# Need a little help

#### draco134

##### New member
HI, I have a math problem that i just don't know how to work it out. Any help to try to understand it would be greatly appreciated....

the question is : According to survey data, 62% of housholds now own a DVD player. If 200 households are contacted at random, what is the probability that no more than 130 of them have a DVD player?

Thank you for helping me out....

#### galactus

##### Super Moderator
Staff member
This is a binomial probability. If you have a calculator, you can just use $$\displaystyle \sum_{k=0}^{130}C(200,k)(\frac{31}{50})^{k})(\frac{19}{50})^{200-k}$$.

But, in the cases where the numbers are rather large, we can use the 'short cut' for binomials.

$$\displaystyle np={\mu}, \;\ (200)(\frac{31}{50})=124$$

$$\displaystyle \sqrt{npq}={\sigma}, \;\ \sqrt{(200)(\frac{31}{50})(\frac{19}{50})}=\frac{\sqrt{1178}}{5}}\approx{6.864}$$

Now, use the z-score with your continuity correction:

$$\displaystyle z=\frac{130.5-124}{6.864}\approx{0.947}$$

Look it up in the table. It should be close to the first method at the top.