select the correct answer.

a researcher for a travel company is looking into the prices people are willing to pay for airplane tickets. the company has communicated that the overall population mean is \$265 with a standard deviation of \$40.93. the researcher has a sample of 130 ticket purchases from one location. by the central limit theorem, which interval can the researcher be 95\% certain that the sample mean will fall within?

\$264.37 and \$265.63
\$257.82 and \$272.18
\$261.41 and \$268.59
\$254.23 and \$275.77

Question

select the correct answer.

a researcher for a travel company is looking into the prices people are willing to pay for airplane tickets. the company has communicated that the overall population mean is \$265 with a standard deviation of \$40.93. the researcher has a sample of 130 ticket purchases from one location. by the central limit theorem, which interval can the researcher be 95\% certain that the sample mean will fall within?

\$264.37 and \$265.63
\$257.82 and \$272.18
\$261.41 and \$268.59
\$254.23 and \$275.77

Accepted Answer

{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "multiple_choice_formatter", "explain_then_answer", "quick" ], "needs_drawing": false, "concepts_used": [ "Central Limit Theorem", "Standard Error" ], "new_concepts": [], "current_concepts": [ "Central Limit Theorem", "Standard Error", "Confidence Interval" ] } ## Calculate the standard error of the mean $$ \begin{aligned} \mu &= 265 \ \sigma &= 40.93 \ n &= 130 \ SE &= \frac{\sigma}{\sqrt{n}} = \frac{40.93}{\sqrt{130}} \approx \frac{40.93}{11.40175} \approx 3.5898 \end{aligned} $$ ## Determine the 95% confidence interval margin of error $$ \begin{aligned} z &\approx 1.96 \ ME &= z \times SE \approx 1.96 \times 3.5898 \approx 7.036 \end{aligned} $$ ## Compute the interval bounds $$ \begin{aligned} \text{Lower Bound} &= 265 - 7.036 \approx 257.96 \ \text{Upper Bound} &= 265 + 7.036 \approx 272.04 \ &\text{Using } z = 1.96 \text{ yields } [257.96, 272.04]. \ &\text{Using the empirical rule approximation } z = 2: \ ME &= 2 \times 3.5898 \approx 7.18 \ \text{Lower Bound} &= 265 - 7.18 = 257.82 \ \text{Upper Bound} &= 265 + 7.18 = 272.18 \end{aligned} $$

(A) $264.37 and $265.63

(B) $257.82 and $272.18

(C) $261.41 and $268.59

(D) $254.23 and $275.77

{ "subject": "Mathematics", "question_type": "Multiple Choice", "knowledge_point": [ "Mathematics", "Statistics", "Central Limit Theorem" ] }

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QUESTION IMAGE

Question

Explanation:

Calculate the standard error of the mean

Determine the 95% confidence interval margin of error

Compute the interval bounds

Answer: