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gizmo_the_mogwai [7]
3 years ago
13

If x is a binomial random variable, compute the mean, the standard deviation, and the variance for each of the following cases:

Mathematics
1 answer:
gogolik [260]3 years ago
4 0

Answer:

(a) \mu_X=1.2, \sigma_X=0.8485, \sigma^{2} _X=0.72

(b) \mu_X=4, \sigma_X=0.8944, \sigma^{2} _X=0.8

(c) \mu_X=1.5, \sigma_X=1.0247, \sigma^{2} _X=1.05

(d) \mu_X=3.2, \sigma_X=0.8, \sigma^{2} _X=0.64

Step-by-step explanation:

A binomial random variable counts how often a particular event occurs in a fixed number of tries or trials.

For a binomial random variable X, we can use these formulas for mean, standard deviation, and variance:

\mu_X=np

\sigma_X=\sqrt{np(1-p)}

\sigma^{2} _X=np(1-p)

Applying the above formulas we get,

(a) n = 3, p = 0.4

\mu_X=3\cdot 0.4=1.2

\sigma_X=\sqrt{3\cdot 0.4(1-0.4)}=0.8485

\sigma^{2} _X=3\cdot \:0.4\left(1-0.4\right)=0.72

(b) n = 5, p = 0.8

\mu_X=5\cdot 0.8=4

\sigma_X=\sqrt{5\cdot 0.8(1-0.8)}=0.8944

\sigma^{2} _X=5\cdot \:0.8\left(1-0.8\right)=0.8

(c) n = 5, p = 0.3

\mu_X=5\cdot 0.3=1.5

\sigma_X=\sqrt{5\cdot 0.3(1-0.3)}=1.0247

\sigma^{2} _X=5\cdot \:0.3\left(1-0.3\right)=1.05

(d) n = 4, p = 0.8

\mu_X=4\cdot 0.8=3.2

\sigma_X=\sqrt{4\cdot 0.8(1-0.8)}=0.8

\sigma^{2} _X=4\cdot \:0.8\left(1-0.8\right)=0.64

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-----------------------------------------------------------

Work Shown:

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