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.

Answers

Answer 1

Answer:

Answer:

$\lambda \geq 6.63835$

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 $X \sim Poisson(\lambda)$

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

$f(x)=(e^(-\lambda) \lambda^x)/(x!) , x=0,1,2,3,4,...$

And f(x)=0 for other case.

For this distribution the expected value is the same parameter $\lambda$

$E(X)=\mu =\lambda$

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

$P(X\geq 2)=1-P(X<2)=1-P(X\leq 1)=1-[P(X=0)+P(X=1)]$

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

$P(X=0)=(e^(-\lambda) \lambda^0)/(0!)=e^(-\lambda)$

$P(X=1)=(e^(-\lambda) \lambda^1)/(1!)=\lambda e^(-\lambda)$

And replacing we have this:

$P(X\geq 2)=1-[P(X=0)+P(X=1)]=1-[e^(-\lambda) +\lambda e^(-\lambda)[]$

$P(X\geq 2)=1-e^(-\lambda)(1+\lambda)$

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

$P(X\geq 2)=1-e^(-\lambda)(1+\lambda)\geq 0.99$

And now we can solve for $\lambda$

$0.01 \geq e^(-\lambda)(1+\lambda)$

Applying natural log on both sides we have:

$ln(0.01) \geq ln(e^(-\lambda)+ln(1+\lambda)$

$ln(0.01) \geq -\lambda+ln(1+\lambda)$

$\lambda-ln(1+\lambda)+ln(0.01) \geq 0$

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:

$x_(n+1)=x_n -(f(x_n))/(f'(x_n))$

Where :

$f(x_n)=\lambda -ln(1+\lambda)+ln(0.01)$

$f'(x_n)=1-(1)/(1+\lambda)$

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

$\lambda \geq 6.63835$

Answer 2

Answer:

Final answer:

The problem pertains to Poisson Distribution in probability theory, focusing on finding the smallest mean (λ) such that the probability of having at least two chocolate chips in a cookie is more than 0.99. This involves solving an inequality using the formula for Poisson Distribution.

Explanation:

This problem pertains to the Poisson Distribution, often used in probability theory. In particular, we're looking at the number of events (in this case, the number of chocolate chips) that occur within a fixed interval. Here, the interval under study is a single cookie. The question requires us to find the smallest value of λ (the mean value of the distribution) such that the probability of getting at least two chocolate chips in a cookie is more than 0.99.

Using the formula for Poisson Distribution, the probability of finding k copies of an event is given by:

P(X=k) = λ^k * exp(-λ) / k!

The condition here is that the probability of finding at least 2 copies is more than 0.99. Therefore, you formally need to solve the inequality:

P(X>=2) = 1 - P(X=0) - P(X=1) > 0.99

Substituting the values of P(X=0) and P(X=1) from our standard formula, you will need to calculate and find the smallest value of λ that satisfies this inequality.

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List all the factor pairs for 48 make a tabletop to help

Answers

1 and 48 are a factor pair of 48 since 1 x 48= 48

2 and 24 are a factor pair of 48 since 2 x 24= 48

3 and 16 are a factor pair of 48 since 3 x 16= 48

4 and 12 are a factor pair of 48 since 4 x 12= 48

6 and 8 are a factor pair of 48 since 6 x 8= 48

8 and 6 are a factor pair of 48 since 8 x 6= 48

12 and 4 are a factor pair of 48 since 12 x 4= 48

16 and 3 are a factor pair of 48 since 16 x 3= 48

24 and 2 are a factor pair of 48 since 24 x 2= 48

48 and 1 are a factor pair of 48 since 48 x 1= 48

Compute 17÷2 enter your answer using remainder notation

Answers

Final answer:

To compute 17 ÷ 2 using remainder notation, divide 17 by 2 and find the quotient and remainder. The quotient is 8 and the remainder is 1.

Explanation:

To compute 17 ÷ 2 using remainder notation, you divide 17 by 2 and find the quotient and remainder. In this case, the quotient is the whole number part of the division and the remainder is the leftover part. When you divide 17 by 2, the quotient is 8 and the remainder is 1. Therefore, the answer is 8 with a remainder of 1.

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Answer:8.5

Step-by-step explanation:

Answers

COMPLETE QUESTION:

Chris went on a vacation for a week and asked his brother Paul to feed his old cat Charlie. But Paul is forgetful, and Chris is 70% sure Paul will forget to feed his cat. Without food, Charlie will die with probability 0.5. With food, he will die with probability 0.03. Chris came back from vacation and found Charlie alive. What is the probability that Paul forgot to feed Charlie (round off to third decimal place)?

Answer:

The probability that Paul forgot to feed charlie is 0.546

Step-by-step explanation:

Lets denote F the event 'Paul forgot to feed Charlie', and L the even 'Charlie is alive', we have

P(F) = 0.7

P(L|F) = 1-0.5 = 0.5

P(L|F^c) = 1-P(L^c|F^c) = 1-0.03 = 0.97

We want to calculate P(F|L). We will use Bayesformula at the start and the theoremoftotalprobability to calculate P(L).

Given that Charlie is alive, the probability that Paul forgot to feed charlie is 0.546.

Answer:

P = 0.546

Step-by-step explanation:

Hi,

This is a question of conditional probability, which means to find probability of a situation given that another event has already occured:

$P(A|B) = (P(B|A) P(A))/(P(B)) = (P(A \cap B))/(P(B))$

In this question, we need to find the probability of Charlie being alive if not fed, with the data given below:

$P(Paul\ forgets)= 0.70\nP(Paul\ feeds) = 0.30\nP( Charlie\ dies\ given\ that\ Paul\ forgets) = 0.50\nP( Charlie\ dies\ given\ that\ Paul\ feeds) = 0.03\n$