# 统计与概率代写|STATISTICS AND PROBABILITY II MATH254 University of Liverpool Assignment

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## Instructions:

It’s good to hear that it covers both discrete and continuous random variables, as well as important distributions like the geometric, exponential, and normal distributions.

I also appreciate that the module highlights the importance of measure theory and probability as a foundation for this material. This should help students gain a deeper understanding of the underlying concepts and be better prepared to apply them in real-world situations.

Overall, it sounds like a valuable module for students interested in actuarial science, financial mathematics, statistics, and the physical sciences.

Corrupted by their power, the judges running the popular game show America’s Next Top Mathematician have been taking bribes from many of the contestants. Each episode, a given contestant is either allowed to stay on the show or is kicked off.

If the contestant has been bribing the judges she will be allowed to stay with probability 1. If the contestant has not been bribing the judges, she will be allowed to stay with probability $1 / 3$.

Suppose that $1 / 4$ of the contestants have been bribing the judges. The same contestants bribe the judges in both rounds, i.e., if a contestant bribes them in the first round, she bribes them in the second round too (and vice versa).
(a) If you pick a random contestant who was allowed to stay during the first episode, what is the probability that she was bribing the judges?

(a) We first compute $P\left(S_1\right)$ using the law of total probability.
$$P\left(S_1\right)=P\left(S_1 \mid B\right) P(B)+P\left(S_1 \mid H\right) P(H)=1 \cdot \frac{1}{4}+\frac{1}{3} \cdot \frac{3}{4}=\frac{1}{2} .$$
We therefore have (by Bayes’ rule) $P\left(B \mid S_1\right)=P\left(S_1 \mid B\right) \frac{P(B)}{P\left(S_1\right)}=1 \cdot \frac{1 / 4}{1 / 2}=\frac{1}{2}$.

(b) If you pick a random contestant, what is the probability that she is allowed to stay during both of the first two episodes?

(b) Using the tree we have the total probability of $S_2$ is
$$P\left(S_2\right)=\frac{1}{4}+\frac{3}{4} \cdot \frac{1}{3} \cdot \frac{1}{3}=\frac{1}{3}$$

(c) If you pick random contestant who was allowed to stay during the first episode, what is the probability that she gets kicked off during the second episode?

(c) We want to compute $P\left(L_2 \mid S_1\right)=\frac{P\left(L_2 \cap S_1\right)}{P\left(S_1\right)}$.
From the calculation we did in part (a), $P\left(S_1\right)=1 / 2$. For the numerator, we have (see the tree)
$$P\left(L_2 \cap S_1\right)=P\left(L_2 \cap S_1 \mid B\right) P(B)+P\left(L_2 \cap S_1 \mid H\right) P(H)=0 \cdot \frac{1}{3}+\frac{2}{9} \cdot \frac{3}{4}=\frac{1}{6}$$
Therefore $P\left(L_2 \mid S_1\right)=\frac{1 / 6}{1 / 2}=\frac{1}{3}$.

# 概率学和统计学代写|STATISTICS AND PROBABILITY I MATH253 University of Liverpool Assignment

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Assignment-daixieTM为您提供利物浦大学University of Liverpool STATISTICS AND PROBABILITY I MATH253概率学和统计学代写代考辅导服务！

## Instructions:

It is true that data analysis has become an essential part of research in many fields, including medicine, pharmacology, and biology. Statistical methods are used to help researchers make sense of large amounts of data and to draw meaningful conclusions from their findings. In addition, data analysis is also important in many other fields, including finance, consultancy, and the public sector.

To perform statistical analysis on real data sets, researchers often use statistical software packages. These packages allow researchers to perform complex statistical analyses quickly and accurately, and to visualize their results in meaningful ways. Some popular statistical software packages include R, SAS, and SPSS.

When performing statistical analysis, it is important to have a strong understanding of the underlying statistical techniques being employed. This includes understanding concepts such as probability, hypothesis testing, and regression analysis. It is also important to understand how to apply these techniques appropriately to real-world data sets.

Interpreting the results of statistical analyses is also a crucial part of the process. Researchers must be able to communicate their findings effectively to others, including colleagues, stakeholders, and the general public. This requires a strong understanding of statistical concepts as well as effective communication skills.

Overall, statistical analysis is a critical tool for researchers and professionals in many fields. By applying statistical techniques appropriately and interpreting their results effectively, researchers can gain valuable insights into their data and make informed decisions based on their findings.

You roll a fair six sided die repeatedly until the sum of all numbers rolled is greater than 6 . Let $X$ be the number of times you roll the die. Let $F$ be the cumulative distribution function for $X$. Compute $F(1), F(2)$, and $F(7)$.

(15) $F(1)$ : Since you never get more than 6 on one roll we have $F(1)=0$.
\begin{aligned} & F(2)=P(X=1)+P(X=2): \ & P(X=1)=0 \ & P(X=2)=P(\text { total on } 2 \text { dice }=7,8,9,10,11,12)=\frac{21}{36}=\frac{7}{12} . \end{aligned}
$F(7)$ : The smallest total on 7 rolls is 7 , so $F(7)=1$.

(15) Suppose $X$ is a random variable with cdf
$$F(x)= \begin{cases}0 & \text { for } x<0 \\ x(2-x) & \text { for } 0 \leq x \leq 1 \\ 1 & \text { for } x>1\end{cases}$$
(a) Find $E(X)$.
(b) Find $P(X<0.4)$.

(a) $f(x)=F^{\prime}(x)=2-2 x$ on $[0,1]$. Therefore
\begin{aligned} E(X) & =\int_0^1 x f(x) d x \ & =\int_0^1 2 x-2 x^2 d x \ & =x^2-\left.\frac{2}{3} x^3\right|_0 ^1 \ & =\frac{1}{3} \end{aligned}
(b) $P(X \leq 0.4)=F(0.4)=0.4(2-0.4)=0.4(1.6)=0.64$.

A political poll is taken to determine the fraction $p$ of the population that would support a referendum requiring all citizens to be fluent in the language of probability and statistics.
(a) Assume $p=0.5$. Use the central limit theorem to estimate the probability that in a poll of 25 people, at least 14 people support the referendum.
Let $X \sim \operatorname{binomial}(25,0.5)=$ the number supporting the referendum.
$$E(X)=12.5, \quad \operatorname{Var}(X)=25 \cdot \frac{1}{4}=\frac{25}{4}, \quad \sigma_X=\frac{5}{2} .$$
Standardizing and using the CLT we have $Z=\frac{X-12.5}{5 / 2} \approx \mathrm{N}(0,1)$ Therefore,
$$P(X \geq 14)=P\left(\frac{X-12.5}{5 / 2} \geq \frac{14-12.5}{5 / 2}\right) \approx P(Z \geq 0.6)=\Phi(-0.6)=0.2743 .$$
where the last probability was looked up in the $Z$-table.