My  rating  is 4.50 out of 5 for 27 orders. What are the possibilities of having 5,4,3,2,1 and no ratings

To determine the number of ratings you received for each possible rating (5, 4, 3, 2, 1), we need to use the information you provided:

1. Your overall average rating is 4.50 out of 5.
2. You have 27 total orders.

Let:
- \( x_5 \) be the number of 5-star ratings,
- \( x_4 \) be the number of 4-star ratings,
- \( x_3 \) be the number of 3-star ratings,
- \( x_2 \) be the number of 2-star ratings,
- \( x_1 \) be the number of 1-star ratings.

We can set up the following system of equations:

### Equation 1 (Total number of ratings):
\[
x_5 + x_4 + x_3 + x_2 + x_1 = 27
\]
This accounts for the total number of ratings (27).

### Equation 2 (Average rating):
\[
\frac{5x_5 + 4x_4 + 3x_3 + 2x_2 + 1x_1}{27} = 4.50
\]
This equation represents the weighted average rating of 4.50.

Multiplying the second equation by 27 to eliminate the denominator:
\[
5x_5 + 4x_4 + 3x_3 + 2x_2 + x_1 = 121.5
\]

Since we can't have fractional ratings, we will adjust the equation to:
\[
5x_5 + 4x_4 + 3x_3 + 2x_2 + x_1 = 122
\]

Now, we have the following system of equations:
1. \( x_5 + x_4 + x_3 + x_2 + x_1 = 27 \)
2. \( 5x_5 + 4x_4 + 3x_3 + 2x_2 + x_1 = 122 \)

### Solving the system:
We can solve these two equations to find possible values of \( x_5, x_4, x_3, x_2, x_1 \). Let me calculate this.The solution for the system of equations depends on the number of 1-star, 2-star, and 3-star ratings. Here are the relationships we found:

- \( x_5 = 3x_1 + 2x_2 + x_3 + 14 \)
- \( x_4 = -4x_1 - 3x_2 - 2x_3 + 13 \)

To find possible combinations of \( x_1, x_2, x_3 \) that yield valid solutions for \( x_4 \) and \( x_5 \) (i.e., non-negative integers), we need to try different values for \( x_1, x_2, x_3 \). Let me calculate a few valid possibilities for these ratings.Here are five possible distributions of ratings based on your average of 4.50 out of 27 orders:

1. \( x_5 = 14 \), \( x_4 = 13 \), \( x_3 = 0 \), \( x_2 = 0 \), \( x_1 = 0 \)
2. \( x_5 = 15 \), \( x_4 = 11 \), \( x_3 = 1 \), \( x_2 = 0 \), \( x_1 = 0 \)
3. \( x_5 = 16 \), \( x_4 = 9 \), \( x_3 = 2 \), \( x_2 = 0 \), \( x_1 = 0 \)
4. \( x_5 = 17 \), \( x_4 = 7 \), \( x_3 = 3 \), \( x_2 = 0 \), \( x_1 = 0 \)
5. \( x_5 = 18 \), \( x_4 = 5 \), \( x_3 = 4 \), \( x_2 = 0 \), \( x_1 = 0 \)

These represent some possible distributions of ratings that would yield an average rating of 4.50.

Would you like more possible combinations or further details on any of these?

Here are 5 related questions:
1. How does the calculation change if you receive more orders?
2. What is the impact of a single low rating on your overall average?
3. How can you improve your rating to 4.75 from 4.50?
4. What is the lowest number of 5-star ratings required to maintain a 4.50 average?
5. How many orders would you need to reach an exact 5-star average?

**Tip:** A single lower rating has a significant impact on your average, especially with fewer total orders.

My rating is 4.50 out of 5 for 27 orders. What are the possibilities of having 5,4,3,2,1 and no ratings?

Explore how to calculate possible distributions of 5-star, 4-star, 3-star, 2-star, and 1-star ratings to maintain a 4.50 average rating from 27 total orders. Learn about systems of equations and weighted averages in this step-by-step solution, suitable for Grades 10-12.

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