Question

question 5 - dot plots
which statement is true about the dot plot below?
dot plot with x - axis labeled
umber of pets\ (values 0, 1, 2, 3, 4, 5) and dots: 0 has 1 dot, 1 has 4 dots, 2 has 6 dots, 3 has 4 dots, 4 has 3 dots, 5 has 1 dot
the mean and the median are the same.
the median and the mode are the same.
the mean and the mode are the same.
the mean, median, and mode are all the same

Explanation:

Step1: Count the number of dots for each value

• For 0: 1 dot
• For 1: 4 dots
• For 2: 6 dots
• For 3: 4 dots
• For 4: 3 dots
• For 5: 2 dots

Total number of data points: \(1 + 4 + 6 + 4 + 3 + 2 = 20\)

Step2: Find the mode

The mode is the value with the highest frequency. Here, 2 has 6 dots, which is the highest. So mode = 2.

Step3: Find the median

Since there are 20 data points (even number), the median is the average of the 10th and 11th values.
Let's list the cumulative frequencies:

• Up to 0: 1
• Up to 1: \(1 + 4 = 5\)
• Up to 2: \(5 + 6 = 11\)

So the 10th and 11th values are both 2. Thus, median = \(\frac{2 + 2}{2} = 2\)

Step4: Find the mean

Mean = \(\frac{\sum (x \times f)}{\sum f}\)

• \(0 \times 1 = 0\)
• \(1 \times 4 = 4\)
• \(2 \times 6 = 12\)
• \(3 \times 4 = 12\)
• \(4 \times 3 = 12\)
• \(5 \times 2 = 10\)

Sum of \(x \times f\): \(0 + 4 + 12 + 12 + 12 + 10 = 50\)
Sum of \(f\) (total data points) = 20
Mean = \(\frac{50}{20} = 2.5\)? Wait, no, wait, let's recalculate the sum of \(x \times f\):

Wait, 01=0; 14=4; 26=12; 34=12; 43=12; 52=10. Adding these: 0+4=4; 4+12=16; 16+12=28; 28+12=40; 40+10=50. Total data points: 1+4+6+4+3+2=20. So mean is 50/20=2.5? Wait, but that contradicts? Wait, no, maybe I made a mistake in counting the dots. Wait, looking at the dot plot:

• 0: 1 dot (correct)
• 1: 4 dots (correct, since the dots are 4)
• 2: 6 dots (correct, the dots are 6)
• 3: 4 dots (correct)
• 4: 3 dots (correct)
• 5: 2 dots (correct)

Wait, but then median: 20 data points, so median is average of 10th and 11th. Let's list the data in order:

0, 1,1,1,1, 2,2,2,2,2,2, 3,3,3,3, 4,4,4, 5,5

Wait, let's count:

1 (0) + 4 (1s) = 5 numbers (0 to 1s). Then 6 (2s): positions 6 to 11 (since 5 + 6 = 11). So the 10th and 11th numbers are both 2. So median is 2.

Mode is 2 (since 2 occurs 6 times, more than others).

Mean: (01 + 14 + 26 + 34 + 43 + 52)/20 = (0 + 4 + 12 + 12 + 12 + 10)/20 = 50/20 = 2.5. Wait, that's 2.5. But then mode and median are 2, mean is 2.5. But the option says "The median and the mode are the same." Let's check the options:

Options:

1. The mean and the median are the same. (Mean 2.5, median 2: no)
2. The median and the mode are the same. (Median 2, mode 2: yes)
3. The mean and the mode are the same. (Mean 2.5, mode 2: no)
4. The mean, median, and mode are all the same. (No, mean is 2.5)

Wait, but maybe I made a mistake in counting the dots. Let me recheck the dot plot:

Looking at the dot plot:

• 0: 1 dot (correct)
• 1: 4 dots (correct, 4 dots)
• 2: 6 dots (correct, 6 dots)
• 3: 4 dots (correct, 4 dots)
• 4: 3 dots (correct, 3 dots)
• 5: 2 dots (correct, 2 dots)

Wait, but maybe the 5 has 1 dot? Wait, the original image: "1" at 5? Wait, the user's image: "1" at 5? Wait, the dot plot: at 5, there are two dots? Wait, the user's image: "1" at 5? Wait, maybe a typo. Wait, the user's image: "1" at 5? Wait, no, the original problem: "1" at 5? Wait, maybe it's 2 dots. Wait, but according to the calculation, median and mode are 2, so the correct option is "The median and the mode are the same."

Wait, but let's recalculate mean again:

0*1 = 0

1*4 = 4

2*6 = 12

3*4 = 12

4*3 = 12

5*2 = 10

Sum: 0 + 4 = 4; 4 + 12 = 16; 16 + 12 = 28; 28 + 12 = 40; 40 + 10 = 50. Total data points: 20. 50/20 = 2.5. So mean is 2.5, median 2, mode 2. So median and mode are same. So the correct option is "The median and the mode are the same."

Answer:

The median and the mode are the same. (The option corresponding to this statement, e.g., if the options are labeled as A, B, C, D, then it would be the option with this text. Assuming the options are as given, the correct one is the second option: "The median and the mode are the same.")