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we have 100 families: 10 families have no children, 40 families have 1 child for each one, 30 families have 2 children for each one, 10 families have 3 children for each one and 10 families have 4 children for each one

A- suppose we selected a family randomly, what is the expected number of children in that family? B-suppose we selected a child randomly, what is the expected number of children in his family? is it the same?

I've tried the expectation to solve it but only for "A" , for "B" I didn't know how to solve it

the answer for A is 1.7 children but it is wrong because prof said the correct one is 3.

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  • $\begingroup$ How many children are there all in all? How many of them have no siblings? How many have one sibling? Two? Three? $\endgroup$ – lulu Apr 22 at 17:54
  • $\begingroup$ A) How many children in all? How many families? You're done! $\endgroup$ – David G. Stork Apr 22 at 17:56
  • $\begingroup$ What do you mean by "the answer for A is 1.7 (2) children"? Is this a multiple-choice question, but you forgot to tell us what the options are? $\endgroup$ – TonyK Apr 22 at 18:53
  • $\begingroup$ I meant my answer is 1.7 so Approximately it's 2 kid ( what is the 0.7 of a child is it his arm or leg? so Approximately it is 2 kids) anyways my prof said the correct answer is NOT 2 $\endgroup$ – Nidal Apr 22 at 18:57
  • $\begingroup$ There is no reason that an average number of children has to be a whole number. $\endgroup$ – Ross Millikan Apr 22 at 19:39
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For (B), the mean is $S/T$ where

$S = 10(0^2) + 40(1^2) + 30(2^2) + 10(3^2) + 10(4^2) = 410$

$T = 10(0) + 40(1) + 30(2) + 10(3) + 10(4) = 170$

so the mean is $410/170 = 2.412$

The idea is, for example, a kid from a $3$ child family contributes a count of $3$ to the sum of family sizes, and that is the case for all $10*3$ such children. So those kids contribute a total of $10(3^2)$ to the total. Etc. Denominator is the total number of kids.

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  • $\begingroup$ I solved the problem a different way and got the same final answer as you for (B)(look at the answer above yours), and your solution looks alot more efficient. Although, I still can't seem to understand why you squared the number of children in a group then divided the sum of all that by the total number of children. $\endgroup$ – Allan Henriques Apr 22 at 20:22
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    $\begingroup$ @AllanHenriques imagine asking every one of the children "how many children in your family?" There are, e.g., 10*3=30 kids who would answer "3" so those 30 3's contribute 3*30 =90 = 10(3*3) to the "sum of answers". Etc. The mean in B is the "average answer", i.e. the sum of all those answers divided by the total number of answers, i.e. divided by the total number of children. I think your method is must amount to the same thing except that your final value is exactly 10 times too big, so I'm not sure what you did. $\endgroup$ – Ned Apr 23 at 0:16
  • $\begingroup$ I found the average number of families in the grouping that will be selected, The average number is 24, which is closer to the group with 30 families. The group with 30 families has 2 children per family. But your solution works better because it is more intuitive and understandable $\endgroup$ – Allan Henriques Apr 23 at 0:37
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Intuitively, the expected number of children must be higher when you select a child because you are more likely to select a family with lots of children. In particular, if you select a child, you will never select the families with no children.

For each question, you just add up the product of the number of children and the probability of selecting a family of that size. For A, you have $0.1$ chance of selecting a family with no children, $0.4$ chance of selecting a family with $1$ child and so on. The expected number is then $$0.1\cdot 0 + 0.4 \cdot 1 +0.3 \cdot 2 + 0.1 \cdot 3 + 0.1 \cdot 4=1.7$$

For B, what fraction of the children are in families of $4$ children? That is the probability you select a family of $4$. The approach is the same as part A, but the probabilities are different. Over to you.

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  • $\begingroup$ I think you have a mistake with the numbers. it should be 0.4*1 (not 0.4*4) $\endgroup$ – adhg Apr 22 at 18:19
  • $\begingroup$ @adhg Correct. Fixed. Thanks. $\endgroup$ – Ross Millikan Apr 22 at 18:23
  • $\begingroup$ I did the same, but my prof said it's wrong, he gave me the answer for A and it is 3 children, not 1.7 $\endgroup$ – Nidal Apr 22 at 18:28
  • $\begingroup$ The average for A is clearly less than $3$, so your professor is wrong. I get $\frac {410}{170}\approx 2.412$ for B. $\endgroup$ – Ross Millikan Apr 22 at 19:18
  • $\begingroup$ The mean for B must be bigger than A, since the distribution in B, relative to that in A, is biased toward larger families. In general the distribution in B is called the "size biasing" of A, and it's mean is always bigger (unless A is a constant distribution in which case the means are the same). $\endgroup$ – Ned Apr 22 at 19:32
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I got the same for part A

for part B I made chart and pooled the number of children for each family grouping, then I calculated what percentage of the total child pool a certain family made up, then I multiplied the number of families in the grouping by the percentage, then I added up all the numbers.

*family grouping refers to the number of families with a certain number of children

format: family grouping, # of children in grouping, percentage of total child pool, family grouping multiplied by percentage

10, 0, 0%, 0

40, 40, 23.53%, 9.412

30, 60, 35.29%, 10.587

10, 30, 17.65%, 1.765

10, 40, 23.53%, 2.353

otal 170, 100% 24.117

10<< 24.117-family grouping <30

30-family group= 2 kids

from part a)

1<< 1.7 kids <2

2 kids

so the number of kids is the same in A) and B) even though the probabilities change slightly

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  • $\begingroup$ prof said the correct one is 3 . $\endgroup$ – Nidal Apr 22 at 19:01
  • $\begingroup$ @NidalAlSaqqa Since he is a professor, he is probably more reliable than a single random guy or a highschool student having a different answer. But the fact that 3 people got the same answer of 1.7 while having bias against the answer 1.7 (since the professor said otherwise), our probability of being right is increased drastically. So bring your solution to classmates and then your teacher. If we are wrong after all this, come back and show us why. $\endgroup$ – Allan Henriques Apr 22 at 19:06
  • $\begingroup$ 10<< 24.117-family grouping <30 30-family group= 2 kids _____________ I didn't get what you do ? the total was 24 , then 2 ?? $\endgroup$ – Nidal Apr 22 at 21:02
  • $\begingroup$ @NidalAlSaqqa I found the average number of families in the grouping that will be selected, the average number of families in the selected group is 24.117. A family group with 24.117 families does not exist, and rightly so, because there is never a 100% chance of selecting a certain family. So we need to round the number. This number is closer to 30 families. Thirty corresponds to the group that has 2 children per family, which is the same answer we got in question (A). btw I'm using the << to mean much greater than. $\endgroup$ – Allan Henriques Apr 22 at 21:18
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    $\begingroup$ thanks alot that was helpful , yet I'm waiting for ned to explain what he did , you both got the same result $\endgroup$ – Nidal Apr 22 at 21:28

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