# normal distribution

#### Jellekool

##### New Member
Hi everyone!

I have a question about the answer of question 19a and e. question a: Find the probability that X is greater than 60. Why do they do 0,5 -0,3944 and not 1 - ...?
Question e was: the probability is 0,05 that X is in the symmetric interval about the mean between which two numbers? additional info: mean is 50 and sigma squared is 64. I don't really understand what I have to do here. Can anyone help me out?

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#### Dason

I'm guessing the z-table they use gives the area between 0 and whatever value you want. So if it asked for find P(0 < Z < some_value) then when you look up some_value in the table it will give you that probability directly. There are many different ways these tables get made so you just need to make sure you know what the particular table you're using actually does.

For your second question they're asking what two values are such that 1) they are symetric around the mean (so they have the same distance from the mean) and 2) the probability of being between those two values is .05. So essentially they asking you to solve for 'a' and 'b' in P(a < X < b) = 0.05 where a and b the same distance from the mean. Another way to put that is to solve for c in the following P(50 - c < X < 50 + c) = 0.05

#### Jellekool

##### New Member
thank you very much, but I still don't really understand the second question..

Can someone write it out step by step? #### obh

##### Active Member
symmetric around the 0.5, so half probability to the right and half probability to the left, the total is 0.05

p(z<=Z1) - p(z<=Z2)=0.05

0.5+(0.05/2)=0.525
0.5-(0.05/2)=0.475

p(z<=Z1)=0.525 => Z1 = 0.0627061 => (x1-50)/8=~0.06 => x1=50.48
p(z<=Z2)=0.475 => Z2 = -0.0627061 => (x2-50)/8=~0.06 =>x2=49.52

Inv Z(0.525)= 0.0627061 =>
Inv Z(0.475)= 0.0627061 =>

P( x ≤ 50.501649 ) = 0.525
P( x ≤ 49.498351 ) = 0.475

Last edited:

#### John Chifofo

##### New Member
Let me help with question (e)

The question says that the probability is 0.05 that X is in the symmetrical interval about the mean.

From the above statement, X's middle point is the mean meaning that they have the same distance from the mean. We are therefore safe to say that from the mean(0) to the left the probability is 0.025 and that from the mean(0) to the right the probability is 0.025. 0.025+0.025=0.05.

0.5-0.025= 0.475
When you check on your z-table, 0.475 is in row 0.0 and column 0.06. our z is therefore 0.06.
Since it's in the symmetrical interval about the mean, it's a reflection. We thus have -0.06.
P(-0.06<z<0.06) we then now unstandardize it.

x-50(divided by √64) < 0.06
x-50 < 0.48
x < 50.48

-0.06 < x-50(divided by √64)
-0.48 < x-50
49.52 < x

P(49.52 < x < 50.48)=0.05