Help me with probability

Question

1. Use the following information to answer the next four exercises. Recently, a nurse commented that when a patient calls the medical advice line claiming to have the flu, the chance that he or she truly has the flu (and not just a nasty cold) is only about 4%. Of the next 25 patients calling in claiming to have the flu,

we are interested in how many actually have the flu.

(a) Define the random variable and list its possible values.

(b) State the distribution of X.

(c) Find the probability that at least five of the 25 patients actually have the flu.

(d) On average, for every 25 patients calling in, how many do you expect to have the flu?

2. Approximately 8% of students at a local high school participate in after-school sports all four years of high school. A group of 60 seniors is randomly chosen. Of interest is the number that participated in after-school sports all four years of high school.

1. (a) In words, define the random variable X.
2. (b) List the values that X may take on.
3. (c) Give the distribution of X, i.e., X ∼ ( , )
4. (d) How many seniors are expected to have participated in after-school sports all four years of high
5. school?
6. (e) Based on numerical values, would you be surprised if none of the seniors participated in after-school sports all four years of high school? Justify your answer numerically.
7. (f) Based on numerical values, is it more likely that four or that five of the seniors participated in after-school sports all four years of high school? Justify your answer numerically.

Answer 1

Note: all my code is in R Markown

1a. Because we are interested in, the "success rate" of patients who actually have the flu, we can use the binomial distribution as an appropriate model for this problem. Note: there may be other families that could be used in this situation, like the beta for example, but the binomial is an easy distribution to use and appropriate for this problem. Let's use X as our Random Variable (RV) which follows a Binomial(x; n = 25, p = 0.04), where x is our number of successes(in this problem, how many patients actually have the flu), n is our sample size, and p is our probability of success. So, X can be any discrete number from {0,1,2,...,25}, inclusive. Note: I chose a discrete model because we typically don't think about counting human beings in anything besides integers.

1b. The pmf of our binomial distribution is as follows:

$p_x = \binom{25}{x}(0.04)^x(1-0.04)^{25-x}$

Our support is x = {0,1,2,...,25} inclusive, because we could possibly have 0-25 of those patients actually have the flu.

1c. The probability we are interested in finding can be expressed as follows: P(X ≥ 5). By De Morgan's Law, this is equal to 1-P(X<5). If we add up the values of x=0,1,2,3,4 from our pmf, this will give us our answer.

``` {r}

x = c(0,1,2,3,4)

p_x = choose(25,x)*(0.04)^(x)*(1-0.04)^(25-x)

sum(p_x)

```

sum(p_x)=0.9972

1d. Since we want to find the expected value of our distribution, and the expected value is equal to np for the binomial distribution, the answer is 25*0.04=1. Thus, we would expect one patient to actually have the flu on average for every 25 patients calling in.

2a.Again, we can think of our RV X as following a Binomial distribution. This is because we can define a success as a student who participates in a sport all 4 years of high school. We then have the probability of our success as 0.08, and the sample size we are going to look at is 60 as said in the problem. Note: Let me know if this is what is meant by describe the RV in words.

2b. X can be any integer between 0 and 60 inclusive. (Only integers because the binomial distribution is discrete). This is because we could have 0-60 of these kids in the sample play sports all four years of high school.

2c. The pmf is as follows:

$p_x=\binom{60}{x}(0.08)^x(1-0.08)^{60-x}$

2d. Again, the expectation of the binomial distribution is np. So, E(X)=60*0.08=4.8.

2e. The easiest thing we could do in this problem is to set x=0 and solve for that probability and then make some decision based off of the probability from our pmf.

``` {r}

x = c(0)

p_x = choose(25,x)*(0.04)^(x)*(1-0.04)^(25-x)

p_x

```

p_x(0)=0.3604

2f. One thing we could do for this problem is to solve our pmf for x=4 and 5 and then do some comparisons between the two.

``` {r}

x = c(4,5)

p_x = choose(25,x)*(0.04)^(x)*(1-0.04)^(25-x)

p_x

```

p_x(4)=0.01374

p_x(5)=0.002405