Suppose that f(x) = x/8 for 3 < x <

osi4a2nxk

osi4a2nxk

Answered question

2021-11-16

Suppose that f(x) = x/8 for 3 < x < 5. Determine the following probabilities: a. P(X < 4) b. P(X > 3.5) c. P(4 < X < 5) d. P(X < 4.5) e. P(X < 3.5 or X > 4.5)

Answer & Explanation

Lorraine David

Lorraine David

Beginner2021-11-17Added 13 answers

The probaility density function f of the tandom variable X is given as:
f(x)=x8,3<x<5
Calculate the general probability
P(a<X<b)=18∫abxdx=b2−a216
Apply this formula
P(X<4)=P(3<X<4)=16−916=0.4375
P(X>3.5)=P(3.5<X<5)=25−12.2516=0.797
P(4<X<5)=25−1616=0.5625
P(X<4.5)=20.25−916=0.703
P(X<3.5 or X>4.5)=P(3)

nick1337

nick1337

Expert2023-06-18Added 777 answers

Given that f(x) = x/8 for 3 < x < 5, we can calculate the CDF as follows:
F(x)=∫−∞xf(t)dt
For 3 < x < 5, the function f(x) is a linear function, so we can easily integrate it within that range:
F(x)=∫3xt8dt
Simplifying the integral:
F(x)=18∫3xtdt
F(x)=18[t22]3x
F(x)=18(x22−322)
F(x)=18(x22−92)
Now, let's calculate the probabilities using the CDF we found.
a. P(X < 4)
To find this probability, we substitute x = 4 into the CDF:
P(X<4)=F(4)=18(422−92)
Simplifying the expression:
P(X<4)=18(162−92)
P(X<4)=18(72)
P(X<4)=716
Therefore, P(X<4)=716.
b. P(X > 3.5)
To find this probability, we subtract the probability of X being less than or equal to 3.5 from 1 (since the total probability must add up to 1):
P(X>3.5)=1−P(X≤3.5)
Using the CDF, we can calculate:
P(X>3.5)=1−F(3.5)
P(X>3.5)=1−18(3.522−92)
Simplifying the expression:
P(X>3.5)=1−18(12.252−92)
P(X>3.5)=1−18(3.252)
P(X>3.5)=1−18(138)
P(X>3.5)=1−1364
P(X>3.5)=5164
Therefore, P(X>3.5<br>)=5164.
c. P(4 < X < 5)
To find this probability, we subtract the probability of X being less than or equal to 4 from the probability of X being less than or equal to 5:
P(4<X<5)=F(5)−F(4)
P(4<X<5)=18(522−92)−18(422−92)
Simplifying the expression:
P(4<X<5)=18(252−92)−18(162−92)
P(4<X<5)=18(162)−18(72)
P(4<X<5)=18(82)
P(4<X<5)=18(4)
P(4<X<5)=12
Therefore, P(4<X<5)=12.
d. P(X < 4.5)
To find this probability, we substitute x = 4.5 into the CDF:
P(X<4.5)=F(4.5)=18(4.522−92)
Simplifying the expression:
P(X<4.5)=18(20.252−92)
P(X<4.5)=18(11.252)
P(X<4.5)=18(92+2.252)
P(X<4.5)=18(92+1.125)
P(X<4.5)=18(9+1.1252)
P(X<4.5)=18(10.1252)
P(X<4.5)=18(5.0625)
P(X<4.5)=5.06258
P(X<4.5)=81128
Therefore, P(X<4.5)=81128.
e. P(X < 3.5 or X > 4.5)
To find this probability, we calculate the sum of the probabilities P
(X < 3.5) and P(X > 4.5). Since these two events are mutually exclusive, we can add their probabilities:
P(X<3.5 or X>4.5)=P(X<3.5)+P(X>4.5)
P(X<3.5 or X>4.5)=F(3.5)+(1−F(4.5))
Substituting the CDF values:
P(X<3.5 or X>4.5)=18(3.522−92)+(1−18(4.522−92))
Simplifying the expression:
P(X<3.5 or X>4.5)=18(12.252−92)+(1−18(20.252−92))
P(X<3.5 or X>4.5)=18(3.252)+(1−18(11.252))
P(X<3.5 or X>4.5)=18(138)+(1−18(92+1.125))
P(X<3.5 or X>4.5)=18(138)+(1−18(5.0625))
P(X<3.5 or X>4.5)=1364+(1−0.078125)
P(X<3.5 or X>4.5)=1364+6464−564
P(X<3.5 or X>4.5)=7264
P(X<3.5 or X>4.5)=98
Therefore, P(X<3.5 or X>4.5)=98.
Don Sumner

Don Sumner

Skilled2023-06-18Added 184 answers

a. To find P(X<4), we need to calculate the integral of f(x) from −∞ to 4:
P(X<4)=∫−∞4f(x)dx=∫34x8dx.
b. To find P(X>3.5), we need to calculate the integral of f(x) from 3.5 to ∞:
P(X>3.5)=∫3.5∞f(x)dx=∫3.55x8dx.
c. To find P(4<X<5), we need to calculate the integral of f(x) from 4 to 5:
P(4<X<5)=∫45f(x)dx=∫45x8dx.
d. To find P(X<4.5), we need to calculate the integral of f(x) from −∞ to 4.5:
P(X<4.5)=∫−∞4.5f(x)dx=∫34.5x8dx.
e. To find P(X<3.5 or X>4.5), we need to calculate the sum of the probabilities of P(X<3.5) and P(X>4.5):
P(X<3.5 or X>4.5)=P(X<3.5)+P(X>4.5).
Vasquez

Vasquez

Expert2023-06-18Added 669 answers

Step 1:
a. P(X<4):
To find the probability that X is less than 4, we need to calculate the definite integral of f(x) over the interval (3,4). The probability can be obtained as follows:
P(X<4)=∫34f(x)dx=∫34x8dx
Evaluating the integral:
P(X<4)=[x216]34=4216−3216=1616−916=716
Therefore, P(X<4)=716.
Step 2:
b. P(X>3.5):
To find the probability that X is greater than 3.5, we need to calculate the definite integral of f(x) over the interval (3.5,5). The probability can be obtained as follows:
P(X>3.5)=∫3.55f(x)dx=∫3.55x8dx
Evaluating the integral:
P(X>3.5)=[x216]3.55=5216−3.5216=2516−12.2516=12.7516
Therefore, P(X>3.5)=5164.
Step 3:
c. P(4<X<5):
To find the probability that X lies between 4 and 5, we need to calculate the definite integral of f(x) over the interval (4,5). The probability can be obtained as follows:
P(4<X<5)=∫45f(x)dx=∫45x8dx
Evaluating the integral:
P(4<X<5)=[x216]45=5216−4216=2516−1616=916
Therefore, P(4<X<5)=916.
Step 4:
d. P(X<4.5):
To find the probability that X is less than 4.5, we need to calculate the definite integral of f(x) over the interval (3,4.5). The probability can be obtained as follows:
P(X<4.5)=∫34.5f(x)dx=∫34.5x8dx
Evaluating the integral:
P(X<4.5)=[x216]34.5=4.5216−3216=20.2516−916=11.2516
Therefore, P(X<4.5)=916.
Step 5:
e. P(X<3.5 or X>4.5):
To find the probability that X is less than 3.5 or greater than 4.5, we can calculate the sum of the probabilities P(X<3.5) and P(X>4.5). These probabilities can be obtained using the same approach as in parts (a) and (b):
P(X<3.5)=∫33.5f(x)dx=∫33.5x8dx
P(X>4.5)=∫4.55f(x)dx=∫4.55x8dx
Evaluating the integrals:
P(X<3.5)=[x216]33.5=3.5216−3216=12.2516−916=3.2516
P(X>4.5)=[x216]4.55=5216−4.5216=2516−20.2516=4.7516
Therefore, P(X<3.5 or X>4.5)=P(X<3.5)+P(X>4.5)=3.2516+4.7516=816=12.

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