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Marianna [84]
3 years ago
8

1. Jovi and Amillia are playing a game using the spinner below. Jovi will win the game on his next spin if the arrow lands on a

section labelled 6, 7, or 8. Jovi claims it is likely, but not certain, that he will win the game on the next spin. Explain why Jovi's claim in correct or incorrect. (10 points) 2. Suppose we toss a fair coin 50 times and have 27 heads and 23 tails. A. What is the relative frequency of heads? (5 points) b. What is the theoretical probability of tossing heads? (5 points) c. Are your answers to (a) and (b) the same? If so, explain why. If not, provide a possible explanation for the difference. (10 points)
Mathematics
1 answer:
Lubov Fominskaja [6]3 years ago
8 0

For question number one, Jovi WILL most likely win.

Hope this helps.

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Let the number of chocolate chips in a certain type of cookie have a Poisson distribution. We want the probability that a cookie
ludmilkaskok [199]

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)=\frac{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

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

P(X=0)=\frac{e^{-\lambda} \lambda^0}{0!}=e^{-\lambda}

P(X=1)=\frac{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

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

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x_{n+1}=x_n -\frac{f(x_n)}{f'(x_n)}

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f'(x_n)=1-\frac{1}{1+\lambda}

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

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Answer:

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Generally the sample proportion is  

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