This question is based on the concept of mean and variance. Therefore, the correct option is B, mean age will stay the same but the variance will decrease. Hence, it adds less variance because now there are two children who are 4 as opposed to just one.
Given that:
There are three children in a room, ages three, four, and five.
We have to choose correct option if a four-year-old child enters the room.
According to the question,
Now, calculating the mean of the ages of children.
As we know that,
[tex]Mean = \dfrac{Sum \,of \,the\.\, terms}{Total\, number\, of\, terms}[/tex]
[tex]Mean = \dfrac{3+4+5}{3} = \dfrac{12}{3} = 4[/tex]
If a four-year-old child enters the room then the mean will be,
[tex]Mean = \dfrac{3+4+5+4}{4} = \dfrac{16}{4} = 4[/tex]
Now, calculating the variance,
[tex]Variance \,(S^2)= \dfrac{\sum (x_i - x)^2}{n-1} = \dfrac{\sum (3 - 0)^2}{4-1} = 3 =1.73[/tex]
Therefore, the correct option is B, mean age will stay the same but the variance will decrease. Hence, it adds less variance because now there are two children who are 4 as opposed to just one.
For more details, prefer this link:
https://brainly.com/question/13708253
