Binomial Distribution using normal curve approximation

[lizzpalmer]

Hi

I have 3 questions that all fall under this topic that I need help with.

First Question:

For certain bird species, with approximate assumptions, the number of nests escaping predation has a binomial distribution. Suppose the porbability of success (that is, a nest's escaping predation) is .3 Find the probability that at least half of the 26 nests escape predation.

Second Question:

In 2007-2008, 66% of all undergraduates receive some type of financial aid. Suppose 50 undergraduates are selected at random. Find the probability that at least 35 received financial aid. Find the probability that at most 25 received financial aid.

Third Question:

In 2007-2008, 34% of all undergraduates took out a Federal Stafford Loan. If a random sample of 250 students is selected, find the probability that at most 75 students took out a Federal Stafford Loan.

If someone could help me figure out how to work these problems, it would be greatly appreciated!

Thank you

 

[galactus]

First Question:

For certain bird species, with approximate assumptions, the number of nests escaping predation has a binomial distribution. Suppose the probability of success (that is, a nest's escaping predation) is .3 Find the probability that at least half of the 26 nests escape predation.

When using the binomial approximation, we use \(\displaystyle z=\frac{x-\mu}{\sigma}\)

Where \(\displaystyle \mu=np, \;\ \sigma=\sqrt{np(1-p)}\)

In this case, \(\displaystyle p=.3, \;\ np=(.3)(26)=7.8, \;\ \sigma=\sqrt{26(.3)(.7)}\approx 2.34\)

We use the continuity correction. In this case, it says "at least half". So "at least 13" escape predation". Instead of using x=13, we use x=12.5

\(\displaystyle P(x\geq 12.5)\)

\(\displaystyle z=\frac{12.5-7.8}{2.34}\approx 2.01\)

So, \(\displaystyle P(x\geq 12.5)=P(z\geq 2.01)\)

Looking z=2.01 up in the z table we see it is .9778

But, it asks for "at least 13". So, we go to the right.

This means we have to subtract from 1.

\(\displaystyle 1-.9778=.0222\)

The probability at least half out of 26 nests escape predation is about 2.22%

Now, I showed this one. Try the other two and let us know what you get.

If we compare this to the binomial probability, \(\displaystyle \displaystyle\sum_{k=13}^{26}\binom{26}{k}(.3)^{k}(.7)^{26-k}=.0255\)

Pretty close to what we got before with the approximation.

The reason the binomial approx. is used is when the population/sample is large and it becomes daunting to calculate. In my opinion, in these days of tech, it has become rather obsolete.

One can easily run the last formula through a calculator. But hey, it is good to know.

 

[lizzpalmer]

Is it the same formula for all three questions?

 

[lizzpalmer]

for question number 2 for the first part I got 82.64%

now with the sceond part: at most 25 receive financial aid, Im not sure on the formula bacause I think it would be different that the other two, correct?

 

[lizzpalmer]

at most?

Im still not sure what 'at most' would change in the formula. If anyone could explain I'd appreciate it. Thanks

 

[galactus]

Yes, 'at most' changes from 'at least'.

For #2, at least 35: \(\displaystyle \displaystyle\sum_{k=35}^{50}\binom{50}{k}(.66)^{k}(.34)^{50-k}=.332\)

Using the approximation, we have \(\displaystyle \mu=(50)(.66)=33, \;\ \sqrt{50(.66)(.34)}=3.35\)

Since the problem asks for 'at least 35' we use a continuity correction of 34.5

\(\displaystyle z=\frac{34.5-33}{3.35}\approx .45\)

Look this up in the table and we see .45 corresponds to .6736.

Subtract from 1 and get .3264

Pretty close to the binomial formula at the top.

Now, use the same process with the 'at most' except the continuity correction will add .5 to the x value you find. You do not have to subtract from 1 because of the 'at most'. That is because we come from the left and go right under the normal curve. 'At most' would mean anything from 0 up to, and including, 25.

 

[lizzpalmer]

thanks!