Question

the monthly incomes for 12 randomly selected people, each with a bachelors degree in economics, are shown on the right. complete parts (a) through (c) below. assume the population is normally distributed. 4450.97 4596.67 4366.39 4455.64 4151.92 3727.05 4283.53 4527.91 4407.37 3946.65 4023.24 4221.57 (a) find the sample mean. \\( \overline { x } = 4263.2 \\) (round to one decimal place as needed.) (b) find the sample standard deviation. \\( s = \square \\) (round to one decimal place as needed.)

Explanation:

Step1: Calculate the deviations from the mean

Let \(x_1 = 4450.97\), \(x_2 = 4596.67\), \(x_3 = 4366.39\), \(x_4 = 4455.64\), \(x_5 = 4151.92\), \(x_6 = 3727.05\), \(x_7 = 4283.53\), \(x_8 = 4527.91\), \(x_9 = 4407.37\), \(x_{10}=3946.65\), \(x_{11}=4023.24\), \(x_{12}=4221.57\) and \(\bar{x}=4263.2\).
The deviations are \((x_i - \bar{x})\) for \(i = 1,\cdots,12\).
For example, \(x_1-\bar{x}=4450.97 - 4263.2=187.77\), \(x_2-\bar{x}=4596.67 - 4263.2 = 333.47\), \(x_3-\bar{x}=4366.39 - 4263.2=103.19\), \(x_4-\bar{x}=4455.64 - 4263.2 = 192.44\), \(x_5-\bar{x}=4151.92 - 4263.2=-111.28\), \(x_6-\bar{x}=3727.05 - 4263.2=-536.15\), \(x_7-\bar{x}=4283.53 - 4263.2 = 20.33\), \(x_8-\bar{x}=4527.91 - 4263.2=264.71\), \(x_9-\bar{x}=4407.37 - 4263.2 = 144.17\), \(x_{10}-\bar{x}=3946.65 - 4263.2=-316.55\), \(x_{11}-\bar{x}=4023.24 - 4263.2=-239.96\), \(x_{12}-\bar{x}=4221.57 - 4263.2=-41.63\).

Step2: Square the deviations

\((x_1 - \bar{x})^2=(187.77)^2 = 35257.6729\), \((x_2 - \bar{x})^2=(333.47)^2=111101.2409\), \((x_3 - \bar{x})^2=(103.19)^2 = 10648.1761\), \((x_4 - \bar{x})^2=(192.44)^2=37033.1536\), \((x_5 - \bar{x})^2=(-111.28)^2 = 12383.9984\), \((x_6 - \bar{x})^2=(-536.15)^2=287456.8225\), \((x_7 - \bar{x})^2=(20.33)^2=413.3089\), \((x_8 - \bar{x})^2=(264.71)^2 = 70071.3841\), \((x_9 - \bar{x})^2=(144.17)^2=20784.9889\), \((x_{10}-\bar{x})^2=(-316.55)^2 = 99903.9025\), \((x_{11}-\bar{x})^2=(-239.96)^2=57580.8016\), \((x_{12}-\bar{x})^2=(-41.63)^2 = 1733.0569\).

Step3: Sum the squared deviations

\(\sum_{i = 1}^{12}(x_i-\bar{x})^2=35257.6729+111101.2409 + 10648.1761+37033.1536+12383.9984+287456.8225+413.3089+70071.3841+20784.9889+99903.9025+57580.8016+1733.0569=744468.41\).

Step4: Calculate the sample variance

The sample variance \(s^2=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n - 1}\), where \(n = 12\). So \(s^2=\frac{744468.41}{12-1}=\frac{744468.41}{11}\approx67678.9464\).

Step5: Calculate the sample standard deviation

The sample standard deviation \(s=\sqrt{s^2}\). So \(s=\sqrt{67678.9464}\approx260.2\).

Answer:

\(s = 260.2\)