Assume that when adults with smartphones are randomly​ selected, 51​% use them in meetings or classes. If 11 adult smartphone users are rand

Question

Assume that when adults with smartphones are randomly​ selected, 51​% use them in meetings or classes. If 11 adult smartphone users are randomly​ selected, find the probability that fewer than 5 of them use their smartphones in meetings or classes.

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Everleigh 2 weeks 2021-09-10T09:53:37+00:00 1 Answer 0

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    2021-09-10T09:54:44+00:00

    Answer:

    The probability is 0.2356.

    Step-by-step explanation:

    Let X is the event of using the smartphone in meetings or classes,

    Given,

    The probability of using the smartphone in meetings or classes, p = 51 % = 0.51,

    So, the probability of not using smartphone in meetings or classes, q = 1 – p = 1 – 0.51 = 0.49,

    Thus, the probability that fewer than 5 of them use their smartphones in meetings or classes.

    P(X<5) = P(X=0) + P(X=1) + P(X=2) + P(X=3)+P(X=4)

    Since, binomial distribution formula is,

    P(x) = ^nC_r p^x q^{n-x}

    Where, ^nC_r=\frac{n!}{r!(n-r)!}

    Here, n = 11,

    Hence,  the probability that fewer than 5 of them use their smartphones in meetings or classes

    =^{11}C_0 (0.5)^0 0.49^{11}+^{11}C_1 (0.5)^1 0.49^{10}+^{11}C_2 (0.5)^2 0.49^{9}+^{11}C_3 (0.5)^3 0.49^{8}+^{11}C_4 (0.5)^4 0.49^{7}

    =(0.5)^0 0.49^{11}+11(0.5)0.49^{10} + 55(0.5)^2 0.49^{9}+165 (0.5)^3 0.49^{8} +330(0.5)^4 0.49^{7}

    =0.235596671797

    \approx 0.2356

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