Problem 81: Find the probability of getting five heads and seven tails in 12 flips of a balanced coin.

Solution:

Given that $x=5,n=12,$ and $p=\frac{1}{2}$
Binomial distribution given by
$b(x;n,p)=\left(\begin{array}{c}n\\ x\end{array}\right){p}^{x}(1-p)^{n-x}$
$b(5;12,\frac{1}{2})=\left(\begin{array}{c}12\\ 5\end{array}\right){\frac{1}{2}}^{5}(1-\frac{1}{2})^{12-5}$
$=^{12}C_{5}(\frac{1}{2})^{12} =792(\frac{1}{2})^{12} = 0.19$.

Comments

Problem 85: A manufacturer of external hard drives claims that only 10% of his drives require repairs within the warranty peroid of 12 months.

Problem 85: A manufacturer of external hard drives claims that only 10% of his drives require repairs within the warranty peroid of 12 months. a. If 5 of 20 of his drives required repairs within the first year, does this tend to support or refuse the claim? b. If 3 of 20 of his drives required repairs within the first year, does this tend to support or refuse the claim? Solution: a. If 5 of 20 of his drives required repairs within the first year Given that $n=20,$ and $p=0.3$ $\sum _{x=5}^{20}b(x;n,p)=\sum _{x=5}^{20}b(x;20,0.1)=1-\sum _{x=0}^{4}b(x;20,0.1)$ $ = 1 - [b(0;20,0.1)+b(1;20,0.1)+b(2;20,0.1)+b(3;20,0.1)+b(4;20,0.1)]$ $b(x;n,p)=\left(\begin{array}{c}n\\ x\end{array}\right){p}^{x}(1-p)^{n-x}$ $b(0;20,0.1) = \left(\begin{array}{c}20\\ 0\end{array}\right){0.1}^{0}(1-0.1)^{20-0} = 0.121577$ $b(1;20,0.1) = \left(\begin{array}{c}20\\ 1\end{array}\right){0.1}^{1}(1-0.1)^{20-1} = 0.27017$ $b(2;20,0.1) = \left(\begin{array}{c}20\\ 2\end{array}\right){0.1}^{...

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 $\R^3$.

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.

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!}...

upāyaṁ karoti

Search This Blog