Need Help fast!!!
use the following table to answer the questioin...
X = P(X)
A = 0.30
B= 0.20
C = ?
D = 0.28
Find P(A and D) assuming that A and D are independent.
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jimthompson5910 (jim_thompson5910):
P(A and D) = P(A) * P(D) where events A and D are independent
jimthompson5910 (jim_thompson5910):
So just multiply the corresponding probabilities
OpenStudy (anonymous):
0.30 * 0.28 = 0.084
jimthompson5910 (jim_thompson5910):
you nailed it
OpenStudy (anonymous):
Thanks! :)
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jimthompson5910 (jim_thompson5910):
yw
OpenStudy (anonymous):
what about this one?
you toss 2 coins. find P ( at least 1 tail is tossed)
jimthompson5910 (jim_thompson5910):
What's the sample space look like? Any ideas?
OpenStudy (anonymous):
TT TH HT HH
jimthompson5910 (jim_thompson5910):
good
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jimthompson5910 (jim_thompson5910):
How many have at least one tail?
In other words, how many have one T or two Ts?
OpenStudy (anonymous):
3
jimthompson5910 (jim_thompson5910):
out of how many?
OpenStudy (anonymous):
4
jimthompson5910 (jim_thompson5910):
Good, so
P ( at least 1 tail is tossed) = 3/4
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OpenStudy (anonymous):
but since their are 2 coins, wouldn't it make it 6/8?
jimthompson5910 (jim_thompson5910):
no, there are only 4 possible ways to toss 2 coins
jimthompson5910 (jim_thompson5910):
and there are only 3 ways to get at least one tail
jimthompson5910 (jim_thompson5910):
besides, 6/8 reduces to 3/4 anyway
OpenStudy (anonymous):
oh okay :) for some reason, it keeps giving me the answer, 7/8??
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jimthompson5910 (jim_thompson5910):
really? hmm odd
jimthompson5910 (jim_thompson5910):
are you sure that it isn't 3 coins?
jimthompson5910 (jim_thompson5910):
if you toss 3 coins, then P(at least one tail) = 7/8
OpenStudy (anonymous):
ohhhh, I see now :P
jimthompson5910 (jim_thompson5910):
this is because the total possible ways to toss 3 coins are
HHH
HHT
HTH
HTT
THH
THT
TTH
TTT
and 7 of these total 8 have at least one tail.
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OpenStudy (anonymous):
ahhh, okay :) hmmm what about this?
When tossing 2 coins, P ( at least 1 head) = P(not all tails) True or False?
jimthompson5910 (jim_thompson5910):
Well we know that P ( at least 1 head) = 3/4 from last time right?
OpenStudy (anonymous):
yes
jimthompson5910 (jim_thompson5910):
What about P(not all tails)?
OpenStudy (anonymous):
= 1/4
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jimthompson5910 (jim_thompson5910):
what is P(all tails)?
OpenStudy (anonymous):
0/4?
jimthompson5910 (jim_thompson5910):
Remember, we have
HH
HT
TH
TT
How many of the 4 are all tails?
OpenStudy (anonymous):
4
jimthompson5910 (jim_thompson5910):
HH has all tails?
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OpenStudy (anonymous):
so 3?
jimthompson5910 (jim_thompson5910):
HT is all tails?
OpenStudy (anonymous):
1?
jimthompson5910 (jim_thompson5910):
good, only TT is all tails
jimthompson5910 (jim_thompson5910):
so P(all tails) = 1/4
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jimthompson5910 (jim_thompson5910):
Now we can use this idea to find P(not all tails)
OpenStudy (anonymous):
2/4
jimthompson5910 (jim_thompson5910):
close, but not quite
jimthompson5910 (jim_thompson5910):
P(not all tails) = 1 - P(all tails)
OpenStudy (anonymous):
0?
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jimthompson5910 (jim_thompson5910):
P(not all tails) = 1 - P(all tails)
P(not all tails) = 1 - 1/4
P(not all tails) = ???
OpenStudy (anonymous):
0.75
jimthompson5910 (jim_thompson5910):
or 3/4
jimthompson5910 (jim_thompson5910):
so because P(not all tails) = 3/4 and P ( at least 1 head) = 3/4, we can say that
P ( at least 1 head) = P(not all tails)
OpenStudy (anonymous):
So that answer is indeed True
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jimthompson5910 (jim_thompson5910):
you got it
OpenStudy (anonymous):
Thanks for your help! Really appreciate it! :)
jimthompson5910 (jim_thompson5910):
you're welcome
OpenStudy (anonymous):
I have trouble figuring these types of questions out, could you help? his is an example of which of the following sampling techniques? HELP?
A chef wants to open a new Italian restaurant. He wishes to find out which kinds of pasta (spaghetti, linguini, or penne) will be the most popular with potential customers.
He hires a polling organization to survey the market. The company builds a telephone survey list by selecting one person at random from the first page of the telephone book, who turns out to be the 123rd person on the page. The company then fills out the list by adding the 123rd person on every page of the book.
A.Systematic sample
B.Stratified sample
C.Simple random sample (SRS)
D.Cluster sample
E. Convenience sample
jimthompson5910 (jim_thompson5910):
Any ideas?
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OpenStudy (anonymous):
I'm thinking its systematic because, hes choosing the 123rd person, and then 123td person on every page of the book?
jimthompson5910 (jim_thompson5910):
you got it, very nice
OpenStudy (anonymous):
thanks !
jimthompson5910 (jim_thompson5910):
yw
OpenStudy (anonymous):
How do you find the mean, using a histogram?
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jimthompson5910 (jim_thompson5910):
each value under the histogram represents a value in the data set
the heights of each histogram represent how frequent each value in the data set occurs
making sense so far?
OpenStudy (anonymous):
Yes, and it also says this histogram as a class width of 1
jimthompson5910 (jim_thompson5910):
So you simply multiply each value under the histogram by the corresponding height...then you add up each product
Finally, you divide that last result by the total number of values (ie the sum of the frequencies)
OpenStudy (anonymous):
Okay, so the Height is 5.....
and the width is 10 but there are only 9 being used.
so 1 * 2
3 * 4
4 * 5
5 * 4
6 * 2
7 * 1
8 * 1
9 * 1
10 * 1
jimthompson5910 (jim_thompson5910):
Multiply all those out, add up the results and divide that by the sum of the heights
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OpenStudy (anonymous):
632 / 15
Is this right?
jimthompson5910 (jim_thompson5910):
hmm I'm getting something different, but that could mean that I'm missing something (since I can't see the histogram)
OpenStudy (anonymous):
well, I'll see if I can post it here... but basically it has a height of 5, and a width of 10 but only nine blocks are used.