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Trisaba
 one year ago
P(x> or = x line over + 2 standard deviation) medal and fan
Trisaba
 one year ago
P(x> or = x line over + 2 standard deviation) medal and fan

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Trisaba
 one year ago
Best ResponseYou've already chosen the best response.0i got one more two similar

Trisaba
 one year ago
Best ResponseYou've already chosen the best response.0x line over means there is a line over the x

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[P(x\ge \bar{x}+2\sigma)~~?\] How is \(x\) distributed? If you're given a normal distribution, this is straightforward if you know the empirical rule: https://en.wikipedia.org/wiki/68%E2%80%9395%E2%80%9399.7_rule

Trisaba
 one year ago
Best ResponseYou've already chosen the best response.0it has no labels they want like 68% is 34% + 34%

Trisaba
 one year ago
Best ResponseYou've already chosen the best response.0we have to use the normal ditribution example ones from this line to this

Trisaba
 one year ago
Best ResponseYou've already chosen the best response.04 and 5 sorry wrong numbers

Trisaba
 one year ago
Best ResponseYou've already chosen the best response.0what am i supposed to do

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1441842345311:dw From the rule, you have that \[\begin{cases} D+E=68\\ C+D+E+F=95\\ B+C+D+E+F+G=99.7\\ A+B+C+D+E+F+G+H=1 \end{cases}\] The distribution is symmetric, so you can determine that \[\begin{cases} D=E=34\\ C=F=13.5\\ B=G=2.35\\ A=H=0.15 \end{cases}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So for your first problem, \[P(x\ge \bar{x}+2\sigma)=G+H=\cdots\]

Trisaba
 one year ago
Best ResponseYou've already chosen the best response.0x with the line over it is the mean

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I have \(\bar{x}\) labeled on the graph above in the middle of the curve.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0All I've done is rephrase the problem from \(P(x\ge\bar{x}+2\sigma)\) to the simple operation \(2.35+0.15\). The areas of the regions are probabilities.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The area of the shaded region is the probability you're looking for: dw:1441843152498:dw Are you wondering how I found \(G=2.35\) and \(H=0.15\)?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well, we have \[B+C+D+E+F+G=99.7\] so substituting this into the total probability equation (the last equation in the system above), we get \[\begin{align*}1&=A+B+C+D+E+F+G+H\\ 1&=A+99.7+H\\ 0.3&=A+H\\ 0.3&=A+A\\ 0.3&=2A\\ 0.15&=A=H \end{align*}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\(x\) is completely symbolic here. It's referring to some event or characteristic that occurs with probability modeled by the curve. It can take on any value in the domain of the distribution. Let's say we're talking about the heights of a large group of people. \(\bar{x}\) is the average of height, while \(x\) can be any height that is seen among the group of people. So as an example, the statement \(P(x=5)\) is the same as "the probability that any given person's height is \(5\) (units)". Meanwhile, \(P(x>5)\) means "the probability that any given person's height is greater than \(5\)", and so on. Does that make sense?

Trisaba
 one year ago
Best ResponseYou've already chosen the best response.0so the answer is a group of probabilities?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The answers are the sums of the areas that belong in the desired intervals. I've already shaded the one you're interested in for (4). (5) is just as easy: dw:1441844287564:dw or \(A+B+C+D+E\).

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[P(x\ge \bar{x}+2\sigma)=G+H\]

Trisaba
 one year ago
Best ResponseYou've already chosen the best response.0wait actually i forgot 6 too

Trisaba
 one year ago
Best ResponseYou've already chosen the best response.0please @SithsAndGiggles
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