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Solution: The mean $A_n = \frac{X_1 + ... + X_{80}}{80}$ is approximately normally distributed. Using $n = 80$, $\hat{p} = 5/8$, and $s^2 = 15/64$, we have a 95% confidence interval: $p_l = \hat{p} - 1.96\left(\frac{s}{\sqrt{n}}\right) \approx 0.52$ $p_h = \hat{p} + 1.96\left(\frac{s}{\sqrt{n}}\right) \approx 0.73$

Name: solution the mean an fracx1 x8080 is approximately normally distributed using n 80 hatp 58 and s2 1564 we have a 95 confidence interval pl hatp 196leftfracssqrtnright approx 052 ph hatp 67913
Uploaded: 2024-04-26T05:10:18-08:00
Duration: 1 min 32 s
Channel: Kenny Mesadieu
Description: solution the mean an fracx1 x8080 is approximately normally distributed using n 80 hatp 58 and s2 1564 we have a 95 confidence interval pl hatp 196leftfracssqrtnright approx 052 ph hatp 67913

          Solution: The mean $A_n = \frac{X_1 + ... + X_{80}}{80}$ is approximately normally distributed. Using $n = 80$, $\hat{p} = 5/8$, and $s^2 = 15/64$, we have a 95% confidence interval:
$p_l = \hat{p} - 1.96\left(\frac{s}{\sqrt{n}}\right) \approx 0.52$
$p_h = \hat{p} + 1.96\left(\frac{s}{\sqrt{n}}\right) \approx 0.73$

$solution the mean an fracx1 x8080 is approximately normally distributed using n 80 hatp 58 and s2 1564 we have a 95 confidence interval pl hatp 196leftfracssqrtnright approx 052 ph hatp 67913$

Added by Laura D.

Elementary Statistics a Step by Step Approach

Allan G. Bluman 9th Edition

Instant Answer

Solved by Expert Kenny Mesadieu

04/26/2024

Step 1

625$ $s^2 = \frac{15}{64}$ $s = \sqrt{\frac{15}{64}} = \frac{\sqrt{15}}{8} \approx 0.484$ The 95% confidence interval is given by: $\hat{p} \pm 1.96 \frac{s}{\sqrt{n}}$ Substituting the values, we get: Show more…

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Solution: The mean $A_n = \frac{X_1 + ... + X_{80}}{80}$ is approximately normally distributed. Using $n = 80$, $\hat{p} = 5/8$, and $s^2 = 15/64$, we have a 95% confidence interval: $p_l = \hat{p} - 1.96\left(\frac{s}{\sqrt{n}}\right) \approx 0.52$ $p_h = \hat{p} + 1.96\left(\frac{s}{\sqrt{n}}\right) \approx 0.73$