# Probability distribution question including mean and variance

The lengths of the drive rods produced by a small engineering company are normally distributed with a mean of 118 cm and a standard deviation of $$0.3\ cm$$. Rods that have a length of more than $$118.37\ cm$$ or less than $$117.11\ cm$$ are rejected. Find the percentage of rods that are rejected.

This is a probability distribution question which I am stuck on. I don't know how to approach the question. I tried finding the value of $$p$$ by isolating it from the variance formula but I don't think that is the right approach for this question.

• This is not a binomial distribution problem, this is a normal distribution problem.
– kccu
Apr 8 '19 at 22:18
• Are you familiar with Z-scores? Can you normalize the bounds $118.37$ and $117.11$ into Z-scores, and the use these in conjunction with the standard normal distribution in some way (think area) to find a probability? Apr 8 '19 at 22:20

## 1 Answer

Perhaps this figure will help:

In Mathematica:

1 - Integrate[
PDF[NormalDistribution[118, 0.3], x],
{x, 117.11, 118.37}]

(* 0.110231 *)

• @Sara: In R, the proportion of rejected rods [shaded here(+1)] is computed using code 1 - diff(pnorm(c(117.11,118.37), 118, .3)), which returns 0.1102309. You will get a similar value starting with $1 - P(117.11 < X < 118.37),$ standardizing, and using printed normal tables. Apr 8 '19 at 23:33