Question

the numbers of courses taught per semester by a random sample of university professors are shown in the histogram. make a frequency distribution for the data. then use the table to estimate the sample mean and the sample standard deviation of the data set. make a frequency distribution for the data x f 1 3 2 16 3 22 4 14 the sample mean is x = 2.9 (round to one decimal place as needed.) the sample standard deviation is s = (round to one decimal place as needed.)

Explanation:

Step1: Recall sample - standard - deviation formula

The formula for the sample standard deviation $s=\sqrt{\frac{\sum_{i = 1}^{n}f_i(x_i-\bar{x})^2}{n - 1}}$, where $f_i$ is the frequency, $x_i$ is the data - value, $\bar{x}$ is the sample mean, and $n=\sum_{i = 1}^{n}f_i$. First, calculate $n$:
$n=3 + 16+22 + 14=55$.

Step2: Calculate $(x_i-\bar{x})^2$ and $f_i(x_i-\bar{x})^2$ for each $i$

For $x_1 = 1$, $\bar{x}=2.9$, $(x_1-\bar{x})^2=(1 - 2.9)^2=(-1.9)^2 = 3.61$, and $f_1(x_1-\bar{x})^2=3\times3.61 = 10.83$.
For $x_2 = 2$, $(x_2-\bar{x})^2=(2 - 2.9)^2=(-0.9)^2 = 0.81$, and $f_2(x_2-\bar{x})^2=16\times0.81 = 12.96$.
For $x_3 = 3$, $(x_3-\bar{x})^2=(3 - 2.9)^2=(0.1)^2 = 0.01$, and $f_3(x_3-\bar{x})^2=22\times0.01 = 0.22$.
For $x_4 = 4$, $(x_4-\bar{x})^2=(4 - 2.9)^2=(1.1)^2 = 1.21$, and $f_4(x_4-\bar{x})^2=14\times1.21 = 16.94$.

Step3: Calculate $\sum_{i = 1}^{n}f_i(x_i-\bar{x})^2$

$\sum_{i = 1}^{n}f_i(x_i-\bar{x})^2=10.83+12.96 + 0.22+16.94=40.95$.

Step4: Calculate the sample standard deviation

$s=\sqrt{\frac{40.95}{55 - 1}}=\sqrt{\frac{40.95}{54}}\approx\sqrt{0.7583}\approx0.9$.

Answer:

$0.9$