Find the mode, median and mean of the number of children

[chwala]

Homework Statement

see attached

Relevant Equations

data stats

See attached question and markscheme.

Solutions

Now they give the Mode as $1$ i am not getting this...
My understanding 'maybe its the English used' is that we have $0$ children living in 3 houses, 7 houses have $1$ kid each bringing total number of kids to 7!, 5 houses have $2$ kids each in the household bringing total number of kids here to $10$, in my case the mode ought to be 2. I need some clarity here.
Also looking at this table, 4 kids live in 0 houses and 5 kids also living in 0 houses does not really make sense to me...why not just have (4+5=9) kids live in 0 houses?...

 

[Mark44]

Homework Statement:: see attached
Relevant Equations:: data stats

See attached question and markscheme.

View attachment 298877

Solutions

View attachment 298878
View attachment 298879Now they give the Mode as $1$ i am not getting this...
My understanding 'maybe its the English used' is that we have $0$ children living in 3 houses, 7 houses have $1$ kid each bringing total number of kids to 7!, 5 houses have $2$ kids each in the household bringing total number of kids here to $10$, in my case the mode ought to be 2. I need some clarity here.
Also looking at this table, 4 kids live in 0 houses and 5 kids also living in 0 houses does not really make sense to me...why not just have (4+5=9) kids live in 0 houses?...

I think you're looking at it backwards. 0 children living in 3 houses, means that there are 3 houses with no kids. 4 and 5 kids living in 0 houses means that no houses had exactly 4 or 5 children.

For the mode, it might be helpful to draw a scatter graph of no. of children (horiz. axis) and no. of houses (vert. axis), so that number of houses is a function of the number of children. That would make 1 the mode, although this seems like a very odd problem to me.

You didn't ask about the median, but I agree with the answer of 1.5. For 0 and 1 child, there are a total of 10 houses. For 2 through 6, there are also a total of 10 houses. That would put the median right between 1 and 2, or 1.5 (children).

I haven't figured out how that got a mean of 1.75,

 

[pbuk]

I haven't figured out how that got a mean of 1.75,

Edit: $\frac{\sum \tt{children}{\sum {tt}houses{/tt}} = \frac{35}{20}$
$$ \frac{\sum \tt{children}} {\sum \tt{houses}} = \frac{35}{20} $$

With thanks to @Mark44 and @SammyS for $\LaTeX$ fixes.

 

[Mark44]

$\frac{\sum \tt{children}{\sum {tt}houses{/tt}} = \frac{35}{20}$

Nice. I hadn't given it any more thought since I posted. Needed to get chores done in the yard while it's not rainy.

Try this:

$\frac{\sum \text{children}}{\sum \text{houses}} = \frac{35}{20}$

 

[SammyS]

Actually, the \tt does work. It's short for \texttt which is the "Typewriter" family of font faces.

@pbuk had some other issues with the LaTeX coding.

## \frac{\sum \tt{children}}{\sum \tt{houses}} = \frac{35}{20} ##

gives: $\frac{\sum \tt{children}}{\sum \tt{houses}} = \frac{35}{20}$

- - - and No, I'm not a LaTeX expert. I found the \tt stuff on :

LaTeX Cheat Sheet, a two-page quick reference (by Winston Chang)

 

[Mark44]

It's short for \texttt which is the "Typewriter" family of font faces.

That's a new on on me. Thanks for the tip on the cheat sheet. I'll add it to my stash of LaTeX links.

 

[pbuk]

@pbuk had some other issues with the LaTeX coding.

Yes, and I still seem to be stuck with a broken MathJax in the cache so can't see what I am doing, thanks for sorting for me.

 

[Merlin3189]

Now they give the Mode as 1 i am not getting this... <and following>

Mode is, the most common number of children in a house (see below): 1 child in a house, occurred in 7 houses; no other number of children per house occurred more often. So mode =1 child in a house.

My understanding 'maybe its the English used' is that we have 0 children living in 3 houses, 7 houses have 1 kid each bringing total number of kids to 7!, 5 houses have 2 kids each in the household bringing total number of kids here to 10, in my case the mode ought to be 2. I need some clarity here.
Also looking at this table, 4 kids live in 0 houses and 5 kids also living in 0 houses does not really make sense to me...why not just have (4+5=9) kids live in 0 houses?...

And

For the mode, it might be helpful to draw a scatter graph of no. of children (horiz. axis) and no. of houses (vert. axis), so that number of houses is a function of the number of children. That would make 1 the mode, although this seems like a very odd problem to me.

I think all of this comes down to understanding the actual "experiment". 20 houses were sampled. The metric for each sample was the number of children in the house.
All the statistics are therefore, the number of children in a house (say CpH.) So;
the mode is the CpH that occurred at most sample points.
the median is the CpH of the house where half the sample points had more CpH and half had fewer CpH.
the mean is the CpH you'd get if all children were shared equally over all the sample points (ie. houses).

Mark has therefore got the median right, except that I would say, he should give units for his answer.

You didn't ask about the median, but I agree with the answer of 1.5. For 0 and 1 child, there are a total of 10 houses. For 2 through 6, there are also a total of 10 houses. That would put the median right between 1 and 2, or 1.5 (children).

Median is CpH, where half the houses have more CpH and half the houses have fewer CpH.
10 houses have 2 CpH or more, 10 houses have 1 CpH or fewer.
So the median lies between 1 CpH and 2 CpH - conventionally, I pick the mid point, 1.5 CpH.

Mean, or average number of children per house, is the total number of children (35) divided by the total number of houses sampled (20), so 1.75 CpH. If all the children found, were spread evenly among all the houses sampled, that would be the number of children per house.

The competing notion that the median should be the CpH, where half the children live in houses with more and half in houses with fewer, is IMO wrong here. Our sample points are houses, not children.
Our data, for the 20 sample points (houses), is ; 0,0,0,1,1,1,1,1,1,1,2,2,2,2,2,3,3,3,3,6 children per house, and so by inspection, mode = 1 CpH, median = (1 to 2 CpH )= 1.5 CpH, mean = 35 /20 = 1.75 CpH

If we had instead sampled 35 children and asked, what size household they lived in, and had gotten the same results, we would have had 35 sample points, 1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,2,2,3,3,3,3,3,3,3,3,3,3,3,3,6,6,6,6,6,6 CpH
Note, we get no data from houses which had no children! We are no longer sampling houses. Some houses with multiple children might be sampled multiple times.
Now, the mode is 3 CpH being reported by 12 children. Most children live in houses with 3 children.
The median is, as OP suggested, also 3 CpH. Half of children live in houses with 3 or less children, half live in houses with 3 or more.
The mean is now 99 / 35 = 2.8 CpH. I find this harder to describe, but children in this set have on average 2.8 children living in their house.
These stats are higher, because children in large families all get sampled individually, rather than collectively as part of one household.

This is not unexpected, since this is a different data set, that just happens to get summarised in a similar way when tabulated. Both have validity according to the experimental purpose and protocol. We were looking at a set of houses and wanting statistics for the CpH in that set of houses. The alternative, which might have been of interest to someone studying children rather than houses, would give us statistics for the home family sizes of a set of children. In fact, if the two sets of data had been obtained from exactly the same houses and children, the correct stats would still be different.