0
$\begingroup$

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.

$\endgroup$
  • 3
    $\begingroup$ This is not a binomial distribution problem, this is a normal distribution problem. $\endgroup$ – kccu Apr 8 at 22:18
  • 1
    $\begingroup$ 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? $\endgroup$ – Brian Apr 8 at 22:20
1
$\begingroup$

Perhaps this figure will help:

enter image description here

In Mathematica:

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

(* 0.110231 *)

$\endgroup$
  • $\begingroup$ @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. $\endgroup$ – BruceET Apr 8 at 23:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.