Answer:
a) ![P(X = 1) = 0.38742](https://tex.z-dn.net/?f=P%28X%20%3D%201%29%20%3D%200.38742)
b) ![P(X = 3) = 0.05740](https://tex.z-dn.net/?f=P%28X%20%3D%203%29%20%3D%200.05740)
c) ![P(X = 9) = 0.00000](https://tex.z-dn.net/?f=P%28X%20%3D%209%29%20%3D%200.00000)
d) ![P(X \geq 5) = 0.00163](https://tex.z-dn.net/?f=P%28X%20%5Cgeq%205%29%20%3D%200.00163)
Step-by-step explanation:
For each container, there are only two possible outcomes. Either it is undefilled, or it is not. This means that we can solve this problem using the binomial probability distribution.
Binomial probability distribution:
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
![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)
In which
is the number of different combinatios of x objects from a set of n elements, given by the following formula.
![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)
And p is the probability of X happening.
In this problem
There are 10 containers, so
.
A food-packaging apparatus underfills 10% of the containers, so
.
a) This is P(X = 1)
![P(X = 1) = C_{10,1}.(0.1)^{1}.(0.9)^{9} = 0.38742](https://tex.z-dn.net/?f=P%28X%20%3D%201%29%20%3D%20C_%7B10%2C1%7D.%280.1%29%5E%7B1%7D.%280.9%29%5E%7B9%7D%20%3D%200.38742)
b) This is P(X = 3)
![P(X = 3) = C_{10,3}.(0.1)^{3}.(0.9)^{7} = 0.05740](https://tex.z-dn.net/?f=P%28X%20%3D%203%29%20%3D%20C_%7B10%2C3%7D.%280.1%29%5E%7B3%7D.%280.9%29%5E%7B7%7D%20%3D%200.05740)
c) This is P(X = 9)
![P(X = 9) = C_{10,9}.(0.1)^{9}.(0.9)^{1} = 0.00000](https://tex.z-dn.net/?f=P%28X%20%3D%209%29%20%3D%20C_%7B10%2C9%7D.%280.1%29%5E%7B9%7D.%280.9%29%5E%7B1%7D%20%3D%200.00000)
d) This is
.
Either the number is lesser than five, or it is five or larger. The sum of the probabilities of each event is decimal 1. So:
![P(X < 5) + P(X \geq 5) = 1](https://tex.z-dn.net/?f=P%28X%20%3C%205%29%20%2B%20P%28X%20%5Cgeq%205%29%20%3D%201)
![P(X \geq 5) = 1 - P(X < 5)](https://tex.z-dn.net/?f=P%28X%20%5Cgeq%205%29%20%3D%201%20-%20P%28X%20%3C%205%29)
In which
![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)
![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 = 0) = C_{10,0}.(0.1)^{0}.(0.9)^{10} = 0.34868](https://tex.z-dn.net/?f=P%28X%20%3D%200%29%20%3D%20C_%7B10%2C0%7D.%280.1%29%5E%7B0%7D.%280.9%29%5E%7B10%7D%20%3D%200.34868)
![P(X = 1) = C_{10,1}.(0.1)^{1}.(0.9)^{9} = 0.38742](https://tex.z-dn.net/?f=P%28X%20%3D%201%29%20%3D%20C_%7B10%2C1%7D.%280.1%29%5E%7B1%7D.%280.9%29%5E%7B9%7D%20%3D%200.38742)
![P(X = 2) = C_{10,2}.(0.1)^{2}.(0.9)^{8} = 0.1937](https://tex.z-dn.net/?f=P%28X%20%3D%202%29%20%3D%20C_%7B10%2C2%7D.%280.1%29%5E%7B2%7D.%280.9%29%5E%7B8%7D%20%3D%200.1937)
![P(X = 3) = C_{10,3}.(0.1)^{3}.(0.9)^{7} = 0.05740](https://tex.z-dn.net/?f=P%28X%20%3D%203%29%20%3D%20C_%7B10%2C3%7D.%280.1%29%5E%7B3%7D.%280.9%29%5E%7B7%7D%20%3D%200.05740)
![P(X = 4) = C_{10,4}.(0.1)^{1}.(0.9)^{9} = 0.38742](https://tex.z-dn.net/?f=P%28X%20%3D%204%29%20%3D%20C_%7B10%2C4%7D.%280.1%29%5E%7B1%7D.%280.9%29%5E%7B9%7D%20%3D%200.38742)
So
![P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.34868 + 0.38742 + 0.19371 + 0.05740 + 0.01116 = 0.99837](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%20%3D%200.34868%20%2B%200.38742%20%2B%200.19371%20%2B%200.05740%20%2B%200.01116%20%3D%200.99837)
Finally
![P(X \geq 5) = 1 - P(X < 5) = 1 - 0.99837 = 0.00163](https://tex.z-dn.net/?f=P%28X%20%5Cgeq%205%29%20%3D%201%20-%20P%28X%20%3C%205%29%20%3D%201%20-%200.99837%20%3D%200.00163)
I think it’s b
have a nice day
Answer:
Try as percentages so
7) 37.5%
8) 50%
9) 12.5%
10) 62.5%
11) 87.5%
12) 50%
13) 62.5%
14) 37.5%
15) 100%
Step-by-step explanation:
Answer: The median number of posts made is 8.5
I believe the answer is 90