+34 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.
+6251 \(19 = 52 - 3\cdot 11 = \mu - 3 \sigma\\ 99\% \text{ of the scores fall into }(\mu-3\sigma,\mu + 3\sigma)\\ \text{so }0.01\% \text{ of the scores fall outside that}\\ 0.005\% \text{ above, and }0.005\% \text{ below}\\ \text{so }\approx 0.005\% \text{ scored 19 points or less}\)
.
+34 Thank you so much!!! That makes perfect sense! I have another question if you wouldn't mind helping me.
