question 35\ngiven the sample, 34, 45, 32, 43, 25, 40, and 33, you estimate that the middle 95% of scores is…

question 35\ngiven the sample, 34, 45, 32, 43, 25, 40, and 33, you estimate that the middle 95% of scores is between (round off to two decimal places, as they occur.)\n21.96 and 50.04\n23.04 and 48.96\n28.98 and 43.02\n29.52 and 42.48

question 35\ngiven the sample, 34, 45, 32, 43, 25, 40, and 33, you estimate that the middle 95% of scores is between (round off to two decimal places, as they occur.)\n21.96 and 50.04\n23.04 and 48.96\n28.98 and 43.02\n29.52 and 42.48

Answer

Answer:

D. 29.52 and 42.48

Explanation:

Step1: Calculate mean

$\bar{x}=\frac{34 + 45+32+43+25+40+33}{7}=36$

Step2: Calculate standard - deviation

$s=\sqrt{\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n - 1}}\approx7.24$

Step3: Find interval

For 95% confidence interval with normal - approximation, use $\bar{x}\pm2s$. $36-2\times7.24 = 21.52$, $36 + 2\times7.24=42.48$ (rounding differences may occur in options).