the number of cars sold at a dealership over several weeks is given below.\n14, 23, 31, 29, 33\nwhat is the…

the number of cars sold at a dealership over several weeks is given below.\n14, 23, 31, 29, 33\nwhat is the standard deviation for this set of population data?\nstandard deviation: $sigma=sqrt{\frac{(x_1 - mu)^2+(x_2 - mu)^2+cdots+(x_n - mu)^2}{n}}$\n6.9\n12.4\n15.4\n47.2

the number of cars sold at a dealership over several weeks is given below.\n14, 23, 31, 29, 33\nwhat is the standard deviation for this set of population data?\nstandard deviation: $sigma=sqrt{\frac{(x_1 - mu)^2+(x_2 - mu)^2+cdots+(x_n - mu)^2}{n}}$\n6.9\n12.4\n15.4\n47.2

Answer

Explanation:

Step1: Calculate the mean $\mu$

The data set is $x = {14,23,31,29,33}$, and $N = 5$. $\mu=\frac{14 + 23+31+29+33}{5}=\frac{130}{5}=26$

Step2: Calculate $(x_i-\mu)^2$ for each $x_i$

For $x_1 = 14$: $(14 - 26)^2=(- 12)^2 = 144$ For $x_2 = 23$: $(23 - 26)^2=(-3)^2 = 9$ For $x_3 = 31$: $(31 - 26)^2=(5)^2 = 25$ For $x_4 = 29$: $(29 - 26)^2=(3)^2 = 9$ For $x_5 = 33$: $(33 - 26)^2=(7)^2 = 49$

Step3: Calculate the sum of $(x_i-\mu)^2$

$\sum_{i = 1}^{5}(x_i-\mu)^2=144 + 9+25+9+49=236$

Step4: Calculate the standard - deviation $\sigma$

$\sigma=\sqrt{\frac{\sum_{i = 1}^{N}(x_i-\mu)^2}{N}}=\sqrt{\frac{236}{5}}=\sqrt{47.2}\approx6.9$

Answer:

A. 6.9