A set of chemistry exam scores are normally distributed with a mean of 70 points and a standard deviation of 5 points. What proportion of exam scores are between 68 and 73 points? You may round your answer to four decimal places. please help β

Find the z-score for 68 points and 73 points
z_{1}= \frac{x_{1} - \mu}{\sigma} = \frac{68 - 70}{5} = -\frac{-2}{5} = -0.4
z_{2} = \frac{x_{2}-\mu}{\sigma} = \frac{73-70}{5} = \frac{3}{5} = 0.6
Using the z-table, looking up z_{1} = -0.4, we see that 0.3446 of exam scores are below 68 points.
Using the z-table, looking up z_{2} = 0.6 we see that 0.7257 exam scores are below 73 points.
To find the area between the area between z_{1} and z_{2} we subtract z_{1} from z_{2} to get z_{2}-z_{1} = 0.7257 - 0.3446 = 0.3811
So, 0.3811 proportion of exam scores lie between 68 and 73 points.
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