Let the number of chocolate chips in a certain type of cookie have a poisson distribution. we want the probability that a cookie of this type contains at least two chocolate chips to be greater than 0.99. find the smallest value of the mean that the distribution can take.

Step-by-step explanation:

The Poisson Distribution is "a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time or space if these events occur with a known constant mean rate and independently of the time since the last event".

Let X the random variable that represent the number of chocolate chips in a certain type of cookie. We know that

The probability mass function for the random variable is given by:

And f(x)=0 for other case.

For this distribution the expected value is the same parameter

On this case we are interested on the probability of having at least two chocolate chips, and using the complement rule we have this:

Using the pmf we can find the individual probabilities like this:

And replacing we have this:

And we want this probability that at least of 99%, so we can set upt the following inequality:

And now we can solve for

Applying natural log on both sides we have:

Thats a no linear equation but if we use a numerical method like the Newthon raphson Method or the Jacobi method we find a good point of estimate for the solution.

Using the Newthon Raphson method, we apply this formula:

Where :

Iterating as shown on the figure attached we find a final solution given by:

2(2x + 3) = -4 + 2(2x + 5)

4x + 6 = -4 + 4x + 10

4x + 6 = 4x + 6

4x - 6 + 6 = 4x - 6 + 6

4x = 4x

x = x

unlimited solutions.

i believe it's a.

i'm not sure on this one, most likely correct.