# Statistics Homework Help

• Jan 27th 2008, 03:40 PM
CavalryScout
Statistics Homework Help
Question:
Julie took five test; her scores are as follows:
IQ 110
SAT Math 605
SAT Verbal 590
math Ach. z = 1.3
Science Ach. PR = 84

They want all the z scores. How do you find the Z score for PR?
The Standard Deviation is 15. Mean 100.
• Jan 27th 2008, 04:13 PM
mr fantastic
Quote:

Originally Posted by CavalryScout
Question:
Julie took five test; her scores are as follows:
IQ 110
SAT Math 605
SAT Verbal 590
math Ach. z = 1.3
Science Ach. PR = 84

They want all the z scores. How do you find the Z score for PR?
The Standard Deviation is 15. Mean 100.

$z = \frac{X - \mu}{\sigma}$, where in this case X = PR = 84.
• Jan 27th 2008, 04:26 PM
CavalryScout
Stat help
Can you explain what the symbols stand for?
• Jan 27th 2008, 04:45 PM
mr fantastic
Quote:

Originally Posted by CavalryScout
Can you explain what the symbols stand for?

$\mu$ = mean.
$\sigma$ = standard deviation.

Maybe you're more comfortable with

$
z = \frac{X - \bar{X}}{s.d.}
$
• Jan 27th 2008, 04:55 PM
CavalryScout
PR is percentile and you need a raw score to use that formula according to this formula sheet. Its so confusing.
• Jan 27th 2008, 05:34 PM
mr fantastic
Quote:

Originally Posted by CavalryScout
PR is percentile and you need a raw score to use that formula according to this formula sheet. Its so confusing.

PR is a statistic that presumably follows a normal distribution. You have a value for PR, a mean for PR and a standard deviation for PR. You put those into the formula:

$z = \frac{PR - \bar{PR}}{s.d.} = \frac{84 - 100}{15} = -\frac{16}{15}$.

But for my money, it doesn't make sense to say PR has a mean of 100 ..... I would've thought the mean of a percentile was 50 by definition ......

In which case, $z = \frac{PR - \bar{PR}}{s.d.} = \frac{84 - 50}{15} = \frac{34}{15} = 2.27$, correct to 2 dp.

Maybe someone else will jump in and offer their 2 cents worth ....