W06 Homework

Due date: Wednesday 2/18, 11:59pm

Normal distribution

01

06

Gaussian basics

Find the probability that one observation of a Gaussian variable will yield a value within 1.5 standard deviations of the mean.

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Solution

Solutions - 5180-06

Compute the desired probability:

$$\begin{array}{c}P[-1.5<Z<1.5]=1-2P[Z>1.5]\\ \\ \gg \gg 1-2[1-P[Z<1.5]]\gg \gg 2P[Z<1.5]-1\\ \\ \gg \gg 2(0.9332)-1\gg \gg 0.8664\end{array}$$Link to original

02

07

Morning commute time

My morning commute time is normally distributed, with a mean of 14 minutes, and a standard deviation of 4 minutes. I leave for work every morning at 8:45am and need to arrive by 9:00am.

(a) On any given day, what is the probability that I am late?

(b) On any given day, what is the probability that I reach before 8:55am?

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Solution

Solutions - 5180-07

(a)

(1) Write $X=\sigma Z+\mu$ and substitute into the probability:

$$\begin{array}{c}X=4Z+14\\ \\ P[X>15]\gg \gg P[4Z+14>15]\\ \\ \gg \gg P[Z>\frac{15-14}{4}]\gg \gg P[Z>0.25]\end{array}$$

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(2) Evaluate using the $\mathrm{\Phi }$ table:

$$\begin{array}{c}\gg \gg 1-P[Z<0.25]\\ \\ \gg \gg 1-0.5987\gg \gg 0.4013\end{array}$$

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(b)

(1) Substitute $X=4Z+14$ into the probability:

$$P[X<10]\gg \gg P[4Z+14<10]\gg \gg P[Z<-1]$$

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(2) Evaluate using the $\mathrm{\Phi }$ table:

$$\begin{array}{c}1-P[Z<1]\gg \gg 1-0.8413\gg \gg 0.1587\end{array}$$Link to original

03

04

Normal distribution - test scores

In a large probability theory exam, the scores are normally distributed with a mean of 75 and a standard deviation of 10.

(a) What is the probability that a student scored between 70 and 80

(b) What is the lowest score a student can achieve to be in the top 5%?

(c) What score corresponds to the 25th percentile?

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Solution

Solutions - 5180-04

(a)

(1) Write $X=\sigma Z+\mu$ and substitute into the probability:

$$\begin{array}{c}X=10Z+75\\ \\ P[70\le X\le 80]\gg \gg P[70\le 10Z+75\le 80]\gg \gg P[-\frac{1}{2}\le Z\le \frac{1}{2}]\end{array}$$

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(2) Express in terms of $\mathrm{\Phi }$ and evaluate:

$$\begin{array}{c}\mathrm{\Phi }\left(\frac{1}{2}\right)-\mathrm{\Phi }(-\frac{1}{2})\gg \gg 2\mathrm{\Phi }\left(\frac{1}{2}\right)-1\\ \\ \gg \gg 2(0.6915)-1\gg \gg 0.383\end{array}$$

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(b)

(1) Interpret problem:

Since we wish to find the top $5\%$, we wish to find $z$ such that $\mathrm{\Phi }(z)=0.95$.

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(2) Use lookup table to find $z$:

$$\mathrm{\Phi }(z)=0.95\gg \gg z\approx 1.6449$$

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(3) Solve for $X$ using $X=10Z+75$:

$$X\approx 10(1.6449)+75\gg \gg 91.449$$

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(c)

(1) Interpret problem:

Since we wish to find the 25th percentile, we wish to find $z$ such that $\mathrm{\Phi }(z)=0.25$.

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(2) Use lookup table to find $z$:

$$\mathrm{\Phi }(z)=0.25\gg \gg z\approx -0.6745$$

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(3) Solve for $X$ using $X=10Z+75$:

$$X\approx 10(-0.6745)+75\gg \gg 68.255$$Link to original

04

03

Generalized normal

Let $X$ be generalized normal variable with $\mu =3$ and $\sigma =2$. Using a chart of $\mathrm{\Phi }$ values, find:

(a) $P[2<X<6]$

(b) $c$ such that $F_X(c)=0.67$

(c) $E[X^2]$ (Hint: Use $\mu$ and $\sigma$ to avoid integration.)

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Solution

Solutions - 5180-03

(a)

(1) Write $X=\sigma Z+\mu$ and substitute into the probability:

$$\begin{array}{c}X=2Z+3\\ \\ P[2<X<6]\gg \gg P[2<2Z+3<6]\gg \gg P[-\frac{1}{2}<Z<\frac{3}{2}]\end{array}$$

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(2) Express in terms of $\mathrm{\Phi }$ values:

$$\begin{array}{c}\mathrm{\Phi }\left(\frac{3}{2}\right)-\mathrm{\Phi }(-\frac{1}{2})\gg \gg \mathrm{\Phi }\left(\frac{3}{2}\right)+\mathrm{\Phi }\left(\frac{1}{2}\right)-1\end{array}$$

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(3) Evaluate using a $\mathrm{\Phi }$ table:

$$\begin{array}{c}0.9332+0.6915-1\gg \gg 0.6247\end{array}$$

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(b)

(1) Use the $\mathrm{\Phi }$ table to find $z$ given $\mathrm{\Phi }(z)=0.67$:

$$\mathrm{\Phi }(z)=0.67\gg \gg z\approx 0.4399$$

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(2) Solve for $c$ using $X=2Z+3$:

$$c=2(0.4399)+3\gg \gg 3.8798$$

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(c)

(1) Recall formula for $\text{Var}[X]$ and solve for $E\left[X^2\right]$:

$$\begin{array}{c}\text{Var}[X]=E\left[X^2\right]-{\left(E[X]\right)}^2\\ \\ \gg \gg E\left[X^2\right]=\text{Var}[X]+{\left(E[X]\right)}^2\end{array}$$

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(2) Plug in $\mu =3$ and ${\sigma }^2=4$:

$$E\left[X^2\right]=4+3^2\gg \gg 13$$Link to original

Review problems

05

02

Insurance expected payout

A car insurance analytics team estimates that the cost of repairs per accident is uniformly distributed between $100 and $1500. The manager wants to offer customers a policy that has a $500 deductible and covers all costs above the deductible.

How much is the expected payout per accident?

(Hint: Graph the PDF for the cost of repairs $X$; write a formula for the payout in terms of $X$ using cases; then integrate.)

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Solution

Solutions - 5120-02

(1) State the PDF of $X$ and the payout function $g(X)$:

Since repair cost is uniformly distributed, $f_X(x)=\frac{1}{1400}$ for $100\le x\le 1500$.

The payout with a $500 deductible is:

$$g(x)=\{\begin{array}{cc}0& x\le 500\\ x-500& x>500\end{array}$$

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(2) Compute $E[g(X)]$:

$$\begin{array}{c}E[g(X)]={\int }_{-\infty }^{\infty }g(x)f_X(x)dx\\ \\ \gg \gg \frac{1}{1400}{\int }_{500}^{1500}(x-500)dx\\ \\ \gg \gg \frac{1}{1400}{[\frac{x^2}{2}-500x]|}_{500}^{1500}\gg \gg \approx \$357.14\end{array}$$Link to original

06

05

CDF of derived variable

Suppose $X$ is a continuous $\mathrm{Unif}[\mathrm{1,4}]$ random variable. Let $Y=|X-2|$. Find the CDF of $Y$.

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Solution

Solutions - 5150-05

(1) Express the CDF of $Y$ in terms of $X$:

$$\begin{array}{c}{F}_Y(y)=P[Y\le y]\\ \\ \gg \gg P[|X-2|\le y]\gg \gg P[-y<X-2<y]\\ \\ \gg \gg P[-y+2<X<y+2]\end{array}$$

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(2) Evaluate Case 1, $0<y<1$:

$$\begin{array}{c}{\int }_{-y+2}^{y+2}\frac{1}{3}dx\\ \\ \gg \gg \frac{1}{3}((y+2)-(-y+2))\gg \gg \frac{2y}{3}\end{array}$$

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(3) Evaluate Case 2, $1<y<2$:

$$\begin{array}{c}{\int }_1^{y+2}\frac{1}{3}dx\gg \gg \frac{1}{3}(y+1)\end{array}$$

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(4) State the complete CDF:

$$F_Y(y)=\{\begin{array}{cc}0& y<0\\ 2y/3& 0\le y<1\\ \frac{1}{3}(y+1)& 1\le y<2\\ 1& 2\le y\end{array}$$Link to original

07

06

Octane revenue

The owner of a small gas station has his 1,500 gallon tank of 93-octane gas filled up once at the beginning of each week. The random variable $X$ is the amount of 93-octane the station sells in one week (in thousands of gallons). The PDF of X is shown below.

$$f_X(x)=\{\begin{array}{cc}x& 0\le x\le 1\\ 1& 1<x\le 1.5\end{array}$$

Assuming the station consistently charges $3.00 per gallon for 93-octane and pays $2,000 for the weekly fill-up, find the CDF of $W=3X-2$, the profit the station makes from the 93-octane in a week.

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Solution

Solutions - 5150-06

(1) Express the CDF of $W$ in terms of $X$:

$$\begin{array}{c}{F}_W(w)=P[W\le w]\\ \\ \gg \gg P[3X-2\le w]\gg \gg P[X\le \frac{w+2}{3}]\end{array}$$

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(2) Evaluate Case 1, $-2\le w\le 1$:

$$\begin{array}{c}{\int }_0^{\frac{w+2}{3}}xdx\gg \gg {\frac{1}{2}{x}^2|}_0^{\frac{w+2}{3}}\\ \\ \gg \gg \frac{1}{2}{\left(\frac{w+2}{3}\right)}^2\gg \gg \frac{{(w+2)}^2}{18}\end{array}$$

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(3) Evaluate Case 2, $1\le w\le 2.5$:

$$\begin{array}{c}{\int }_0^{1}xdx+{\int }_1^{\frac{w+2}{3}}1dx\\ \\ \gg \gg {\frac{1}{2}{x}^2|}_0^1+{x|}_1^{\frac{w+2}{3}}\\ \\ \gg \gg \frac{1}{2}+\frac{w+2}{3}-1\gg \gg \frac{w+2}{3}-\frac{1}{2}\end{array}$$

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(4) State the complete CDF:

$$F_W(w)=\{\begin{array}{cc}0& w<-2\\ \frac{{(w+2)}^2}{18}& -2\le w<1\\ \frac{w+2}{3}-\frac{1}{2}& 1\le w<2.5\\ 1& 2.5\le w\end{array}$$Link to original