Suppose that 90 % of people are right-handed. | vedclass

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(B) Let $X$ be the number of right-handed people in a sample of $n = 10$. The probability of a person being right-handed is $p = 0.9$,and the probabil

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$\sum_{r=0}^{6} {^{10}C_r} (0.9)^r (0.1)^{10-r}$

Vedclass · Answer

(B) Let $X$ be the number of right-handed people in a sample of $n = 10$. The probability of a person being right-handed is $p = 0.9$,and the probability of being left-handed is $q = 1 - 0.9 = 0.1$. $X$ follows a binomial distribution $B(n, p) = B(10, 0.9)$. The probability that at most $6$ people are right-handed is $P(X \le 6)$. Using the probability mass function of the binomial distribution,$P(X = r) = {^{n}C_r} p^r q^{n-r}$. Thus,$P(X \le 6) = \sum_{r=0}^{6} {^{10}C_r} (0.9)^r (0.1)^{10-r}$.

Suppose that $90 \%$ of people are right-handed. What is the probability that at most $6$ of a random sample of $10$ people are right-handed?

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