Finding the percentage of students with a higher grade than 63?

I'm having some trouble with a question on my stats assignment, I went to the free tutoring on campus but the notes they made are a little confusing:


Q: Suppose that you have a grade of 63. What percent of students scored higher than you? Please show the procedures of your calculations


Mean: 70
SD: 6.1


So I assumed the procedure would be:


P (x > 63)
P (z > 63 – 70 / 6.1)
P (z > -1.147)


and the tutor on campus said they answer would be, 86.43% (0.8643 X 100 = 86.43%).


I don't know why but I feel like this is wrong? I don't understand how they made the leap from -1.147 to getting the .8643 from our table


This is the table my prof supplied us with in her notes, I used it because our textbook doesn't have a complete table, only one with selected numbers, which isn't much help (just in case the table won't enlarge here is the link to it: http://www.sfu.ca/personal/archives/richards/Table/Pages/Table1.htm)







Thanks!

coreygeorge said:

I'm having some trouble with a question on my stats assignment, I went to the free tutoring on campus but the notes they made are a little confusing:


Q: Suppose that you have a grade of 63. What percent of students scored higher than you? Please show the procedures of your calculations


Mean: 70
SD: 6.1


So I assumed the procedure would be:


P (x > 63)
P (z > 63 – 70 / 6.1)
P (z > -1.147)


and the tutor on campus said they answer would be, 86.43% (0.8643 X 100 = 86.43%).


I don't know why but I feel like this is wrong? I don't understand how they made the leap from -1.147 to getting the .8643 from our table


This is the table my prof supplied us with in her notes, I used it because our textbook doesn't have a complete table, only one with selected numbers, which isn't much help (just in case the table won't enlarge here is the link to it: http://www.sfu.ca/personal/archives/richards/Table/Pages/Table1.htm)

Thanks!

Click to expand...

In my opinion, it is wrong answer.

Do you know the meaning of the table above and how to read it?

Do you know how to do linear interpolation?

coreygeorge said:

I'm having some trouble with a question on my stats assignment, I went to the free tutoring on campus but the notes they made are a little confusing:


Q: Suppose that you have a grade of 63. What percent of students scored higher than you? Please show the procedures of your calculations


Mean: 70
SD: 6.1


So I assumed the procedure would be:


P (x > 63)
P (z > 63 – 70 / 6.1)
P (z > -1.147)


and the tutor on campus said they answer would be, 86.43% (0.8643 X 100 = 86.43%).


I don't know why but I feel like this is wrong? I don't understand how they made the leap from -1.147 to getting the .8643 from our table


This is the table my prof supplied us with in her notes, I used it because our textbook doesn't have a complete table, only one with selected numbers, which isn't much help (just in case the table won't enlarge here is the link to it: http://www.sfu.ca/personal/archives/richards/Table/Pages/Table1.htm)

Thanks!

Click to expand...

To get that solution the z-value has been rounded to -1.1. Note that the "larger area" corresponding to z = 1.1 (ie the area where z>-1.1) is 0.8643.
For more accuracy, you could use a more accurate table (which allows for interpolation) or do a linear interpolation here.
For example if you rounded the z-value to -1.15, and approximation would be halfway between 0.8643 and 0.8849 (the number directly under it).