A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing


  • 5 years ago

A watch company determines that each box of 500 watches has an average of 10 defective watches with standard deviation 3. Suppose that 1000 boxes are produced. Use Chebychev′s Theorem to estimate the number of boxes having between 0 and 20 defective watches. Round the answer to the nearest integer.

  • This Question is Closed
  1. dumbcow
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Chebychev′s Inequality states given standard deviation k no more than 1/k^2 of values will be more than k deviations from mean to find k we see 20 is 10 more than our mean of 10 watches 10 is 10/3 deviations k = 10/3 k^2 = 100/9 1/k^2 = 9/100 9/100 * 1000 = 90 Therefore no more than 90 boxes will have more than 20 defective watches 1000-90=910 910 boxes will have between 0 and 20 defective watches Hope this helps

  2. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...


  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.