the bureau of tobacco and alcohol took a sample of cigarettes and measured the amount of tar (in milligram)…

Explanation:

Step1: Calculate the mean

The mean $\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}$, where $n = 16$ and $x_{i}$ are the data - points. $\sum_{i=1}^{16}x_{i}=16 + 18+8 + 6+5 + 18+12 + 9+17+15+13+15+18+2+11+19=190$ $\bar{x}=\frac{190}{16}=11.875$

Step2: Find the mode

The mode is the most frequently occurring value. In the data - set, 18 appears 3 times, more frequently than any other number. So the mode is 18.

Step3: Calculate the median

First, order the data: 2, 5, 6, 8, 9, 11, 12, 13, 15, 15, 16, 17, 18, 18, 18, 19. Since $n = 16$ (an even number), the median is the average of the $\frac{n}{2}$th and $(\frac{n}{2}+1)$th ordered values. The 8th value is 13 and the 9th value is 15. So the median $M=\frac{13 + 15}{2}=14$

Step4: Calculate the range

The range $R=x_{\max}-x_{\min}$, where $x_{\max}=19$ and $x_{\min}=2$. So $R=19 - 2=17$

Step5: Calculate the variance

The variance $s^{2}=\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}$ $(x_{1}-\bar{x})^{2}=(16 - 11.875)^{2}=4.125^{2}=17.015625$ $(x_{2}-\bar{x})^{2}=(18 - 11.875)^{2}=6.125^{2}=37.515625$ $\cdots$ $(x_{16}-\bar{x})^{2}=(19 - 11.875)^{2}=7.125^{2}=50.765625$ $\sum_{i = 1}^{16}(x_{i}-\bar{x})^{2}=383.875$ $s^{2}=\frac{383.875}{15}\approx25.592$

Step6: Calculate the standard deviation

The standard deviation $s=\sqrt{s^{2}}=\sqrt{25.592}\approx5.059$

Step7: Calculate the inter - quartile range

First, find the first quartile $Q_{1}$ and the third quartile $Q_{3}$. The lower half of the data (first 8 values: 2, 5, 6, 8, 9, 11, 12, 13) has a median $Q_{1}=\frac{6 + 8}{2}=7$ The upper half of the data (last 8 values: 15, 15, 16, 17, 18, 18, 18, 19) has a median $Q_{3}=\frac{17+18}{2}=17.5$ The inter - quartile range $IQR = Q_{3}-Q_{1}=17.5 - 7 = 10.5$

Answer:

Mean: $\bar{x}=11.875$ Mode: 18 Median: $M = 14$ Range: $R = 17$ Variance: $s^{2}\approx25.592$ Standard deviation: $s\approx5.059$ Inter - quartile range: $IQR = 10.5$