Five students have the following scores 30, 30, 50, 60, 80, which statement is true for the data?
a. median = mode
b. mean = 20
c. median = 20
d. mean = median
86. Which set of data do the mean = median = mode?
a. 7, 9, 9, 9, 11
b. 7, 7, 8, 11, 12
c. 7, 7, 7, 8, 11
d. 7, 7, 9, 11, 16
87. What is the probability that the mode will be selected from the array 2, 2, 2, 2, 3, 3, 3, 7?   
a. 0
b. 1/8
c. 3/ 8
d. 4/ 8

### Question 1: Five students have the following scores 30, 30, 50, 60, 80, which statement is true for the data?
Given scores: 30, 30, 50, 60, 80

Let's calculate the mean, median, and mode.

1. **Mode:** The mode is the most frequent number in the data set. Here, 30 appears twice, so the mode is 30.
2. **Median:** To find the median, order the data (which is already ordered) and pick the middle value. Since there are five numbers, the median is the third number:
   \[
   \text{Median} = 50
   \]
3. **Mean:**
   \[
   \text{Mean} = \frac{30 + 30 + 50 + 60 + 80}{5} = \frac{250}{5} = 50
   \]

Now, let's evaluate the statements:
- **a. median = mode:** Median is 50, and Mode is 30. This is **false**.
- **b. mean = 20:** Mean is 50, not 20. This is **false**.
- **c. median = 20:** Median is 50, not 20. This is **false**.
- **d. mean = median:** Both the mean and median are 50. This is **true**.

**Answer:** **d. mean = median**

### Question 2: Which set of data do the mean = median = mode?
Let's evaluate each option:

**a. 7, 9, 9, 9, 11**
- Mean: \( \text{Mean} = \frac{7 + 9 + 9 + 9 + 11}{5} = \frac{45}{5} = 9 \)
- Median: The middle value is 9.
- Mode: 9 is the most frequent value.

Since the mean, median, and mode all equal 9, this option satisfies the condition.

**Answer:** **a. 7, 9, 9, 9, 11**

### Question 3: What is the probability that the mode will be selected from the array 2, 2, 2, 2, 3, 3, 3, 7? 
Given array: 2, 2, 2, 2, 3, 3, 3, 7

- **Mode:** The mode is the number that appears most frequently. Here, 2 appears 4 times, so the mode is 2.
- **Total number of elements:** 8
- **Number of times the mode (2) appears:** 4

**Probability of selecting the mode:**
\[
\text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{4}{8} = \frac{1}{2}
\]

**Answer:** **d. 4/8** (which simplifies to 1/2)

Would you like more details or have any questions?

Here are some related questions you might find interesting:
1. How do you calculate the mean of a dataset with more than one mode?
2. How does adding an outlier affect the mean, median, and mode?
3. What is the significance of the mode in a bimodal distribution?
4. How can the median be used to measure the central tendency in skewed distributions?
5. What happens to the mean and median if all data points are increased by a constant?
6. Can a data set have more than one median? Why or why not?
7. What is the impact of having no mode in a dataset?
8. How do you determine the mode in a continuous data distribution?

**Tip:** The median is resistant to outliers, while the mean is sensitive. This makes the median a better measure of central tendency when dealing with skewed data.

Explore the concepts of mean, median, and mode in statistics through practical examples. This page provides detailed solutions to problems involving the calculation of mean, median, and mode for given datasets.

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