Thanks a lot guys you have no idea how much I appreciated it! 
I have a few more assignments that I could use some help with if anyone knows the answer
4-17) Find z such that P(Z>z) = 0.12
The answer is 1.175
I don't know how to start with this one. First of all, if Z is the standard normal distribution, then what is z? I'm not really sure what they mean when they write Z>z or X>x. Else what should I do?
The Z value 0.12 gives a probability of 0.0478
If I need to find the Z value where the probability is 0.12, then it's between 0.30 (0.1179) and 0.31 (0.1217)
4-39 For a normal random variable with mean 16.5 and standard deviation 0.8, find a point of the distribution such that there is a 0.85 probability that the value of the random variable will be above it.
What should I do here?
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Solved!
4-21. The deviation of a magnetic needle from the magnetic pole in a certain area in northern Canada is a normally distributed random variable with the mean 0 and standard deviation 1. What is the probability that the absolute value of the deviation from the north pole at a given moment will be more than 2.4?
Ok, since mean is 0 and std. dev. is 1, it's a Z distribution. So they are asking "what is the chance of Z>2.4" right? I can then look in my table and find P(Z>2.4) = 0.4918. But since the table measures from 0 to infinity, we have to add 0.5 to our number. So 0.5+0.4918=0.9918. So the area from minus infinity to 2.4 is 99.18%, but we are asked to find the chance of it being larger than 2.4. So we take the inverse of it and find 1-99.18 = 0.0082%. Right now I would say I'm done but the result is supposed to be 0.0162%. I can see that if I multiply my result then I'll get that number, but I don't see why I have to do so Have I done something wrong?
4-33 - Fluctations on a French CAC-40 stock index from march to june 1997 seem to follow a normal distribution with mean 2600 and standard deviation of 50. Find the probability that the CAC-40 stock
will be between 2520 and 2670 on a random day in the period of study.
Here I'll transform X to Z using P(a < X < b) = P (a -μ / σ < Z < b - μ / σ
So 2520-2600/50 = -0.8
and 2670-2600/50 = 1.4
Meaning P(-0.8 < Z < 1.4)
Now this should be easy here on, but I can't get the right result.
I look in my table and find 0.8 = 0.2881 and 1.4 = 0.4192. So by adding the numbers I should get the area between -0.8 and 1.4, right? I get 0.7073 but I should get 0.8644
I have a few more assignments that I could use some help with if anyone knows the answer 4-17) Find z such that P(Z>z) = 0.12
The answer is 1.175
I don't know how to start with this one. First of all, if Z is the standard normal distribution, then what is z? I'm not really sure what they mean when they write Z>z or X>x. Else what should I do?
The Z value 0.12 gives a probability of 0.0478
If I need to find the Z value where the probability is 0.12, then it's between 0.30 (0.1179) and 0.31 (0.1217)
4-39 For a normal random variable with mean 16.5 and standard deviation 0.8, find a point of the distribution such that there is a 0.85 probability that the value of the random variable will be above it.
What should I do here?
------------------------------------------------------------------------------
Solved!
4-21. The deviation of a magnetic needle from the magnetic pole in a certain area in northern Canada is a normally distributed random variable with the mean 0 and standard deviation 1. What is the probability that the absolute value of the deviation from the north pole at a given moment will be more than 2.4?
Ok, since mean is 0 and std. dev. is 1, it's a Z distribution. So they are asking "what is the chance of Z>2.4" right? I can then look in my table and find P(Z>2.4) = 0.4918. But since the table measures from 0 to infinity, we have to add 0.5 to our number. So 0.5+0.4918=0.9918. So the area from minus infinity to 2.4 is 99.18%, but we are asked to find the chance of it being larger than 2.4. So we take the inverse of it and find 1-99.18 = 0.0082%. Right now I would say I'm done but the result is supposed to be 0.0162%. I can see that if I multiply my result then I'll get that number, but I don't see why I have to do so
Have I done something wrong?4-33 - Fluctations on a French CAC-40 stock index from march to june 1997 seem to follow a normal distribution with mean 2600 and standard deviation of 50. Find the probability that the CAC-40 stock
will be between 2520 and 2670 on a random day in the period of study.
Here I'll transform X to Z using P(a < X < b) = P (a -μ / σ < Z < b - μ / σ
So 2520-2600/50 = -0.8
and 2670-2600/50 = 1.4
Meaning P(-0.8 < Z < 1.4)
Now this should be easy here on, but I can't get the right result.
I look in my table and find 0.8 = 0.2881 and 1.4 = 0.4192. So by adding the numbers I should get the area between -0.8 and 1.4, right? I get 0.7073 but I should get 0.8644
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