Homework Help: Urgent help required!

1. Sep 11, 2005

bayan

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 plz.

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

Cheers!

2. Sep 12, 2005

HallsofIvy

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 $$\sqrt{p(1-p)N}$$. 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].

3. Sep 12, 2005

bayan

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

I absloutly hate probability so I am abit lost :(

Thanx for your reply

4. Sep 13, 2005

bayan

anyone knows how I could get an answer?

5. Sep 13, 2005

bayan

can anyone confirm if this is right?

6. Sep 13, 2005

bayan

damnz,

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

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

How does it look? any obvious mistakes?

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

Cheers

Last edited: Sep 13, 2005
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