Question

A standardized​ exam's scores are normally distributed. In a recent​ year, the mean test score was 1491 and the standard deviation was 312. The test scores of four students selected at random are 1910​, 1230​, 2190​, and 1380. Find the​ z-scores that correspond to each value and determine whether any of the values are unusual.

Answer 1

Answer:

$z(1910)=1.3429\\\\z(1230)=-0.9006\\\\z(2190)=2.2404\\\\z(1380)=-0.3558$

Step-by-step explanation:

-Given the population mean is $\mu=1491$and  the standard deviation is $\sigma=312$, we calculate the z-scores using the formula:

$z=\frac{\bar x-\mu}{\sigma}$

#The z-score for 1910 can be calculated as:

$z(1910)=\frac{x-\mu}{\sigma}\\\\=\frac{1910-1491}{312}\\\\=\frac{419}{312}\\\\=1.3429$

#The z-score for 1230can be calculated as:

$z(1230)=\frac{x-\mu}{\sigma}\\\\=\frac{1230-1491}{312}\\\\=\frac{-281}{312}\\\\=-0.9006$

#The z-score for 2190 is calculated as follows:

$z(2190)=\frac{x-\mu}{\sigma}\\\\=\frac{2190-1491}{312}\\\\=\frac{699}{312}\\\\=2.2404$

#The z-score for 1380 is calculated as:

$z(1380)=\frac{x-\mu}{\sigma}\\\\=\frac{1380-1491}{312}\\\\=\frac{-111}{312}\\\\=-0.3558$