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# Modeling Approximately Normal Distributions Question

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2500 students take a college entrance exam. The scores on the exam have an approximately normal distribution  (use the​ 68-95-99 rule) with a mean of 52 points and stand deviation of 11 points.

Estimate the percentage of students scoring 19 points or less.

Apr 21, 2019

### 2+0 Answers

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$$19 = 52 - 3\cdot 11 = \mu - 3 \sigma\\ 99\% \text{ of the scores fall into }(\mu-3\sigma,\mu + 3\sigma)\\ \text{so }0.01\% \text{ of the scores fall outside that}\\ 0.005\% \text{ above, and }0.005\% \text{ below}\\ \text{so }\approx 0.005\% \text{ scored 19 points or less}$$

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Apr 21, 2019
edited by Rom  Apr 21, 2019
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Thank you so much!!! That makes perfect sense! I have another question if you wouldn't mind helping me.

RandiLaine98  Apr 22, 2019
edited by RandiLaine98  Apr 22, 2019