Question

velma is examining the hourly wage of college students. she collects data from 15 juniors, given below. create a histogram and interpret the plot. select he best answer. set xmin to 8, xmax to 20, and xscl to 3. 23 8 23 15 21 8 21 14 19 13 19 14 19 18 22 select the correct answer below. the frequencies decrease as we move to the right of the histogram. the frequencies increase as we move to the right of the histogram. the smallest frequencies are between 17 and 20.

Explanation:

Step1: Determine the classes

Given \( X_{\text{min}} = 8 \), \( X_{\text{max}} = 20 \), and \( X_{\text{scl}} = 3 \), the classes are:
- Class 1: \( 8 - 10 \) (since \( 8+3 - 1=10 \))
- Class 2: \( 11 - 13 \)
- Class 3: \( 14 - 16 \)
- Class 4: \( 17 - 19 \)
- Class 5: \( 20 - 22 \) (but our \( X_{\text{max}} \) is 20, so it's \( 20 - 22 \) but we check data within \( 8 - 20 \))

Step2: Count frequencies for each class

- Class \( 8 - 10 \): Data points are 8, 8. So frequency \( f_1 = 2 \)
- Class \( 11 - 13 \): Data point is 13. So frequency \( f_2 = 1 \)
- Class \( 14 - 16 \): Data points are 15, 14, 14. So frequency \( f_3 = 3 \)
- Class \( 17 - 19 \): Data points are 19, 19, 19, 18. So frequency \( f_4 = 4 \)
- Class \( 20 - 22 \): Data points within \( 20 - 22 \) (but \( X_{\text{max}} = 20 \), our data has 21,23 etc. but we consider up to 20. Wait, original data has 21,23 which are above 20, but we set \( X_{\text{max}} = 20 \), so we ignore data above 20? Wait no, the problem says "Set \( X_{\text{min}} \) to 8, \( X_{\text{max}} \) to 20, and \( X_{\text{scl}} \) to 3". So we consider data in [8,20]. So data above 20 (21,23,21,22) are outside our \( X_{\text{max}} = 20 \), so we exclude them? Wait no, maybe the problem has a typo, but let's proceed with given \( X_{\text{max}} = 20 \). Wait, original data: 23, 8, 23, 15, 21, 8, 21, 14, 19, 13, 19, 14, 19, 18, 22. But we take \( X_{\text{max}} = 20 \), so we consider data ≤20: 8, 8, 15, 14, 19, 13, 19, 14, 19, 18. Wait, no, the 15 juniors: wait the data given is 15 numbers? Wait the data is: 23, 8, 23, 15, 21, 8, 21, 14, 19, 13, 19, 14, 19, 18, 22. That's 15 numbers. But we set \( X_{\text{max}} = 20 \), so data points above 20: 23,23,21,21,22. So we have 15 - 5 = 10 data points within [8,20]? Wait no, let's list data within [8,20]: 8,8,15,14,19,13,19,14,19,18. Wait that's 10? Wait no, 8,8,15,21 (no,21>20),14,19,13,19,14,19,18,22 (no). Wait I must have miscounted. Let's list all 15 data points:
1. 23 (above 20)
2. 8 (in)
3. 23 (above 20)
4. 15 (in)
5. 21 (above 20)
6. 8 (in)
7. 21 (above 20)
8. 14 (in)
9. 19 (in)
10. 13 (in)
11. 19 (in)
12. 14 (in)
13. 19 (in)
14. 18 (in)
15. 22 (above 20)

So data within [8,20]: 8,8,15,14,19,13,19,14,19,18. Wait that's 10? Wait no, 8 (2),15(1),14(2),19(3),13(1),18(1). Total 2+1+2+3+1+1=10? Wait maybe the problem allows Xmax to 20 but data above 20 are included? No, the problem says "Set Xmin to 8, Xmax to 20, and Xscl to 3". So the x - axis is from 8 to 20 with scale 3. So the classes are 8 - 10, 11 - 13, 14 - 16, 17 - 19, 20 - 22 (but 20 - 22 is beyond Xmax? No, Xmax is 20, so 20 - 22 is 20,21,22 but Xmax is 20, so 20 - 22 is up to 20? Maybe the problem has a mistake, but let's proceed with the data within 8 - 20.

Wait, maybe I made a mistake. Let's re - check the classes:

Class 1: 8 - 10 (8 ≤ x < 11, since Xscl = 3, so 8,9,10)
Class 2: 11 - 13 (11 ≤ x < 14)
Class 3: 14 - 16 (14 ≤ x < 17)
Class 4: 17 - 19 (17 ≤ x < 20)
Class 5: 20 - 22 (20 ≤ x < 23)

Now count frequencies with this correct class interval (using exclusive upper limit):

- Class 8 - 10 (8 ≤ x < 11): Data points 8,8. Frequency = 2
- Class 11 - 13 (11 ≤ x < 14): Data point 13. Frequency = 1
- Class 14 - 16 (14 ≤ x < 17): Data points 15,14,14. Frequency = 3
- Class 17 - 19 (17 ≤ x < 20): Data points 19,19,19,18. Frequency = 4
- Class 20 - 22 (20 ≤ x < 23): Data points 21,21,22 (but 21,22 are in this class, but our Xmax is 20? Wait no, Xmax is 20, but the class 20 - 22 includes 20,21,22. So data points 21,21,22 are in this class. So frequency for class 20 - 22: 3 (21,21,22)

Now let's recalculate frequencies with all data (including above 20 but within 20 - 22 class):

- Class 8 - 10: 2
- Class 11 - 13: 1
- Class 14 - 16: 3
- Class 17 - 19: 4
- Class 20 - 22: 3 (21,21,22)

Now, as we move from left (class 8 - 10) to right (class 20 - 22), the frequencies are 2,1,3,4,3. Wait, but let's check the trend. From class 8 - 10 (f=2) to 11 - 13 (f=1) (decrease), then 14 - 16 (f=3) (increase), 17 - 19 (f=4) (increase), 20 - 22 (f=3) (decrease). But the options are about "as we move to the right". Wait maybe the problem considers the main trend. Wait the option "The frequencies increase as we move to the right of the histogram" – let's check the frequencies of the classes in order (8 - 10, 11 - 13, 14 - 16, 17 - 19, 20 - 22) with correct counting (including data above 20 in 20 - 22 class):

Wait no, the original data has 15 points. Let's list all data points and their classes:

1. 23: above 22 (but our Xscl=3, so 20 - 22 is 20,21,22; 23 is in 23 - 25, but we set Xmax=20, so maybe we ignore data above 20. Let's do that. So data within 8 - 20: 8,8,15,14,19,13,19,14,19,18. Wait that's 10 points? No, 8,8,15,14,19,13,19,14,19,18, and also 21,23 are above 20, 22 is above 20. Wait I think the problem has a typo, but let's proceed with the given data and the classes as 8 - 10, 11 - 13, 14 - 16, 17 - 19, 20 - 22 (even if data is above 20, we count them in 20 - 22 as per Xscl=3).

So:

- 8 - 10: 8,8 → 2
- 11 - 13: 13 → 1
- 14 - 16: 15,14,14 → 3
- 17 - 19: 19,19,19,18 → 4
- 20 - 22: 21,21,22 → 3

Now, as we move from the left - most class (8 - 10) to the right - most class (20 - 22), the frequencies are 2,1,3,4,3. But the option "The frequencies increase as we move to the right of the histogram" – let's check the trend from 11 - 13 (f=1) to 14 - 16 (f=3) (increase), 14 - 16 to 17 - 19 (f=4) (increase), 17 - 19 to 20 - 22 (f=3) (decrease). But maybe the problem considers the main part. Wait the other options:

- "The frequencies decrease as we move to the right" – no, because from 11 - 13 to 17 - 19, frequencies increase.
- "The smallest frequencies are between 17 and 20" – no, frequency between 17 - 19 is 4, which is not small.
- "The frequencies increase as we move to the right" – let's see the classes in order (8 - 10, 11 - 13, 14 - 16, 17 - 19). The frequencies are 2,1,3,4. Ignoring the last class (20 - 22) which has data above 20 (maybe the problem expects us to ignore data above 20). So from 8 - 10 (f=2) to 11 - 13 (f=1) (decrease), then 11 - 13 to 14 - 16 (f=3) (increase), 14 - 16 to 17 - 19 (f=4) (increase). So overall, from 11 - 13 to 17 - 19, frequencies increase. And if we consider the general trend (ignoring the dip at 11 - 13), the frequencies are increasing as we move towards the right (from lower classes to higher classes within 8 - 20).

Step3: Analyze the options

- Option 1: "The frequencies decrease as we move to the right" – False, because from 11 - 13 to 17 - 19, frequencies increase.
- Option 2: "The frequencies increase as we move to the right" – True, because the frequencies in the classes 11 - 13 (f=1), 14 - 16 (f=3), 17 - 19 (f=4) show an increasing trend as we move right.
- Option 3: "The smallest frequencies are between 17 and 20" – False, frequency between 17 - 19 is 4, which is not the smallest (smallest is 1 in 11 - 13).

Answer:

The frequencies increase as we move to the right of the histogram. (The option corresponding to this statement)