The cross-section area of plastic tubing for use in pulmonary resuscitators in normally distributed with $\mu = 12.5mm^{2} $ and $\sigma = 0.2 mm^{2}$ . When the area is less than $12 mm^{2}$ or larger than $13 mm^{2}$, the tube does not fit properly. If the tubes are shipped in boxes of $1000$, how many wrong-sized tubes per box can doctors expect to find?
Attempt: we need to find the area when it is less than $12 mm^{2}$ or larger than $13 mm^{2}$, the tube does not fit properly.
Then using the above information and finding the area under the standard normal distribution together with the normal table , we have $P(X < 12 $ or $ X > 13) = P(X <12) + P(X > 13) = P(\frac{X-\mu}{\sigma} < \frac{12-\mu}{\sigma}) + 1 - P(\frac{X-\mu}{\sigma} \leq \frac{13-\mu}{\sigma}) = P(Z < \frac{12-12.5}{0.2}) + [1 - P({Z} \leq \frac{13-12.5}{0.2})] = P(X < -2.5) + 1 - P(Z \leq 2.5) = 0.0062 + 1 - 0.9938 = 0.0124 $
Thus the number of wrong-sized tubes per box that doctors can expect to find is the probability times $100$ Thus, $100*0.0124 = 12.4$
Can anyone please verify this? Any help/feeback would really help. Thank you,
The answer is about right, but there are two errors that happen to cancel!
The probability a tube is bad (too big or too small) is roughly $2(0.0062)$, that is, $0.0124$. And you need to multiply by $1000$, not by $100$, giving expected value about $12.4$.