Problem 86: Human error is given as the reason for 75% of all accidents in a plant. Use the formula for the binomial distribution to find the probability that human error will be given as the reaosn for two of the next four accidents.

Solution:

a. Given that $x=2, n=4,$ and $p=0.75$
$b(x;n,p)=\left(\begin{array}{c}n\\ x\end{array}\right){p}^{x}(1-p)^{n-x}$
$b(2;4,0.75) = \left(\begin{array}{c}4\\ 2\end{array}\right){0.75}^{2}(1-0.75)^{4-2}$
$ = ^{4}C_{2}{0.75}^{2}(0.25)^{2}= 6({0.75}^{2})(0.25^{2}) = 0.2109.$

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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?

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 98: If 6 of 18 new buildings in a city violate the building code, what is the probability that a building inspector, who randomly selects 4 of the new buildings for inspection, will catch
a. none of the buildings that violate the building code?
b. 1 of the new buildings that violate the building code?
c. 2 of the new buildings that violate the building code?
d. at least 3 of the new buildings that violate the building code?

Problem 98: If 6 of 18 new buildings in a city violate the building code, what is the probability that a building inspector, who randomly selects 4 of the new buildings for inspection, will catch a. none of the buildings that violate the building code? b. 1 of the new buildings that violate the building code? c. 2 of the new buildings that violate the building code? d. at least 3 of the new buildings that violate the building code? Solution: a. none of the buildings that violate the building code? Given that $x=0,n=4,a=6, N=18$ Hypergeometric distribution, $h(x;n,a,N)=\frac{\left(\begin{array}{c}a\\ x\end{array}\right)\left(\begin{array}{c}N-a\\ n-x\end{array}\right)}{\left(\begin{array}{c}N\\ n\end{array}\right)}$ $P(x=0) = h(0;4,6,18)=\frac{\left(\begin{array}{c}6\\ 0\end{array}\right)\left(\begin{array}{c}18-6\\ 4-0\end{array}\right)}{\left(\begin{array}{c}18\\ 4\end{array}\right)} = 0.16176.$ b. 1 of the new buildings that violate the building code? Given ...

Problem 101: Among the 300 employees of a company, 240 are union members, while the others are not, If 8 of the employees are chosen to serve on the administrative committee, find the probability that 5 of them will be union member while the others are not.
a. the formula for the hypergeometric distribution;
b. the formula for the binomial distribution as an approximation.

Problem 101: Among the 300 employees of a company, 240 are union members, while the others are not, If 8 of the employees are chosen to serve on the administrative committee, find the probability that 5 of them will be union member while the others are not. a. the formula for the hypergeometric distribution; b. the formula for the binomial distribution as an approximation. Solution: a. the formula for the hypergeometric distribution Given that $x=5,n=8,a=240, N=300$ Hypergeometric distribution, $h(x;n,a,N)=\frac{\left(\begin{array}{c}a\\ x\end{array}\right)\left(\begin{array}{c}N-a\\ n-x\end{array}\right)}{\left(\begin{array}{c}N\\ n\end{array}\right)}$ $P(x=5) = h(5;8,240,300)=\frac{\left(\begin{array}{c}240\\ 5\end{array}\right)\left(\begin{array}{c}300-240\\ 8-5\end{array}\right)}{\left(\begin{array}{c}300\\ 8\end{array}\right)} = 0.1470.$ b. the formula for the binomial distribution as an approximation? Given that $x=5,n=8,p = \frac{240}{300}$ $P(x= 5) =...

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