How to find a percentage from standard deviation and mean?

Question

Tommy Wait’s times between pitches is normally distributed with a mean of 36 sec. and a standard deviation of 2.5 sec. What percentage of his times between his pitches are 30 and 34 seconds? Please show the steps so I can understand how to do it. Thank you!

Answer 1

For this problem, you will be using the Z-table to look up probabilities. We know the standard normal distribution has a mean of 0 and standard deviation of 1, but the normal distribution of times given in this problem does not match that for the standard normal (since mean = 36 and sd = 2.5). 

The good news is we can convert any wait time value to a Z value... all we need to do is use the following relationship:  Z = ( x - μ ) / σ.

The probability  P ( 30 < X < 34 ) needs to be converted to a P ( ? < Z < ? ) to determine it.  So we need to take those end points of the interval, 30 and 34, and convert them both to Z-scores. 

For 30, Z = ( 30 - 36 ) / 2.5  = -2.4. 

For 34, Z = ( 34 - 36 ) / 2.5 = -0.8.

So P ( 30 < X < 34 ), when converted to Z values, is P ( -2.4 < Z < -0.8 ).

Use your Z table.  The area we seek is between -0.8 and -2.4, but since the table is written in terms of total area from negative infinity up to a Z value, we need to first find the area to the left of -0.8 and then subtract out the area to the left of -2.4, which will leave us just the area in between -0.8 and -2.4.  The Greek letter phi represents the area to the left of a Z value: P ( - infinity < Z < x ) = Φ(x).

Mathematically, P ( -2.4 < Z < -0.8 ) = Φ(-0.8) - Φ(-2.4).  Looking at the Z table, Φ(-0.8) = 0.2119, and Φ(-2.4) = 0.0082.  The probability is 0.2119 - 0.0082, or 0.2037.