
#1
Mar212, 03:39 PM

P: 216

TITLE: Normal distributions (sorry)
Question 1: Working: Now the answer for sigma i got correct, but μ i got incorrect for some reason. Could anyone explain where i've gone wrong for μ? Question 2: Workings: All i need help is with part c and d: for c) i done 1  0.6463 to get the probability of the two shaded bits, but where do i go to get the part they are asking for? for d) i have no idea, any help would be great. thanks. 



#2
Mar212, 05:20 PM

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Well, one obvious error is that you have both [itex](25\mu)/\sigma[/itex] and [itex](64\mu)/\sigma[/itex] equal to 0.67. One of them should be positive. In any case, the normal distribution is symmertric so it should be clear that the mean will be exactly between the two "quartiles".
The two shaded bits? I assume you mean the one shaded bit and the other outlieing area. That, of course, is 1 .6463. To find the two parts separately, use the fact, as you are given, that upper piece has twice the area of the lower piece. 



#3
Mar312, 02:06 AM

P: 216

for question 2) I still dont get how to get part C. 



#4
Mar312, 02:48 PM

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High school Normal distribution two questionsRGV 



#5
Mar312, 03:05 PM

P: 216





#6
Mar312, 04:14 PM

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The distance between the mean and the 25th percentile is the same as the distance between the mean and the 75th percentile, because of symmetry. Therefore, the mean is the average of the 25th and 75th percentiles. RGV 



#7
Mar312, 05:48 PM

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#8
Mar312, 06:03 PM

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RGV 



#9
Mar312, 06:19 PM

P: 216

for c i done (10.6463)/3 as there is 3 parts, with the shaded region being the smallest, which gets me the correct value. For D I've standardized it but i'm not sure what value i should put it against... 



#10
Mar412, 11:35 AM

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Never mind trying to understand whether (45  mu) / sigma is 0.67, just tell me: what do YOU think the value of (45  mu)/sigma should be? Remember, read the whole question carefully before answering. RGV 



#11
Mar512, 03:17 PM

P: 216

thanks for continuing to help. 



#12
Mar612, 11:26 AM

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Your "equation" P(z>(45mu)/sigma) = 0.75 is wrong, but would be OK if you replaced the 0.75 by ___ ? (I'm leaving it to you to fill in the blank.) Always, always, draw a picture. RGV 



#13
Mar612, 11:43 AM

P: 216

by 0.25? so P(Z>(45mu)/sigma) = 0.25 ? Is this correct?




#14
Mar612, 11:50 AM

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RGV 



#15
Mar612, 11:52 AM

P: 216




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