Please Help! @Loser66 @Vandreigan @iPwnBunnies
With what you know about probabilities, which one makes the most sense for question 1?
C
The survey revealed that 20/50 liked fall, while only 6/50 liked winter. So, I probably wouldn't say that there is the same amount. That's quite a large difference within 50 data points!
D, i didnt read c carefully
I'd agree with D :)
yeh
For question 2, I'm not familiar with that graphing style. I need to find out exactly what it means (specifically, what the box represents. I get the impression that it represents some number of standard deviations, but I need to know for sure). Stay with me. Hopefully this won't take but a moment.
So, you may want to get some clarification on this one, but I'd go with B. The girls values show more variation, as illustrated by the larger range. However, they show a lower standard deviation (if you've covered this), as they are more focused around one region.
ok
The next one reads like they want you to list all possible combinations of heads and tails that can be made by tossing 4 coins at once. Since you are tossing them all at once, you lose track of which quarter is which (order doesn't matter), so it's a bit easier than before. 4H 3H 1T etc. If they want you to keep track of each quarter, it would look like: HHHH HHHT HHTH HTHH THHH etc. through all the possible combinations.
ok
i got 10 combinations
Which way are you doing it? Either way, 10 shouldn't be your answer.
a list
what did you get
If order doesn't matter: 5 If order does matter: 16
ok
for 16 what would be the probability
of what? All we've done is count the total number of combinations possible.
oh part b
So, how many different ways can the quarters fall to leave two heads and two tails? Example: HHTT, HTHT, etc.
4
HHTT HTHT THHT HTTH THTH TTHH
6 ways, with a total of 16 combinations. Which gives us what probability?
6 out of 16
Reduce it, and you have your answer :)
3/8
number 4
Hmm, some terminology that I'm not familiar with. One second.
ok
I can't find anything solid for that one. I've never heard of a "mean average distribution" before.
I'd suggest trying to get the attention of someone else to help you on what's left. Unfortunately, I have to go. Best of luck to you!
k
can u just help with number 5 part a
a and b
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