1 select all the distribution shapes for which it is most often appropriate to use the mean.
a bell - shaped
b bimodal
c skewed
d symmetric
e uniform

2 for which distribution shape is it usually appropriate to use the median when summarizing the data?
a bell - shaped
b skewed
c symmetric
d uniform

3 the number of writing instruments in some teachers’ desks is displayed in the dot plot.
dot plot with x - axis labeled number of writing instruments from 4 to 13, with dots at 5 (1), 6 (1), 7 (1), 8 (1), 9 (3), 10 (4), 11 (2), 12 (1)
which is greater, the mean or the median?
explain your reasoning, using the shape of the distribution.

Question

1 select all the distribution shapes for which it is most often appropriate to use the mean.
a bell - shaped
b bimodal
c skewed
d symmetric
e uniform

2 for which distribution shape is it usually appropriate to use the median when summarizing the data?
a bell - shaped
b skewed
c symmetric
d uniform

3 the number of writing instruments in some teachers’ desks is displayed in the dot plot.
dot plot with x - axis labeled number of writing instruments from 4 to 13, with dots at 5 (1), 6 (1), 7 (1), 8 (1), 9 (3), 10 (4), 11 (2), 12 (1)
which is greater, the mean or the median?
explain your reasoning, using the shape of the distribution.

Accepted Answer

### Question 1
# Brief Explanations:
The mean is most appropriate for symmetric distributions (including bell - shaped, symmetric, and uniform as bell - shaped is a type of symmetric, and uniform is also symmetric in a sense). Bimodal and skewed distributions are better summarized with median as mean is affected by outliers or multiple modes. So bell - shaped (A), symmetric (D), and uniform (E) are correct.
# Answer:
A. bell - shaped, D. symmetric, E. uniform

### Question 2
# Brief Explanations:
Skewed distributions have outliers that affect the mean. The median is resistant to outliers, so for skewed (B) distributions, median is usually appropriate. Bell - shaped, symmetric, and uniform distributions are better summarized with mean.
# Answer:
B. skewed

### Question 3
# Explanation:
## Step 1: Analyze the distribution shape
First, we count the number of dots (frequency) for each value:
- 5: 1 dot
- 6: 1 dot
- 7: 1 dot
- 8: 1 dot
- 9: 3 dots
- 10: 4 dots
- 11: 2 dots
- 12: 1 dot

The distribution is symmetric (the left and right sides of the peak at 10 are relatively balanced). For a symmetric distribution, the mean and median are approximately equal? Wait, no, let's calculate mean and median.

## Step 2: Calculate the median
First, find the total number of data points. Sum the frequencies: $1 + 1+1 + 1+3 + 4+2 + 1=14$. Since $n = 14$ (even), the median is the average of the $\frac{n}{2}=7$th and $\frac{n}{2}+ 1 = 8$th values.

Let's list the data in order: 5, 6, 7, 8, 9, 9, 9, 10, 10, 10, 10, 11, 11, 12.

The 7th value is 9 and the 8th value is 10. So median $=\frac{9 + 10}{2}=9.5$.

## Step 3: Calculate the mean
Mean $=\frac{\sum_{i = 1}^{n}x_{i}f_{i}}{n}$, where $x_{i}$ is the value and $f_{i}$ is the frequency.

$$
\begin{align*}
\sum_{i = 1}^{n}x_{i}f_{i}&=(5\times1)+(6\times1)+(7\times1)+(8\times1)+(9\times3)+(10\times4)+(11\times2)+(12\times1)\
&=5 + 6+7 + 8+27+40+22+12\
&=5 + 6=11;11 + 7 = 18;18+8 = 26;26 + 27=53;53+40 = 93;93+22 = 115;115+12 = 127
\end{align*}
$$

Mean $=\frac{127}{14}\approx9.07$? Wait, no, I must have miscalculated. Wait, 5*1 = 5, 6*1=6 (total 11), 7*1 = 7 (18), 8*1=8 (26), 9*3 = 27 (53), 10*4=40 (93), 11*2 = 22 (115), 12*1=12 (127). $n = 1 + 1+1+1 + 3+4+2+1=14$. So mean $=\frac{127}{14}\approx9.07$? But median is 9.5. Wait, maybe my frequency count is wrong. Let's re - count the dots:

Looking at the dot plot:

- 5: 1 dot
- 6: 1 dot
- 7: 1 dot
- 8: 1 dot
- 9: 3 dots (so 3)
- 10: 4 dots (so 4)
- 11: 2 dots (so 2)
- 12: 1 dot (so 1)

Total: 1+1+1+1 + 3+4+2+1=14. Correct.

Wait, maybe the distribution is symmetric? Wait, the left side (5,6,7,8) has 4 data points, the right side (11,12) has 3 data points, and the middle (9,10) has 7 data points. Wait, maybe it's approximately symmetric. Wait, but when I calculated mean is approximately 9.07 and median is 9.5? That can't be. Wait, no, I think I made a mistake in the order of the data.

Wait, the data points are: 5, 6, 7, 8, 9, 9, 9, 10, 10, 10, 10, 11, 11, 12. Let's list them in order:

Position 1: 5

Position 2: 6

Position 3: 7

Position 4: 8

Position 5: 9

Position 6: 9

Position 7: 9

Position 8: 10

Position 9: 10

Position 10: 10

Position 11: 10

Position 12: 11

Position 13: 11

Position 14: 12

Ah! I made a mistake in the 7th and 8th positions. The 7th position is 9, the 8th position is 10. So median is $\frac{9 + 10}{2}=9.5$.

Now calculate the mean again:

$$
\begin{align*}
\sum x_{i}f_{i}&=5\times1+6\times1 + 7\times1+8\times1+9\times3+10\times4+11\times2+12\times1\
&=5 + 6+7 + 8+27+40+22+12\
&=5+6 = 11;11 + 7=18;18 + 8 = 26;26+27 = 53;53+40 = 93;93+22 = 115;115+12 = 127
\end{align*}
$$

Mean $=\frac{127}{14}\approx9.07$? Wait, that's less than median. But the distribution looks symmetric? Wait, no, maybe the dot plot is symmetric around 9.5? Wait, 5 and 12: 5 + 12 = 17, 6 and 11: 6+11 = 17, 7 and 10: 7 + 10=17, 8 and 9: 8 + 9 = 17. Oh! So it is symmetric. Then why is the mean less than median? Wait, no, my calculation of the mean must be wrong.

Wait, 5*1 = 5, 6*1=6 (total 11), 7*1 = 7 (18), 8*1=8 (26), 9*3 = 27 (53), 10*4=40 (93), 11*2 = 22 (115), 12*1=12 (127). 127 divided by 14: 14*9 = 126, so 127/14=9 + 1/14≈9.07. But median is 9.5. Wait, this is a contradiction. Wait, no, the number of data points: let's count the dots again.

Looking at the dot plot:

- 5: 1 dot

- 6: 1 dot

- 7: 1 dot

- 8: 1 dot

- 9: 3 dots (so 3)

- 10: 4 dots (so 4)

- 11: 2 dots (so 2)

- 12: 1 dot (so 1)

Total: 1+1+1+1+3 + 4+2+1 = 14. Correct.

Wait, maybe the dot plot is not symmetric. Wait, the left tail (5,6,7,8) has 4 points, the right tail (11,12) has 3 points. So it's slightly skewed left? Wait, no, left tail is longer? Wait, 5,6,7,8: 4 points, right tail 11,12: 2 points? Wait, no, 11 has 2 dots, 12 has 1 dot, so 3 points. Left tail: 5,6,7,8: 4 points. So left tail is longer. So the distribution is skewed left. In a left - skewed distribution, the mean is less than the median. Wait, but earlier I thought it was symmetric. My mistake. So the distribution is skewed left? Wait, no, the peak is at 10, and the left side has more spread? Wait, no, let's list the data in order and see the shape.

Data in order: 5, 6, 7, 8, 9, 9, 9, 10, 10, 10, 10, 11, 11, 12.

The left side (values less than the median) has values 5,6,7,8,9,9,9 (7 values) and the right side (values greater than the median) has 10,10,10,10,11,11,12 (7 values). Wait, median is between 9 and 10, so the number of values less than median: positions 1 - 7 (5,6,7,8,9,9,9) and values greater than median: positions 8 - 14 (10,10,10,10,11,11,12). So it is symmetric. Then why is the mean less than median? Wait, no, I think I miscalculated the mean.

Wait, 5+6+7+8+9+9+9+10+10+10+10+11+11+12. Let's add them step by step:

5 + 6 = 11; 11+7 = 18; 18+8 = 26; 26+9 = 35; 35+9 = 44; 44+9 = 53; 53+10 = 63; 63+10 = 73; 73+10 = 83; 83+10 = 93; 93+11 = 104; 104+11 = 115; 115+12 = 127. Yes, that's correct. 127 divided by 14 is approximately 9.07. Median is 9.5. So median is greater than mean? But the question is which is greater, mean or median. Wait, maybe my analysis of the distribution shape is wrong.

Wait, the dot plot: the number of dots at 5,6,7,8 is 1 each, at 9 is 3, at 10 is 4, at 11 is 2, at 12 is 1. So the distribution is symmetric around 9.5 (since 5 and 12, 6 and 11, 7 and 10, 8 and 9 are symmetric pairs). In a symmetric distribution, mean and median should be equal. But here they are not. Wait, maybe the dot plot has a different number of dots. Wait, maybe I miscounted the dots at 9 and 10. Let's look again: at 9, there are 3 dots? Or 4? Wait, the original dot plot: "at 9: three dots, at 10: four dots, at 11: two dots, at 12: one dot". Wait, maybe at 9 there are 4 dots? Let's re - examine the problem statement.

The dot plot: "5: 1, 6:1,7:1,8:1,9:3,10:4,11:2,12:1". Wait, maybe the user made a typo, or I misread. Alternatively, maybe the distribution is symmetric and my calculation is wrong. Wait, 14 data points. The median is the average of the 7th and 8th terms. Let's list the data with frequencies:

- 5: 1 (position 1)

- 6: 1 (position 2)

- 7: 1 (position 3)

- 8: 1 (position 4)

- 9: 3 (positions 5,6,7)

- 10: 4 (positions 8,9,10,11)

- 11: 2 (positions 12,13)

- 12: 1 (position 14)

Ah! Here's the mistake. The 7th position is 9 (since positions 5,6,7 are 9), and the 8th position is 10 (position 8 is 10). So median is $\frac{9 + 10}{2}=9.5$.

Now calculate the mean:

Sum of values: $5\times1+6\times1 + 7\times1+8\times1+9\times3+10\times4+11\times2+12\times1$

$=5 + 6+7 + 8+27+40+22+12$

$=5+6 = 11;11 + 7=18;18 + 8 = 26;26+27 = 53;53+40 = 93;93+22 = 115;115+12 = 127$

Mean $=\frac{127}{14}\approx9.07$

Wait, so median (9.5) is greater than mean (≈9.07). But the distribution is symmetric? No, because the frequency of 9 is 3 and 10 is 4, and the left side has 4 values (5,6,7,8) and the right side has 3 values (11,12) plus 4 values of 10. Wait, maybe the distribution is slightly skewed left. In a left - skewed distribution, mean < median. So the median is greater than the mean.

# Answer:
The median is greater than the mean. The distribution is approximately symmetric (or slightly left - skewed). In a symmetric distribution, mean and median are close, but due to the slight left - skew (more values on the left with lower numbers), the mean is pulled down slightly, making the median greater than the mean. (Or, if we consider the exact calculation, median = 9.5 and mean≈9.07, so median > mean)

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QUESTION IMAGE

Question

Explanation:

Question 1

Step 1: Analyze the distribution shape

Step 2: Calculate the median

Step 3: Calculate the mean

Answer:

Question 2