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for the following set of data, find the population standard deviation, …

Question

for the following set of data, find the population standard deviation, to the nearest thousandth.
69, 13, 51, 63, 59, 85, 26, 20, 19

Explanation:

Step1: Calculate the mean

$\bar{x}=\frac{69 + 13+51+63+59+85+26+20+19}{9}=\frac{405}{9} = 45$

Step2: Calculate the squared - differences

$(69 - 45)^2=24^2 = 576$
$(13 - 45)^2=(-32)^2 = 1024$
$(51 - 45)^2=6^2 = 36$
$(63 - 45)^2=18^2 = 324$
$(59 - 45)^2=14^2 = 196$
$(85 - 45)^2=40^2 = 1600$
$(26 - 45)^2=(-19)^2 = 361$
$(20 - 45)^2=(-25)^2 = 625$
$(19 - 45)^2=(-26)^2 = 676$

Step3: Calculate the variance

$\sigma^{2}=\frac{576+1024+36+324+196+1600+361+625+676}{9}=\frac{5418}{9}=602$

Step4: Calculate the standard deviation

$\sigma=\sqrt{602}\approx24.536$

Answer:

$24.536$