Question

lesson 7 standard deviation do now: find the interquartile range of the data (iqr) 1. 60 - 20 = 40 0 10 20 30 40 50 60 70 80 90 miles driven each day in august 1. the shoe - size for all the pairs of shoes in a persons closet are recorded 7 7 7 7 7 7 a) what is the mean of the data? 7 b) what is the standard deviation of the data? 7 2. the mean of data set a is 43.5 and the standard deviation is 7.4. the mean of data set b is 12.8 and the standard deviation is 4.1 a) which data set shows greater variability? explain your reasoning. b) what differences would you expect to see when comparing the dot plots of the two data sets?

Explanation:

Step1: Recall inter - quartile range formula

The inter - quartile range (IQR) is $Q_3 - Q_1$, where $Q_3$ is the third quartile and $Q_1$ is the first quartile. From the box - and - whisker plot, if we assume the left end of the box is $Q_1 = 20$ and the right end of the box is $Q_3=60$, then $IQR=Q_3 - Q_1$.

Step2: Calculate IQR

$IQR = 60 - 20=40$.

For the shoe - size data:

Part a

Step1: Recall mean formula

The mean $\bar{x}=\frac{\sum_{i = 1}^{n}x_i}{n}$. Here, $x_1=x_2=\cdots=x_6 = 7$ and $n = 6$. So $\bar{x}=\frac{7 + 7+7+7+7+7}{6}=\frac{42}{6}=7$.

Part b

Step1: Recall standard deviation formula

The standard deviation $s=\sqrt{\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n - 1}}$ (for a sample, for a population $\sigma=\sqrt{\frac{\sum_{i = 1}^{n}(x_i-\mu)^2}{n}}$). Since $x_i = 7$ for all $i$ and $\bar{x}=7$, then $(x_i-\bar{x})=0$ for all $i$. So $s=\sqrt{\frac{0+0+0+0+0+0}{n - 1}} = 0$.

For the two data - sets:

Part a

Step1: Understand variability and standard deviation

The standard deviation is a measure of variability. A larger standard deviation indicates greater variability. Data set A has a standard deviation of $7.4$ and data set B has a standard deviation of $4.1$. Since $7.4>4.1$, data set A shows greater variability.

Part b

Step1: Analyze dot - plots based on mean and standard deviation

The dot - plot of data set A will be more spread out around its mean of $43.5$ compared to the dot - plot of data set B around its mean of $12.8$. The dots in the dot - plot of data set A will be more dispersed, while the dots in the dot - plot of data set B will be more clustered around the mean value.

Answer:

1. IQR = 40

a. Mean of shoe - size data = 7
b. Standard deviation of shoe - size data = 0

1. a. Data set A shows greater variability because its standard deviation ($7.4$) is greater than the standard deviation of data set B ($4.1$).

b. The dot - plot of data set A will be more spread out around the mean of $43.5$ compared to the dot - plot of data set B which will be more clustered around the mean of $12.8$.