Question

listed below are measured amounts of caffeine (mg per 12oz of drink) obtained from each of 14 brands. find the range, variance, and standard deviation for the given sample data. include appropriate units in the results. are the statistics obtained in one sample representative of the population of all cans of the same 14 brands consumed?
32 41 58 50 0 35 51 39 60 56 50 46 0 0
the range of the sample data is 60.0 mg per 12oz of drink.
(round to one decimal place as needed.)
the standard deviation of the sample data is 21.7 mg per 12oz of drink.
(round to one decimal place as needed.)
the variance of the sample data is
(round to one decimal place as needed)

Explanation:

Step1: Calculate the mean

First, find the sum of the data values: \(32 + 41+58 + 50+0 + 35+51+39+60+56+50+46+0 + 0=518\)
The number of data points \(n = 14\)
The mean \(\bar{x}=\frac{518}{14}\approx37.0\)

Step2: Calculate the variance

The formula for the sample variance \(s^{2}=\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}\)
\((32-37)^{2}=(-5)^{2}=25\)
\((41 - 37)^{2}=4^{2}=16\)
\((58-37)^{2}=21^{2}=441\)
\((50 - 37)^{2}=13^{2}=169\)
\((0-37)^{2}=(-37)^{2}=1369\)
\((35 - 37)^{2}=(-2)^{2}=4\)
\((51-37)^{2}=14^{2}=196\)
\((39-37)^{2}=2^{2}=4\)
\((60 - 37)^{2}=23^{2}=529\)
\((56-37)^{2}=19^{2}=361\)
\((50-37)^{2}=13^{2}=169\)
\((46-37)^{2}=9^{2}=81\)
\((0 - 37)^{2}=(-37)^{2}=1369\)
\((0-37)^{2}=(-37)^{2}=1369\)

\(\sum_{i = 1}^{14}(x_{i}-\bar{x})^{2}=25+16 + 441+169+1369+4+196+4+529+361+169+81+1369+1369=6758\)

\(s^{2}=\frac{6758}{14 - 1}=\frac{6758}{13}\approx519.8\)

Answer:

The variance of the sample data is \(519.8\) mg² per \(12\) oz of drink.