I am fairly new to statistics and I would appreciate any help on this problem.
The length of a large batch of 1-inch nails has a normal distribution with μ = 1 and σ = 0.08
If a nail has a length < 0.9 it's too short, and if it has a length of > 1.15 it's too long.
If we pick 10 nails randomly, what's the probability that exactly one nail is too short and one nail is too long?
There are three categories of nails, too short, too long, and acceptable. The probabilities of drawing each are given by \begin{align*} p_s &= \Phi_{\mu,\sigma^2}(.9) \\ p_l &= 1 - \Phi_{\mu,\sigma^2}(1.15) \\ p_a &= \Phi_{\mu,\sigma^2}(1.15) - \Phi_{\mu,\sigma^2}(.9) \end{align*} Converting to the CDF of a standard normal, we have \begin{align*} \frac{X - \mu}{\sigma} \stackrel{(d)}{=} N(0,1) \Rightarrow \frac{X - 1}{.08} \stackrel{(d)}{=} N(0,1) \end{align*} So our probabilities can be written \begin{align*} p_s &= \Phi(-.1 / .08) \approx .106 \\ p_l &= 1 - \Phi(.15/.08) \approx .03 \\ p_a &= \Phi(.15/.08) - \Phi(-.1/.08) \approx .864 \end{align*} Thus the probability that we get exactly 1 short nail, 1 long, and the rest acceptable is \begin{align*} Pr = \frac{10!}{1! 1! 8!} p_s p_l p_a^{8} \approx 90 \cdot (0.106) \cdot (0.03) \cdot (0.864)^8 = 0.0897 \end{align*}