the ages (in years) of a random sample of shoppers at a gaming store are shown. determine the range, mean…

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: $\approx3.18$