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klasskru [66]
2 years ago
8

1.Were your two estimates the same? Did the mean length of all the straws in the bag change in between selecting the two samples

? Explain your reasoning.
2.The actual mean length of all of the straws in the bag is about 2.37 inches. How do your estimates compare to this mean length?

3.If you repeated the same process again but you selected a larger sample (such as 10 or 20 straws, instead of just 5), would your estimate be more accurate? Explain your reasoning.





Remember that sample 1 mean= 4


Sample 2 mean = 3.8
Mathematics
2 answers:
RUDIKE [14]2 years ago
6 0
14.8, 16/3 and 16.2637
wariber [46]2 years ago
5 0
1. 14.8
2. 16/3
3. 16.2637
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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
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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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Step-by-step explanation:

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Answer:

$1800

Step-by-step explanation:

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Help find the answer to this question
marusya05 [52]
First, find the area of the entire traingle using the formula bh/2.
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