Assume that when adults with smartphones are randomly selected, 64% use them in meetings or classes. If 9 adult smartphone users are randomly selected, find the probability that exactly 2 of them use their smartphones in meetings or classes.
Solution:
"p = 64 \\% = 0.64"
n = 9
"X\\sim Bin(n,p)"
"P(X=x) = \\frac{n!}{x!(n-x)!} \\times p^x \\times (1-p)^{n-x} \\\\\n\nP(X=2) = \\frac{9!}{2!(9-2)!} \\times 0.64^2 \\times (1-0.64)^{9-2} \\\\\n= 0.01155"
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