Respuesta :
Answer:
Explained below.
Step-by-step explanation:
The probability density function of lifetime in months of a transistor in a certain application is:
[tex]f(x)=0.4\cdot e^{-0.4x};x>0[/tex]
The probability density function suggests that the random variable X follows a exponential function with parameter λ = 0.4.
(a)
Compute the mean lifetime as follows:
[tex]\mu=\frac{1}{\lambda}=\frac{1}{0.40}=2.5[/tex]
Thus, the mean lifetime is 2.5 months.
(b)
Compute the standard deviation of the lifetimes as follows:
[tex]\sigma=\sqrt{\frac{1}{\lambda^{2}}}=\sqrt{\frac{1}{0.40^{2}}}=2.5[/tex]
Thus, the standard deviation of the lifetimes is 2.5 months.
(c)
The cumulative distribution function of the lifetime is:
[tex]CDF=1-e^{-0.40x}[/tex]
(d)
Compute the probability that the lifetime will be less than 6 months as follows:
[tex]P(X<6)=\int\limits^{6}_{0} {0.40\cdot e^{-0.40x}} \, dx[/tex]
[tex]=0.40\times [\frac{e^{-0.40x}}{-0.40}]^{6}_{0}\\\\=-1\times [e^{-0.40\times6}-e^{-0.40\times0}]\\\\=-0.090718+1\\\\=0.909282\\\\\approx 0.9093[/tex]
Thus, the probability that the lifetime will be less than 6 months is 0.9093.
(e)
Compute the 60th percentile of the lifetime as follows:
[tex]P(X<x)=0.60\\\\\int\limits^{x}_{0} {0.40\cdot e^{-0.40x}} \, dx=0.60\\\\0.40\times [\frac{e^{-0.40x}}{-0.40}]^{x}_{0}=0.60\\\\-1\times [e^{-0.40\times x}-e^{-0.40\times0}]=0.60\\\\1-e^{-0.40x}=0.60\\\\e^{-0.40x}=0.40\\\\-0.40x=\ln(0.40)\\\\-0.40x=-0.916291\\\\x=2.2907275\\\\x\approx 2.3[/tex]
Thus, the 60th percentile of the lifetime is 2.3 months.
