STATS HELP PLEASE; MORE DIFFICULT THEN IT IS?

[asanz89]

Hey guys!

I am working on a math problem for my statistics class and I am stuck on this question. I feel like I might be making it more difficult then it is so if someone can please give some input and/or help that would be greatly appreciated. Ill show what I have done also so far.

Question:

Car color preference change over the years and according to the particular model that the customer selects. In a recent year, 10% of a ceratin brand of luxury cars sold are black. If 25 cars sold are selected randomly for a customer satifaction survey, find the probabilities:

a) Exactly 5 cars are black.

so I used the binomial formula..
n=25 x=5 p=0.10 q=0.90
p(X=5)= n!/x!(n-x)! * p^x * q^n-x and I got 0.0646
Just wondering if I have used the correcct approach and if I need to use the normal distribution table or not.

b)at least 4 cars are black.

for this I did P( x greater than/or equal to 4) = (x=0)+(x=1)+(x=2)+(x=3)+(x=4)

Is this also the correct approach to use?

Help would be greatly appreciated, thank you.

-Anthony

 

[galactus]

Car color preference change over the years and according to the particular model that the customer selects. In a recent year, 10% of a ceratin brand of luxury cars sold are black. If 25 cars sold are selected randomly for a customer satifaction survey, find the probabilities:

a) Exactly 5 cars are black.

Yes, use the binomial. You are correct. \(\displaystyle \binom{25}{5}(\frac{1}{10})^{5}(\frac{9}{10})^{20}\approx .0646\)

b)at least 4 cars are black.

You can find the probability that no more than three are not black and subtract from 1:

\(\displaystyle 1-\sum_{k=0}^{3}\binom{25}{k}(\frac{1}{10})^{k}(\frac{9}{10})^{25-k}\)

or

\(\displaystyle \sum_{k=4}^{25}\binom{25}{k}(\frac{1}{10})^{k}(\frac{9}{10})^{25-k}\)

If you are using the normal distribution for these, then use the continuity correction.

\(\displaystyle z=\frac{x-{np}}{\sqrt{np(1-p)}}\)

Since it says 'at least', subtract the result from 1.

\(\displaystyle 1-\frac{3.5-2.5}{1.5}\approx .2525\)

You really do not have to use the normal distribution for the binomial here because the sample is not that large, only 25.

If it were several hundred, then it is a good idea. This was used back before calculators onstead of summing up an enormous number of binomial probabilities.

 

[asanz89]

Thanks Galactus.

Not sure if you'd be able to verify this for me, if not somebody please lol.

diameters of douglas furs are normally distributed with mean of 4 inches and standard deviation of 1.5 inches.

I've been asked to take into consideration that the tree stand will expand up to a diamter of 6 inches. If a randomly selected tree is chosen, what is the probability that the tree will not fit in your christmas tree stand?

at first I thought I could do 1-P(4<x<6), in which I got an answer of 0.5918. Now I'm also thinking it could be done by P(x>6)?