Need a little help

[draco134]

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]

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.