mean and standard deviation of a binomial distribution
DESCRIPTION
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. - PowerPoint PPT PresentationTRANSCRIPT
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Lesson 10.2
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μ (mean)= npWhere n = number of trials, p = probability of
success
σ2 (variance) = npq where q = probability of failure
σ ( standard deviation) =
![Page 3: Mean and Standard Deviation of a binomial distribution](https://reader036.vdocuments.pub/reader036/viewer/2022082711/5681324e550346895d98c8a5/html5/thumbnails/3.jpg)
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:
0q2 + 2pq + 2p2
0 1q2
1 2pq
2 1p2
2p(q+p)2p
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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
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