Binomial Theorem > Normal Approximation to solve a binomial problem

## Normal Approximation: Overview

When n * p and n * q are greater than 5, you can use the normal approximation to solve a binomial distribution problem. This article shows you how to solve those types of problem using the

**continuity correction factor**.

## Normal Approximation: Example #1 (Video)

## Normal Approximation: Example#2

Sixty two percent of 12th graders attend school in a particular urban school district. If a sample of 500 12th grade children are selected, find the probability that at least 290 are actually enrolled in school.

**Step 1:** *Determine p,q, and n:*

p is defined in the question as 62%, or 0.62

To find q, subtract p from 1: 1 – 0.62 = 0.38

n is defined in the question as 500

**Step 2:*** Determine if you can use the normal distribution*:

n * p = 310 and n * q = 190. These are both larger than 5.

**Step 3:** *Find the mean, μ by multiplying n and p*:

n * p = 310

**Step 4:** *Multiply step 3 by q *:

310 * 0.38 = 117.8.

**Step 5:** *Take the square root of step 4 to get the standard deviation, σ*:

sqrt(117.8)=10.85

**Note**: The formula for the standard deviation for a binomial is &sqrt;(n*p*q).

**Step 6:** *Write the problem using correct notation*:

P(X≥290)

**Step 7:** *Rewrite the problem using the continuity correction factor*:

P (X ≥ 290-0.5)= P (X ≥ 289.5)

**Step 8:** *Draw a diagram with the mean in the center*. Shade the area that corresponds to the probability you are looking for. We’re looking for X ≥ 289.5, so…

**Step 9:*** Find the z-score*.

You can find this by subtracting the mean (μ) from the probability you found in step 7, then dividing by the standard deviation (σ): (289.5 – 310) / 10.85 = -1.89

**Step 10:** *Look up the z-value in the z-table: *

The area for -1.819 is 0.4706.

**Step 11:** *Add .5 to your answer in step 10 to find the total area pictured:*

0.4706+ 0.5 = 0.9706.

That’s it! The probability is .9706, or 97.06%.

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Prof,or anyone who knows, is there a way to do this problem using my ti-89 titanium calculator?

If n*p and p*q are not greater than or equal to 5, what do you do then?

oh my goodness, i feel like kissing you right now! thankyou thankyou thankyou! i will be buying the book for my end semester exam!!!

God bless you!

Professor, there is something I don’t get in this example – how can p*q be >5? since they are both probabilities, this means they are maximum 1 each (and not in the same time :), so p*q can’t be >1 ever (in fact, I guess the max p*q =.25)

If n*p is less than 30 or in this case <5, then instead of normal distribution, you would be using t-distribution as the central limit theorem does not apply. t-distribution is similar but you would use the t-table chart and find the t-value instead of the z-value and work it the same way after

rc we are talking about n*p or n*q being greater than 5

prof plz tell me the procedure to solve this problem by using normal distribution

if p(x>a)=0.4995 then find the value of a

its a very good example ,thnks professor…:)

Hello. In step 11 is it necessary to add 0.5 to the calculated area?

Lianne,

Yes — if you are using the z-tables on this website.

Regards.

Stephanie

Wouldn’t it just be easier to find the z-score of -1.89

And then subtract it. So it would be 1-.0294

The last two steps confuse me greatly.

Eddie,

There’s more than one way to get the answer in stats. This is the way I taught my students, but do whatever works for you :)

Stephanie

Well I do have a question how would we find the t-table and will the answer be correct since my z table stops at 3.49

z- score of -1.819 is 0.0294 and why did you add 0.5 in that. we can simply subtract this z-value from 1 and we will get the answer as z-score is always calculate CDF (area to the left of the curve). Since we are interested in knowing area to the right of the curve.

The z-score isn’t always the whole area to the left of the curve. Sometimes it’s the area between the mean and the score (as in this case). You have to add .5 to account for the area to the left of the mean.

hi prof.plz guide me when should we use correction factor and when not we need to use this?how can v differentiate?…….

You use it when you have a discrete function and you want to use a continuous function (usually a binomial to a normal or, more rarely, a Poisson with a large λ to a normal). These are the only two cases I can think of where you would want to use it…

Excuse me, I want to ask you something.

Actually i did an exercise about this topic.

And I got the answer

a. P=77.29% by binomial

b. P=76.94% by normal approximation

The question is how do I comment on answer (b) to exact percentage found in answer (a)?

Lim,

I’m not sure what you mean by “The question is how do I comment on answer (b) to exact percentage found in answer (a)?”. Could you rephrase your question?

Thanks.

I mean “how to comment on (b) to exact percentage found in answer (a)”?

This exercise requires to calculate the percentage of probability by two ways. Question (a) calculate by binomial, Question (b) calculate by normal approximation plus compare and explain and comment why it is similar or equal to (a). But I can only calculate and cannot comment on it.

OK, if I have this right, you have found (a), your binomial, and (b) the normal approximation. What are your results? I’m assuming they are different: the question is probably asking you to say what the difference is and why (i.e. because you’re using the two different methods).

OMG….. at first it was kind of difficult…… But i got it now.. thanks a lot

Dear Professor,

Both examples are very clear up to the point of z tables.

I noticed you used the A-3 Table for both examples on this page. What is not clear to me is what criteria you use to determine what column on the Z table can be used. It is not clear to me why you used Column D for question 1 and column A for example 2. Thanks for explaining why.

Hi, Eric,

Thanks for your question. I’m using this table. Not sure what table you’re looking at that has columns D and A?