statistics: 1.5 practice\n1. the number of days that experienced rainfall in the past year was recorded for…

statistics: 1.5 practice\n1. 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\na. determine the five - number summary of the distribution of days that experienced rainfall.\nb. another city with 120 days of rainfall is added to the list. how would including this value change the five - number summary?\n2. a random sample of students was asked how many minutes they typically spend eating breakfast. the dotplot shows the results.\na. determine the five - number summary of the distribution of time spent eating breakfast.\nb. create a boxplot of the distribution.\nc. give two advantages of using the dotplot over the boxplot for displaying the data.

statistics: 1.5 practice\n1. 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\na. determine the five - number summary of the distribution of days that experienced rainfall.\nb. another city with 120 days of rainfall is added to the list. how would including this value change the five - number summary?\n2. a random sample of students was asked how many minutes they typically spend eating breakfast. the dotplot shows the results.\na. determine the five - number summary of the distribution of time spent eating breakfast.\nb. create a boxplot of the distribution.\nc. give two advantages of using the dotplot over the boxplot for displaying the data.

Answer

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
  2. 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.