Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

If the probability of a defective fuse from a manufactuning unit is $2\%$ in a box of $200$ fuses find the probability that more than $3$ fuses are defective$ [e^{-4}=0.0183].$

Can you answer this question?

1 Answer

0 votes
  • A random variable $X$ is said to have a poisson distribution of the probability mass function of $X$ is
  • $P(X=x)=\large\frac{e^{\Large -\lambda }\lambda^x}{x!}$$\qquad (x=0,1,2........$ for some $\lambda > 0)$
  • Constants of a poisson distribution :
  • Mean=Variance=$\lambda$
  • A continuous random variable $X$ is said to follow a normal distribution with parameter $\mu$ and $\sigma$ (or $\mu$ and $\sigma^2$) if the probability density function is
  • $f(x)=\large\frac{1}{\sigma \sqrt{2\pi}}$$e^{-\large\frac{1}{2}(\frac{x-\mu}{\sigma})^2};-\infty < x < \infty,-\infty< \mu <0$ and $\sigma > 0$
  • $X\sim N(\mu,\sigma)$
  • Constants of a normal distribution :
  • Mean =$\mu$,variance =$\sigma^2$,standard deviation =$\sigma$
Step 1:
Let $X$ be the random variable denoting the number of defective fuses in a box of 200 fuses.
Probability of a defective fuse P=0.02
$\quad=200\times 0.02$
$\therefore X\sim P(4) $
$\Rightarrow P(X=x)=\large\frac{e^{-4}4^4}{4!}$$\qquad x=0,1,2......$
Step 2:
Probability that more than 3 fuses are defective.
$P(X > 3)=1-P(X \leq 3)=1-e^{-4}[\large\frac{4^0}{0!}+\frac{4^1}{1!}+\frac{4^2}{2!}+\frac{4^3}{3!}]$
answered Sep 18, 2013 by sreemathi.v

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App