Which of the following statements regarding the probability density function of the normal distribution $N(1,2^2)$ is false:

We need to find the one false statement, and go through the claims one at a time.

1: Yes, the total area under the curve is one, since this is true for all probability distributions, cf. the definition of a continuous
density function

2: Yes. The mean value is one, and we have that $1^2 = 1$.

3: No, the standard deviation is two and the variance is four.

4: Yes, the distribution is symmetric around the mean value

5: Yes, the normal density is defined between $-\infty$ and $\infty$.

So, the correct answer is 3.

Question 2 of 5

Calls come into a telephone exchange at random at a rate of 5 per minute.

What is the probability that exactly 10 calls come during a given 2-minute period?

We are dealing with a discrete probability distribution here, and counts which are not upper bounded. Thus it is natural to use the Poisson distribution, book Chapter 2.
Define a random variable $X \sim Pois(5)$, where $X$ denotes the number of calls during some given minute. Thean we see that $Y \sim Pois(10)$ must denote the number of calls during some given two minute interval. We now need to find $P(Y=10)$:
\(P(Y=10) = \frac{10^{10}}{10!}e^{-10}= 0.125\)
Or in R:

Two brothers are evenly matched when playing chess. One day they decide to play until one or the other has won three games in a row.

Ignoring draws(ties), what is the probability that they play exactly three games?

The probability that the brothers end up playing precisely three matches is equal to the probability that whoeever wins the first match also wins the next two matches. The probability of either brother winning some given match is $\frac{1}{2}$. Thus the probability that the brother to win the first match also wins the next two matches is $\frac{1}{2}^2 = \frac{1}{4}$.
\medskip

Or differently put: It is like a binomial distribution with $n=3$ and $p=0.5$: The event asked about can be expressed as a function of what happens in the first three mathces, since they definitely will play at least three matches. E.g.: Let $X$ be the number of times brother A wins among the first three matches (still assuming that draws do not occur). Then the event we are asked about is.
\(P(\mbox{Brother A wins 3 or 0 times})=P(X=0)+P(X=3)=1/8 + 1/8 = 1/4\)

So, the correct answer is 1.

Question 4 of 5

Two brothers are evenly matched when playing chess. One day they decide to play until one or the other has won three games in a row.

Ignoring draws(ties), what is the probability that they play exactly four games?

There are two options

Brother 1 wins the first match and brother 2 wins the next three matches.

The probability of this is (the probability that brother 1 wins the first match) times (the probability that brother 2 wins the next three matches)
\(= 1/2 \cdot (1/2)^3 = (1/2)^4\)
Brother 2 wins the first match and brother 1 wins the next three matches. As above the probability of this is also $(1/2)^4$.

In both ways four matches are played so the probability of four matches being played is:
\((1/2)^4 + (1/2)^4 = (1/2)^3,\)
and the correct answer is hence 2.

Question 5 of 5

Let X be normally distributed with mean $36$ and variance $9$.

$P(X \geq 30)$ equals approximately:

We have that $X \sim N(36, 3^2)$, let $Z \sim N(0,1)$ and calculate

Or in R: (in three different ways)<br>
<pre><code>
> 1-pnorm(30,m=36, sd=3)
[1] 0.9772499
> 1-pnorm(-2)
[1] 0.9772499
> pnorm(2)
[1] 0.9772499
</code></pre>