Math Professors Bowling: Normalcdf Probability Analysis

[rowdy3]

One summer night at Bellair Lanes, a group of math professors went bowling. In true form, they decided to calculate their mean bowling score and standard deviation to use for the statistics problems. Here are the results: u(mean) 88, o(standard dev.)15.
Find the probability that if a math professor is selected, his or her score will be greater than 100.
100-88 / 15 = .8 normal cdf(.8,4) =.2118

Find the probability that, if a math professor is selected, his or her score will be between 50 and 100.
50-88 / 15 = -2.53 100-88 / 15 =.8 normal cdf(-2.53,.8)= .7824.

Did I do the problems right? Thanks

 

[HallsofIvy]

Why "cdf(.8, 4)" for the first one? Where did the "4" come from?

 

[statdad]

Unless the problem states that the scores resemble a normal distribution your calculation isn't justified. If your instructor left that comment out, it's pretty sloppy.