Exercise
Jun 27'24
Answer
Solution: D
The number of heads is approximately normally distributed with mean 200 and variance 400 *1/2 * 1/2 = 100. 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}{10}[/math] and [math]\frac{x}{10} [/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 = 12.8[/math].