53% 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 little confidence in newspapers is (a) exactly five, (b) at least six, and (c) less than four.

(a) $p(5) = 0.242$ (round to three decimal places as needed.)
(b) $p(x \ge 6) = 0.462$ (round to three decimal places as needed.)

Question

53% 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 little confidence in newspapers is (a) exactly five, (b) at least six, and (c) less than four.

(a) \$p(5) = 0.242\$ (round to three decimal places as needed.)
(b) \$p(x \ge 6) = 0.462\$ (round to three decimal places as needed.)

(a) $p(5) = 0.242$ (round to three decimal places as needed.)
(b) $p(x \ge 6) = 0.462$ (round to three decimal places as needed.)

Accepted Answer

{ "quality": "clear", "question_count": 3, "skills_matched": [ "step_cot", "explain_then_answer", "quick" ], "needs_drawing": false, "concepts_used": [], "new_concepts": [ "Binomial Distribution", "Binomial Probability Formula", "Cumulative Binomial Probability" ], "current_concepts": [ "Binomial Distribution", "Binomial Probability Formula", "Cumulative Binomial Probability" ] } ## Define the binomial parameters $$ \begin{aligned} &n = 10\ &p = 0.53\ &q = 1 - p = 0.47\ &X \sim \text{Binomial}(10, 0.53) \end{aligned} $$ ## Calculate the probabilities for each part $$ \begin{aligned} &\text{(a) } P(X = 5) = \binom{10}{5} (0.53)^5 (0.47)^5 \approx 252 \times 0.041819 \times 0.022935 \approx 0.2417 \approx 0.242\ &\text{(b) } P(X \ge 6) = \sum_{k=6}^{10} \binom{10}{k} (0.53)^k (0.47)^{10-k}\ &P(X = 6) \approx 0.2272\ &P(X = 7) \approx 0.1468\ &P(X = 8) \approx 0.0621\ &P(X = 9) \approx 0.0156\ &P(X = 10) \approx 0.0018\ &P(X \ge 6) \approx 0.2272 + 0.1468 + 0.0621 + 0.0156 + 0.0018 = 0.4535 \approx 0.454\ &\text{(c) } P(X < 4) = \sum_{k=0}^{3} \binom{10}{k} (0.53)^k (0.47)^{10-k}\ &P(X = 0) \approx 0.0005\ &P(X = 1) \approx 0.0060\ &P(X = 2) \approx 0.0303\ &P(X = 3) \approx 0.0911\ &P(X < 4) \approx 0.0005 + 0.0060 + 0.0303 + 0.0911 = 0.1279 \approx 0.128 \end{aligned} $$ ## Summarize the rounded values $$ \begin{aligned} &\text{(a) } P(5) = 0.242\ &\text{(b) } P(X \ge 6) = 0.454\ &\text{(c) } P(X < 4) = 0.128 \end{aligned} $$ ## Question 1 (a) $P(5) = 0.242$ ## Question 2 (b) $P(X \ge 6) = 0.454$ ## Question 3 (c) $P(X < 4) = 0.128$ { "subject": "Mathematics", "question_type": "Multi-part", "knowledge_point": [ "Mathematics", "Statistics", "Binomial Distribution" ] }

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QUESTION IMAGE

Question

Explanation:

Define the binomial parameters

Calculate the probabilities for each part

Summarize the rounded values

Answer:

Question 1

Question 2

Question 3