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.

Answer 1

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

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

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

$=0.235596671797$

$\approx 0.2356$