Y varies as t. if y = 6 when t= 4

Mathematics

1 answer:

jonny [76]3 years ago

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A.
$\mathbb P(X>6)=\mathbb P(X=7)+\mathbb P(X=8)+\mathbb P(X=9)$
$=\dbinom97(0.4)^7(1-0.4)^{9-7}+\dbinom98(0.4)^8(1-0.4)^{9-8}+\dbinom99(0.4)^9(1-0.4)^{9-9}$
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b.
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c.
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$\mathbb P(2$
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f.
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g, h.
For $X\sim\mathcal B(n,p)$ , recall that $\mathbb 
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$\mu_X=9(0.4)=3.6$
${\sigma_X}^2=9(0.4)(0.6)=2.16$

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