Question

question 35
given 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.)
21.96 and 50.04
23.04 and 48.96
28.98 and 43.02
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).

Answer:

D. 29.52 and 42.48