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.

Answer:[tex]\lambda \geq 6.63835[/tex]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 [tex]X \sim Poisson(\lambda)[/tex]The probability mass function for the random variable is given by:[tex]f(x)=\frac{e^{-\lambda} \lambda^x}{x!} , x=0,1,2,3,4,...[/tex]And f(x)=0 for other case.For this distribution the expected value is the same parameter [tex]\lambda[/tex][tex]E(X)=\mu =\lambda[/tex]On this case we are interested on the probability of having at least two chocolate chips, and using the complement rule we have this:[tex]P(X\geq 2)=1-P(X<2)=1-P(X\leq 1)=1-[P(X=0)+P(X=1)][/tex]Using the pmf we can find the individual probabilities like this:[tex]P(X=0)=\frac{e^{-\lambda} \lambda^0}{0!}=e^{-\lambda}[/tex][tex]P(X=1)=\frac{e^{-\lambda} \lambda^1}{1!}=\lambda e^{-\lambda}[/tex]And replacing we have this:[tex]P(X\geq 2)=1-[P(X=0)+P(X=1)]=1-[e^{-\lambda} +\lambda e^{-\lambda}[][/tex][tex]P(X\geq 2)=1-e^{-\lambda}(1+\lambda)[/tex]And we want this probability that at least of 99%, so we can set upt the following inequality:[tex]P(X\geq 2)=1-e^{-\lambda}(1+\lambda)\geq 0.99[/tex]And now we can solve for [tex]\lambda[/tex][tex]0.01 \geq e^{-\lambda}(1+\lambda)[/tex]Applying natural log on both sides we have:[tex]ln(0.01) \geq ln(e^{-\lambda}+ln(1+\lambda)[/tex][tex]ln(0.01) \geq -\lambda+ln(1+\lambda)[/tex][tex]\lambda-ln(1+\lambda)+ln(0.01) \geq 0[/tex]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:[tex]x_{n+1}=x_n -\frac{f(x_n)}{f'(x_n)}[/tex]Where :[tex]f(x_n)=\lambda -ln(1+\lambda)+ln(0.01)[/tex][tex]f'(x_n)=1-\frac{1}{1+\lambda}[/tex]Iterating as shown on the figure attached we find a final solution given by:[tex]\lambda \geq 6.63835[/tex]