A lab printer’s warm-up time \(X\) is uniformly distributed between 20 and 40 seconds. What is \(P(X\le 30)\)?
Model \(X\sim \text{Unif}[a=20,\ b=40]\)
Given: a=20, b=40. Find: \(P(X\le 30)\).
Compute: CDF on [a,b] is \(F(x)=\dfrac{x-a}{b-a}\). So \(F(30)=\dfrac{30-20}{40-20}=\dfrac{10}{20}=\mathbf{0.5000}\).
=(30-20)/(40-20)
A shuttle arrives uniformly between 0 and 15 minutes after you arrive. What is \(P(5\le X\le 12)\)?
Model \(X\sim \text{Unif}[0,15]\)
Compute: \(P(5\le X\le 12)=\dfrac{12-5}{15-0}=\dfrac{7}{15}=\mathbf{0.4667}\).
=(12-5)/(15-0)
Assume page lengths \(X\) are \(\text{Unif}[280,320]\) words. Find the mean and variance.
Model \(X\sim \text{Unif}[a=280,\ b=320]\)
Compute: \(\operatorname{E}[X]=(a+b)/2=\mathbf{300}\). \(\operatorname{Var}(X)=(b-a)^2/12=(40^2)/12=\mathbf{133.33}\).
Mean: =(280+320)/2 Var: =(320-280)^2/12
For \(X\sim \text{Unif}[2,6]\), find the PDF value \(f_X(4)\).
Model \(X\sim \text{Unif}[2,6]\)
Compute: PDF on [a,b] is \(f(x)=1/(b-a)=1/(6-2)=\mathbf{0.25}\).
=1/(6-2)
Time to failure \(X\) (in years) is exponential with mean 5 years. What is the rate \(\lambda\)? What is \(P(X>3)\)?
Model \(X\sim \mathrm{Exp}(\lambda)\)
Given: mean=5 ⇒ \(\lambda=1/5=\mathbf{0.2}\).
Tail: \(P(X>t)=e^{-\lambda t}\Rightarrow P(X>3)=e^{-0.2\cdot 3}=\mathbf{0.5488}\).
=EXP(-0.2*3) or =1-EXPON.DIST(3, 0.2, TRUE)
Waiting time (minutes) is \(\mathrm{Exp}(\lambda=0.5)\). Find \(P(X\le 2)\) and \(P(X>2)\).
Model \(X\sim \mathrm{Exp}(0.5)\)
CDF: \(F(x)=1-e^{-\lambda x}\Rightarrow F(2)=1-e^{-1}=\mathbf{0.6321}\).
Tail: \(P(X>2)=e^{-1}=\mathbf{0.3679}\).
=EXPON.DIST(2, 0.5, TRUE) and =1-EXPON.DIST(2, 0.5, TRUE)
With \(\lambda=0.25\) per minute, find the 90th percentile wait time \(x_{0.90}\).
Model \(X\sim \mathrm{Exp}(0.25)\)
Quantile: \(x_p=F^{-1}(p)=-\tfrac{1}{\lambda}\ln(1-p)\Rightarrow x_{0.90}= -\tfrac{1}{0.25}\ln(0.1)=\mathbf{9.21}\,\text{min}\).
=-LN(1-0.9)/0.25
If mean time is 4 minutes, what is \(P(X<1)\)?
Model mean=4 ⇒ \(\lambda=1/4=0.25\)
CDF: \(F(1)=1-e^{-0.25\cdot 1}=\mathbf{0.2212}\).
=EXPON.DIST(1, 0.25, TRUE)
Let cycle life \(X\) be lognormal with \(\ln X\sim \mathcal{N}(\mu=6.5,\,\sigma=0.4)\). Find \(P(X\le 1000)\).
Model \(\ln X\sim \mathcal{N}(6.5,\,0.4^2)\)
Compute: \(P(X\le 1000)=P(\ln X\le \ln 1000)=\Phi\!\Big(\dfrac{\ln 1000-6.5}{0.4}\Big)\).
\(\ln 1000=6.9078\Rightarrow z=(6.9078-6.5)/0.4=1.0195\Rightarrow P=\mathbf{0.8469}\) (approx).
=LOGNORM.DIST(1000, 6.5, 0.4, TRUE)
Assume \(\ln X\sim \mathcal{N}(\mu=2.0,\,\sigma=0.5)\) for download time \(X\) in minutes. Find the median and the 90th percentile.
Facts For lognormal, median = \(e^{\mu}\), and \(x_p=\exp(\mu+\sigma z_p)\).
Median: \(e^{2.0}=\mathbf{7.389}\) min.
90th pct: \(z_{0.90}=1.2816\Rightarrow x_{0.90}=\exp(2+0.5\cdot1.2816)=\mathbf{15.96}\) min.
Median: =EXP(2)
90th: =EXP(2 + 0.5*NORM.S.INV(0.90))
If \(\ln X\sim \mathcal{N}(3.2,\,0.6^2)\) for cost multiplier \(X\), what is \(P(X>40)\)?
Model \(\ln X\sim \mathcal{N}(3.2,\,0.6^2)\)
Tail: \(P(X>40)=1- P(\ln X\le \ln 40)\).
\(\ln 40=3.6889\Rightarrow z=(3.6889-3.2)/0.6=0.8148\Rightarrow P=1-\Phi(0.8148)=\mathbf{0.2070}\) (approx).
=1 - LOGNORM.DIST(40, 3.2, 0.6, TRUE)
Let \(\ln X\sim \mathcal{N}(\mu=1.5,\,\sigma=0.3)\). Find \(x\) such that \(P(X\le x)=0.80\).
Quantile \(x_p=\exp(\mu+\sigma z_p)\)
\(z_{0.80}=0.8416\Rightarrow x=\exp(1.5+0.3\cdot0.8416)=\mathbf{5.18}\) (approx).
=EXP(1.5 + 0.3*NORM.S.INV(0.80))