Answer:
Option A would be your answer
Step-by-step explanation:
Had this question on Achieve3000
<span> 2 student tickets
</span><span>=1.50+1.50+2.50+2.50+2.50=10.50</span>
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: the second one
Step-by-step explanation:
