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\centerline{MONT 105Q -- Mathematical Journeys}
\centerline{Problem Set 4 -- Using the Standard Normal Table}
\centerline{{\bf due}: March 30, 2016}
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\noindent
Use the standard normal table to the following questions.  
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\noindent
A)  Let $Z$ be a standard normal.  
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\item{1)} Find $P(Z > .57)$.  
\item{2)} Find $P(Z < -1.3)$.
\item{3)} Find $P(.37 < Z < 1.20)$.  
\item{4)} Find $P(-.55 < Z < 1.30)$.
\item{5)} Find $c$ such that $P(Z > c) = .01$. (You'll probably want to ``interpolate.'')
\item{6)} How would your answer to 5) change if it said:  Find $c$ such that $P(|Z| > c) = .01$?
(Note: $|Z| > c$ says: the absolute value of $Z$ is greater than $c$.  That can happen if
either $Z > c$ or $Z < -c$.)
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\noindent
B) $Y$ is normally distributed with mean $10$ and SD $3$.  Find
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\item{1)}  $P(Y < 9)$.
\item{2)}  $P(Y > 11.3)$.
\item{3)}  $P(7 < Y \le 16)$.  How does this answer compare with the ``rule of thumb''
given by Wheelan in the graph on page 26 of {\it Naked Statistics}?)
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\noindent 
C) For this question, you will need to combine what we learned before about
the Central Limit Theorem with our new techniques based on the standard normal
table.  Let $X$ be the maximal oxygen intake of a human on a treadmill where
the measurements are made in milliliters of oxygen per minute per kilogram of weight.
Assume that for a particular population of patients the population mean of
$X$ is $\mu = 54.030$ and the population standard deviation is $\sigma = 5.8$.
Let $\overline{X}$ be the sample mean of a random sample of size $n = 50$ from 
this population.
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\item{1)}  Find $P(52.761 \le \overline{X} \le 54.453)$.
\item{2)}  (Without calculation, and explain):  Would the answer to part 1) be
larger or smaller if the sample size was $n = 100$?
\item{3)}  Suppose 10 different random samples of size $n = 50$ are taken 
from this population.  What is the probability that exactly 7 of them will have
$52.761 \le \overline{X} \le 54.453$? 
\bye
