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Problem 89: The probability that the noise level of a wide-band amplifier will exceed 2 dB is 0.05. Use Table 1 or software to find the probabilities that among 12 such amplifiers the noise level of (a) one will exceed 2 dB; (b) at most two will exceed 2 dB; (c) two or more will exceed 2 dB

Problem 89: The probability that the noise level of a wide-band amplifier will exceed 2 dB is 0.05. Use Table 1 or software to find the probabilities that among 12 such amplifiers the noise level of (a) one will exceed 2 dB; (b) at most two will exceed 2 dB; (c) two or more will exceed 2 dB

Solution:


a. one will exceed 2 dB;
Given that $n=12,$ and $p=0.05$
$P(x=1) = b(1;12,0.05)$
$b(x;n,p)=\left(\begin{array}{c}n\\ x\end{array}\right){p}^{x}(1-p)^{n-x}$
$P(x=1) = b(1;12,0.05) = \left(\begin{array}{c}12\\ 1\end{array}\right){0.05}^{1}(1-0.05)^{12-1} = 0.34128 $
b. at most two will exceed 2 dB;
Given that $n=12,$ and $p=0.05$
$P(x\le 2) = \sum _{x=0}^{2}b(x;n,p)=\sum _{x=0}^{2}b(x;12,0.05)$
$=b(0;12,0.05)+b(1;12,0.05)+b(2;12,0.05)$
$b(x;n,p)=\left(\begin{array}{c}n\\ x\end{array}\right){p}^{x}(1-p)^{n-x}$
$b(0;12,0.05) = \left(\begin{array}{c}12\\ 0\end{array}\right){0.05}^{0}(1-0.05)^{12-0} = 0.54036 $
$b(1;12,0.05) = \left(\begin{array}{c}12\\ 1\end{array}\right){0.05}^{1}(1-0.05)^{12-1} = 0.34128$
$b(2;12,0.05) = \left(\begin{array}{c}12\\ 2\end{array}\right){0.05}^{2}(1-0.05)^{12-2} = 0.09872$
$P(x\le 2) = b(0;12,0.05)+b(1;12,0.05)+b(2;12,0.05) =0.98043 $
c. two or more will exceed 2 dB
Given that $n=15,$ and $p=0.7$
$P(x\ge 2) = \sum _{x=2}^{12}b(x;n,p)=\sum _{x=2}^{12}b(x;12,0.05)$
$ = 1 -\sum _{x=0}^{2}b(x;12,0.05) = 1-[b(0;12,0.05)+b(1;12,0.05)]$
$b(x;n,p)=\left(\begin{array}{c}n\\ x\end{array}\right){p}^{x}(1-p)^{n-x}$
$b(0;12,0.05) = \left(\begin{array}{c}12\\ 0\end{array}\right){0.05}^{0}(1-0.05)^{12-0} = 0.54036 $
$b(1;12,0.05) = \left(\begin{array}{c}12\\ 1\end{array}\right){0.05}^{1}(1-0.05)^{12-1} = 0.34128$
$P(x\ge 2) = 1-[b(0;12,0.05)+b(1;12,0.05)] = 1-0.8816 = 0.1184$

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