Please help me (Probability Distribution)

r1terrell23

New member
Joined
Mar 15, 2006
Messages
3
I have a ten question quiz and I keep missing one question. I suck at Binomial Distribution. I set the problem up right but I keep getting the wrong answer. Here it is.

Suppose that 55% of all adults with children favor school voucher programs. A sample of 8 adults with children is taken. What is the probability that either 5 or 6 of the adults favor school vouchers?
 
\(\displaystyle \L
\left( \begin{array}{l}
8 \\
5 \\
\end{array} \right)\left( {.55} \right)^5 \left( {.45} \right)^3 + \left( \begin{array}{l}
8 \\
6 \\
\end{array} \right)\left( {.55} \right)^6 \left( {.45} \right)^2\)
 
pka said:
\(\displaystyle \L
\left( \begin{array}{l}
8 \\
5 \\
\end{array} \right)\left( {.55} \right)^5 \left( {.45} \right)^3 + \left( \begin{array}{l}
8 \\
6 \\
\end{array} \right)\left( {.55} \right)^6 \left( {.45} \right)^2\)

I got.010191
 
There's a pattern to the binomial distribution.

\(\displaystyle (p+q)^{8}=p^{8}+8p^{7}q+28p^{6}q^{2}+56p^{5}q^{3}+70p^{4}q^{4}+56p^{3}q^{5}+28p^{2}q^{6}+8pq^{7}+q^{8}\)

p=.55 and q=.45

Now, suppose you wanted to know the probability that 6 or

more favor vouchers. You'd use the parts of the distribution which

includes 6,7, and 8.

\(\displaystyle 28p^{6}q^{2}+8p^{7}q+p^{8}=28(.55)^{6}(.45)^{2}+8(.55)^{7}(.45)+(.55)^{8}=.22\)

Now, to do 5 or 6. Use \(\displaystyle 56p^{5}q^{3}+28p^{6}q^{2}\)

See how that works?. Use the parts of the expansion you need. It's a cool formula to know for probability. I always thought so anyway.
 
galactus said:
There's a pattern to the binomial distribution.

\(\displaystyle (p+q)^{8}=p^{8}+8p^{7}q+28p^{6}q^{2}+56p^{5}q^{3}+70p^{4}q^{4}+56p^{3}q^{5}+28p^{2}q^{6}+8pq^{7}+q^{8}\)

p=.55 and q=.45

Now, suppose you wanted to know the probability that 6 or

more favor vouchers. You'd use the parts of the distribution which

includes 6,7, and 8.

\(\displaystyle 28p^{6}q^{2}+8p^{7}q+p^{8}=28(.55)^{6}(.45)^{2}+8(.55)^{7}(.45)+(.55)^{8}=.22\)

Now, to do 5 or 6. Use \(\displaystyle 56p^{5}q^{3}+28p^{6}q^{2}\)

See how that works?. Use the parts of the expansion you need. It's a cool formula to know for probability. I always thought so anyway.


.413775?
 
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