given the sample, 34, 45, 32, 43, 25, 40, and 33, what is the standard deviation?\ns = sqrt{\frac{sum (x…

given the sample, 34, 45, 32, 43, 25, 40, and 33, what is the standard deviation?\ns = sqrt{\frac{sum (x - \bar{x})^2}{n - 1}}\n(round off to two decimal places, as they occur.)\no 6\no 6.48\no 6.5\no 7.02

given the sample, 34, 45, 32, 43, 25, 40, and 33, what is the standard deviation?\ns = sqrt{\frac{sum (x - \bar{x})^2}{n - 1}}\n(round off to two decimal places, as they occur.)\no 6\no 6.48\no 6.5\no 7.02

Answer

Answer:

B. 6.48

Explanation:

Step1: Calculate mean

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

Step2: Calculate squared - differences

$(34 - 36)^2+(45 - 36)^2+(32 - 36)^2+(43 - 36)^2+(25 - 36)^2+(40 - 36)^2+(33 - 36)^2=2^2 + 9^2+4^2+7^2+11^2+4^2+3^2=4 + 81+16+49+121+16+9 = 296$

Step3: Calculate standard deviation

$s=\sqrt{\frac{296}{7 - 1}}=\sqrt{\frac{296}{6}}\approx6.48$