Skip to main content

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?

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, $h(x;n,a,N)=\frac{\left(\begin{array}{c}a\\ x\end{array}\right)\left(\begin{array}{c}N-a\\ n-x\end{array}\right)}{\left(\begin{array}{c}N\\ n\end{array}\right)}$
$P(x=2) = h(2;4,20,30)=\frac{\left(\begin{array}{c}20\\ 2\end{array}\right)\left(\begin{array}{c}30-20\\ 4-2\end{array}\right)}{\left(\begin{array}{c}30\\ 4\end{array}\right)} = 0.311987.$

c. at least 3 will need a speed test
Given that $n=4,a=6, N=13$
Hypergeometric distribution, $h(x;n,a,N)=\frac{\left(\begin{array}{c}a\\ x\end{array}\right)\left(\begin{array}{c}N-a\\ n-x\end{array}\right)}{\left(\begin{array}{c}N\\ n\end{array}\right)}$
$P(x\ge 3) = P(x=3) + P(x=4)$
$P(x=3) = h(3;4,20,30)=\frac{\left(\begin{array}{c}20\\ 3\end{array}\right)\left(\begin{array}{c}30-20\\ 4-3\end{array}\right)}{\left(\begin{array}{c}30\\ 4\end{array}\right)} = 0.41592.$

$P(x=4) = h(4;4,20,30)=\frac{\left(\begin{array}{c}20\\ 4\end{array}\right)\left(\begin{array}{c}30-20\\ 4-4\end{array}\right)}{\left(\begin{array}{c}30\\ 4\end{array}\right)} = 0.1767.$

$P(x\ge 3) = P(x=3) + P(x=4) = 0.41592+0.1767 = 0.5927$

Labels:

Comments

Post a Comment

Popular posts from this blog

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 ...
Post a Comment
Read more

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 $x=10, \lambda = \alphaT = 6\dot 2 = 12$ $f(10;12 )=\frac{{12 }^{10}{e}^{-12 }}{10!} = 0.104837$
Post a Comment
Read more

Problem 56: Let $S = \{x \in \R^3 : x = (r-2s,3r+s,s), r,s \in \R \}$. Establish that $S$ is a subspace of $\R^3$.

Problem 56: Let $S = \{x \in \R^3 : x = (r-2s,3r+s,s), r,s \in \R \}$. Establish that $S$ is a subspace of $\R^3$. Solution: $S = \{x \in \R^3 : x = (r-2s,3r+s,s), r,s \in \R \}$. S is certainly nonempty. Let $x, y \in S$. Then for some $r, s, u,v \in \R$, $x = (r-2s,3r+s,s)$ and $y = (u-2v,3u+v,v)$. Hence, $x + y = (r-2s,3r+s,s) + (u-2v,3u+v,v)$ $= (r-2s+u-2v,3r+s+3u+v,s+v) = ((r+u)-2(s+v),3(r+u)+(s+v),(s+v))$ $= (k-2l,3k+l,l)$, where $k = r + u, l=s+v$. Consequently, $S$ is closed under addition. Further, if $c \in \R$, then $cx = c(r-2s,3r+s,s) = (c(r-2s),c(3r+s),cs) $ $= (cr-2cs,3cr+cs,cs) = (a-2b,3a+b,b)$, where $a = cr,b=cs$. Therefore $S$ is also closed under scalar multiplication. It follows that $S$ is a subspace of $\R^3$.
Post a Comment
Read more