that is between 75 and 85).

2)Find the percentage of students whose scores are equal to or higher than 90.

My firstsolution here was from using theStandard Normal Table in the textbook.FYI.

N= 100; X= 90; mean = 75; standard deviation = 10

a) Convert 90 to a Z score: Z = (90-75)/10 = +1.5

b) Look up 1 in the standard normal table (Column A) and find the corresponding area in Column C. The value(in the percentage format)is 6.68.

c)Thus, the percentage of students whose scoresare equal to or higher than 90 is6.68%.

Using thegivenwebsite is much easier.

http://davidmlane.com/hyperstat/z_table.html

3)Find thenumberof the students whose scores range between 70 and 90.

The followings show how to directly use thewebsite to solve this problem:http://davidmlane.com/hyperstat/z_table.html

N=100; mean = 75; standard deviation = 10.

We need to first find the proportion of the students whose scores are between 70 and 90.Enter the information needed at the above website and we will get the probability value (the shaded area) that is 0.6247.

The number of students who belong to this category:0.6247*100 = 62.47

4)How manystudents fail the test if the passing grade is 60?

7.The GSS dataset provides the following statistics for the average years of education for lower-, working-, middle-, and upper-class respondents, and their associated standard deviation. (Note:All the information is given in the following table.You do NOT need the SPSS to solve this problem.)

Social Class	Mean	SD	N
Lower class	11	3.5	100
Working class	12	2.5	500
Middle class	14	3.0	500
Upper class	15	3.5	50

1). Assuming that the years of education is normally distributed in the population, how many percentages ofworking-classrespondents have 12 to 16 years of education?

2). Assuming that the years of education is normally distributed in the population, how manymiddle-classrespondents have 16 to 20 years of education?

3) Assuming that the years of education is normally distributed in the population, what is the probability that anupper-classrespondent will have less than 12 years of education?

8.Usethegsslab.savtoanswerexamine the distribution of "agekdbrn" (Respondent's age when 1^stchild born).

1)Split your output by gender. Use a histogramwith a normal curveforthe distribution ofmen's ages when 1^stchild was born. (Use the variable"agekdbrn".)

2)Calculate the mean, standard deviation, and skewness for "agekdbrn" in thismalesample. (Include the descriptive stats tablewhere you find your answer.)