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Suppose that an unfair weighted coin has a probability of 0.6 of getting heads when

the coin is flipped. Assuming that the coin is flipped ten times and that successive
coin flips are independent of one another, what is the probability that the number
of heads is within one standard deviation of the mean?

User BlueBadger
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1 Answer

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The answer is 0.6659 or 66.59%

To find the probability that the number of heads is within one standard deviation of the mean, we need to calculate the mean and standard deviation of the binomial distribution.

The mean (μ) of a binomial distribution is given by n * p, where n is the number of trials and p is the probability of success (getting a head in this case). In this case, n = 10 (number of coin flips) and p = 0.6.

μ = n * p = 10 * 0.6 = 6

The standard deviation (σ) of a binomial distribution is given by sqrt(n * p * (1 - p)). Let's calculate the standard deviation:

σ = sqrt(n * p * (1 - p))
= sqrt(10 * 0.6 * (1 - 0.6))
= sqrt(10 * 0.6 * 0.4)
= sqrt(2.4 * 0.4)
= sqrt(0.96)
≈ 0.9798

Now, we need to calculate the range within one standard deviation of the mean. The lower bound will be μ - σ, and the upper bound will be μ + σ.

Lower bound = 6 - 0.9798 ≈ 5.0202
Upper bound = 6 + 0.9798 ≈ 6.9798

To find the probability that the number of heads is within one standard deviation of the mean, we calculate the cumulative probability of getting 5, 6, or 7 heads. We can use the binomial cumulative distribution function or a calculator that provides binomial probabilities.

P(5 ≤ X ≤ 7) = P(X = 5) + P(X = 6) + P(X = 7)

Using the binomial cumulative distribution function or a calculator, we can find the probabilities associated with each value:

P(X = 5) ≈ 0.2007
P(X = 6) ≈ 0.2508
P(X = 7) ≈ 0.2144

Now, let's sum up these probabilities:

P(5 ≤ X ≤ 7) ≈ 0.2007 + 0.2508 + 0.2144
≈ 0.6659

Therefore, the probability that the number of heads is within one standard deviation of the mean is approximately 0.6659, or 66.59%.
User Ashtonium
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