Math Professors Bowling: Normalcdf Probability Analysis

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SUMMARY

The discussion centers on the probability analysis of bowling scores among math professors using the normal cumulative distribution function (normalcdf). The mean score is established at 88 with a standard deviation of 15. Calculations reveal that the probability of a professor scoring above 100 is approximately 0.2118, while the probability of scoring between 50 and 100 is approximately 0.7824. Concerns are raised regarding the assumption of normal distribution and the justification for using the value "4" in the normalcdf function.

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  • Familiarity with the normal cumulative distribution function (normalcdf)
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rowdy3
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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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