Originally Posted by

**lax600** The average height for players on a certain basketball team is 73 inches, with a standard deviation of 6 inches. What percent of players' hgeights are:

a.) between 70 and 73 inches?

I did:

$\displaystyle z1 = 70 - 73/6$

$\displaystyle z1 = -3/6$

$\displaystyle z1 = -0.5$

$\displaystyle z1 = 0.3085$

$\displaystyle z2 = 73 - 73/6$ this work is redundant

$\displaystyle z2 = 0/6$

$\displaystyle z2 = 0$

$\displaystyle z2 = 0.5000$

$\displaystyle 0.3085 + 0.5000 = 0.8135$

1 - 0.8135 = 0.1915 which is 19.15%

So 19.15% of the people are between 70 and 73 inches. But my question is there a faster simpler way of solving this?