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2 years ago

Please need help with this assignment… thx

Mathematics

1 answer:

Citrus2011 [14]2 years ago

5 0

Answer:

1.length = 15, width = 12, height = 13

2. 13 × 12 × 15 = 2340

Step-by-step explanation:

3 ÷ 1/4 = 12

3 3/4 = 15/4

15/4 ÷ 1/4 = 15

3 1/4 ÷ 1/4 = 13

length = 15

width = 12

height = 13

13 × 12 × 15 = 2340

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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: