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D. Translation between N(0,1) variables and N(?,?) variables We have the following NISNID ("Normal Is Standard Normal In Disguise") Fact, which we present without proof (but which is not hard to show, using the above formula for the N(?,?) pdf f??(2)): \frac{Z - ?}{?} NISNID Fact. If Z is N(?,?), then is N(0,1). In stats texts, you will typically find tables of N(0,1) variables, but not other N(?,?) variables. Now we know why: we can compute probabilities associated with N(?,?) random variables if all we know are probabilities associated with N(0,1) random variables. Here's an example showing how. Example. Suppose Z is N(8,1.5). Find P(5 < z < 11). Solution. Since, in this case, ? = 8 and ? = 1.5, we have P(5 < z < 11) = P(\frac{5 - 8}{1.5} < \frac{z - 8}{1.5} < \frac{11 - 8}{1.5}) = P(-2 < \frac{z - 8}{1.5} < 2) = 0.955. The last step is by the NISNID Fact, and by exercise B1(b) above. Using the strategy of the above example (and using part B above where necessary), complete the following exercises. Exercise D1. Suppose Z is N(-2,0.3). Find P(-2.3 < z < -1.7). Exercise D2. Suppose Z is N(2,2). Find P(-1.92 < z < 5.92).

          D. Translation between N(0,1) variables and N(?,?) variables
We have the following NISNID ("Normal Is Standard Normal In Disguise") Fact, which we present
without proof (but which is not hard to show, using the above formula for the N(?,?) pdf f??(2)):
\frac{Z - ?}{?}
NISNID Fact. If Z is N(?,?), then  is N(0,1).
In stats texts, you will typically find tables of N(0,1) variables, but not other N(?,?) variables.
Now we know why: we can compute probabilities associated with N(?,?) random variables if all
we know are probabilities associated with N(0,1) random variables.
Here's an example showing how.
Example. Suppose Z is N(8,1.5). Find P(5 < z < 11).
Solution. Since, in this case, ? = 8 and ? = 1.5, we have
P(5 < z < 11) = P(\frac{5 - 8}{1.5} < \frac{z - 8}{1.5} < \frac{11 - 8}{1.5})
= P(-2 < \frac{z - 8}{1.5} < 2) = 0.955.
The last step is by the NISNID Fact, and by exercise B1(b) above. Using the strategy of the
above example (and using part B above where necessary), complete the following exercises.
Exercise D1. Suppose Z is N(-2,0.3). Find P(-2.3 < z < -1.7).
Exercise D2. Suppose Z is N(2,2). Find P(-1.92 < z < 5.92).

D. Translation between N(0,1) variables and N(?,?) variables
We have the following NISNID ("Normal Is Standard Normal In Disguise") Fact, which we present
without proof (but which is not hard to show, using the above formula for the N(?,?) pdf f??(2)):
(Z - ?)/(?)
NISNID Fact. If Z is N(?,?), then is N(0,1).
In stats texts, you will typically find tables of N(0,1) variables, but not other N(?,?) variables.
Now we know why: we can compute probabilities associated with N(?,?) random variables if all
we know are probabilities associated with N(0,1) random variables.
Here's an example showing how.
Example. Suppose Z is N(8,1.5). Find P(5 < z < 11).
Solution. Since, in this case, ? = 8 and ? = 1.5, we have
P(5 < z < 11) = P((5 - 8)/(1.5) < (z - 8)/(1.5) < (11 - 8)/(1.5))
= P(-2 < (z - 8)/(1.5) < 2) = 0.955.
The last step is by the NISNID Fact, and by exercise B1(b) above. Using the strategy of the
above example (and using part B above where necessary), complete the following exercises.
Exercise D1. Suppose Z is N(-2,0.3). Find P(-2.3 < z < -1.7).
Exercise D2. Suppose Z is N(2,2). Find P(-1.92 < z < 5.92).

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