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GRE Subject Test: Math : Discrete Distributions

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Example Questions

Example Question #101 : Other Topics

X is a continuously and uniformly distributed on the interval (0,50). Find the Expected Value (E[x]) and Variance (Var(x)) of X.

Possible Answers:

E[x]=25 $\displaystyle E[x]=25$

$\displaystyle Var(x)=2500$

$\displaystyle E[x]=4.167$

$\displaystyle Var(x)=208.33$

$\displaystyle E[x]=4.167$

$\displaystyle Var(x)=1250$

E[x]=25 $\displaystyle E[x]=25$

$\displaystyle Var(x)=200.166$

E[x]=25 $\displaystyle E[x]=25$

$\displaystyle Var(x)=208.33$

Correct answer:

E[x]=25 $\displaystyle E[x]=25$

$\displaystyle Var(x)=208.33$

Explanation:

Because x is a continuous uniform random variable the expected value and variance can be found with the following formulas:

$E[x]=\frac{a+b}{2}$ $\displaystyle E[x]=\frac{a+b}{2}$

$Var(x)=\frac{(b-a)^2}{12}$ $\displaystyle Var(x)=\frac{(b-a)^2}{12}$

X is uniform on (a,b). In this case a is 0 and b is 50. Plugging the values of a and b into the given formulas will give the answers:

$E[x]=\frac{a+b}{2}=\frac{0+50}{2}=25$ $\displaystyle E[x]=\frac{a+b}{2}=\frac{0+50}{2}=25$

$Var(x)=\frac{(b-a)^2}{12}=\frac{2500}{12}=208.33$ $\displaystyle Var(x)=\frac{(b-a)^2}{12}=\frac{2500}{12}=208.33$

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Example Question #102 : Other Topics

A fair coin is tossed 15 times. What is the probability of observing less than 3 heads?

Possible Answers:

0.0037 $\displaystyle 0.0037$

0.0032 $\displaystyle 0.0032$

0.5 $\displaystyle 0.5$

0.125 $\displaystyle 0.125$

0.0139 $\displaystyle 0.0139$

Correct answer:

0.0037 $\displaystyle 0.0037$

Explanation:

This problem uses the Binomial Distribution: $\binom{n}{k}p^k(1-p)^{n-k}$ $\displaystyle \binom{n}{k}p^k(1-p)^{n-k}$

For this problem n is the number of trials, or 15. Because the problem stated that the coin was a fair coin the probability of heads is one half, or .5.

The binomial distribution is a discrete distribution so the expression x<3 has to be broken down.

$\displaystyle Pr(x< 3)=Pr(x=0)+Pr(x=1)+Pr(x=2)$

$Pr(x=0)=\begin{pmatrix} 15\\0 \end{pmatrix} (.5)^0(1-.5)^{15}=.000031$ $\displaystyle Pr(x=0)=\begin{pmatrix} 15\\0 \end{pmatrix} (.5)^0(1-.5)^{15}=.000031$

$Pr(x=1)=\begin{pmatrix} 15\\1 \end{pmatrix} (.5)^1(1-.5)^{14}=.00046$ $\displaystyle Pr(x=1)=\begin{pmatrix} 15\\1 \end{pmatrix} (.5)^1(1-.5)^{14}=.00046$

$Pr(x=2)=\begin{pmatrix} 15\\2 \end{pmatrix} (.5)^2(1-.5)^{13}=.0032$ $\displaystyle Pr(x=2)=\begin{pmatrix} 15\\2 \end{pmatrix} (.5)^2(1-.5)^{13}=.0032$

Adding the probabilities will give the final answer.

$\displaystyle .000031+.0046+.0032=.0037$

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GRE Subject Test: Math : Discrete Distributions

Study concepts, example questions & explanations for GRE Subject Test: Math

All GRE Subject Test: Math Resources

Example Questions

Example Question #101 : Other Topics

Example Question #102 : Other Topics

All GRE Subject Test: Math Resources

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