Question

compute the range and sample standard deviation for strength of the concrete (in psi)
3960, 4140, 3300, 3200, 2920, 3860, 4140, 4040
the range is 1220 psi.
s = psi (round to one decimal place as needed.)

Explanation:

Step1: Calculate the mean

The formula for the mean \(\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}\).
Here \(n = 8\), and \(x=\{3960,4140,3300,3200,2920,3860,4140,4040\}\).
\(\sum_{i=1}^{8}x_{i}=3960 + 4140+3300+3200+2920+3860+4140+4040=30560\).
\(\bar{x}=\frac{30560}{8}=3820\).

Step2: Calculate \((x_{i}-\bar{x})^{2}\)

For \(x_1 = 3960\): \((3960 - 3820)^{2}=140^{2}=19600\).
For \(x_2 = 4140\): \((4140 - 3820)^{2}=320^{2}=102400\).
For \(x_3 = 3300\): \((3300 - 3820)^{2}=(- 520)^{2}=270400\).
For \(x_4 = 3200\): \((3200 - 3820)^{2}=(-620)^{2}=384400\).
For \(x_5 = 2920\): \((2920 - 3820)^{2}=(-900)^{2}=810000\).
For \(x_6 = 3860\): \((3860 - 3820)^{2}=40^{2}=1600\).
For \(x_7 = 4140\): \((4140 - 3820)^{2}=320^{2}=102400\).
For \(x_8 = 4040\): \((4040 - 3820)^{2}=220^{2}=48400\).

Step3: Calculate the sum of \((x_{i}-\bar{x})^{2}\)

\(\sum_{i = 1}^{8}(x_{i}-\bar{x})^{2}=19600+102400+270400+384400+810000+1600+102400+48400 = 1739200\).

Step4: Calculate the sample variance \(s^{2}\)

The formula for the sample variance \(s^{2}=\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}\).
Here \(n = 8\), so \(s^{2}=\frac{1739200}{8 - 1}=\frac{1739200}{7}\approx248457.14\).

Step5: Calculate the sample standard deviation \(s\)

\(s=\sqrt{s^{2}}=\sqrt{248457.14}\approx498.4\).

Answer:

\(498.4\)