statistic help?

Question

1. In the spinner, Sectors A, B,D and E have central angles of 45 degrees, while sectors C and F have central angels of 90.
a. If the spinner is spun once, what is the probability that it will land in either sector A or sector C?
b. Suppose the spinner is spun 6 times. Construct a histogram for the probability distribution P, where P (n) is the probability that the spinner lands in sector A or sector C n times.

2. Suppose the fpllowing experiment is conducted: A die with "s" sides marked 1 through "s" is tossed "n" times and the number of times a 1 is tossed is recorded/ After many repetitions of the experiment, it is found that the number of 1s tossed has a mean of 33 and a standard deviation of 5.5
a. If the die is assumed to be fair, what is the most likely number of sides it has?
b. If the die is assumed to be fair, how many times "n" was it tossed in each experiment?

Answer 1

1. if the spinner is spun once, the probability that it will land in either A or C would be 135/360 and you can reduce that.
If the spinner is spun six times you'd work that out as a binomial experiement with p=135/360 and n = 6.

The formula for binomial probability is
A probability formula for Bernoulli trials. The probability of achieving exactly k successes in n trials is shown below.
http://www.mathwords.com/b/binomial_probability_formula.htm
Formula:
probability of k successes in n trials is C(n,k) p^k q^N9-k)
n = number of trials
k = number of successes
n – k = number of failures
p = probability of success in one trial
q = 1 – p = probability of failure in one trial

So for example your probability table would look like this:
k = number of times spinner lands in A or C in 6 trials
p(k) = probabilty of that many times occurring
k ..........0............1...........2..........3...........4..........5...........6
p(k)..0.0596..0.21458..0.32186 etc

see also http://en.wikipedia.org/wiki/Binomial_distribution

2. To get this, you'd have to treat it as a binomial experiment, with a "1" as success and any other result as failure. Then you'd work backwards from the definitions of mean and standard deviation for a binomial experiment.
In a binomial experiment, the mean is np and the standard deviation is sqrt npq, which can also be written sqrt (meanxq) or sqrt (meanx(1-p))
So plugging in what we know..
5.5 = sqrt (33q) square both sides
30.25 = 33q
divide each side by 33
30.35 / 33 = q ~= .9167 = probability of anything other than a 1
p = 1-.9167 ~= .0833 = probability of getting a 1
Assuming the die is fair, the die has 1/p sides so p works out to 12.
Now, with mean = 33 = np and p ~= (1/12), solve for n and you get n = 33/p = 33/(1/12) = 33x12 = 396 tosses of the die
Now try this yourself and see that you can do it too.

Answer 2

1. a.
P(A+C) = (45 + 90)/360 = 3/8 = 0.375 = p

b.
q = 1 - p = 5/8 = 0.625
0 0.625^6
1 6(0.625^5)(0.375)
2 15(0.625^4)(0.375^2)
3 20(0.625^3)(0.375^3)
4 15(0.625^2)(0.375^4)
5 6(0.625)(0.375^5)
6 0.375^6

0, 0.05960
1, 0.21458
2, 0.32187
3, 0.25749
4, 0.11587
5, 0.02780
6, 0.00278

np = 33
npq = 30.25
q = 30.25/33 = 0.91667
p = 0.08333 = 1/12
a. 12 sides
b. n = 396

Answer 3

Find the probability of no successes in six trials of a binomial experiment in which the probability of success is 30%. Round to the nearest tenth of a percent.