The lengths of a large stock of titanium rods follow a normal distribution with a mean (x) of 440 mm and a standard deviation (o) of 1 mm. What is the percentage of rods whose lengths lie between 438 mm and 441 mm?

To find the percentage of titanium rods whose lengths lie between 438 mm and 441 mm, we can use the properties of the normal distribution.

We are given the following parameters for the normal distribution of the rod lengths:
Mean ($\mu$) = 440 mm
Standard deviation ($\sigma$) = 1 mm

We want to find the probability (or percentage) that a randomly selected rod's length, $X$, falls in the interval:
$P(438\le X\le 441)$

First, we standardize the values of 438 mm and 441 mm by converting them into standard normal z-scores using the formula:
$Z=\frac{X-\mu }{\sigma }$

For $X_1=438$ mm:
$Z_1=\frac{438-440}{1}=-2$

For $X_2=441$ mm:
$Z_2=\frac{441-440}{1}=1$

So, we need to find:
$P(-2\le Z\le 1)=\Phi \left(1\right)-\Phi (-2)$

Using standard normal distribution table values:
$\Phi \left(1\right)\approx 0.8413$
$\Phi (-2)\approx 0.0228$

Now, we calculate the difference:
$P(-2\le Z\le 1)=0.8413-0.0228=0.8185$

To convert this probability to a percentage, we multiply by 100:
$0.8185\times 100\%=81.85\%$

Therefore, the percentage of titanium rods whose lengths lie between 438 mm and 441 mm is 81.85%.