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Mathematics 12 Online
OpenStudy (roo777x):

Find the average mean? http://prnt.sc/dk06yu

OpenStudy (roo777x):

@3mar i know how to find the average (mean) but im not sure how on this one because the results never match up with the answer options.

OpenStudy (3mar):

What do you think?

OpenStudy (roo777x):

2 - 2.5 was the closest number i was getting

OpenStudy (roo777x):

@3mar i added all the numbers (equaling 1) and found the average (divided by 5) and the result was 0.2. Then i found the average of just the ones who HAD siblings and left out 0. But it still wasnt right

OpenStudy (3mar):

"Then i found the average of just the ones who HAD siblings and left out 0." What so you mean by that?

OpenStudy (3mar):

What do x and p(x) stand for?

OpenStudy (roo777x):

i didnt include the decimal of those who did not have siblings

satellite73 (satellite73):

good idea multiply and add

OpenStudy (roo777x):

multiply what?

satellite73 (satellite73):

the probabilities by the numbers

satellite73 (satellite73):

\[1\times 0.6+2\times 0.05+3\times 0.15+4\times 0.1\]

OpenStudy (roo777x):

@satellite73 1.55 - would it mean B or C?

satellite73 (satellite73):

dunno i cannot read the mind of the person who made the question, but given the \(4+\) at the end of the table, i would round up

satellite73 (satellite73):

i would not bet more than $7 on that answer though

OpenStudy (mathmale):

Does it tell you anything that the greatest probability by far is 0.6? Under what number (of siblings) do you find that? Notice the prob. to the left of 0.6 is 0.1 and that to the right of 0.6 is 0.05. What do you suppose the graph of these probabilities would look like? Would it be possible to infer the mean of all this data just by looking at the given table or the graph that you would draw?

OpenStudy (roo777x):

i dont know, ill try

OpenStudy (roo777x):

@mathmale i did but i dont think it means anyhting

satellite73 (satellite73):

is it not clear how to do this?

satellite73 (satellite73):

if not that is fine, tell me and i will show you how to do it

OpenStudy (roo777x):

i found the average, though i guess not the right way because i cant get the correct result. So yeah if you could thatd be great

satellite73 (satellite73):

ok lets imagine that there are 100 students in the class

OpenStudy (mathmale):

There is a simple procedure for finding the average value; it consists of multiplying each possible sibling count (0, 1, 2, .. ) by the probability that appears underneath each such count. Then you add the products together. Unfortunately, that 4+ creates a problem, since that 4+ could represent 4, 5, 6, 7, ... On the other hand... the value 1 appears many times more often than does any other value. Seems to me that that tells us immediately that the average number of siblings is 1 or very close to 1.

satellite73 (satellite73):

i messed up it is \[1\\ 0.6\] so \(60\%\) of the students have on sibling

satellite73 (satellite73):

what is \(60\%\) of 100? (don't think too hard)

OpenStudy (roo777x):

3/5 = 0.6

satellite73 (satellite73):

yeah i was asking what \(60\%\) of 100 was, not what .6 is as a decimal

satellite73 (satellite73):

i see the problem i just computed it and got 1.55 too, which is not an answer choice so you have to choose between 1.5 and 2

satellite73 (satellite73):

i.e. round up or down like i said, i would round up but that is just a guess

OpenStudy (roo777x):

exactly, thats where im at

OpenStudy (roo777x):

thank you @satellite73 and @mathmale

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