f(x)=2/x³ if x>1 and f(x)=0 elsewhere.
The diameter of a particle of contamination (in micrometers) is modeled with the probability density function
Determine P(4 <X<8) and P(X<2)
AP(4<X<8)=0.75 and P(X-2) 3/8
P(4<X<8) 3/64 and P(X2)-3/4
None of the other three answers.
CP(4<X<8)=1/2 and P(X2)=4.6875-10-2
The Correct Answer and Explanation is :
We are given a probability density function f(x)f(x) that defines the probability distribution for the diameter of a particle of contamination: f(x)={2×3,if x>10,elsewheref(x) = \begin{cases} \frac{2}{x^3}, & \text{if } x > 1 \\ 0, & \text{elsewhere} \end{cases}
Step 1: Understand the probability density function (PDF)
The PDF f(x)f(x) is defined for values of x>1x > 1. This means that the random variable XX representing the particle diameter takes values greater than 1. The function 2×3\frac{2}{x^3} describes the probability density for these values of xx.
For a probability density function, the total probability must be equal to 1. Thus, we can verify that the given PDF is valid by checking the integral over its domain: ∫1∞f(x) dx=1\int_{1}^{\infty} f(x) \, dx = 1
Let’s calculate this integral: ∫1∞2×3 dx=[−1×2]1∞=0−(−1)=1\int_{1}^{\infty} \frac{2}{x^3} \, dx = \left[ -\frac{1}{x^2} \right]_{1}^{\infty} = 0 – \left( -1 \right) = 1
Since this equals 1, the PDF is valid.
Step 2: Calculate P(4<X<8)P(4 < X < 8)
To find P(4<X<8)P(4 < X < 8), we need to compute the integral of the PDF from 4 to 8: P(4<X<8)=∫482×3 dxP(4 < X < 8) = \int_{4}^{8} \frac{2}{x^3} \, dx
We solve the integral: ∫2×3 dx=−1×2\int \frac{2}{x^3} \, dx = -\frac{1}{x^2}
So, evaluating from 4 to 8: P(4<X<8)=[−1×2]48=−182+142=−164+116=116−164P(4 < X < 8) = \left[ -\frac{1}{x^2} \right]_{4}^{8} = -\frac{1}{8^2} + \frac{1}{4^2} = -\frac{1}{64} + \frac{1}{16} = \frac{1}{16} – \frac{1}{64}
To simplify: 116=464,464−164=364\frac{1}{16} = \frac{4}{64}, \quad \frac{4}{64} – \frac{1}{64} = \frac{3}{64}
Thus, P(4<X<8)=364P(4 < X < 8) = \frac{3}{64}.
Step 3: Calculate P(X<2)P(X < 2)
Now, let’s calculate P(X<2)P(X < 2), which is the integral of f(x)f(x) from 1 to 2: P(X<2)=∫122×3 dxP(X < 2) = \int_{1}^{2} \frac{2}{x^3} \, dx
Using the same integral calculation: P(X<2)=[−1×2]12=−122+112=−14+1=1−14=34P(X < 2) = \left[ -\frac{1}{x^2} \right]_{1}^{2} = -\frac{1}{2^2} + \frac{1}{1^2} = -\frac{1}{4} + 1 = 1 – \frac{1}{4} = \frac{3}{4}
Thus, P(X<2)=34P(X < 2) = \frac{3}{4}.
Step 4: Compare with the given options
- P(4<X<8)=364P(4 < X < 8) = \frac{3}{64}, which matches one of the options.
- P(X<2)=34P(X < 2) = \frac{3}{4}, which also matches one of the options.
Thus, the correct answer is:
C) P(4 < X < 8) = 3/64 and P(X < 2) = 3/4
This explanation shows how we calculated the probabilities using the PDF and demonstrated that the given answer options match our results.