probability problem

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ruthlathe

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I am having trouble with this problem, and if I can get guidance to just one of the answers I got wrong, I can probably figure out the rest.
A study conducted by the AC Nielsen Company found that 59% of Americans can name the Three Stooges, but only 17% can identify at least three Supreme Court Justices.
a. In a random sample of 30 Americans, what is the probablilty that no more than 5 can identify at least three Supreme Court justices?

My work: using Cn,r X p X q, where n = 30 and r = 5, p = .17, q = 1-.17 = .83:
30C5(.17)to the 5th power (.83) to the 25th power
= .1919, which is wrong, but I don't know if I am using the formula incorrectly, or if I am using an incorrect formula.
b, which I got right: in a random sample of 50 Americans, what is the probablilty that 40 cannot identify at least three Supreme Court justices: 50C40 (.83)to the 40th power X (.17) to the 10th power
= .1200

c. another incorrect answer: in a random sample of 100 Americans, what is the probablity that at least half can name the Three Stooges?
My work: using Cn,r X p X q, where n = 100, r = 50, p = .59, q = .41:

100C50 (.59) to the 50th power (.41) to the 50th power
= .0153
d. another incorrect answer: In a random sample of 100 Americans, what is the probability that between 60 and 70 (including 60 and 70) can name the Three Stooges?
My work: {100C59 X (.59) to the 59th power (.41) to the 41st power} - {100C70(.59) to the 70th power (.41) to the 30th power}
= 0.0744
 
I think I was applying the wrong formula. Can I run this by you, though? I finally used the binomial cdf function of my T83, so
a. binomcdf(30,.59.5) = 2.4424 to the -6
b. i already got that correct
c. at least half of 100 is greater than or equal to 50, so I used:
1 - binomcdf(100, .59.50) = .9572
d. 1 - { binomcdf9100,.59, 70) - binomcdf(100,.59,59) = .5467
 
ruthlathe said:
I am having trouble with this problem, and if I can get guidance to just one of the answers I got wrong, I can probably figure out the rest.
A study conducted by the AC Nielsen Company found that 59% of Americans can name the Three Stooges, but only 17% can identify at least three Supreme Court Justices.
a. In a random sample of 30 Americans, what is the probablilty that no more than 5 can identify at least three Supreme Court justices?

My work: using Cn,r X p X q, where n = 30 and r = 5, p = .17, q = 1-.17 = .83:
30C5(.17)to the 5th power (.83) to the 25th power
= .1919, which is wrong, but I don't know if I am using the formula incorrectly, or if I am using an incorrect formula.

With this equation you are finding the probability that exactly 5 can identify all the supreme court justices.

Now you need to find - for no more than 5:

the probability that exactly 4 can identify all the supreme court justices.

the probability that exactly 3 can identify all the supreme court justices.

the probability that exactly 2 can identify all the supreme court justices.

the probability that exactly 1 can identify all the supreme court justices.

the probability that exactly 0 can identify all the supreme court justices.

Then add those all up.....

b, which I got right: in a random sample of 50 Americans, what is the probablilty that 40 cannot identify at least three Supreme Court justices: 50C40 (.83)to the 40th power X (.17) to the 10th power
= .1200

c. another incorrect answer: in a random sample of 100 Americans, what is the probablity that at least half can name the Three Stooges?
My work: using Cn,r X p X q, where n = 100, r = 50, p = .59, q = .41:

100C50 (.59) to the 50th power (.41) to the 50th power
= .0153
d. another incorrect answer: In a random sample of 100 Americans, what is the probability that between 60 and 70 (including 60 and 70) can name the Three Stooges?
My work: {100C59 X (.59) to the 59th power (.41) to the 41st power} - {100C70(.59) to the 70th power (.41) to the 30th power}
= 0.0744
Click to expand...
 
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