A manufacturer produces a certain type of components that has a length of $2.1 \,\mathrm{cm}$. the distribution of this components is normal with mean $2.15 \,\mathrm{cm}$ and sd $0.4 \,\mathrm{cm}$. There are $50$ pieces in a bag. Find the probability that a randomly selected bag will contain more than $30\%$ of the components with length longer than $2.1 \,\mathrm{cm}$. Thank You

  • $\begingroup$ What have you tried so far? Where exactly are you stuck on? Here are some hints: (1) Compute the probability that a component has a length > 2.1 cm. (2) Combine this with the binomial distribution. $\endgroup$ – Cavents Mar 20 '16 at 13:27
  • $\begingroup$ yes i have tried. The probability is P((2.1-2.15)/0.4) = P(Z>-0.125) = 0.54975. I think it is very stupid to combine this with the binomial distribution that i need to calculate 15times. $\endgroup$ – Nobi Mar 20 '16 at 13:36
  • $\begingroup$ No, it's not. It's a typical problem. $\endgroup$ – Em. Mar 20 '16 at 13:47
  • $\begingroup$ And you're expected to show your thoughts and attempts in every post. Formatting tips here. $\endgroup$ – Em. Mar 20 '16 at 13:47

The probability that one piece is longer than 2.1 cm is

$$P(X> 2.1)=1-P(X\leq 2.1)=1-\Phi\left( \frac{x-\mu}{\sigma} \right)=1-\Phi\left( \frac{2.1-2.15}{0.4} \right)=1-\Phi\left( \frac{-0.05}{0.4} \right)=1-\Phi(-0.125)$$

$\Phi(Z )$ is the cdf of the standard normal distribution. Looking at the table we get for $Z=-0.12$ a value of $0.45224$ and for $Z=-0.13$ a value of $0.44828$.

A linear interpolation gives $\Phi(-0.125)=\frac{0.45224+0.44828}{2}=0.45026\approx 45\%$

Now you can use the binomial distribution to find the probability that more than $30\%$ of the components have a length longer than 2.1cm.

$P(Y > 15)=1-P(Y\leq 15)=1-\sum_{k=0}^{15}{50 \choose k} 0.55^k\cdot 0.45^{50-k}$

It is time consuming to calculate this. You can do an approximation by using the standard normal distribution.

$P(Y\leq 15)=\Phi\left(\frac{x+0.5-\mu}{\sigma} \right)$ $=\Phi\left(\frac{15+0.5-0.55\cdot 50}{\sqrt{50\cdot 0.55\cdot 0.45}} \right)=\Phi(-3,41)=0.00032=0.32\%$

  • $\begingroup$ Consider providing hints and constructive questions instead of complete solutions. $\endgroup$ – Em. Mar 20 '16 at 13:53

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