Answer: 3.5 integers can be in decimals too. and any number counts as an integer except for 0
Answer:
a. E(x) = 3.730
b. c = 3.8475
c. 0.4308
Step-by-step explanation:
a.
Given
0 x < 3
F(x) = (x-3)/1.13, 3 < x < 4.13
1 x > 4.13
Calculating E(x)
First, we'll calculate the pdf, f(x).
f(x) is the derivative of F(x)
So, if F(x) = (x-3)/1.13
f(x) = F'(x) = 1/1.13, 3 < x < 4.13
E(x) is the integral of xf(x)
xf(x) = x * 1/1.3 = x/1.3
Integrating x/1.3
E(x) = x²/(2*1.13)
E(x) = x²/2.26 , 3 < x < 4.13
E(x) = (4.13²-3²)/2.16
E(x) = 3.730046296296296
E(x) = 3.730 (approximated)
b.
What is the value c such that P(X < c) = 0.75
First, we'll solve F(c)
F(c) = P(x<c)
F(c) = (c-3)/1.13= 0.75
c - 3 = 1.13 * 0.75
c - 3 = 0.8475
c = 3 + 0.8475
c = 3.8475
c.
What is the probability that X falls within 0.28 minutes of its mean?
Here we'll solve for
P(3.73 - 0.28 < X < 3.73 + 0.28)
= F(3.73 + 0.28) - F(3.73 + 0.28)
= 2*0.28/1.3 = 0.430769
= 0.4308 -- Approximated
Answer:
I am a 4th grader I don't know
Step-by-step explanation:
Answer:
x+74=180
x=180-74
x=106
x=4y+22
106-22=4y
84/4=y
21=y
y=21
Step-by-step explanation:
Answer:
a

b

Step-by-step explanation:
From the question we are told that
The number of students in the class is N = 20 (This is the population )
The number of student that will cheat is k = 3
The number of students that he is focused on is n = 4
Generally the probability distribution that defines this question is the Hyper geometrically distributed because four students are focused on without replacing them in the class (i.e in the generally population) and population contains exactly three student that will cheat.
Generally probability mass function is mathematically represented as

Here C stands for combination , hence we will be making use of the combination functionality in our calculators
Generally the that he finds at least one of the students cheating when he focus his attention on four randomly chosen students during the exam is mathematically represented as

Here




Hence


Generally the that he finds at least one of the students cheating when he focus his attention on six randomly chosen students during the exam is mathematically represented as

![P(X \ge 1) =1- [ \frac{^{k}C_x * ^{N-k}C_{n-x}}{^{N}C_n}]](https://tex.z-dn.net/?f=P%28X%20%20%5Cge%201%29%20%3D1-%20%5B%20%20%5Cfrac%7B%5E%7Bk%7DC_x%20%2A%20%5E%7BN-k%7DC_%7Bn-x%7D%7D%7B%5E%7BN%7DC_n%7D%5D%20)
Here n = 6
So
![P(X \ge 1) =1- [ \frac{^{3}C_0 * ^{20 -3}C_{6-0}}{^{20}C_6}]](https://tex.z-dn.net/?f=P%28X%20%20%5Cge%201%29%20%3D1-%20%5B%20%20%5Cfrac%7B%5E%7B3%7DC_0%20%2A%20%5E%7B20%20-3%7DC_%7B6-0%7D%7D%7B%5E%7B20%7DC_6%7D%5D%20)
![P(X \ge 1) =1- [ \frac{^{3}C_0 * ^{17}C_{6}}{^{20}C_6}]](https://tex.z-dn.net/?f=P%28X%20%20%5Cge%201%29%20%3D1-%20%5B%20%20%5Cfrac%7B%5E%7B3%7DC_0%20%2A%20%5E%7B17%7DC_%7B6%7D%7D%7B%5E%7B20%7DC_6%7D%5D%20)
![P(X \ge 1) =1- [ \frac{1 * 12376}{38760}]](https://tex.z-dn.net/?f=P%28X%20%20%5Cge%201%29%20%3D1-%20%5B%20%20%5Cfrac%7B1%20%20%2A%20%2012376%7D%7B38760%7D%5D%20)

