Question

the state test scores for 12 randomly selected high school seniors are shown on the right. complete parts (a) through (c) below. assume the population is normally distributed. (a) find the sample mean. \\( \overline { x } = 909.3 \\) (round to one decimal place as needed.) (b) find the sample standard deviation. \\( s = \square \\) (round to one decimal place as needed)

Explanation:

Step1: Recall the formula for sample standard deviation

The formula for the sample standard deviation \(s=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}}\), where \(n\) is the sample size, \(x_{i}\) are the data points, and \(\bar{x}\) is the sample mean. Here \(n = 12\) and \(\bar{x}=909.3\).

Step2: Calculate \((x_{i}-\bar{x})^{2}\) for each data point

For \(x_1 = 1425\): \((1425 - 909.3)^{2}=(515.7)^{2}=265946.49\)
For \(x_2 = 1228\): \((1228 - 909.3)^{2}=(318.7)^{2}=101569.69\)
For \(x_3 = 985\): \((985 - 909.3)^{2}=(75.7)^{2}=5730.49\)
For \(x_4 = 692\): \((692 - 909.3)^{2}=(- 217.3)^{2}=47219.29\)
For \(x_5 = 730\): \((730 - 909.3)^{2}=(-179.3)^{2}=32148.49\)
For \(x_6 = 830\): \((830 - 909.3)^{2}=(-79.3)^{2}=6288.49\)
For \(x_7 = 724\): \((724 - 909.3)^{2}=(-185.3)^{2}=34336.09\)
For \(x_8 = 747\): \((747 - 909.3)^{2}=(-162.3)^{2}=26341.29\)
For \(x_9 = 544\): \((544 - 909.3)^{2}=(-365.3)^{2}=133444.09\)
For \(x_{10}=621\): \((621 - 909.3)^{2}=(-288.3)^{2}=83116.89\)
For \(x_{11}=1442\): \((1442 - 909.3)^{2}=(532.7)^{2}=283769.29\)
For \(x_{12}=944\): \((944 - 909.3)^{2}=(34.7)^{2}=1204.09\)

Step3: Sum up \((x_{i}-\bar{x})^{2}\)

\(\sum_{i = 1}^{12}(x_{i}-\bar{x})^{2}=265946.49+101569.69 + 5730.49+47219.29+32148.49+6288.49+34336.09+26341.29+133444.09+83116.89+283769.29+1204.09\)
\(\sum_{i = 1}^{12}(x_{i}-\bar{x})^{2}=1021122.1\)

Step4: Calculate the sample standard deviation

\(s=\sqrt{\frac{1021122.1}{12 - 1}}=\sqrt{\frac{1021122.1}{11}}\approx\sqrt{92829.2818}\approx304.7\)

Answer:

\(s = 304.7\)