The mean of a distribution is the sum of the data elements divided by the count of the dataset.
<em>The mean of the distribution is 4</em>
The complete table is given as
![\left[\begin{array}{cc}People & Frequency &0 - 2 & 5 & 3 - 5 & 25 & 6 - 8 & 5\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7DPeople%20%26%20Frequency%20%260%20-%202%20%26%205%20%26%203%20-%205%20%26%2025%20%26%206%20-%208%20%26%205%5Cend%7Barray%7D%5Cright%5D)
The complete question requires that, we calculate the mean of the dataset
First, we calculate the class midpoint
This is the average of the class interval
<u />
<u>For interval 0 - 2, </u>

<u>For interval 3 - 5,</u>

<u>For interval 6 - 8</u>

So, the table becomes
![\left[\begin{array}{ccc}People & x & Frequency &0 - 2 &1 & 5 & 3 - 5 & 4& 25 & 6 - 8& 7 & 5\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7DPeople%20%26%20x%20%26%20Frequency%20%260%20-%202%20%261%20%26%205%20%26%203%20-%205%20%26%204%26%2025%20%26%206%20-%208%26%207%20%26%205%5Cend%7Barray%7D%5Cright%5D)
The mean is then calculated as:

This gives



Hence, the mean of the distribution is 4
Read more about mean at:
brainly.com/question/17060266
The estimated amount of people in Clairette, Texas is roughly 55-65 people
This is fairly easy.
60x30=1800
Just add up the zeros from both numbers and do 6x3.
So therefore there will be 18 hundreds sections since there is nothing on the tens digit and nothing on the ones digit.
Hope this helps!
53, and 9. Are the answers.