# Math Help - Normal Distribution

1. ## Normal Distribution

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:

$z1 = 70 - 73/6$
$z1 = -3/6$
$z1 = -0.5$
$z1 = 0.3085$

$z2 = 73 - 73/6$
$z2 = 0/6$
$z2 = 0$
$z2 = 0.5000$

$0.3085 + 0.5000 = 0.8135$
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?

2. 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:

$z1 = 70 - 73/6$
$z1 = -3/6$
$z1 = -0.5$
$z1 = 0.3085$

$z2 = 73 - 73/6$ this work is redundant
$z2 = 0/6$
$z2 = 0$
$z2 = 0.5000$

$0.3085 + 0.5000 = 0.8135$
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?

By definition for a normal distribution the mean is the mode and median value, so there was no need to calculate the probability of someone being less then 73 inches.

Bobak

3. i need to find those who were in between 70 and 73 inches

4. 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:

$z1 = 70 - 73/6$
$z1 = -3/6$
$z1 = -0.5$
$z1 = 0.3085$

$z2 = 73 - 73/6$
$z2 = 0/6$
$z2 = 0$
$z2 = 0.5000$

$0.3085 + 0.5000 = 0.8135$
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?
Yes. As bobak already mentioned, 73 is the mean so you know that z2 = 0 without having to do the calculation you've done to get it. Then the required probability is 0.5 - 0.3085 = 0.1915 etc.