excans:359e4b78f2: Difference between revisions
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(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...") |
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'''Solution: | '''Solution: D''' | ||
The number of heads is approximately normally distributed with mean 200 and variance | 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 10<sup>th</sup> percentile of a standard normal variable is approximately equal to -1.2816, this gives <math>x = 12.8</math>. | ||
Latest revision as of 13:20, 28 November 2024
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].