In a 5 star rating, how can i achieve a rating of 4.44 in the least amount of responses? The responses can only be 1, 2, 3, 4, or 5.

Guest Jul 31, 2017

edited by
Guest
Jul 31, 2017

#5**+1 **

Interesting question! I would not consider this a "simple" question. It requires one to think about it!

Before we attempt this problem, we have to understand how the rating system works. Generally, the rating system simply takes the average of all the responses' ratings. Therefore, we can create a formula for it.

\(f(\# \hspace{1mm}\text{of responses})=\frac{x_1+x_2+x_3...x_{\# \hspace{1mm}\text{of responses}}}{\# \hspace{1mm}\text{of responses}}=4.44\)

I will create a table, first:

# of responses | Formula | Solve for numerator | ||||

2 | \(\frac{x_1+x_2}{2}=4.44\) | \({x_1+x_2}=8.88\) | ||||

3 | \(\frac{x_1+x_2+x_3}{3}=4.44\) | \({x_1+x_2+x_3}=13.32\) | ||||

4 | \(\frac{x_1+x_2...+x_4}{4}=4.44\) | \({x_1+x_2...+x_4}=17.76\) | ||||

5 | \(\frac{x_1+x_2...+x_5}{5}=4.44\) | \({x_1+x_2...+x_5}=22.2\) | ||||

6 | \(\frac{x_1+x_2...+x_6}{6}=4.44\) | \({x_1+x_2...+x_6}=26.64\) | ||||

7 | \(\frac{x_1+x_2...+x_7}{7}=4.44\) | \({x_1+x_2...+x_7}=31.08\) | ||||

8 | \(\frac{x_1+x_2...+x_8}{8}=4.44\) | \({x_1+x_2...+x_8}=35.52\) | ||||

9 | \(\frac{x_1+x_2...+x_9}{9}=4.44\) | \({x_1+x_2...+x_9}=39.96\) | ||||

10 | \(\frac{x_1+x_2...+x_{10}}{10}=4.44\) | \({x_1+x_2...+x_{10}}=44.4\) | ||||

n | \(\frac{x_1+x_2...+x_n}{n}=4.44\) | \({x_1+x_2...+x_n}=4.44n\) |

We have tried the first 10 responses. Will 10 responses ever give an average of 4.44? No. How do I know? We need the integers from 1 to 5 to equal a decimal number. That is impossible. Let's try the first one in the table, 2.

\(x_1+x_2=8.88\)

If the numbers we can substitute into x are integers, it is impossible to find 2 integers that add up to 8.88 because you cannot add two integers and get a decimal. The same logic can be used for the above ones. Now we must hunt for the least number for *n* such that 4.44*n* is an integer. I have effectively changed the question now. I will use a fraction to help me out with this:

\(4.44\) | I am going to convert 4.44 into a fraction. 4.44 extends until the hundredth place, so put it over 100. |

\(4+\frac{44}{100}\) | Now, I only care about the 44/100, so I will forget about the 4. |

\(\frac{44}{100}\div\frac{4}{4}=\frac{11}{25}\) | This fraction is irreducible. |

But what does this mean? The denominator says the least amount of response, 25. How do I know this? Well, what is 4.44*25=111.

Now the last question we must ask ourselves is can we make 111 with 25 responses?

Yes, we can! You can have 21 5-star responses and 2 2-star responses and 2 1-star responses.

TheXSquaredFactor Jul 31, 2017

#1**0 **

In a 5 star rating, how can i achieve a rating of 4.44 in the least amount of responses? The responses can only be 1, 2, 3, 4, or 5.

4.44 .... I thought it was repeater :(

If it was repeater, and yes I know it is not, then the rating wouldbe 42/9 so you would need a multiple of 9 responses.

The minimum would be 9

Melody Jul 31, 2017

#6**+1 **

Wait a second! If the desired rating was \(4.\overline{44}\), then the numbers would have to add to 40, right?

\(4\frac{4}{9}\) | Convert this into an improper fraction. |

\(\frac{9*4+4}{9}=\frac{36+4}{9}=\frac{40}{9}\) | |

\(\frac{40}{9}=4.\overline{44}\)

\(\frac{42}{9}=4.\overline{66}\neq4.\overline{44}\)

5,5,5,5,5,5,5,4,1 would give the desired result.

TheXSquaredFactor
Jul 31, 2017

#3**+1 **

If it is EXACTLY 4.44 then that is 4 and 11/25

So it would be a minimum of 25 responses.

Melody Jul 31, 2017

#5**+1 **

Best Answer

Interesting question! I would not consider this a "simple" question. It requires one to think about it!

Before we attempt this problem, we have to understand how the rating system works. Generally, the rating system simply takes the average of all the responses' ratings. Therefore, we can create a formula for it.

\(f(\# \hspace{1mm}\text{of responses})=\frac{x_1+x_2+x_3...x_{\# \hspace{1mm}\text{of responses}}}{\# \hspace{1mm}\text{of responses}}=4.44\)

I will create a table, first:

# of responses | Formula | Solve for numerator | ||||

2 | \(\frac{x_1+x_2}{2}=4.44\) | \({x_1+x_2}=8.88\) | ||||

3 | \(\frac{x_1+x_2+x_3}{3}=4.44\) | \({x_1+x_2+x_3}=13.32\) | ||||

4 | \(\frac{x_1+x_2...+x_4}{4}=4.44\) | \({x_1+x_2...+x_4}=17.76\) | ||||

5 | \(\frac{x_1+x_2...+x_5}{5}=4.44\) | \({x_1+x_2...+x_5}=22.2\) | ||||

6 | \(\frac{x_1+x_2...+x_6}{6}=4.44\) | \({x_1+x_2...+x_6}=26.64\) | ||||

7 | \(\frac{x_1+x_2...+x_7}{7}=4.44\) | \({x_1+x_2...+x_7}=31.08\) | ||||

8 | \(\frac{x_1+x_2...+x_8}{8}=4.44\) | \({x_1+x_2...+x_8}=35.52\) | ||||

9 | \(\frac{x_1+x_2...+x_9}{9}=4.44\) | \({x_1+x_2...+x_9}=39.96\) | ||||

10 | \(\frac{x_1+x_2...+x_{10}}{10}=4.44\) | \({x_1+x_2...+x_{10}}=44.4\) | ||||

n | \(\frac{x_1+x_2...+x_n}{n}=4.44\) | \({x_1+x_2...+x_n}=4.44n\) |

We have tried the first 10 responses. Will 10 responses ever give an average of 4.44? No. How do I know? We need the integers from 1 to 5 to equal a decimal number. That is impossible. Let's try the first one in the table, 2.

\(x_1+x_2=8.88\)

If the numbers we can substitute into x are integers, it is impossible to find 2 integers that add up to 8.88 because you cannot add two integers and get a decimal. The same logic can be used for the above ones. Now we must hunt for the least number for *n* such that 4.44*n* is an integer. I have effectively changed the question now. I will use a fraction to help me out with this:

\(4.44\) | I am going to convert 4.44 into a fraction. 4.44 extends until the hundredth place, so put it over 100. |

\(4+\frac{44}{100}\) | Now, I only care about the 44/100, so I will forget about the 4. |

\(\frac{44}{100}\div\frac{4}{4}=\frac{11}{25}\) | This fraction is irreducible. |

But what does this mean? The denominator says the least amount of response, 25. How do I know this? Well, what is 4.44*25=111.

Now the last question we must ask ourselves is can we make 111 with 25 responses?

Yes, we can! You can have 21 5-star responses and 2 2-star responses and 2 1-star responses.

TheXSquaredFactor Jul 31, 2017