upāyaṁ karoti

Problem 108: If the probability density of a random variable is given by $f(x)= \begin{array}{cc} \Bigg\{ & \begin{array}{cc} kx^3 & 0\le x\le1 \\ 0 & elsewhere \end{array} \end{array} $ find the value k and the probability that the random variable takes on a value a. betweeen $\frac{1}{4}$ and $\frac{3}{4}$; b. greater than $\frac{2}{3}.$ Solution: To find k, we must integrate $f(x)$ from $x=0$ to $x=1$ and set it equal to 1. Thus $\int _{0}^{1}f(x)dx=1\Rightarrow \int _{0}^{1}k{x}^{3}dx=1\Rightarrow k\frac{{x}^{4}}{4}\Big|_0^1 =1 \Rightarrow k/4=1 \Rightarrow k =4$ a. betweeen $\frac{1}{4}$ and $\frac{3}{4}$; $P(0.25\le X \le 0.75) = \int _{0.25}^{0.75}4{x}^{3}dx = x^4\Big|_{0.25}^{0.75} = 0.75^4-0.25^4 = 0.3125$ b. greater than $\frac{2}{3}.$ $P(X > \frac{2}{3}) = \int _{\frac{2}{3}}^{\1}4{x}^{3}dx = x^4\Big|_{\frac{2}{3}}^{1} =1^4 - \frac{2}{3}^4 = 0.802469$

Problem 107: In a given city, 6% of all drivers get at least one parking ticket per year. Use the Poisson approximation to the binomial distribution to determine the probabilities that among 80 drivers randomly chosen in this given city: a. 4 will get at least one parking ticket in any given year, b. At least 3 will get at least at least one parking ticket in any given year, c. Anywhere from 3 to 6, inclusive will get at least one parking ticket in any given year. Solution: Poisson distribution, $f(x;\lambda )=\frac{{\lambda }^{x}{e}^{-\lambda }}{x!}$ a. 4 will get at least one parking ticket in any given year Given that $x=4, n=80, p=0.06, np = 4.8$ $f(4;4.8)=\frac{{4.8}^{4}{e}^{-4.8}}{4!} = 0.182029$ b. At least 3 will get at least at least one parking ticket in any given year Given that $x\ge 3, n=80, p=0.06, np = 4.8$ $P(x\ge 3) = 1-P(x $P(x=0) = f(0;4.8)=\frac{{4.8}^{0}{e}^{-4.8 }}{0!} = 0.00823$ $P(x=1) = f(1;4.8)=\frac{{4.8}^{1}{e}^{-4.8 }}{1!}...

Problem 106: A computing system manager states that the rate of interruptions to the internet service is 0.2 per week. Use the Poisson distribution to find the probability of a. one interruption in 3 weeks b. at least two interruptions in 5 weeks c. at most one interruption in 15 weeks Solution: Poisson distribution, $f(x;\lambda )=\frac{{\lambda }^{x}{e}^{-\lambda }}{x!}$ a. one interruption in 3 weeks Given that $x=1, \lambda = \alphaT = 0.2\dot 3 = 0.6$ $f(1;0.6 )=\frac{{0.6}^{1}{e}^{-0.6}}{1!} = 0.329$ b. at least two interruptions in 5 weeks Given that $x\ge 2, \lambda = \alphaT = 0.2dot 5 = 1$ $P(x\ge 2) = 1-P(x $P(x=0) = f(0;1)=\frac{{1}^{0}{e}^{-1 }}{0!} = 0.367879$ $P(x=1) = f(1;1)=\frac{{1}^{1}{e}^{-1 }}{1!} = 0.367879$ $P(x\ge 2) = 1-[P(x=0)+P(x=1)] = 1-(0.367879+0.367879) = 1-0.7357 = 0.264241.$ c. at most one interruption in 15 weeks Given that $x\le 1, \lambda = \alphaT = 0.2dot 15 = 3$ $P(x\le 1) = [P(x=0)+P(x=1)]$ $P(x=0) = f...

Problem 105: If a bank receives on the average α = 6 bad checks per day, what are the probabilities that it will receive (a) 4 bad checks on any given day? (b) 10 bad checks over any 2 consecutive days? Solution: Poisson distribution, $f(x;\lambda )=\frac{{\lambda }^{x}{e}^{-\lambda }}{x!}$ a. 4 bad checks on any given day Given that $x=4, \lambda = \alphaT = 6\dot 1 = 6$ $f(4;6 )=\frac{{6 }^{4}{e}^{-6 }}{4!} = 0.134$ b. 10 bad checks over any 2 consecutive days Given that $x=10, \lambda = \alphaT = 6\dot 2 = 12$ $f(10;12 )=\frac{{12 }^{10}{e}^{-12 }}{10!} = 0.104837$

Problem 104: Suppose that, next month, the quality control division will inspect 30 units. Among these, 20 will undergo a speed test and 10 will be tested for current flow. If an engineer is randomly assigned 4 units, what are the probabilities that a. none of them will need a speed test? b. only 2 will need a speed test? c. at least 3 will need a speed test? Solution: a. none of them will need a speed test Given that $x=0,n=4,a=20, N=30$ Hypergeometric distribution, $h(x;n,a,N)=\frac{\left(\begin{array}{c}a\\ x\end{array}\right)\left(\begin{array}{c}N-a\\ n-x\end{array}\right)}{\left(\begin{array}{c}N\\ n\end{array}\right)}$ $P(x=0) = h(0;4,20,30)=\frac{\left(\begin{array}{c}20\\ 0\end{array}\right)\left(\begin{array}{c}30-20\\ 4-0\end{array}\right)}{\left(\begin{array}{c}30\\ 4\end{array}\right)} = 0.007663.$ b. only 2 will need a speed test Given that $x=0,n=4,a=6, N=13$ Hypergeometric distribution, $h(x;n,a,N)=\frac{\left(\begin{array}{c}a\\ x\end{array}...

Problem 103: Among the 13 countries that an international trade federation is considering for their next 4 annual conferences, 6 are in Asia. To avoid arguments, the selection is left to chance. If none of the countries can be selected more than once, what are the probabilities that (a) all the conferences will be held in Asia? (b) none of the conferences will be held in Asia? Solution: a. all the conferences will be held in Asia Given that $x=4,n=4,a=6, N=13$ Hypergeometric distribution, $h(x;n,a,N)=\frac{\left(\begin{array}{c}a\\ x\end{array}\right)\left(\begin{array}{c}N-a\\ n-x\end{array}\right)}{\left(\begin{array}{c}N\\ n\end{array}\right)}$ $P(x=4) = h(4;4,6,13)=\frac{\left(\begin{array}{c}6\\ 4\end{array}\right)\left(\begin{array}{c}13-6\\ 4-4\end{array}\right)}{\left(\begin{array}{c}13\\ 4\end{array}\right)} = 0.020979.$ b. none of the conferences will be held in Asia Given that $x=0,n=4,a=6, N=13$ Hypergeometric distribution, $h(x;n,a,N)=\frac{\left...

Problem 102: Amaker of specialized instruments receives shipments of 24 circuit boards. Suppose one shipment contains 4 that are defective. An engineer selects a random sample of size 4. What are the probabilities that the sample will contain a. 0 defective circuit boards? b. 1 defective circuit board ? c. 2 or more defective circuit boards? Solution: a. 0 defective circuit boards Given that $x=0,n=4,a=4, N=24$ Hypergeometric distribution, $h(x;n,a,N)=\frac{\left(\begin{array}{c}a\\ x\end{array}\right)\left(\begin{array}{c}N-a\\ n-x\end{array}\right)}{\left(\begin{array}{c}N\\ n\end{array}\right)}$ $P(x=0) = h(0;4,4,24)=\frac{\left(\begin{array}{c}4\\ 0\end{array}\right)\left(\begin{array}{c}24-4\\ 4-0\end{array}\right)}{\left(\begin{array}{c}24\\ 4\end{array}\right)} = 0.456.$ b. 1 defective circuit board Given that $x=1,n=4,a=4, N=24$ Hypergeometric distribution, $h(x;n,a,N)=\frac{\left(\begin{array}{c}a\\ x\end{array}\right)\left(\begin{array}{c}N-a\\ n-x...