Math Professors Bowling: Normalcdf Probability Analysis

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A group of math professors calculated their mean bowling score as 88 with a standard deviation of 15. They found the probability of a professor scoring above 100 to be approximately 21.18% using the normal cumulative distribution function (normal cdf). For scores between 50 and 100, the probability was calculated at about 78.24%. Concerns were raised regarding the justification of using the normal distribution without explicit confirmation of the data's normality. The discussion highlights the importance of proper statistical assumptions in probability calculations.
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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
 
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Why "cdf(.8, 4)" for the first one? Where did the "4" come from?
 
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.
 

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