An important factor in solid missile fuel is the particle size distrib

Ashley Bell

Ashley Bell

Answered question

2021-12-24

An important factor in solid missile fuel is the particle size distribution. Significant problems occur if the particle sizes are too large. From production data in the past, it has been determined that the particle size (in micrometers) distribution is characterized by f(x)=3x4,for x>1,f(x)=0, elsewhere.
a) Verify that this is a valid density function.
b) Evalute F(x).
c) What is the probability that a random particle from the manufactured fuel exceeds 4 micrometers?

Answer & Explanation

Cheryl King

Cheryl King

Beginner2021-12-25Added 36 answers

Step 1
a) In order to verify that f is a valid density function, we need to check if the condition f(x)dx=1 is satisfied.
f(x)dx=13x4dx
=(1x3)1
=0(113)(1)
=1
In (1) we used the fact:
(1x3)=0
which is true because of:
(1x3)=limx1x3=0.
The last equality is obtained from the fact:
limxx3=
Therefore, f truly is a valid density function
Step 2
b) Let X be the random variable which represents the particle size in micrometers.
By the definition of the cumulative distribution function F, for
x>1
we have: F(x)=P(Xx)
=xf(t)dt
=1x3t4dt
=(1t3)1x
=1x3(113)
=11x3
Step 3
c) We need to find the probability P(X>4).
P(X>4)=1P(X4)
=1F(4)
=1(1143)
=11+164
=164
habbocowji

habbocowji

Beginner2021-12-26Added 22 answers

a) To verify whether f(x) is a density function.
We know that, if f(x) is a density function then,
xf(x)dx=1
Consider,
xf(x)dx=13x4dx
=31x4dx
=3[x4+14+1]1
=33[1x3]1
=1[13113]
=1[11]
=1[01](1=0)
=1
Therefore, we get.
1f(x)dx=1
Hence, given f(x) is a density function.
b) To find F(x), i.e, distribution function.
We know that, distribution is obtained as,
F(t)=0xf(t)dt
F(t)=1x3t4dt
=3[t4+14+1]1x
=33[1t3]1x
=1[1x3113]
=1[1x31]
=11x3
Hence, F(x)=11x3;x>1
c) To find the probability that a random particle from the manufactured fuel exceeds 4 micrometres
Let X: size of a random particle from the manufactured fuel
We need to find P(X>4), which is obtained as,
P(X>4)=4f(x)dx
=43x4dx

karton

karton

Expert2021-12-30Added 613 answers

a) Solution: To verify it is a valid density function, we take the integral and see that it is one. I.e.
f(x)dx=13x4dx=(x3)1=0(1)=1
b) Solution: To evaluate F(x) we use the definition of CDF of a continuous RV,
F(x)=P(Xx)=xf(t)dt
Because f is piecewise, we have
F(x)={1x3x>10elsewhere
c) What is the probability that a random particle from the manufactured fuel exceeds 4 micrometers?
Solution:We want to find P(X>4)=1P(X4). This can be done quickly with the CDF. That is,
P(X>4)=1P(X4)=1F(4)=1(143)=43

xleb123

xleb123

Skilled2023-06-12Added 181 answers

a) To verify that the given function is a valid density function, we need to check two conditions:
1. The function must be non-negative for all values of x:
f(x)=3x4 for x>1 and f(x)=0 elsewhere. Since x4 is always positive for x>0, multiplying it by 3 will also yield a positive value. Therefore, f(x) is non-negative for all values of x.
2. The integral of the function over its entire domain must be equal to 1:
f(x)dx=13x4dx
Evaluating the integral, we have:
13x4dx=[33x3]1=limx33x3(33)=0(11)=1
Since the integral of the function over its domain is equal to 1, the function is a valid density function.
b) To evaluate F(x), which represents the cumulative distribution function (CDF), we need to integrate the density function from to x:
F(x)=xf(t)dt=1x3t4dt
Evaluating the integral, we have:
F(x)=[t3]1x=1x3(113)=11x3 for x>1 and F(x)=0 for x1.
c) The probability that a random particle from the manufactured fuel exceeds 4 micrometers is given by P(X>4), where X represents the particle size. Since the density function is zero for x1, we only need to evaluate the CDF at x=4:
P(X>4)=1F(4)=1(1143)=164
Therefore, the probability that a random particle from the manufactured fuel exceeds 4 micrometers is 164.
fudzisako

fudzisako

Skilled2023-06-12Added 105 answers

a) To verify if the given function f(x)=3x4 is a valid density function, we need to check if it satisfies the properties of a density function.
The properties of a valid density function are as follows:
1. f(x)0 for all x.
2. The integral of f(x) over its entire range is equal to 1.
Let's check these properties for f(x)=3x4:
1. For all x>1, f(x)=3x4, which is always positive. Therefore, f(x)0 for all x.
2. To check the integral, we need to evaluate f(x)dx and see if it equals 1:
f(x)dx=13x4dx
Using the power rule of integration, we can integrate 3x4 as follows:
13x4dx=[x3]1=[1x3]1=lima1a3(113)=0(1)=1
The integral of f(x) over its entire range is equal to 1. Hence, f(x)=3x4 is a valid density function.
b) To evaluate F(x), the cumulative distribution function (CDF) of f(x), we need to integrate f(x) from to x:
F(x)=xf(t)dt=1x3t4dt
Integrating 3t4 with respect to t gives:
F(x)=[t3]1x=1x3(113)=11x3
Therefore, the cumulative distribution function F(x) is 11x3.
c) The probability that a random particle from the manufactured fuel exceeds 4 micrometers is given by P(X>4), where X represents the random variable representing particle size.
To calculate this probability, we can use the cumulative distribution function F(x) as follows:
P(X>4)=1P(X4)=1F(4)
Substituting x=4 into the cumulative distribution function, we get:
P(X>4)=1F(4)=1(1143)=164
Therefore, the probability that a random particle from the manufactured fuel exceeds 4 micrometers is 164.

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