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\centerline{MONT 105Q -- Mathematical Journeys}
\centerline{Discussion -- Using the Standard Normal Table}
\centerline{March 21, 2016}
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\noindent
{\it Background}
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Right-hand tail probabilities for a standard normal random 
variable (i.e. normal distribution with $\mu = 0$, $\sigma^2 = 1$) 
are given in the table on the attached sheet.  For instance, 
The entry for $z = 1.43$ is $.0764$.  This means that 
$$P(Z > 1.43) = .0764.$$  
From this we could also derive facts such as 
$$P(Z \le 1.43) = 1 - .0764 = .9236.$$
Moreover, by the symmetry of the standard normal curve, if $c > 0$, we can 
see 
$$P(0 < Z < c) = {1\over 2} - P(Z > c) = P(-c < Z < 0).$$
Hence, for instance, 
$$\eqalign{
P(-1.0 < Z < 2.0) &= P(-1 < Z \le 0) + P(0 < Z < 2)\cr
&= \left({1\over 2} - P(Z > 1)\right) + \left({1\over 2} - P(Z > 2)\right)\cr
&= (.5 - .1587) + (.5 - .0228)\cr
&= .3413 + .4772 = .8185.}$$
For today's discussion problems, the following key fact will be needed too.
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\noindent
{\it Key Fact:}  
If $Y$ is normal with mean $\mu$ and
standard deviation $\sigma$, then 
$$Z = {Y - \mu \over \sigma}$$
is standard normal, and the table can be applied to $Z$. 
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\noindent 
{\it For Example:}  Say we know that $Y$ has a normal distribution with mean 
$\mu = 12$ and SD $\sigma = 2$.  Suppose we want to know
the probability $P(Y > 11.3)$.  We can determine this from
the Key Fact as follows:
$$\eqalign{Y > 11.3 &\Leftrightarrow Z = {Y - 12\over 2} > {11.3 - 12\over 2} = -.35,\ {\rm so}\cr
P(Y > 11.3) &= P(Z > -.35)\cr
           &= 1 - P(Z > +.35)\ {\rm by\ symmetry}\cr
           &= 1 - .3632\ {\rm from\ table}\cr
           &= .6368\cr}$$
In today's discussion, you will practice
using the standard normal table to answer questions about normally distributed
quantities following similar methods.  
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\noindent
{\it Discussion Questions}
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\noindent
A)  Let $Z$ be a standard normal.  
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\item{1)} Find $P(Z > 1.29)$.  
\item{2)} Find $P(Z < 1.0)$.  
\item{3)} Find $P(-2.13 < Z < -0.56)$.
\item{4)} Find $c$ such that $P(Z > c) = .05$. (You'll need to ``interpolate'' here. Do 
you see why?)
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\noindent
B) $Y$ is normally distributed with mean $6$ and SD $4$.  Find
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\item{1)}  $P(Y < 7)$.
\item{2)}  $P(Y > 8.5)$.
\item{3)}  $P(5 < Y \le 8)$.
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\noindent 
C)  SlimMints are sold in two-packs with a stated label weight of 
20.4 grams.  The actual weights of the packages are normally
distributed with mean $\mu = 21.37$ and SD $\sigma = .4$.
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\item{1)}  Let $Y$ be the weight of a single package selected
at random from the production line.  What is the probability $P(Y > 22.07)$?
\item{2)}  Suppose that 15 packages are selected independently.
Let $X$ be the number that weigh less than 21 grams.  What
is $P(X = 2) = $ the probability that exactly two have weight less than 21 grams?
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\noindent
We'll compare notes and I'll give the correct answers at the end of the class.
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\bye
