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)=3x−4,for x&gt;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?

Question

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)=3x−4,for x&amp;gt;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?

Cheryl King · Accepted Answer

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=∫1∞3x−4dx  =(−1x3)1∞  =0−(−113)(1)  =1  In (1) we used the fact:  (−1x3)∞=0  which is true because of:  (−1x3)∞=limx⇒∞−1x3=0.  The last equality is obtained from the fact:  limx⇒∞x3=∞  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&amp;gt;1  we have: F(x)=P(X≤x)  =∫∞xf(t)dt  =∫1x3t−4dt  =(−1t3)1x  =−1x3−(−113)  =1−1x3  Step 3  c) We need to find the probability P(X&amp;gt;4).  P(X&amp;gt;4)=1−P(X≤4)  =1−F(4)  =1−(1−143)  =1−1+164  =164

habbocowji · Accepted Answer

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=∫1∞3x−4dx =3∫1∞x−4dx =3[x−4+1−4+1]1∞ =3−3[1x3]1∞ =−1[1∞3−113] =−1[1∞−1] =−1[0−1](∴1∞=0) =1 Therefore, we get. ∫1∞f(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)=∫1x3t−4dt =3[t−4+1−4+1]1x =3−3[1t3]1x =−1[1x3−113] =−1[1x3−1] =1−1x3 Hence, F(x)=1−1x3;x&amp;gt;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&amp;gt;4), which is obtained as, P(X&amp;gt;4)=∫4∞f(x)dx =∫4∞3x−4dx

karton · Accepted Answer

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=∫1∞3x−4dx=(−x−3)1∞=−0−(−1)=1 b) Solution: To evaluate F(x) we use the definition of CDF of a continuous RV, F(x)=P(X≤x)=∫−∞xf(t)dt Because f is piecewise, we have F(x)={1−x−3x&amp;gt;10elsewhere c) What is the probability that a random particle from the manufactured fuel exceeds 4 micrometers? Solution:We want to find P(X&amp;gt;4)=1−P(X≤4). This can be done quickly with the CDF. That is, P(X&amp;gt;4)=1−P(X≤4)=1−F(4)=1−(1−4−3)=4−3