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MaRussiya [10]
2 years ago
10

You move out into the country and you notice every Spring there are more and more Deer Fawns that appear. You decide to try and

predict how many Fawns there will be for the up coming Spring. You collect data to, to help estimate Fawn Count for the upcoming Spring season. You collect data on over the past 10 years.
x1 = Adult Deer Count
x2 = Annual Rain in Inches
x3 = Winter Severity

Where Winter Severity Index:
1 = Warm
2 = Mild
3 = Cold
4 = Freeze
5 = Severe

Required:
Interpret the slope(s) of the significant predictors for Fawn Count (if there are any).
Mathematics
1 answer:
marusya05 [52]2 years ago
3 0

Answer: hello your question is incomplete attached below is the missing part

answer : There is no slope

Step-by-step explanation:

To interpret the slope we plot a graph of Winter severity index vs Fawn count

slope = Δy / Δx  = 0

attached below is the graph

There no slope because the graph is a curve and not a straight line graph

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Wires manufactured for use in a computer system are specified to have resistances between 0.11 and 0.13 ohms. The actual measure
Marina CMI [18]

Answer:

a) P(0.11

And we can find this probability with this difference and with the normal standard table or excel:

P(-1.11

b) P(0.11 < \bar X < 0.13)

And we can use the z score defined by:

z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}

And using the limits we got:

z = \frac{0.11-0.12}{\frac{0.009}{\sqrt{4}}}= -2.22

z = \frac{0.13-0.12}{\frac{0.009}{\sqrt{4}}}= 2.22

And we want to find this probability:

P(-2.22< Z

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Part a

Let X the random variable that represent the resitances of a population, and for this case we know the distribution for X is given by:

X \sim N(0.12,0.009)  

Where \mu=0.12 and \sigma=0.009

We are interested on this probability

P(0.11

And the best way to solve this problem is using the normal standard distribution and the z score given by:

z=\frac{x-\mu}{\sigma}

If we apply this formula to our probability we got this:

P(0.11

And we can find this probability with this difference and with the normal standard table or excel:

P(-1.11

Part b

We select a sample size of n =4. And since the distribution for X is normal then we know that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

And we want this probability:

P(0.11 < \bar X < 0.13)

And we can use the z score defined by:

z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}

And using the limits we got:

z = \frac{0.11-0.12}{\frac{0.009}{\sqrt{4}}}= -2.22

z = \frac{0.13-0.12}{\frac{0.009}{\sqrt{4}}}= 2.22

And we want to find this probability:

P(-2.22< Z

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Answer:

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