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\centerline{Mathematics 375 -- Probability and Statistics 1}
\centerline{Discussion 3 -- Using the Standard Normal Table}
\centerline{October 26, 2009}
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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 $\mu = 0$, $\sigma = 1$) 
are given in the tables printed inside the front cover and on
page 792 of the text.  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 $Z$.  
In today's discussion, you will practice
using the table to answer questions about normally distributed
quantities.
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\noindent
{\it Discussion Questions}
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A)  Let $Z$ be a standard normal.  
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\item{1)} Find $P(-2.13 < Z < -0.56)$.
\item{2)} Find $c$ such that $P(Z > c) = .05$
\item{3)} Find $c$ such that $P(|Z| < c) = .75$.
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\noindent
B) $Y$ is normally distributed with mean $6$ and variance
$16$.  Find
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\item{1)}  $P(Y < 7)$.
\item{2)}  $P(5 < Y \le 8)$.
\item{3)}  $P(|Y - 4| < 1)$.
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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 variance $\sigma^2 = .16$.
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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 the probability $P(X \le 2)$.
\item{3)}  In order to cut costs, the manufacturer of SlimMints
wants to change the production process to reduce the 
actual mean weight $\mu$, while keeping the same 20.4 gram
stated label weight.  Market research finds that
customers will not notice this if the actual weight of a package
is less than the stated label weight no more than $20\%$ of the time.
What is the smallest value of $\mu$ for which $P(Y < 20.4) \le .20$?
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\noindent
{\it Assignment}
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Group writeups due at end of class.
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