Question

the ages (in years) of a random sample of shoppers at a gaming store are shown. determine the range, mean, variance, and standard deviation of the sample data set. 12, 18, 23, 15, 17, 17, 19, 16, 13, 19 the range is (simplify your answer.)

Explanation:

Step1: Find the range

Range = Max - Min. Max = 23, Min = 12. So Range = 23 - 12 = 11.

Step2: Calculate the mean

Mean $\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}$. Here $n = 10$, $\sum_{i=1}^{10}x_{i}=12 + 18+23+15+17+17+19+16+13+19=169$. So $\bar{x}=\frac{169}{10}=16.9$.

Step3: Calculate the variance

The formula for sample - variance $s^{2}=\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}$.
$(12-16.9)^{2}=(-4.9)^{2}=24.01$, $(18 - 16.9)^{2}=(1.1)^{2}=1.21$, $(23-16.9)^{2}=(6.1)^{2}=37.21$, $(15-16.9)^{2}=(-1.9)^{2}=3.61$, $(17-16.9)^{2}=(0.1)^{2}=0.01$, $(17-16.9)^{2}=(0.1)^{2}=0.01$, $(19-16.9)^{2}=(2.1)^{2}=4.41$, $(16-16.9)^{2}=(-0.9)^{2}=0.81$, $(13-16.9)^{2}=(-3.9)^{2}=15.21$, $(19-16.9)^{2}=(2.1)^{2}=4.41$.
$\sum_{i = 1}^{10}(x_{i}-16.9)^{2}=24.01+1.21+37.21+3.61+0.01+0.01+4.41+0.81+15.21+4.41 = 90.9$.
$s^{2}=\frac{90.9}{9}=10.1$.

Step4: Calculate the standard deviation

The standard - deviation $s=\sqrt{s^{2}}=\sqrt{10.1}\approx3.18$.

Answer:

Range: 11
Mean: 16.9
Variance: 10.1
Standard deviation: $\approx3.18$