Revision as of 03:28, 27 June 2024 by Admin (Created page with "'''Solution: B''' The number of heads is approximately normally distributed with mean 200 and variance 200 *1/2 * 1/2 = 50. Then the probability that the number of heads is between <math>200-x </math> and <math>200 + x </math> is the probability that a standard normal variable is between <math>\frac{-x}{\sqrt{50}}</math> and <math>\frac{x}{\sqrt{50}} </math>, which must equal 0.8. Since the 10<sup>th</sup> percentile of a standard normal variable is approximately equal...")
Exercise
ABy Admin
Jun 27'24
Answer
Solution: B
The number of heads is approximately normally distributed with mean 200 and variance 200 *1/2 * 1/2 = 50. Then the probability that the number of heads is between [math]200-x [/math] and [math]200 + x [/math] is the probability that a standard normal variable is between [math]\frac{-x}{\sqrt{50}}[/math] and [math]\frac{x}{\sqrt{50}} [/math], which must equal 0.8. Since the 10th percentile of a standard normal variable is approximately equal to -1.2816, this gives [math]x = 9.0622[/math].