Revision as of 23:56, 26 June 2024 by Admin (Created page with "A balanced coin is flipped 400 times. Determine the number <math>x</math> such that the probability that the number of heads is between <math>200- x</math> and <math>200 + x</math> is approximately .80. '''References''' {{cite web |url=https://math.dartmouth.edu/~prob/prob/prob.pdf |title=Grinstead and Snell’s Introduction to Probability |last=Doyle |first=Peter G.|date=2006 |access-date=June 6, 2024}}")
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ABy Admin
Jun 27'24

Exercise

A balanced coin is flipped 400 times. Determine the number [math]x[/math] such that the probability that the number of heads is between [math]200- x[/math] and [math]200 + x[/math] is approximately .80.

References

Doyle, Peter G. (2006). "Grinstead and Snell's Introduction to Probability" (PDF). Retrieved June 6, 2024.

ABy Admin
Jun 27'24

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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