Answer to Question #95832 in Statistics and Probability for Ari

Question

Q2
You are to take a multiple-choice exam consisting of 100 questions with five possible responses to each. Suppose that you have not studied and so must guess (select one of the five answers in a completely random fashion) on each question. Let random variable X represent the number of correct responses on the exam.
a) Specify the probability distribution of X
b) What is your expected number of correct responses?
c) What are the values of the variance and standard deviation of X?
d) What is the probability that you will get exactly the expected number of correct responses?

Q3
The number of elementary particles, recorded by a device in a space vehicle during a one-day flight, has a Poisson distribution with a mean of 3.0 particles. Find the probability that during a one-day flight there will be:
(a) no particle recorded
(b) at least 4 particles recorded

Answer 1

Question 2.

a) $X$ is defined as the number of success (correct answers) in a series of independent experiments (answering the questions "at random") thus $X \sim {\text{Bin(}}n,p{\text{)}}$ - binomial distribution with $n = 100$ and $p = \frac{1}{5}$ where probability mass function is

$${p_X}(k) = \binom{n}{k}{p^k}{(1 - p)^{n-k}}=\binom{100}{k}{(\frac{1}{5})^k}{(\frac{4}{5})^{100-k}}$$

where $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ - binomial coefficient

b) We need to find a formula for expected value $\mathbb{E} X$ . The easiest way to do it - represents our random variable $X$ as a sum of other random variables (independent) $X = \sum\limits_{i = 1}^n {{Y_i}}$ where ${{Y_i}}$ is defined as a result of one experiment (yes-no). Its probability mass function is

$${p_Y}(k) = \begin{cases} 1-p, & k=0 \\ p, & k=1 \end{cases}$$

So it is easy to find its expected value

$$\mathbb{E} Y = \sum\limits_{k = 0}^1 {k{p_Y}(k)} = 0 \cdot (1 - p) \cdot 0 + 1 \cdot p = p$$

And now we can find the expected value of binomial distribution as (we use that expected value of sum is the sum of expected values)

$$\mathbb{E} X = \mathbb{E} \sum\limits_{i = 1}^n {{Y_i}} = \sum\limits_{i = 1}^n {\mathbb{E} {Y_i}} = \sum\limits_{i = 1}^n p = np$$

Let's calculate it in our case

$$\mathbb{E} X = np = 100 \cdot \frac{1}{5} = 20$$

c) We can find the variance using the same way as in b). We shall use that

$$\mathbb{D} Y = \mathbb{E}({Y^2})- {(\mathbb{E} Y)^2}$$ $$\mathbb{E} ({Y^2}) = \sum\limits_{k = 0}^1 {{k^2}{p_Y}(k)} = 0 \cdot (1 - p) \cdot 0 + 1 \cdot p = p$$ $$\mathbb{D} Y = \mathbb{E} {X^2} - {(\mathbb{E} X)^2}=p - {p^2} = p(1 - p)$$

And now let's use that variance of sum is sum of variances in case of independent (or uncorrelated, more explicitly) random variables

$$\mathbb{D} X = \mathbb{D} \sum\limits_{i = 1}^n {{Y_k}} = \sum\limits_{i = 1}^n {\mathbb{D} {Y_k}} = \sum\limits_{i = 1}^n {p(1 - p)} = np(1 - p)$$

Let's do the calculations

$$\mathbb{D} X = 100 \cdot \frac{1}{5} \cdot \frac{4}{5} = 16$$

Standard deviation is the square root of variance

$${\sigma _X} = \sqrt {\mathbb{D} X} = 4$$

d) We need to calculate $P(X = 20) = {p_X}(20)=\binom{100}{20}{p^{20}}{(1 - p)^{80}}$ but it can't be done wihout computer. We can use De Moivre–Laplace theorem which claims ${\text{Bin(}}n,p{\text{)}} \to N(np,np(1 - p))$ as $n \to + \infty$ and $p$ remains fixed. So in means that

$$\binom{n}{k}{p^k}{(1 - p)^{n-k}} \to \frac{1}{{\sqrt {2\pi np(1 - p)} }}{e^{ - \frac{{{{(k - np)}^2}}}{{2np(1 - p)}}}}$$

and now we can do the calculations

$$P(x = 20) \approx \frac{1}{{\sqrt {2\pi \cdot 100 \cdot 0.2 \cdot 0.8} }}{e^{ - \frac{{20 - 100 \cdot 0.2}}{{2 \cdot 100 \cdot 0.2 \cdot 0.8}}}} \approx 0.0997$$

Question 3.

a) The probability mass function for Poisson distribution with mean ${\lambda}$ has the form

$$p(k) = \frac{{{\lambda ^k}{e^{ - \lambda }}}}{{k!}}$$

Thus probability that zero particles will be recorded is

$$P(X = 0) = p(0) = \frac{{{3^0}{e^{ - 3}}}}{{0!}} = {e^{ - 3}} \approx 0.049787$$

b) We need to find the probability $P(X \geqslant 4)$ . We shall use that $\sum\limits_{k = 0}^n {\frac{{{\lambda ^k}{e^{ - \lambda }}}}{{k!}}} = 1$ thus

$$P(X \geqslant 4) = \sum\limits_{k = 4}^n {\frac{{{\lambda ^k}{e^{ - \lambda }}}}{{k!}}} = 1 - \sum\limits_{k = 0}^3 {\frac{{{\lambda ^k}{e^{ - \lambda }}}}{{k!}}}$$ $$P(X \geqslant 4) = 1 - \frac{{{3^0}{e^{ - 3}}}}{{0!}} - \frac{{{3^1}{e^{ - 3}}}}{{1!}} - \frac{{{3^2}{e^{ - 3}}}}{{2!}} - \frac{{{3^3}{e^{ - 3}}}}{{3!}}$$

Do the calculations

$$P(X \geqslant 4) = 1 - 13{e^{ - 3}} \approx 0.352768$$