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].

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