1. A set of exam scores are reported as X values and z-scores. On this exam a score of X = 75 corresponds to a z-score of z = +1.00, and X = 60 corresponds to a z-score of z = –0.50. What are the values for the mean and standard deviation for this exam?

Question

1.	A set of exam scores are reported as X values and z-scores. On this exam a score of X = 75 corresponds to a z-score of z = +1.00, and X = 60 corresponds to a z-score of z = –0.50. What are the values for the mean and standard deviation for this exam?

anonymous · Answer

IT IS NOT Z=+1.00 IT IS Z=+100

debbieg · Answer

z=100? are you sure? highly unlikely. z=1.00 makes much more sense.

kropot72 · Answer

This can be solved using simultaneous equations, based on the following formula:
$$z=\frac{X-\mu}{\sigma}$$
From the first given data we get
$$1=\frac{75-\mu}{\sigma}$$
which can be rearranged to
$$\sigma=75-\mu\ ............(1)$$
From the second given data we get
$$-0.5=\frac{60-\mu}{\sigma}$$
$$-0.5\sigma=60-\mu\ ............(2)$$
Now you just need to solve the system of equations (1) and (2).

anonymous · Answer

YES, it is +1.00

debbieg · Answer

Ok then. Like @kropot72 said - you can set up a system of 2 equations. The z-score is just telling you the number of standard deviations above the mean that the raw score is.

So if  X = 75 corresponds to a z-score of z = +1.00, and X = 60 corresponds to a z-score of z = –0.50, then:

$\Large 75=\mu + \sigma$
$\Large 60=\mu - 0.5 \sigma$

anonymous · Answer

I am still confused! How do I solve for the mean and s.d. of those. I am lost!

debbieg · Answer

You have a system of 2 equations in 2 unknowns. You can solve using either elimination (easy here, since you can just subtract one equation from the other) or substitution (solve one equation for one of the variables, and substitute that back into the other equation, then solve for the remaining variable).

debbieg · Answer

The other, less "technical" way to think of it is this:

75 is one stnd dev above the mean, right?
and 60 is 1/2 of a stnd dev below the mean.

Thus, the DISTANCE between 75 and 60, is 1.5 standard deviations.

Hence:

$\Large 15=1.5 \sigma$

Now solve THAT for $\sigma$, and you have your standard deviation.

To get your mean, go back to what I said above:
75 is one stnd dev above the mean
and 60 is 1/2 of a stnd dev below the mean.

That's really what the system of 2 equations in 2 unknowns does, also.... but this is just a less technical way to think about it.

anonymous · Answer

Thank you so much! That really helped me!

debbieg · Answer

you're welcome. :)

anonymous · Answer

So are my (u) things 65 and 55?! Do I have two means?