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trapecia [35]
3 years ago
15

An important factor in solid missile fuel is the particle size distribution. Significant problems occur if the particle sizes ar

e too large. From production data in the past, it has been determined that the particle size (in micrometers) distribution is characterized by 3x−4, x > 1, f(x) = 0, elsewhere. (a) Verify that this is a valid density function. (b) Evaluate F(x). (c) What is the probability that a random particle from the manufactured fuel exceeds 4 micrometers?
Mathematics
1 answer:
baherus [9]3 years ago
7 0

Answer:

b) F(x)=1-x^{-3}

c) 0.0156

Step-by-step explanation:

Let's call X : '' Particle size (in micrometers) ''

X is a random variable

The distribution function for X is :

f(x)=3x^{-4}  \\

For x > 1

f(x) = 0, elsewhere

For a) the condition for f(x) to be a valid density function is that

the integral between -∞ and ∞ of f(x) must be equal to 1.

For the integral I change ∞ for j so ∞ = j

Also for the integral I change -∞ for k so k = -∞

\int\limits^j_k {f(x)} \, dx =1

\int\limits^j_1 {3x^{-4} } \, dx =\\

3\frac{j^{-3}}{-3} -3\frac{1^{-3}}{-3} =1

Then f(x) is a valid density function

b) To find F(x) we must integrate between -∞ and x the function f(t)

We calculate f(t) changing x by t in the f(x) function

x = t \\f(x) =3x^{-4} \\f(t)=3t^{-4}

Now we integrate

\int\limits^x_1 {3t^{-4} } \, dt =-x^{-3}+1

F(x)=1-x^{-3}

c) P(X>4)

P(X>4) = 1 - P(X≤4)

P(X>4) = 1 - F(4)

P(X>4) = 1 - (1-4^{-3})=4^{-3}=0.0156

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