For each of the binomial distributions, compute the mean, variance, and standard deviation. n=600, p=0.52

Konciljev56
2022-09-13
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mercuross8

Answered 2022-09-14
Author has **16** answers

For the binomical distribution we calculate mean using a equation,

$$\mu =np$$

where p is the probability and n is the sample size. Since n=600 and p=0.52 then we obtain,

$$\mu =600\ast 0.52=312$$

For the binomical distribution we calculate standard deviation using a equation,

$$\sigma =\sqrt{npq}$$

where p is the probability, n is the sample size, and q=1-p since n=600, p=0.52 therefore q=1-0.52=0.48 So, when substitute we get,

$$\sigma =\sqrt{600\ast 0.52\ast 0.48}=\sqrt{149.76}=12.24$$

$$\mu =np$$

where p is the probability and n is the sample size. Since n=600 and p=0.52 then we obtain,

$$\mu =600\ast 0.52=312$$

For the binomical distribution we calculate standard deviation using a equation,

$$\sigma =\sqrt{npq}$$

where p is the probability, n is the sample size, and q=1-p since n=600, p=0.52 therefore q=1-0.52=0.48 So, when substitute we get,

$$\sigma =\sqrt{600\ast 0.52\ast 0.48}=\sqrt{149.76}=12.24$$

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