Question

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

Answer 1

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

Answer 2

Yes, but is there such thing as P(X>29.2)=-0.9345?

Answer 3

no, probabilities must be within 0 and 1

Answer 4

i get
m = 42
s = 8.47

Answer 5

that's right.. How did you get that?

Answer 6

by solving the system of equations

you should get
Z1 = 2.018
Z2 = -1.51
right

Answer 7

Wait, how did you get Z2?

Answer 8

looked up 1-.9345

Answer 9

which is 0.0655... But I don't have that value on my distribution table?

Answer 10

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

Answer 11

It starts at 0.5?
And I can't use excel in my exam, so... how can I find the value otherwise?

Answer 12

it doesn't show probabilities less than 0.5 ??

Answer 13

No, it doesn't.

Answer 14

ok look up .9345 and flip the sign of the z-value

Answer 15

No

Answer 16

It's 1.51

Answer 17

So, it'd be -1.51?

Answer 18

yeah, you can do that because the distribution is symmetric

Answer 19

Ah Ok. Thanks... So from there, what are the two equations which result?

Answer 20

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

Answer 21

Ok. Thanks!!!

Answer 22

yw