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3 years ago

Find the mean and range.

2 answers:

6 0

Hey there!

$18,23,10,39,22,17,16,15 \\ \\ \\ Mean: 20 \\ \\ \\ Range: 29 \\ \\ \\ \\ \\ Answer: Option B.$

Good luck on your assignment and enjoy your day!

~ $LoveYourselfFirst:)$

photoshop1234 [79]3 years ago

4 0

Hello!

Let's review what mean and range is.

REVIEW

Mean is calculated by adding all the numbers and dividing by how many numbers there are.

Range is calculated by subtracting the lowest number from the greatest number, also known as finding the difference

_____________________________________________________________

Mean

18 + 23 + 10 + 39 + 22 + 17 + 16 + 15 = 160

There are 8 numbers.

Divide 160 by 8 to get the mean

160 ÷ 8 = 20

The mean is 20

___________________________________________________________

Range

The highest number is 39, while the lowest number is 10. Subtract 10 from 39 to get the range.

39 - 10 = 29

The range is 29

Your answer is the 2nd option.

I hope you find this useful! :)

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3 years ago

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:

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

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:

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