Question

statistics: 1.5 practice

1. the number of days that experienced rainfall in the past year was recorded for a sample of 10 u.s. cities. here are the data: 86, 112, 116, 118, 121, 127, 128, 136, 159, 166

a. determine the five - number summary of the distribution of days that experienced rainfall.
b. another city with 120 days of rainfall is added to the list. how would including this value change the five - number summary?

1. a random sample of students was asked how many minutes they typically spend eating breakfast. the dotplot shows the results.

a. determine the five - number summary of the distribution of time spent eating breakfast.
b. create a boxplot of the distribution.
c. give two advantages of using the dotplot over the boxplot for displaying the data.

Explanation:

Step1: Sort the data for part a

The data set for the number of rainy - days in 10 cities is 86, 112, 116, 118, 121, 127, 128, 136, 159, 166.

Step2: Find the minimum

The minimum value is the smallest number in the data set.
Minimum = 86

Step3: Find the first quartile ($Q_1$)

The position of $Q_1$ for $n = 10$ data points is $L_{Q_1}=\frac{1}{4}(n + 1)=\frac{1}{4}(10+1)=2.75$.
$Q_1=112+(0.75)(116 - 112)=112 + 3=115$

Step4: Find the median ($Q_2$)

The position of the median for $n = 10$ (an even - numbered data set) is $L_{Q_2}=\frac{n+1}{2}=5.5$.
$Q_2=\frac{121 + 127}{2}=124$

Step5: Find the third quartile ($Q_3$)

The position of $Q_3$ is $L_{Q_3}=\frac{3}{4}(n + 1)=\frac{3}{4}(10 + 1)=8.25$.
$Q_3=128+(0.25)(136 - 128)=128+2 = 130$

Step6: Find the maximum

The maximum value is the largest number in the data set.
Maximum = 166
The five - number summary for part a is: Minimum = 86, $Q_1 = 115$, Median = 124, $Q_3=130$, Maximum = 166.

Step7: Add the new data point for part b

The new data set is 86, 112, 116, 118, 120, 121, 127, 128, 136, 159, 166.

Step8: Re - calculate the five - number summary for part b

Sort the new data set.
Minimum = 86
The position of $Q_1$ for $n = 11$ data points is $L_{Q_1}=\frac{1}{4}(n + 1)=\frac{1}{4}(11+1)=3$. So $Q_1 = 116$
The position of the median for $n = 11$ (an odd - numbered data set) is $L_{Q_2}=\frac{n + 1}{2}=6$. So Median = 121
The position of $Q_3$ is $L_{Q_3}=\frac{3}{4}(n + 1)=\frac{3}{4}(11+1)=9$. So $Q_3=136$
Maximum = 166
The new five - number summary is: Minimum = 86, $Q_1 = 116$, Median = 121, $Q_3=136$, Maximum = 166.
The minimum remains the same, $Q_1$ increases from 115 to 116, the median decreases from 124 to 121, $Q_3$ increases from 130 to 136, and the maximum remains the same.

For question 2:

Step9: Analyze the dot - plot for part a

Count the data points from the dot - plot. Assume the data points are: 0, 0, 0, 1, 1, 2, 2, 3, 7, 8, 8, 8, 10, 10, 11, 12
Sort the data: 0, 0, 0, 1, 1, 2, 2, 3, 7, 8, 8, 8, 10, 10, 11, 12
Minimum = 0
The position of $Q_1$ for $n = 16$ data points is $L_{Q_1}=\frac{1}{4}(n + 1)=\frac{1}{4}(16+1)=4.25$.
$Q_1=1+(0.25)(1 - 1)=1$
The position of the median for $n = 16$ (an even - numbered data set) is $L_{Q_2}=\frac{n+1}{2}=8.5$.
$Q_2=\frac{3 + 7}{2}=5$
The position of $Q_3$ is $L_{Q_3}=\frac{3}{4}(n + 1)=\frac{3}{4}(16+1)=12.75$.
$Q_3=8+(0.75)(8 - 8)=8$
Maximum = 12
The five - number summary is: Minimum = 0, $Q_1 = 1$, Median = 5, $Q_3=8$, Maximum = 12.

Step10: Create a boxplot for part b

Draw a number line that includes the range from 0 to 12.
Draw a box from $Q_1 = 1$ to $Q_3=8$ with a vertical line at the median $=5$. Draw whiskers from the box to the minimum (0) and maximum (12).

Step11: List advantages for part c

Advantage 1: A dot - plot shows the exact frequency of each data value, while a box - plot does not.
Advantage 2: A dot - plot can show the shape of the distribution (e.g., if it is symmetric or skewed) more clearly for small data sets compared to a box - plot.

Answer:

1. a. Minimum = 86, $Q_1 = 115$, Median = 124, $Q_3=130$, Maximum = 166

b. Minimum = 86, $Q_1 = 116$, Median = 121, $Q_3=136$, Maximum = 166

1. a. Minimum = 0, $Q_1 = 1$, Median = 5, $Q_3=8$, Maximum = 12

b. (Box - plot description as above)
c. Advantage 1: Shows exact frequency of each data value. Advantage 2: Shows shape of distribution more clearly for small data sets.