I feel my adults with smart phones are randomly selected 64% use them in meetings or classes if five adults smart phone users are randomly selected find the probability that exactly 3 of them use their smart phones and meetings our classes
Solution;
Given;
"P=0.64"
"n=5"
"x=3"
Then;
"q=1-p=1-0.64=0.36"
Using binomial distribution;
"P(X=x)=\\frac{n!}{(n-x)!x!}p^xq^{n-x}"
By direct substitution;
"P(X=3)=\\frac{5!}{2!\u00d73!}\u00d70.64^3\u00d70.36^2"
"P(X=3)=0.3397"
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