• #todo/school : Finish the STAT HW 📅 2024-10-12 ✅ 2024-10-12

4.12

Proof
Say $S$ is the event that someone comes to a full stop, so $P(S)=.25$. We are given $n=20$ suggesting a binomial model.

a.

b.

c.

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4.14

Proof
Say the event $S$ is the event a purchasing customer wants the paperback version. So then $P(S)=.6=p$ with varying $n$ per part (this is a binomial distribution).

a. Here $n=10$:

b. Still $n=10$:

Thus:

Thus:

c. All customers will get what they want if $X>7$ doesn't happen, along with $X<3$. So we want to know:

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4.24

Proof
a. We require that $X=12$ to all agree on a guilty verdict, or $X=0$ for a non-guilty verdict. Hence:

Similarly:

b. Clearly when $p\to 1$ then $P(\text{guilty})$ is higher, making $P(\text{not guilty})$ lower. On the opposite, when $p\to 0$ then $P(\text{not guilty})$ is higher, making $P(\text{guilty})$ lower. This makes sense since if there's a high probability an individual juror would come to one verdict or the other, then all jurors voting unanimously for one or the other would have to favor that same verdict.

c. We require that now $X\ge 9$ so then:

Note that this is greater than the answer from (a), mainly due to the fact that we have a $1-$ out in the front, and typically $p^{12}$ will be a very small value.
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4.66

Proof
Here we have a negative binomial distribution. Here $p=.2$ and $X$ is the number of individuals administered.

a. Here $r=1$ so:

b. The experiment happening 5 times is exactly the same as when the experiment has 4 non-adverse reaction subjects. That's because $SSSSF$ is the only possible event of this happening. This is the same probability from (a), so it's $\boxed{{\displaystyle .08192}}$.

c.

d.

For counting the number of people without and adverse reaction $Y$, that would be $Y=X-1$, which is linear, so then:

e. Here:

So then:

Then:

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4.68

Proof
First, notice that we can say that $3\le X\le 5$ since if it's any less then there's not even 3 children in the family, and any more means that there were already 3 children of the same sex as it was. Thus, we can consider each case separately. All of these are negative binomial where $r=3$ and varying $p$.

For $X_{\text{boy}}$:

For $X_{\text{girl}}$:

Checking, adding the two gives 1, as expected. In general:

Giving the following Probability Distribution, PMF:

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Question 6

Proof
This is a negative binomial distribution with $r=3$ and $p=0.062$. Let $X$ denote the number of scans required.

a.

b. This number is just $Y=X-1$ since we just ignore the last scan. That way it's before the target is detected for the full 3 times. Then ${\mu }_Y={\mu }_X-1=\boxed{{\displaystyle 47.3871\text{ scans}}}$.

c. Now this is a normal binomial distribution with $n=30$ and $p=.062$ still. Then if $X^{\prime }$ is the number of times we detect the target then:

d. Now this is back to being a negative binomial distribution, where $r=1$ and $p=.062$. Then:

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4.38

Proof
a.

b.

c. Since successive weekdays are independent, then:

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4.44

Proof
a. Here $t=1\text{ year}$ so then $\mu =4.8\text{ rainstorms}$. As such:

b. Now $t=3\text{ years}$ so $\mu =14.4\text{ rainstorms}\Rightarrow \sigma =\sqrt{\mu }\approx 3.7947\text{ rainstorms}$. Then that means that we want to see $\mu +\sigma \approx 18.195\text{rainstorms}$. So:

c. In general:

We know that $P(X\ge 1)=.999$ so then:

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4.46

Proof
Here $\lambda =5$ where $t$ is in hours.

a. $t=1$ so then since $\mu =\lambda t=5$:

b.

c. Here now $t=.75$ so $\mu =3.75$, so

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4.90

Proof
Here $\mu =\lambda A$ where $A$ is the yardage. Then snice $A=\pi r^2$ then here $\mu =2\pi r^2\text{ grasshoppers}$. Thus:

This requires that:

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