a chemical company sells a specialized indust...
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. $\sigma = 0.55$