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Mean and Standard Deviation of a binomial distribution

Mean and Standard Deviation of a binomial distribution. Lesson 10.2. Formulas. μ (mean)= np Where n = number of trials, p = probability of success σ 2 (variance) = npq where q = probability of failure σ ( standard deviation) =. Example 1. 0q 2 + 2pq + 2p 2. 2p(q+p). 2p.

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Mean and Standard Deviation of a binomial distribution

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  1. Mean and Standard Deviation of a binomial distribution Lesson 10.2

  2. Formulas μ (mean)= np Where n = number of trials, p = probability of success σ2 (variance) = npq where q = probability of failure σ ( standard deviation) =

  3. Example 1 0q2 + 2pq + 2p2 2p(q+p) 2p Suppose a dart player has probability p of hitting the bull’s-eye with a single dart, and all attempts are independent. Prove that the expected number of bull’s eyes the player will hit in two attempts is 2p. Binomial distribution:

  4. Example 2 In the binomial distribution of tossing a far coin 250 times and counting the number of heads, what is the mean and standard deviation? Mean: np 250 * .5 = 125 Sd: (npq).5 (250 * .5 * .5).5 7.91

  5. Example 3 For some binomial probability distribution, μ = 45 and σ = 6, find n and p.

  6. homework Pages 638 – 639 1 - 11

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