Respuesta :
Answer:
0.9704 = 97.04% probability of getting a cookie with at least 3 chips
Step-by-step explanation:
We only have the mean number of chocolate chips. So we use the Poisson distribution to solve this question.
Poisson distribution:
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
In which
x is the number of sucesses
e = 2.71828 is the Euler number
[tex]\mu[/tex] is the mean in the given time interval.
A well-mixed cookie dough will produce cookies with a mean of 7 chocolate chips apiece.
This means that [tex]\mu = 7[/tex]
What is the probability of getting a cookie with at least 3 chips?
Either we have less than 3 chips, or we have at least 3 chips. The sum of the probabilities of these events is decimal 1.
[tex]P(X < 3) + P(X \geq 3) = 1[/tex]
We want [tex]P(X \geq 3)[/tex]. Then
[tex]P(X \geq 3) = 1 - P(X < 3)[/tex]
In which
[tex]P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)[/tex]
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-7}*7^{0}}{(0)!} = 0.0009[/tex]
[tex]P(X = 1) = \frac{e^{-7}*7^{1}}{(1)!} = 0.0064[/tex]
[tex]P(X = 2) = \frac{e^{-7}*7^{2}}{(2)!} = 0.0223[/tex]
[tex]P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0009 + 0.0064 + 0.0223 = 0.0296[/tex]
[tex]P(X \geq 3) = 1 - P(X < 3) = 1 - 0.0296 = 0.9704[/tex]
0.9704 = 97.04% probability of getting a cookie with at least 3 chips
