Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing

This Question is Closed

satellite73Best ResponseYou've already chosen the best response.0
3 are brown and so evidently 6 are not brown use the binomial \[\binom{9}{3}(.21)^3(.79)^6\]
 one year ago

dhruv_khatkarBest ResponseYou've already chosen the best response.0
i think you can use the formula (n x) . p^x . q ^(nx)
 one year ago

yashar806Best ResponseYou've already chosen the best response.0
what about the second one?
 one year ago

kropot72Best ResponseYou've already chosen the best response.0
For the third one you use the fact that the binomial distribution can be approximated by the normal distribution when n is large and p is neither small nor near 1.\[\mu=np=175\times 0.21\] \[\sigma=\sqrt{np(1p)}=\sqrt{175\times 0.21\times 0.79}\] Then you find the required probability by means of a standard normal distribution table.
 one year ago

kropot72Best ResponseYou've already chosen the best response.0
Basically you use the same method for the fourth one as for the third, but change the values of n and p accordingly.
 one year ago

kropot72Best ResponseYou've already chosen the best response.0
You need to find one value of zscore for the third one and two values of zscore for the fourth.
 one year ago

yashar806Best ResponseYou've already chosen the best response.0
because there two values 17% and 24%
 one year ago
See more questions >>>
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...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 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.