Question

find a) the mean and b) the standard deviation on an examination in which grades of 70 and 88 correspond to standard scores of -0.6 and 1.4 respectively.

Explanation:

Step1: Recall the z - score formula

The z - score formula is \(z=\frac{x - \mu}{\sigma}\), where \(x\) is the raw score, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.
For \(x = 70\) and \(z=-0.6\), we have \(-0.6=\frac{70 - \mu}{\sigma}\), which can be rewritten as \(70-\mu=- 0.6\sigma\) or \(\mu=70 + 0.6\sigma\).
For \(x = 88\) and \(z = 1.4\), we have \(1.4=\frac{88-\mu}{\sigma}\), which can be rewritten as \(88-\mu=1.4\sigma\).

Step2: Substitute \(\mu\) from the first equation into the second equation

Substitute \(\mu = 70+0.6\sigma\) into \(88-\mu=1.4\sigma\).
We get \(88-(70 + 0.6\sigma)=1.4\sigma\).
First, simplify the left - hand side: \(88 - 70-0.6\sigma=1.4\sigma\).
\(18-0.6\sigma=1.4\sigma\).
Add \(0.6\sigma\) to both sides: \(18=1.4\sigma+0.6\sigma\).
Since \(1.4\sigma+0.6\sigma=(1.4 + 0.6)\sigma=2\sigma\), then \(\sigma=\frac{18}{2}=9\).

Step3: Find the mean \(\mu\)

Substitute \(\sigma = 9\) into \(\mu=70 + 0.6\sigma\).
\(\mu=70+0.6\times9\).
\(\mu=70 + 5.4=75.4\).

Answer:

a) The mean \(\mu = 75.4\).
b) The standard deviation \(\sigma=9\).