Using the binomial distribution, it is found that there is a:
a) 0.9298 = 92.98% probability that at least 8 of them passed.
b) 0.0001 = 0.01% probability that fewer than 5 passed.
For each student, there are only two possible outcomes, either they passed, or they did not pass. The probability of a student passing is independent of any other student, hence, the binomial distribution is used to solve this question.
<h3>What is the binomial probability distribution formula?</h3>
The formula is:
![P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20C_%7Bn%2Cx%7D.p%5E%7Bx%7D.%281-p%29%5E%7Bn-x%7D)
![C_{n,x} = \frac{n!}{x!(n-x)!}](https://tex.z-dn.net/?f=C_%7Bn%2Cx%7D%20%3D%20%5Cfrac%7Bn%21%7D%7Bx%21%28n-x%29%21%7D)
The parameters are:
- x is the number of successes.
- n is the number of trials.
- p is the probability of a success on a single trial.
In this problem:
- 90% of the students passed, hence
.
- The professor randomly selected 10 exams, hence
.
Item a:
The probability is:
![P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10)](https://tex.z-dn.net/?f=P%28X%20%5Cgeq%208%29%20%3D%20P%28X%20%3D%208%29%20%2B%20P%28X%20%3D%209%29%20%2B%20P%28X%20%3D%2010%29)
In which:
![P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20C_%7Bn%2Cx%7D.p%5E%7Bx%7D.%281-p%29%5E%7Bn-x%7D)
![P(X = 8) = C_{10,8}.(0.9)^{8}.(0.1)^{2} = 0.1937](https://tex.z-dn.net/?f=P%28X%20%3D%208%29%20%3D%20C_%7B10%2C8%7D.%280.9%29%5E%7B8%7D.%280.1%29%5E%7B2%7D%20%3D%200.1937)
![P(X = 9) = C_{10,9}.(0.9)^{9}.(0.1)^{1} = 0.3874](https://tex.z-dn.net/?f=P%28X%20%3D%209%29%20%3D%20C_%7B10%2C9%7D.%280.9%29%5E%7B9%7D.%280.1%29%5E%7B1%7D%20%3D%200.3874)
![P(X = 10) = C_{10,10}.(0.9)^{10}.(0.1)^{0} = 0.3487](https://tex.z-dn.net/?f=P%28X%20%3D%2010%29%20%3D%20C_%7B10%2C10%7D.%280.9%29%5E%7B10%7D.%280.1%29%5E%7B0%7D%20%3D%200.3487)
Then:
![P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10) = 0.1937 + 0.3874 + 0.3487 = 0.9298](https://tex.z-dn.net/?f=P%28X%20%5Cgeq%208%29%20%3D%20P%28X%20%3D%208%29%20%2B%20P%28X%20%3D%209%29%20%2B%20P%28X%20%3D%2010%29%20%3D%200.1937%20%2B%200.3874%20%2B%200.3487%20%3D%200.9298)
0.9298 = 92.98% probability that at least 8 of them passed.
Item b:
The probability is:
![P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)](https://tex.z-dn.net/?f=P%28X%20%3C%205%29%20%3D%20P%28X%20%3D%200%29%20%2B%20P%28X%20%3D%201%29%20%2B%20P%28X%20%3D%202%29%20%2B%20P%28X%20%3D%203%29%20%2B%20P%28X%20%3D%204%29)
Using the binomial formula, as in item a, to find each probability, then adding them, it is found that:
![P(X < 5) = 0.0001](https://tex.z-dn.net/?f=P%28X%20%3C%205%29%20%3D%200.0001)
Hence:
0.0001 = 0.01% probability that fewer than 5 passed.
You can learn more about the the binomial distribution at brainly.com/question/24863377
Answer:
Roses= $8.4
Daises= $15.6
Step-by-step explanation:
Let represent the daises and let r represent d roses
r + d= 24....equation 1
0.25r + 0.90d= 16.40.......equation 2
r= 25-d
Substitute 25-d for r in equation 2
0.25(25-d) + 0.90d= 16.40
6.25-0.25d+0.90d= 16.40
6.25+0.65d= 16.40
0.65d= 16.40-6.25
0.65d= 10.15
d= 10.15/0.65
d = 15.6
Sub 15.6 for d in equation 1
r+d= 24
r+15.6= 24
r= 24-15.6
= 8.4
Price of daises is $15.6
Roses is $8.4
Answer:
2.1/////_____\\\\\\
Step-by-step explanation:
hope this helps
You need to provide a homie wit some more details lol