SAC Probability Help: Find the Probability of 15-20 Students

  • Thread starter Thread starter bayan
  • Start date Start date
  • Tags Tags
    Probability
Click For Summary

Homework Help Overview

The discussion revolves around a probability problem related to a binomial distribution, specifically focusing on the likelihood of a certain number of students receiving apprenticeships from a larger group. The original poster is attempting to find the probability that between 15 and 20 students out of 110 will get an apprenticeship, given a success probability of 0.15.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the use of the normal approximation for the binomial distribution due to the large sample size. There are questions about the interpretation of the interval for the probability calculation and the application of the normal distribution.

Discussion Status

Some participants have provided guidance on using the normal approximation and have shared calculations related to the mean and standard deviation. There is an ongoing exploration of methods, with no explicit consensus reached on the final approach or calculations.

Contextual Notes

The original poster expresses frustration with the problem and mentions a time constraint, indicating a sense of urgency in finding a solution. There are also indications of confusion regarding the various methods available in textbooks.

bayan
Messages
202
Reaction score
0
Hi guys.

There was a question in my SAC which I still have an hour to finish, that I had no Idea about.

The question was like following.

5 students graduated from a school which has a apprenticeship probability of .15.

I answered a) b) and c) which was to find the probability of none getting an apprenticeship and one getting it and at least one gettingit.


In part D it say using Binomial method find the probability that 15 to 20 students (Inclusive) out of 110 will get apprenticeship.


Can someone help me please.

I have made no progress from getting any info from the question :(

Cheers!
 
Physics news on Phys.org
I suspect your time is up by now but with numbers as large as 110 students, I would be inclined to use the normal approximation. For large N, a binomial distribution with probability of "success" p can be approximated by a normal distribution with mean pN and standard deviation [tex]\sqrt{p(1-p)N}[/tex]. Here p= 0.15 so the normal approximation has mean 16.5 and standard deviation 3.75 (approx). Since the normal distribution is continuous we interpret "between 15 to 20 students" as the interval [14.5, 20.5].
 
HallsofIvy said:
I suspect your time is up by now but with numbers as large as 110 students, I would be inclined to use the normal approximation. For large N, a binomial distribution with probability of "success" p can be approximated by a normal distribution with mean pN and standard deviation [tex]\sqrt{p(1-p)N}[/tex]. Here p= 0.15 so the normal approximation has mean 16.5 and standard deviation 3.75 (approx). Since the normal distribution is continuous we interpret "between 15 to 20 students" as the interval [14.5, 20.5].


Can you please clarify how I can get the answer? I took a look at my textbookbut there are many diffrent methods of doing it :(

I absloutly hate probability so I am abit lost :(

Thanx for your reply
 
damnz,

I accidently deleted the other post
here is it again :(

[tex]X ~ N (16.5,14.06) Pr(15 < Z < 20)[/tex]
[tex]Pr((\frac{15-16.5}{3.75})(\frac{20-16.5}{3.75}))[/tex]
[tex]Pr(-.4 < Z < .933)[/tex]
[tex]Pr(Z < .933)- Pr(Z > .4)[/tex]
[tex]Pr(Z <.933) - (1-Pr(Z < .4))[/tex]
[tex]1.5-1.25=.25[/tex]

How does it look? any obvious mistakes?

i only have another 5 hours left before I have that class again.

Cheers
 
Last edited:

Similar threads

Replies
28
Views
2K
  • · Replies 17 ·
Replies
17
Views
3K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 11 ·
Replies
11
Views
3K
Replies
29
Views
3K
Replies
5
Views
2K
Replies
9
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
9
Views
3K