Can you see if you can get an answer to this question? X is distributed normally, P(X>59.1)=0.0218 and P(X>29.2)=0.9345. Find the mean and standard deviation of the distribution to 3 s.f

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Can you see if you can get an answer to this question? X is distributed normally, P(X>59.1)=0.0218 and P(X>29.2)=0.9345. Find the mean and standard deviation of the distribution to 3 s.f

Mathematics
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Because its P(X>a) you need to do 1-P to get correct Z-values Z1 = (59.1- m)/s Z2 = (29.2 -m)/s you will have to solve system of 2 equations using substitution
Yes, but is there such thing as P(X>29.2)=-0.9345?
no, probabilities must be within 0 and 1

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i get m = 42 s = 8.47
that's right.. How did you get that?
by solving the system of equations you should get Z1 = 2.018 Z2 = -1.51 right
Wait, how did you get Z2?
looked up 1-.9345
which is 0.0655... But I don't have that value on my distribution table?
just round it to the closest one.. 0.066 if you have excel, use "Norminv" function, it will give you the z-values as well
It starts at 0.5? And I can't use excel in my exam, so... how can I find the value otherwise?
it doesn't show probabilities less than 0.5 ??
No, it doesn't.
ok look up .9345 and flip the sign of the z-value
No
It's 1.51
So, it'd be -1.51?
yeah, you can do that because the distribution is symmetric
Ah Ok. Thanks... So from there, what are the two equations which result?
2.018 = (59.1- m)/s -1.51 = (29.2 -m)/s solve one for s --> s = (59.1-m)/2.018 plug that into 2nd equation to find m
Ok. Thanks!!!
yw

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