Respuesta :
Answer:
(a) [tex]P(X=10) = 0.2070[/tex]
(b) [tex]P(X\geq 10) = 0.3798[/tex]
Step-by-step explanation:
Note that in this problem we have an initial population N = 50, of which 30 fulfill a certain characteristic "m" (belong to the second section). Then, from the population N, a sample of size n = 15 is selected and it is desired to know how many comply with the desired characteristic (second section).
So
Let X be the number of projects in the second section, then X is a discrete random variable that can be modeled by a hypergeometric distribution.
(a)
Therefore, to answer question (a) we use the following equation presented in the attached image:
Where:
[tex]N = 50\\m = 30\\n = 15\\X = 10[/tex]
Then:
[tex]P (X = 10) = \frac{\frac{30!}{10!(30-10)!}\frac{20!}{5!(20-5)!}}{\frac{50!}{15!(50-15)!}}\\\\\\P(X=10) = 0.2070[/tex]
(b)
For part (b) we have:
[tex]P(X\geq10) = 1-P(X<10)\\\\P(X\geq10)= 1- \sum_{X=0}^{9}\frac{\left[\begin{array}{cc}30&\\X&\end{array}\right]\left[\begin{array}{cc}50-&30&\\15-&X&\end{array}\right]}{\left[\begin{array}{cc}50&\\15&\end{array}\right]}[/tex]
[tex]P(X\geq 10) = 1-0.6202 \\\\P(X\geq 10) = 0.3798[/tex]
You can use the binomial distribution to find out the value of needed probability.
The needed probabilities are:
- [tex]P(X = 10) = 0.0167[/tex]
- [tex]P(X \geq 10) = 0.2339[/tex]
How to find that a given condition can be modeled by binomial distribution?
Binomial distributions consists of n independent Bernoulli trials.
Bernoulli trials are those trials which end up randomly either on success (with probability p) or on failures( with probability 1- p = q (say))
Suppose we have random variable X pertaining binomial distribution with parameters n and p, then it is written as
The probability that out of n trials, there'd be x successes is given by
[tex]P(X =x) = \: ^nC_xp^x(1-p)^{n-x}[/tex]
How to model the situation given with binomial distribution?
Consider those first 15 projects.
Since the instructor had arranged the projects randomly and the projects can be either from the first section or second section(and only one of them), thus, each project is independent of each other for being either from first section (call it failure) or from second section (call it success)
The probability that a project will belong to first section is
[tex]\dfrac{20}{20+30} = \dfrac{2}{5} = 0.4[/tex]
(since there are 20 students from first section and total students of both the sections are having total of 50 students)
Since we call it failure when the project belongs to first section, thus, we have:
[tex]0.4 = q = 1-p\\p = 1-0.4 = 0.6[/tex]
Thus, probability of success(probability of a project to be from second section) is 0.6
Thus, if we let random variable X track the count of successes out of 15 for the given condition, then we have
[tex]X \sim B(n,p)\\\\X \sim B(15, 0.6)[/tex]
Using the probability function of binomial distribution, we get:
a) Probability that exactly 10 of the projects out of 15 projects selected is
[tex]P(X = 10) = \: ^{15}C_{10}(0.6)^{10}(0.4)^{5} = 273 \times 0.006 \times 0.01024 \\P(X = 10) \approx 0.0167[/tex]
b) Probability that at least 10 of these selected projects are from the second section is
[tex]\begin{aligned } P(X \geq 10) &= P(X = 10) + (P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15) )\\&= 0.0167 + ( \: ^{15}C_{11}(0.6)^{11}(0.4)^{4} + \: ^{15}C_{12}(0.6)^{12}(0.4)^{3} + \: ^{15}C_{13}(0.6)^{13}(0.4)^{2}\\ + \: ^{15}C_{14}(0.6)^{14}(0.4)^{1} + \: ^{15}C_{15}(0.6)^{15}(0.4)^{0} )\\&= 0.0167 + (0.1267 + 0.0634 + 0.0219 + 0.0047 + 0.0005)\\&\approx 0.2339\\\end{aligned}[/tex]
Thus,
The needed probabilities are:
- [tex]P(X = 10) = 0.0167[/tex]
- [tex]P(X \geq 10) = 0.2339[/tex]
Learn more about binomial distribution here:
https://brainly.com/question/13609688
