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

xleb123 · Accepted Answer

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)=3x−4 for x&amp;gt;1 and f(x)=0 elsewhere. Since x−4 is always positive for x&amp;gt;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=∫1∞3x−4dxEvaluating the integral, we have:∫1∞3x−4dx=[−33x3]1∞=limx→∞−33x3−(−33)=0−(−11)=1Since 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=∫1x3t−4dtEvaluating the integral, we have:F(x)=[−t−3]1x=−1x3−(−113)=1−1x3 for x&amp;gt;1 and F(x)=0 for x≤1.c) The probability that a random particle from the manufactured fuel exceeds 4 micrometers is given by P(X&amp;gt;4), where X represents the particle size. Since the density function is zero for x≤1, we only need to evaluate the CDF at x=4:P(X&amp;gt;4)=1−F(4)=1−(1−143)=164Therefore, the probability that a random particle from the manufactured fuel exceeds 4 micrometers is 164.

fudzisako · Accepted Answer

a) To verify if the given function f(x)=3x−4 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&#039;s check these properties for f(x)=3x−4:1. For all x&amp;gt;1, f(x)=3x−4, 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=∫1∞3x−4dxUsing the power rule of integration, we can integrate 3x−4 as follows:∫1∞3x−4dx=[−x−3]1∞=[−1x3]1∞=lima→∞−1a3−(−113)=0−(−1)=1The integral of f(x) over its entire range is equal to 1. Hence, f(x)=3x−4 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=∫1x3t−4dtIntegrating 3t−4 with respect to t gives:F(x)=[−t−3]1x=−1x3−(−113)=1−1x3Therefore, the cumulative distribution function F(x) is 1−1x3.c) The probability that a random particle from the manufactured fuel exceeds 4 micrometers is given by P(X&amp;gt;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&amp;gt;4)=1−P(X≤4)=1−F(4)Substituting x=4 into the cumulative distribution function, we get:P(X&amp;gt;4)=1−F(4)=1−(1−143)=164Therefore, the probability that a random particle from the manufactured fuel exceeds 4 micrometers is 164.

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