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

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The discussion focuses on calculating the probability that 15 to 20 students out of 110 will secure an apprenticeship, given a success probability of 0.15. The normal approximation to the binomial distribution is recommended due to the large sample size, with a mean of 16.5 and a standard deviation of approximately 3.75. The interval for 15 to 20 students is interpreted as [14.5, 20.5] for the normal distribution. The user attempts to calculate the probability using Z-scores and expresses concern over potential mistakes in their calculations. The urgency of the situation is emphasized, as the user has limited time before their class.
bayan
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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!
 
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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].
 
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 \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].


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 :(

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
 
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