# Homework 1 for Stat Inference

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Brian Caffo
Johns Hopkins Bloomberg School of Public Health

## About these slides

• These are some practice problems for Statistical Inference Quiz 1
• They were created using slidify interactive which you will learn in Creating Data Products
• Please help improve this with pull requests here (https://github.com/bcaffo/courses)

Consider influenza epidemics for two parent heterosexual families. Suppose that the probability is 15% that at least one of the parents has contracted the disease. The probability that the father has contracted influenza is 10% while that the mother contracted the disease is 9%. What is the probability that both contracted influenza expressed as a whole number percentage? Watch a video solution

1. 15%
2. 10%
3. 9%
4. 4%

$A = Father$, $P(A) = .10$, $B = Mother$, $P(B) = .09$ $P(A\cup B) = .15$,

$P(A\cup B) = P(A) + P(B) - P(AB)$ thus $.15 = .10 + .09 - P(AB)$

.10 + .09 - .15

 0.04


A random variable, $X$, is uniform, a box from $0$ to $1$ of height $1$. (So that its density is $f(x) = 1$ for $0\leq x \leq 1$.) What is its median expressed to two decimal places? Watch a video solution.

1. 1.00
2. 0.75
3. 0.50
4. 0.25

The median is the point so that 50% of the density lies below it.

This density looks like a box. So, notice that $P(X \leq x) = width\times height = x$. We want $.5 = P(X\leq x) = x$.

You are playing a game with a friend where you flip a coin and if it comes up heads you give her $X$ dollars and if it comes up tails she gives you $Y$ dollars. The odds that the coin is heads is $d$. What is your expected earnings? Watch a video solution.

1. $-X \frac{d}{1 + d} + Y \frac{1}{1+d}$
2. $X \frac{d}{1 + d} + Y \frac{1}{1+d}$
3. $X \frac{d}{1 + d} - Y \frac{1}{1+d}$
4. $-X \frac{d}{1 + d} - Y \frac{1}{1+d}$

The odds that you lose on a given round is given by $p / (1 - p) = d$ which implies that $p = d / (1 + d)$.

You lose $X$ with probability $p = d/(1 +d)$ and you win $Y$ with probability $1-p = 1/(1 + d)$. So your answer is $-X \frac{d}{1 + d} + Y \frac{1}{1+d}$

A random variable takes the value -4 with probability .2 and 1 with probability .8. What is the variance of this random variable? Watch a video solution.

1. 0
2. 4
3. 8
4. 16

This random variable has mean 0. The variance would be given by $E[X^2]$ then.

$E[X] = 0$ $Var(X) = E[X^2] = (-4)^2 * .2 + (1)^2 * .8$

-4 * .2 + 1 * .8

 0

(-4)^2 * .2 + (1)^2 * .8

 4


If $\bar X$ and $\bar Y$ are comprised of $n$ iid random variables arising from distributions having means $\mu_x$ and $\mu_y$, respectively and common variance $\sigma^2$ what is the variance $\bar X - \bar Y$? Watch a video solution of this problem.

1. 0
2. $2\sigma^2/n$
3. $\mu_x - \mu_y$
4. $2\sigma^2$

Remember that $Var(\bar X) = Var(\bar Y) = \sigma^2 / n$.

$Var(\bar X - \bar Y) = Var(\bar X) + Var(\bar Y) = \sigma^2 / n + \sigma^2 / n$

Let $X$ be a random variable having standard deviation $\sigma$. What can be said about $X /\sigma$? Watch a video solution of this problem.

1. Nothing
2. It must have variance 1.
3. It must have mean 0.
4. It must have variance 0.

$Var(aX) = a^2 Var(X)$

$Var(X / \sigma) = Var(X) / \sigma^2 = 1$

If a continuous density that never touches the horizontal axis is symmetric about zero, can we say that its associated median is zero? Watch a video solution.

1. Yes
2. No.
3. It can not be determined given the information given.

This is a surprisingly hard problem. The easy explanation is that 50% of the probability is below 0 and 50% is above so yes. However, it is predicated on the density not being a flat line at 0 around 0. That's why the caveat that it never touches the horizontal axis is important.

Consider the following pmf given in R

p <- c(.1, .2, .3, .4)
x <- 2 : 5


What is the variance expressed to 1 decimal place? Watch a solution to this problem.

1. 1.0
2. 4.0
3. 6.0
4. 17.0

The variance is $E[X^2] - E[X]^2$

sum(x ^ 2 * p) - sum(x * p) ^ 2

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