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?

The correct option is 81.85%.

Step-by-step Explanation:

We are given that the lengths of the titanium rods follow a normal distribution with the following parameters:
Mean ($\mu$) = 440 mm
Standard deviation ($\sigma$) = 1 mm

We want to find the percentage of rods whose lengths ($X$) lie between 438 mm and 441 mm. This can be written as finding the probability:

$P(438\le X\le 441)$

Step 1: Convert the X-values into standard normal variable (Z-scores)
The formula to convert a raw score $X$ to a standard score $Z$ is:

$Z=\frac{X-\mu }{\sigma }$

For the lower limit, $X=438$ mm:

$Z_1=\frac{438-440}{1}=-2$

For the upper limit, $X=441$ mm:

$Z_2=\frac{441-440}{1}=1$

Thus, we need to find the probability that $Z$ is between $-2$ and $1$:

$P(-2\le Z\le 1)$

Step 2: Find the cumulative probabilities from the standard normal table
Using standard normal distribution values (Z-table):
The cumulative probability for $Z=1$ is:

$P(Z<1)\approx 0.8413$

The cumulative probability for $Z=-2$ is:

$P(Z<-2)\approx 0.0228$

Step 3: Calculate the probability between the Z-scores

$P(-2\le Z\le 1)=P(Z<1)-P(Z<-2)$

$P(-2\le Z\le 1)=0.8413-0.0228=0.8185$

Step 4: Convert to percentage
Multiply the probability by 100 to get the percentage:

$0.8185\times 100\%=81.85\%$

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