assume the average amount of caffeine consumed daily by adults is normally distributed with a mean of 220 mg and a standard deviation of 47 mg. determine the percent of adults who consume less than 220 mg of caffeine daily.

% of adults consume less than 220 mg of caffeine daily.
(round to two decimal places as needed.)

Question

assume the average amount of caffeine consumed daily by adults is normally distributed with a mean of 220 mg and a standard deviation of 47 mg. determine the percent of adults who consume less than 220 mg of caffeine daily.

% of adults consume less than 220 mg of caffeine daily.
(round to two decimal places as needed.)

Accepted Answer

{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "blank_filling_formatter", "explain_then_answer", "quick" ], "needs_drawing": false, "concepts_used": [ "Standard Normal Distribution" ], "new_concepts": [], "current_concepts": [ "Standard Normal Distribution", "Z-Score Calculation" ] } ## Calculate the z-score $$ \begin{aligned} \mu &= 220 \ \sigma &= 47 \ X &= 220 \ z &= \frac{X - \mu}{\sigma} = \frac{220 - 220}{47} = 0 \end{aligned} $$ ## Find the cumulative probability $$ \begin{aligned} P(X < 220) &= P(Z < 0) \ P(Z < 0) &= 0.5000 \end{aligned} $$ ## Convert to percentage $$ \begin{aligned} \text{Percentage} &= 0.5000 \times 100\% = 50\% \ \text{Percentage} &= 50.00\% \end{aligned} $$ Assume the average amount of caffeine consumed daily by adults is normally distributed with a mean of 220 mg and a standard deviation of 47 mg. Determine the percent of adults who consume less than 220 mg of caffeine daily. 50% of adults consume less than 220 mg of caffeine daily. (Round to two decimal places as needed.) { "subject": "Mathematics", "question_type": "Fill-in-the-blank", "knowledge_point": [ "Mathematics", "Statistics", "Standard Normal Distribution" ] }

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Question

Explanation:

Calculate the z-score

Find the cumulative probability

Convert to percentage

Answer: