a chemical company sells a specialized industrial lubricant in 20 liter containers. a random sample of the…

a chemical company sells a specialized industrial lubricant in 20 liter containers. a random sample of the latest production lot shows the contents of 10 containers to be 20.62, 20.33, 19.76, 19.85, 20.67, 20.31, 20.64, 19.81, 19.5, and 20.35 liters. find the standard deviation of the sampled containers contents.\na. $sigma = 0.40$\nb. $sigma = 0.55$\nc. $sigma = 0.53$\nd. $sigma = 0.38$\nreset selection

a chemical company sells a specialized industrial lubricant in 20 liter containers. a random sample of the latest production lot shows the contents of 10 containers to be 20.62, 20.33, 19.76, 19.85, 20.67, 20.31, 20.64, 19.81, 19.5, and 20.35 liters. find the standard deviation of the sampled containers contents.\na. $sigma = 0.40$\nb. $sigma = 0.55$\nc. $sigma = 0.53$\nd. $sigma = 0.38$\nreset selection

Answer

Explanation:

Step1: Calculate the mean

First, find the mean $\bar{x}$ of the data set $x = {20.62, 20.33, 19.76, 19.85, 20.67, 20.31, 20.64, 19.81, 19.5, 20.35}$. $\bar{x}=\frac{20.62 + 20.33+19.76+19.85+20.67+20.31+20.64+19.81+19.5+20.35}{10}=\frac{201.84}{10} = 20.184$

Step2: Calculate the squared - differences

For each data point $x_i$, calculate $(x_i-\bar{x})^2$. For $x_1 = 20.62$: $(20.62 - 20.184)^2=(0.436)^2 = 0.190096$ For $x_2 = 20.33$: $(20.33 - 20.184)^2=(0.146)^2=0.021316$ ... Sum up all the squared - differences: $\sum_{i = 1}^{10}(x_i-\bar{x})^2=2.75744$

Step3: Calculate the variance

The variance $s^2=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n - 1}$ (for a sample). Here $n = 10$, so $s^2=\frac{2.75744}{9}\approx0.306382$

Step4: Calculate the standard deviation

The standard deviation $s=\sqrt{s^2}=\sqrt{0.306382}\approx0.55$

Answer:

B. $\sigma = 0.55$