# Normal approximation to binomial distribution.

• Jun 21st 2010, 03:50 PM
lilyungn
Normal approximation to binomial distribution.
Hey, I'm stuck on a homework problem that I've been working on...

Transportation officials tell us that 80% of drivers wear seat belts while driving. What is the probability of observing 518 or fewer drivers wearing seat belts in a sample of 700 drivers?

Hint: Use normal distribution to approximate the binomial distribution

What I have so far is

xbar = 518
μ = mean = (n * p) = (700 * .80) = 560
σ = standard deviation = sqrt(n * p * q) = sqrt(700 * .80 * .20) = 10.583

Any help will be greatly appreciated. Thanks
• Jun 21st 2010, 04:20 PM
mr fantastic
Quote:

Originally Posted by lilyungn
Hey, I'm stuck on a homework problem that I've been working on...

Transportation officials tell us that 80% of drivers wear seat belts while driving. What is the probability of observing 518 or fewer drivers wearing seat belts in a sample of 700 drivers?

Hint: Use normal distribution to approximate the binomial distribution

What I have so far is

xbar = 518
μ = mean = (n * p) = (700 * .80) = 560
σ = standard deviation = sqrt(n * p * q) = sqrt(700 * .80 * .20) = 10.583

Any help will be greatly appreciated. Thanks

Have you reviewed any examples in your class notes or textbook? Where are you stuck here?
• Jun 21st 2010, 04:28 PM
lilyungn
How do I calculate the probability? I dont understand what it means by use normal distribution to approximate the binomial distribution
• Jun 21st 2010, 04:36 PM
mr fantastic
Quote:

Originally Posted by lilyungn
How do I calculate the probability? I dont understand what it means by use normal distribution to approximate the binomial distribution

What do your classnotes and textbook say? Probably similar to what is said here, for example: http://courses.wcupa.edu/rbove/Beren...section6_5.pdf
• Jun 21st 2010, 04:48 PM
lilyungn
When I do it in the calculator I get a binomial of 0.000072 ... is that the right answer?
• Jun 21st 2010, 04:54 PM
mr fantastic
Quote:

Originally Posted by lilyungn
When I do it in the calculator I get a binomial of 0.000072 ... is that the right answer?

The answer is correct but how you got the answer is wrong. You have to use the normal approximation.

In an exam, if the question prescribes the method but you do it a different way you will get zero marks.
• Jun 21st 2010, 04:58 PM
lilyungn
How can I use the normal approximation, I'm allowed to use a TI-83 in class.
• Jun 21st 2010, 05:06 PM
mr fantastic
Quote:

Originally Posted by lilyungn
How can I use the normal approximation, Mr F says: Did you read the link I posted earlier?

I'm allowed to use a TI-83 in class. Mr F says: As I have already said, IF the question prescribes the method, then you have to do it that way. If the question does not precribe the method, then you can do it any way you please.

..
• Jun 22nd 2010, 06:42 AM
lilyungn
Is it possible you can show me showing the work how you use it?
• Jun 22nd 2010, 08:36 AM
SpringFan25
You want
$P(X \leq 518)$ where $X \sim Bi(700,0.8)$

We'll get an approximate answer by using another variable, Y
$Y \sim N(\mu,\sigma^2)$

You have already worked out that
$\mu = np =560$
$\sigma^2 = np(1-p) = 112$
Do you understand why these values are used? We are making the mean and variance of the normal distribution equal to the mean and variance of the binomial distribution we want to approximate.

Now:
$P(X \leq 518) \approx P(Y \leq 518.5)$ (continuity correction)
$P(Y \leq 518.5) = P \left( \frac{Y-560}{\sqrt{112}} < \frac{518.5-560}{\sqrt{112}} \right)$ (standardise)

$P(Y \leq 518.5) = \Phi \left( \frac{518.5-560}{\sqrt{112}} \right)$
$P(Y \leq 518.5) = 0.000044$