assume the hold time of callers to a cable company is normally distributed with a mean of 4.1 minutes and a standard deviation of 0.8 minute. determine the percent of callers who are on hold between 3.6 minutes and 5.2 minutes.

click here to view page 1 of the standard normal distribution table.
click here to view page 2 of the standard normal distribution table.

$\square$% (round to two decimal places as needed.)

Question

assume the hold time of callers to a cable company is normally distributed with a mean of 4.1 minutes and a standard deviation of 0.8 minute. determine the percent of callers who are on hold between 3.6 minutes and 5.2 minutes.

click here to view page 1 of the standard normal distribution table.
click here to view page 2 of the standard normal distribution table.

$\square$% (round to two decimal places as needed.)

Accepted Answer

{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "blank_filling_formatter", "explain_then_answer", "tutor", "mastery_adaptive_brevity" ], "needs_drawing": true, "concepts_used": [ "Normal Distribution", "Z-score Definition", "Standard Normal Table" ], "new_concepts": [], "current_concepts": [ "Normal Distribution", "Z-score Definition", "Standard Normal Table" ] } ## Identify the given parameters Using the *Normal Distribution* knowledge point $$ \begin{aligned} &\mu = 4.1\ &\sigma = 0.8\ &X_1 = 3.6,\quad X_2 = 5.2 \end{aligned} $$ ## Calculate the Z-scores Using the *Z-score Definition* knowledge point $$ \begin{aligned} &z_1 = \frac{3.6 - 4.1}{0.8} = \frac{-0.5}{0.8} = -0.625 \approx -0.63\ &z_2 = \frac{5.2 - 4.1}{0.8} = \frac{1.1}{0.8} = 1.375 \approx 1.38 \end{aligned} $$ ## Find the cumulative probabilities Using the *Standard Normal Table* knowledge point $$ \begin{aligned} &P(Z < 1.38) = 0.9162\ &P(Z < -0.63) = 0.2643 \end{aligned} $$ ## Calculate the final percentage Using the *Normal Distribution* knowledge point $$ \begin{aligned} &P(-0.63 < Z < 1.38) = 0.9162 - 0.2643 = 0.6519\ &\text{Percentage} = 0.6519 \times 100\% = 65.19\% \end{aligned} $$ Assume the hold time of callers to a cable company is normally distributed with a mean of 4.1 minutes and a standard deviation of 0.8 minute. Determine the percent of callers who are on hold between 3.6 minutes and 5.2 minutes. 65.19% (Round to two decimal places as needed.) { "elements": [ { "type": "functiongraph", "params": [ { "js": "Math.exp(-0.5 * Math.pow((x - 4.1) / 0.8, 2)) / (0.8 * Math.sqrt(2 * Math.PI))", "latex": "\frac{1}{0.8\sqrt{2\pi}} e^{-\frac{1}{2}\left(\frac{x-4.1}{0.8}\ ight)^2}" }, 1.0, 7.2 ], "properties": { "strokeColor": "#8C55F2", "strokeWidth": 2, "name": "Normal Curve" } }, { "type": "inequality", "params": [ { "js": "Math.exp(-0.5 * Math.pow((x - 4.1) / 0.8, 2)) / (0.8 * Math.sqrt(2 * Math.PI))", "latex": "y \le f(x)" }, [4.1, 0.2] ], "properties": { "inverse": false, "strict": false, "fillColor": "rgba(140, 85, 242, 0.3)" } }, { "type": "line", "params": [ [3.6, 0], [3.6, 0.5] ], "properties": { "strokeColor": "#F2557F", "strokeWidth": 1.5, "dash": 2, "name": "x = 3.6" } }, { "type": "line", "params": [ [5.2, 0], [5.2, 0.5] ], "properties": { "strokeColor": "#F2557F", "strokeWidth": 1.5, "dash": 2, "name": "x = 5.2" } } ], "timestamps": [0.5] } { "subject": "Mathematics", "question_type": "Fill-in-the-blank", "knowledge_point": [ "Mathematics", "Statistics", "Normal Distribution" ] }

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QUESTION IMAGE

Question

Explanation:

Identify the given parameters

Calculate the Z-scores

Find the cumulative probabilities

Calculate the final percentage

Answer: