Question

jin recorded the ages (in years) for a sample of 4 zebras from a wildlife preserve. here are the ages: 2,9,14,15 the mean of these ages is $\bar{x}=10$ years. what is the standard deviation? round your answer to two decimal places. $s_xapproxsquare$ years show calculator

Explanation:

Step1: Recall sample - standard - deviation formula

The formula for the sample standard deviation $s_x=\sqrt{\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n - 1}}$, where $x_i$ are the data points, $\bar{x}$ is the sample mean, and $n$ is the sample size. Here, $n = 4$ and $\bar{x}=10$.

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

For $x_1 = 2$: $(2 - 10)^2=(-8)^2 = 64$.
For $x_2 = 9$: $(9 - 10)^2=(-1)^2 = 1$.
For $x_3 = 14$: $(14 - 10)^2=(4)^2 = 16$.
For $x_4 = 15$: $(15 - 10)^2=(5)^2 = 25$.

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

$\sum_{i = 1}^{4}(x_i-\bar{x})^2=64 + 1+16 + 25=106$.

Step4: Calculate the sample standard deviation

$s_x=\sqrt{\frac{106}{4 - 1}}=\sqrt{\frac{106}{3}}\approx\sqrt{35.333\cdots}\approx5.94$.

Answer:

$5.94$