Question

the following are a students five exam scores: 77,72,63,62,68. calculate the mean: $\bar{x}=$. calculate the standard deviation, the long way. fill in the values in the table below. using $s = sqrt{\frac{sum(x - \bar{x})^2}{n - 1}}$, what is the standard deviation? (take the total from the table, divide by $n - 1$ and then find the square root) round your answer to 2 decimal places.

Explanation:

Step1: Calculate the mean

The data set is \(77,72,63,62,68\). The formula for the mean \(\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}\), where \(n = 5\) and \(x_{i}\) are the data - points.
\(\bar{x}=\frac{77 + 72+63+62+68}{5}=\frac{342}{5}=68.4\)

Step2: Calculate \((x-\bar{x})\) for each data - point

For \(x = 77\): \(77−68.4 = 8.6\)
For \(x = 72\): \(72−68.4 = 3.6\)
For \(x = 63\): \(63−68.4=-5.4\)
For \(x = 62\): \(62−68.4=-6.4\)
For \(x = 68\): \(68−68.4=-0.4\)

Step3: Calculate \((x - \bar{x})^2\) for each data - point

For \(x = 77\): \((77 - 68.4)^2=(8.6)^2 = 73.96\)
For \(x = 72\): \((72 - 68.4)^2=(3.6)^2 = 12.96\)
For \(x = 63\): \((63 - 68.4)^2=(-5.4)^2 = 29.16\)
For \(x = 62\): \((62 - 68.4)^2=(-6.4)^2 = 40.96\)
For \(x = 68\): \((68 - 68.4)^2=(-0.4)^2 = 0.16\)

Step4: Calculate the sum of \((x-\bar{x})^2\)

\(\sum_{i = 1}^{5}(x_{i}-\bar{x})^2=73.96 + 12.96+29.16+40.96+0.16=157.2\)

Step5: Calculate the standard deviation

The formula for the sample standard deviation is \(s=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^2}{n - 1}}\), with \(n = 5\).
\(s=\sqrt{\frac{157.2}{4}}=\sqrt{39.3}\approx6.27\)

Answer:

Mean: \(68.4\)
Standard deviation: \(6.27\)