Question

ii
a class of 20 students was surveyed to determine the number of pets each student owned. the data are represented in the dot plot below.

(1) which statement about the data is correct?
(1) the mean and the median are the same.
(2) the median and the mode are the same.
(3) the mean and the mode are the same.
(4) the mean, median, and mode are all the same.

(4) the range of ages of the swim team is smaller than the range of ages of the track team.

Explanation:

Step1: Count the number of dots for each number of pets

• For 0 pets: 1 dot
• For 1 pet: 4 dots
• For 2 pets: 7 dots
• For 3 pets: 4 dots
• For 4 pets: 3 dots
• For 5 pets: 2 dots

Step2: Calculate the total number of students (data points)

Total dots = \(1 + 4 + 7 + 4 + 3 + 2 = 21\)? Wait, the problem says 20 students. Maybe a typo, but let's proceed with the dot counts. Wait, maybe I miscounted. Let's check again:

0: 1

1: 4 (so 1*4)

2: 7 (2*7)

3: 4 (3*4)

4: 3 (4*3)

5: 2 (5*2)

Total data points: \(1 + 4 + 7 + 4 + 3 + 2 = 21\). But the problem says 20. Maybe a mistake, but let's continue.

Step3: Calculate the mean

Mean = \(\frac{(01) + (14) + (27) + (34) + (43) + (52)}{21}\)

Calculate numerator: \(0 + 4 + 14 + 12 + 12 + 10 = 52\)

Mean = \(\frac{52}{21} \approx 2.476\)

Step4: Find the median

Since there are 21 data points, the median is the 11th value (since \((21 + 1)/2 = 11\)). Let's order the data:

• 0: 1 (1st)
• 1: 4 (2nd - 5th)
• 2: 7 (6th - 12th)

So the 11th value is 2. So median = 2.

Step5: Find the mode

The mode is the number with the most dots, which is 2 (7 dots).

Step6: Compare mean, median, mode

• Mean ≈ 2.476
• Median = 2
• Mode = 2

Wait, but the options are about which are the same. Wait, maybe I miscounted the total number of students. Let's check again. Maybe the dot plot has:

0: 1

1: 4

2: 6

3: 4

4: 3

5: 2

Total: 1+4+6+4+3+2=20. Ah, maybe I miscounted 2's dots. Let's re - count the dot plot:

Looking at the dot plot:

• 0: 1 dot

• 1: 4 dots (so 4 students)

• 2: Let's see, the dots for 2: maybe 6? Wait, the original problem says 20 students. Let's assume the correct counts are:

0:1, 1:4, 2:6, 3:4, 4:3, 5:2. Total: 1 + 4+6 + 4+3 + 2=20.

Now recalculate:

Mean: \(\frac{0*1 + 1*4 + 2*6 + 3*4 + 4*3 + 5*2}{20}=\frac{0 + 4+12 + 12 + 12 + 10}{20}=\frac{50}{20}=2.5\)

Median: For 20 data points, the median is the average of the 10th and 11th values.

Order of data:

• 0:1 (1st)

• 1:4 (2nd - 5th)

• 2:6 (6th - 11th)

• 3:4 (12th - 15th)

• 4:3 (16th - 18th)

• 5:2 (19th - 20th)

10th value: 2 (since 6th - 11th are 2s), 11th value: 2. So median=\(\frac{2 + 2}{2}=2\)

Mode: The number with the most dots is 2 (6 dots? Wait, no, if total is 20, and 0:1,1:4,2:7,3:4,4:3,5:2, total is 21. There is a contradiction. But assuming the correct total is 20, and the mode is 2 (most frequent), median is 2, and let's recalculate mean with 20 data points (correcting 2's count to 6):

Mean=\(\frac{0*1 + 1*4 + 2*6 + 3*4 + 4*3 + 5*2}{20}=\frac{0 + 4+12 + 12 + 12 + 10}{20}=\frac{50}{20}=2.5\)

Median: average of 10th and 11th. 10th is 2, 11th is 2, so median = 2.

Mode: 2 (since 6 or 7 is the highest count for 2).

Now, median = 2, mode = 2. Mean is 2.5. So the median and mode are the same. So option (2) "The median and the mode are the same" is correct.

Answer:

(2) The median and the mode are the same.