PRECALC: Suppose you conduct a random experiment that has the binomial probability distribution.?

Supposed the probability that outcome C occurs on any one repetition is 0.4. Let P(x) be the probability that outcome C occurs xtimes in 5 repetitions.


a) Calculate P(x) for each value of x in the domain.


b) Find the mathematically expected value of x.


c) Show that the mathematically expected value of x is equal to 0.4


d) If the probability that C occurs on any one repetition is b and the probability that C does not occur on one repetition is a=1-b, prove that in 5 trials, the expected value of x is 5b.


e) From what you have observed in this problem, make a conjecture about the mathematically expected balue of x in n repetitions, if the probability that C occurs on any one repetition is b.


f) If you plant 100 seeds, each of which has a probability of 0.71 of germinating, how many seeds would you epxect to germinate?





I have no clue what any of these questions are asking. Can anyone help explain or solve any?


Thank you!!!

PRECALC: Suppose you conduct a random experiment that has the binomial probability distribution.?
a) Each repetition has possible outcomes. Success here called ‘outcome C’


and failure ie ‘not outcome C’. Usually they are denoted as outcome ‘1’ and ‘0’. The number ‘1’ is usually used for a success and ‘0’ for a failure.





We are given probability of C = .4. This means that probability of not C = .6


since the two probabilities (for 1 and 0) add to 1. ie .4 + .6 = 1





Now, P(x) = the probability of x successes out of 5 repetitions.


The possible values for x are 0,1,2,3,4,5. We can’t have C occur


Less than 0 time and we can’t have more than 5 times. (since only


5 repetitions).





The value for P(x) is the usual binomial distribution.


P(x) = 5Cx * (.4) ^ x * (.6) ^ * (5-x)





So we need to calculate this for each value 0 to 4.


For example


P(0) = 5C0 * (.4) ^ 0 * (.6) ^ (5-0) = .6^5 = .0776


P(1) = 5C1 * (.4) ^1 * (.6) ^ (4) = 5 * .4 * .6 ^ 4 = .2592


etc





b) From the definition of expected value, this =


0 * P(0) + 1*P(1) + 2*P(2) + 3*P(3) + 4*P(4) + 5*P(5)





c)


You can plug in the numbers from (a) and calculate, the value should be 5*.4 = 2


(ie not =4 which was stated).





d) the proof can be done in a number of ways. The simplest way is


to use (c) above. We saw that the definition of the expected value with


N=5 and prob = .4 gave 5*.4 = 2. So, the same calculation with any prob = b


gives 5*b.





(there are formal mathematical proofs of this fact, but that seems to be beyond


the level of your course)





e) Generalize from 5*prob to n*b = expected value





f) Expected number to germinate is just the expected value N*P(1)


ie 100*.71 = 71
Reply:X has the binomial distribution with n = 5 trials and success probability p = 0.4





In general, if X has the binomial distribution with n trials and a success probability of p then


P[X = x] = n!/(x!(n-x)!) * p^x * (1-p)^(n-x)


for values of x = 0, 1, 2, ..., n


P[X = x] = 0 for any other value of x.





The probability mass function is derived by looking at the number of combination of x objects chosen from n objects and then a total of x success and n - x failures.


Or, in other words, the binomial is the sum of n independent and identically distributed Bernoulli trials.





X ~ Binomial( n , p )





the mean of the binomial distribution is n * p = 2


the variance of the binomial distribution is n * p * (1 - p) = 1.2


the standard deviation is the square root of the variance = √ ( n * p * (1 - p)) = 1.095445





The Probability Mass Function, PMF,


f(X) = P(X = x) is:





P( X = 0 ) = 0.07776


P( X = 1 ) = 0.2592


P( X = 2 ) = 0.3456


P( X = 3 ) = 0.2304


P( X = 4 ) = 0.0768


P( X = 5 ) = 0.01024








== -- == -- == -- ==





Let Xb be the number of seeds that germinate. Xb has the binomial distribution with n = 100 trials and success probability p = 0.71





In general, if X has the binomial distribution with n trials and a success probability of p then


P[Xb = x] = n!/(x!(n-x)!) * p^x * (1-p)^(n-x)


for values of x = 0, 1, 2, ..., n


P[Xb = x] = 0 for any other value of x.





To use the normal approximation to the binomial you must first validate that you have more than 10 expected successes and 10 expected failures. In other words, you need to have n * p %26gt; 10 and n * (1-p) %26gt; 10.





Some authors will say you only need 5 expected successes and 5 expected failures to use this approximation. If you are working towards the center of the distribution then this condition should be sufficient. However, the approximations in the tails of the distribution will be weaker especially if the success probability is low or high. Using 10 expected successes and 10 expected failures is a more conservative approach but will allow for better approximations especially when p is small or p is large.





In this case you have:


n * p = 100 * 0.71 = 71 expected success


n * (1 - p) = 100 * 0.29 = 29 expected failures





We have checked and confirmed that there are enough expected successes and expected failures. Now we can move on to the rest of the work.





If Xb ~ Binomial(n, p) then we can approximate probabilities using the normal distribution where Xn is normal with mean μ = n * p, variance σ² = n * p * (1-p), and standard deviation σ





Xb ~ Binomial(n = 100 , p = 0.71 )


Xn ~ Normal( μ = 71 , σ² = 20.59 )


Xn ~ Normal( μ = 71 , σ = 4.537621 )





I have noted two different notations for the Normal distribution, one using the variance and one using the standard deviation. In most textbooks and in most of the literature, the parameters used to denote the Normal distribution are the mean and the variance. In most software programs, the standard notation is to use the mean and the standard deviation.





The probabilities are approximated using a continuity correction. We need to use a continuity correction because we are estimating discrete probabilities with a continuous distribution. The best way to make sure you use the correct continuity correction is to draw out a small histogram of the binomial distribution and shade in the values you need. The continuity correction accounts for the area of the boxes that would be missing or would be extra under the normal curve.





P( Xb %26lt; x) ≈ P( Xn %26lt; (x - 0.5) )


P( Xb %26gt; x) ≈ P( Xn %26gt; (x + 0.5) )


P( Xb ≤ x) ≈ P( Xn ≤ (x + 0.5) )


P( Xb ≥ x) ≈ P( Xn ≥ (x - 0.5) )


P( Xb = x) ≈ P( (x - 0.5) %26lt; Xn %26lt; (x + 0.5) )


P( a ≤ Xb ≤ b ) ≈ P( (a - 0.5) %26lt; Xn %26lt; (b + 0.5) )


P( a ≤ Xb %26lt; b ) ≈ P( (a - 0.5) %26lt; Xn %26lt; (b - 0.5) )


P( a %26lt; Xb ≤ b ) ≈ P( (a + 0.5) %26lt; Xn %26lt; (b + 0.5) )


P( a %26lt; Xb %26lt; b ) ≈ P( (a + 0.5) %26lt; Xn %26lt; (b - 0.5) )





In the work that follows Xb has the binomial distribution, Xn has the normal distribution and Z has the standard normal distribution.





Remember that for any normal random variable Xn, you can transform it into standard units via: Z = (Xn - μ ) / σ