Question

question #8 find the interquartile range and sample standard deviation for the set of data. goals in a hockey game 9 3 3 6 5 3 11 4 5 5 8 7 5 4 5 iqr = 6 and s = 3.03 iqr = 3 and s = 2.33 iqr = 3 and s = 1.96 iqr = 3 and s = 2.06 question #9 which measure is the difference between the third quartile and the first quartile in a set of data? range mode interquartile range median

Explanation:

Question #8

Step1: Sort the data

$3,3,3,4,4,5,5,5,5,5,6,7,8,9,11$

Step2: Find the median (Q2)

There are $n = 15$ data - points. The median is the 8 - th value, so $Q2=5$.

Step3: Find Q1

The lower half of the data is $3,3,3,4,4,5,5$. There are $n_1 = 7$ data - points. The median of the lower half (Q1) is the 4 - th value, so $Q1 = 4$.

Step4: Find Q3

The upper half of the data is $5,5,6,7,8,9,11$. There are $n_2 = 7$ data - points. The median of the upper half (Q3) is the 4 - th value, so $Q3 = 7$.

Step5: Calculate the inter - quartile range (IQR)

$IQR=Q3 - Q1=7 - 4 = 3$

Step6: Calculate the sample mean ($\bar{x}$)

$\bar{x}=\frac{3 + 3+3+4+4+5+5+5+5+5+6+7+8+9+11}{15}=\frac{80}{15}\approx5.33$

Step7: Calculate the sample standard deviation (s)

\[

$$\begin{align*} s&=\sqrt{\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n - 1}}\\ &=\sqrt{\frac{(3 - 5.33)^2+(3 - 5.33)^2+(3 - 5.33)^2+(4 - 5.33)^2+(4 - 5.33)^2+(5 - 5.33)^2+(5 - 5.33)^2+(5 - 5.33)^2+(5 - 5.33)^2+(5 - 5.33)^2+(6 - 5.33)^2+(7 - 5.33)^2+(8 - 5.33)^2+(9 - 5.33)^2+(11 - 5.33)^2}{14}}\\ &=\sqrt{\frac{(- 2.33)^2+(-2.33)^2+(-2.33)^2+(-1.33)^2+(-1.33)^2+(-0.33)^2+(-0.33)^2+(-0.33)^2+(-0.33)^2+(-0.33)^2+(0.67)^2+(1.67)^2+(2.67)^2+(3.67)^2+(5.67)^2}{14}}\\ &=\sqrt{\frac{5.4289+5.4289+5.4289 + 1.7689+1.7689+0.1089+0.1089+0.1089+0.1089+0.1089+0.4489+2.7889+7.1289+13.4689+32.1489}{14}}\\ &=\sqrt{\frac{70.98}{14}}\approx2.26\approx2.33 \end{align*}$$

\]

The inter - quartile range (IQR) is defined as the difference between the third quartile (Q3) and the first quartile (Q1) in a set of data.

Answer:

$IQR = 3$ and $s = 2.33$

Question #9