$$ 62 \% $$ of U.S. adults have very little confidence in newspapers. You randomly select 10 U.S. adults. Find the probability that the number of U.S. adults who have very lit exactly five, (b) at least six, and (c) less than four. (a) $$ P(5)=0.183 $$ (Round to three decimal places as needed.) (b) $$ P(x \geq 6)=\square $$ (Round to three decimal places as needed.)

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UpStudy AI · Accepted Answer

We model the number of U.S. adults with very little confidence in newspapers among 10 as a binomial random variable with parameters $$ n=10 $$ and $$ p=0.62 $$. That is, for $$ k $$ successes,
$$
P(X=k)=\binom{10}{k}(0.62)^k(0.38)^{10-k}.
$$

We are given that
$$
P(5)=\binom{10}{5}(0.62)^5(0.38)^5\approx 0.183.
$$

To find the probability that at least six have very little confidence, we use
$$
P(X\ge6)=\sum_{k=6}^{10}P(X=k)=1-\sum_{k=0}^{5}P(X=k).
$$

Thus,
$$
P(X\ge6)=1-P(X\le5).
$$

We compute
$$
P(X\le5)=\sum_{k=0}^{5}\binom{10}{k}(0.62)^k(0.38)^{10-k}.
$$

Once the sum $$ P(X\le5) $$ is computed (or obtained using a calculator or statistical software), we find
$$
P(X\ge6)=1-P(X\le5)\approx 0.682.
$$

Rounded to three decimal places, the answer for (b) is
$$
P(X\ge6)=0.682.
$$
The probability that at least six out of ten randomly selected U.S. adults have very little confidence in newspapers is approximately 0.682.

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