Question

3.2 measures of spread: standard deviation and variance
concrete formulated for use in commercial structures is designed to withstand pressures of 4000 pounds per square inch. the data set below represents the strength of 20 randomly selected samples of concrete. click on the data button shown below to display the data and copy and paste it into a statistical software package. then compute the mean and standard deviation rounding your answers to 1 decimal place.
data
strength
4444
4085
4078
4444
4001
4322
4345
4398
4175
4319
4211
4239
4133
3922
4494
4172
4263
3911
3935
3976

Explanation:

Step1: Calculate the sum of data

Let the data - set be \(x_1,x_2,\cdots,x_{20}\). The sum \(S=\sum_{i = 1}^{20}x_i=4444 + 4085+4078+4444+4001+4322+4345+4398+4175+4319+4211+4239+4133+3922+4494+4172+4263+3911+3935+3976 = 83714\)

Step2: Calculate the mean \(\bar{x}\)

The formula for the mean of a sample is \(\bar{x}=\frac{\sum_{i = 1}^{n}x_i}{n}\), where \(n = 20\). So \(\bar{x}=\frac{83714}{20}=4185.7\)

Step3: Calculate the squared - differences \((x_i-\bar{x})^2\)

For example, when \(x_1 = 4444\), \((x_1-\bar{x})^2=(4444 - 4185.7)^2=(258.3)^2 = 66718.89\). Do this for all \(i\) from \(1\) to \(20\) and sum them up. Let \(Q=\sum_{i = 1}^{20}(x_i - \bar{x})^2\).
\(Q=(4444 - 4185.7)^2+(4085 - 4185.7)^2+(4078 - 4185.7)^2+(4444 - 4185.7)^2+(4001 - 4185.7)^2+(4322 - 4185.7)^2+(4345 - 4185.7)^2+(4398 - 4185.7)^2+(4175 - 4185.7)^2+(4319 - 4185.7)^2+(4211 - 4185.7)^2+(4239 - 4185.7)^2+(4133 - 4185.7)^2+(3922 - 4185.7)^2+(4494 - 4185.7)^2+(4172 - 4185.7)^2+(4263 - 4185.7)^2+(3911 - 4185.7)^2+(3935 - 4185.7)^2+(3976 - 4185.7)^2\)
\(Q = 340740.2\)

Step4: Calculate the variance \(s^2\)

The formula for the sample variance is \(s^2=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n - 1}\), with \(n = 20\). So \(s^2=\frac{340740.2}{19}\approx17933.7\)

Step5: Calculate the standard deviation \(s\)

The standard deviation \(s=\sqrt{s^2}\), so \(s=\sqrt{17933.7}\approx133.9\)

Answer:

Mean: \(4185.7\), Standard - deviation: \(133.9\)