# Normal Approximation Binomial

The probability that an appliance is in repair is 0.5. If an apartment complex has 100 such appliances, what is the probability that at least 60 will be in repair? Use the normal approximation to the binomial.

I calculated n*p = 100*.5=mean I calculated the standard deviation to be 100*.5 (1-.5) = 25 then the squareroot of 25 is 5 for the standard deviation

For the z score I did $Z = \frac{60 - 50}{5} = 2$. Probability associated with that z-score is .9772.

I'm not sure if I did the z score correctly for normal approximation? Also having difficulty in determining if .9772 is my answer or should I subract that from 1??

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+1 for a good attempt on your question. :) – user61752 Mar 8 '13 at 3:56

Use normal with mean $np$ and variance $np(1-p)$.

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A good rule of thumb to see if you can approximate your binomial distribution with your normal, is if you multiply your n-value and probability together ($np$), and if $np \geq 35$ (which it is 'cause you got 50),. then you can definitely approximate your binomial distribution with a normal distribution.

As far as I can tell, what you did seems to be correct. Your Z-value is right, and you could always check your probability by plugging into a calculator, if you have a distribution tab on your calculator. Since you want to know the X = the number of appliances in need of repair, $P(X \geq 60) = 1-P(X < 60)$ and see if that Probability matches. And you know your upper limit will have to be 100, since that's the n-value. Will check answer when I find my new batteries for my calculator.

Note: If you used the Standard Normal Table, then that table takes the area under the curve to the left of the line (your lower limit). Since you want the values greater than 60, then you will want to take $1-$ your probability.

Edit: I finally borrowed a calculator. I used the binomial distribution cdf option on the TI-89. $P(X \in [60,100]) = .28443966363$. Hope this helps! I am also not sure where you are supposed to round your answer to, so I didn't round it for you.

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The Normal Approximation to the Binomial Distribution can be Defined as:

$P$($a \leq X \leq b$)$\approx \phi(\frac{b+\frac{1}{2}-\mu}{\sigma})-\phi(\frac{a-\frac{1}{2}-\mu}{\sigma})$

So to find what you're looking for is

$P(X \geq a)= 1-\phi(\frac{a-\frac{1}{2}-\mu}{\sigma})$

The rest is just computing values.

$a=60$

$\mu=np=(100)(0.5)=50$,

$\sigma=\sqrt{npq}=\sqrt{(100)(0.5)(0.5)}=\sqrt{25}=5$

Now put the values in for $\phi$

$P(X \geq 60)= 1- \phi (\frac{60-\frac{1}{2}-50}{5})$ $P(X \geq 60)=1-\phi(1.9)$

then from looking in the Normal Table I found the value of $\phi (1.9)\approx 0.8621$

$P(X \geq 60)=1-0.8621=0.1379$