1. Derby collects samples of insects in the forest and in a marsh to count the number of insects in each sample and then releases the insects. The number of insects in each sample from the two locations is provided below.
Forest 15 13 14 9 16 11 15 14 15 14
Marsh 17 14 15 17 16 18 17 19 13 16
What is the median and the interquartile range for each of the two locations? Round to one decimal place.
The median number of insects in the forest samples is---- and the interquartile range is---
the median number of insects in the marsh samples is --- and the interquartile range is---
The median number of insects in the forest samples is 14.5 and the interquartile range is 3.
The median number of insects in the marsh samples is 16.5 and the interquartile range is 2.5.
1. An urgent care center has two campuses, A and B. The doctors at the two campuses are comparing the wait time in minutes for customers. The amount of time was measured for each patient at both campuses. The results are provided in the table below.
Campus A Campus B
9 24 15 11 21 14
19 13 30 22 31 5
16 11 27 17 19 21
7 17 22 26 28 24
21 18 4 37 33 8
Justify the measure of center and measure of spread that could be used to compare the two data sets.
A. The median and interquartile range could be used because the distribution for Campus A is skewed right with an outlier of 30
minutes and the distribution for Campus B is roughly symmetric with no outliers.
B. The median and interquartile range could be used. Although there are no outliers, the distribution for Campus A is skewed left, and the distribution for Campus B is skewed right.
C. The mean and standard deviation could be used because both of the distributions are roughly symmetric, and they do not have outliers.
D.The mean and standard deviation could be used. Both of the distributions are roughly symmetric, and only Campus B has an outlier of 37
minutes.
2. Two data sets are being compared. One data set is skewed left, and the minimum value is an outlier. Another data set is symmetric about its peak about the center of the distribution with no outliers. Explain whether the mean and standard deviation should be used to compare the measures of center and measures of spread.
A. The mean and standard deviation should not be used because one of the distributions is skewed left and has an outlier.
B. The mean and standard deviation should not be used because both of the distributions are skewed left, and both data sets have at least one outlier.
C. The mean and the standard deviation should be used because the mean and the standard deviation are not affected by outliers.
D. The mean and the standard deviation should be used because one of the distributions is symmetric with no outliers.
3. Noah is on a cross country team. He is analyzing the amount of time in minutes and seconds it takes for him and his teammates to complete a 5-kilometer race and a 20-kilometer race. The results are shown in the accompanying table.
5-km Race 17:34 17:52 18:06 16:44 17:21 17:38 18:32 17:07 16:58 17:56
20-km Race 72:41 71:54 72:20 70:36 75:42 80:36 73:14 74:22 71:26 74:39
Interpret how the shape of the times for the 5-kilometer race compares to the shape of the times for the 20-kilometer race.
A. The distribution for the 5-kilometer race is roughly symmetric, and the distribution for the 20-kilometer race is skewed right. The time for one member of the team to complete the 20-kilometer race is unusually high.
B. The distribution of the 5-kilometer race is skewed right, and the distribution for the 20-kilometer race is roughly symmetric. The time for one member of the team to complete the 5-kilometer race is unusually high.
C. The distributions for the 5-kilometer race and 20-kilometer race are skewed left. The times for one member of the team to complete the 5-kilometer race and 20-kilometer race are unusually low for the respective races.
D.The distributions for the 5-kilometer race and 20-kilometer race are both symmetric, and there are no outliers. The times for the 5-kilometer race peak between 17 and 18 minutes, while the times for the 20-kilometer race peak between 71 and 75 minutes.
1. A. The median and interquartile range could be used because the distribution for Campus A is skewed right with an outlier of 30 minutes and the distribution for Campus B is roughly symmetric with no outliers.
2. A. The mean and standard deviation should not be used because one of the distributions is skewed left and has an outlier.
3. A. The distribution for the 5-kilometer race is roughly symmetric, and the distribution for the 20-kilometer race is skewed right. The time for one member of the team to complete the 20-kilometer race is unusually high.
1. Kaitlyn measures the upload speed in megabits per second of her home broadband internet connection during peak hours and off-peak hours. The results are provided in the accompanying table.
Peak Off-Peak
5.83 5.14 6.05 6.80 7.14 6.98
3.54 5.77 5.45 7.27 5.90 6.62
5.81 6.19 4.82 6.08 6.37 6.59
4.65 5.58 5.30 7.41 6.48 6.26
5.96 6.37 5.79 6.31 6.84 6.55
5.06 4.95 5.61 7.03 6.45 6.77
Examine the results of Kaitlyn's test to determine the statements that compare the medians and interquartile ranges of the data sets to each other in terms of this situation. Select the two correct answers.
A. The median upload speed during peak hours is 5.595 megabits per second, which is less than the median upload speed during off-peak hours, 6.605 megabits per second. The upload speeds during peak hours is slower on average compared to off-peak times.
B. The median upload speed during peak hours is 6.37 megabits per second, which is greater than the median upload speed during off-peak hours, 6.08 megabits per second. The upload speeds during peak hours is faster on average compared to off-peak times.
C. The interquartile range is 0.69 megabit per second for the upload speed during peak hours, which is greater than the interquartile range during off-peak hours, 0.67 megabit per second. So the spread of the upload speeds during off-peak hours is about the same as during peak hours.
D. The interquartile range is 0.77 megabit per second for the upload speed during peak hours, which is greater than the interquartile range during off-peak hours, 0.61 megabit per second. So the upload speeds are closer to each other during off-peak hours than during peak hours.
E. The median upload speed during peak hours is 5.905 megabits per second, which is less than the median upload speed during off-peak hours, 7.005 megabits per second. The upload speeds during peak hours is slower on average compared to off-peak times.
A. The median upload speed during peak hours is 5.595 megabits per second, which is less than the median upload speed during off-peak hours, 6.605 megabits per second. The upload speeds during peak hours is slower on average compared to off-peak times.
D. The interquartile range is 0.77 megabit per second for the upload speed during peak hours, which is greater than the interquartile range during off-peak hours, 0.61 megabit per second. So the upload speeds are closer to each other during off-peak hours than during peak hours.
To find the median and interquartile range for each location, we will first need to sort the data in ascending order.
For the forest sample:
15 13 14 9 16 11 15 14 15 14
Sorting the data in ascending order:
9 11 13 14 14 14 15 15 15 16
To find the median, we need to find the middle number. Since we have an even number of data points (10), we need to find the average of the two middle numbers, which are 14 and 15.
Median = (14 + 15) / 2 = 29 / 2 = 14.5
To find the interquartile range, we need to find the first quartile (Q1) and the third quartile (Q3).
The first quartile (Q1) is the median of the lower half of the data. In this case, the lower half is: 9 11 13 14.
Q1 = (11 + 13) / 2 = 24 / 2 = 12
The third quartile (Q3) is the median of the upper half of the data. In this case, the upper half is: 15 15 16.
Q3 = (15 + 15) / 2 = 30 / 2 = 15
The interquartile range is the difference between Q3 and Q1.
Interquartile range = Q3 - Q1 = 15 - 12 = 3
Therefore, the median number of insects in the forest samples is 14.5, and the interquartile range is 3.
For the marsh sample:
17 14 15 17 16 18 17 19 13 16
Sorting the data in ascending order:
13 14 15 16 16 17 17 17 18 19
To find the median, we need to find the middle number. Since we have an even number of data points (10), we need to find the average of the two middle numbers, which are 16 and 17.
Median = (16 + 17) / 2 = 33 / 2 = 16.5
To find the interquartile range, we need to find the first quartile (Q1) and the third quartile (Q3).
The first quartile (Q1) is the median of the lower half of the data. In this case, the lower half is: 13 14 15 16.
Q1 = (14 + 15) / 2 = 29 / 2 = 14.5
The third quartile (Q3) is the median of the upper half of the data. In this case, the upper half is: 17 17 17 18 19.
Q3 = (17 + 17) / 2 = 34 / 2 = 17
The interquartile range is the difference between Q3 and Q1.
Interquartile range = Q3 - Q1 = 17 - 14.5 = 2.5
Therefore, the median number of insects in the marsh samples is 16.5, and the interquartile range is 2.5.