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  1. #1
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    binomial

    50 (a) $\displaystyle b(3;8,0.6) = \binom{8}{3}(0.6)^{3}(0.4)^{5} - \binom{8}{2}(0.6)^{2}(0.4)^{6} $

    (b) $\displaystyle b(5;8,0.6) = \binom{8}{5}(0.6)^{5}(0.4)^{3} - \binom{8}{4}(0.6)^{4}(0.4)^{4} $

    (c) $\displaystyle P(3 \leq X \leq 5) = \binom{8}{5}(0.6)^{5}(0.4)^{3} - \binom{8}{2}(0.6)^{2}(0.4)^{6} $

    (d) $\displaystyle P(1 \leq X) = \sum_{x=1}^{12} \binom{12}{x}(0.1)^{x}(0.9)^{12-x} $

    59. (a) $\displaystyle B(4;10,0.3) = 0.850 $
    (b) $\displaystyle b(4;10,0.3) = B(4;10,0.3) - B(3;10,0.3) = 0.2 $
    (c) $\displaystyle b(6;10,0.7) = B(6;10,0.7) - B(5;10,0.7) = 0.2 $
    (d) $\displaystyle P(2 \leq X \leq 4) = B(4;10,0.3) - B(1;10,0.3) = 0.701 $
    (e) $\displaystyle P(2 \leq X) = 1- P(X \leq 1) = 1 - B(1;10, 0.3) = 0.851 $
    (f) $\displaystyle P(X \leq 1) = B(1;10, 0.7) = 0 $

    (g) $\displaystyle P(2<X<6) = B(5;10,0.3) - B(2,10,0.3) = 0.570$

    60. The long-run percentage of defectives is $\displaystyle 5 \% $. Also $\displaystyle X \sim \text{Bin}(25,.05) $.
    (a) $\displaystyle P(X \leq 2) = B(2;25, .05) = 0.873 $
    (b) $\displaystyle P(X \geq 5) = 1-P(X \leq 4) = 1-B(4;25,.05) = 1-0.993 = 0.007 $
    (c) $\displaystyle P(1 \leq X \leq 4) = B(4;25,.05) - B(0;25,.05) $

    these look right?
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by shilz222 View Post
    50 (a) $\displaystyle b(3;8,0.6) = \binom{8}{3}(0.6)^{3}(0.4)^{5} - \binom{8}{2}(0.6)^{2}(0.4)^{6} $
    By definition:

    (a) $\displaystyle b(3;8,0.6) = \binom{8}{3}(0.6)^{3}(0.4)^{5} $

    RonL
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by shilz222;106361 (c) [tex
    P(3 \leq X \leq 5) = \binom{8}{5}(0.6)^{5}(0.4)^{3} - \binom{8}{2}(0.6)^{2}(0.4)^{6} [/tex]
    If $\displaystyle X\sim B(8,0.6)$:

    $\displaystyle P(3 \leq X \leq 5) = \binom{8}{5}(0.6)^{5}(0.4)^{3} +\binom{8}{4}(0.6)^{4}(0.4)^{4}+\binom{8}{3}(0.6)^ {3}(0.4)^{5} $

    RonL
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