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2 years ago

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A drawer of loose socks contains 2 red socks, 2 green socks, and 6 white socks. Which best describes how to determine the probab

ility of pulling out a white sock, not replacing it, and pulling out another white sock?

2 answers:

kenny6666 [7]2 years ago

7 0

Well, before you draw out the first sock, you had 60% chance of getting a white sock since six of the ten socks are white! (6/10)

So, if you were to not replace the first drawn white sock, your percentage goes to 55.5556%! This is because you now have 5/9 socks remaining in the drawer being white!

Vikki [24]2 years ago

7 0

Answer:

The probability that the first sock is white is 6/10 and that the second sock is white is 5/9, so the probability of choosing a pair of white socks is 30/90 = 1/3! Hope that helps.

Step-by-step explanation:

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The time it takes to transmit a file always depends on the file size. Suppose you transmitted 30 files, with the average size of

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Answer:

Y = 0.009042 + 0.0002457X

Y = 0.1073 seconds

Step-by-step explanation:

In the given problem we have two variables: the transmission time and average size of file.

Y = transmission time

X = average file size

The linear regression model is given by

Y = a + bX

The slope b is given by

b = correlation coefficient*(SDy/SDx)

Where SDy is the standard deviation of average transmittance time and SDx is the standard deviation of average file size.

b = 0.86(0.01/35)

b = 0.0002457

The y-intercept a is given by

a = y - bx

a = 0.04 - (0.0002457)126

a = 0.04 - 0.030958

a = 0.009042

Therefore, the linear regression model is

Y = 0.009042 + 0.0002457X

Predict the time it will take to transmit a 400 Kbyte file.

Substitute X = 400 in the regression model

Y = 0.009042 + 0.0002457(400)

Y = 0.009042 + 0.09828

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2 years ago

A machine produces parts that are either defect free (90%), slightly defective (3%), or obviously defective (7%). Prior to shipm

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Answer:

(a) 0.0686

(b) 0.9984

(c) 0.0016

Step-by-step explanation:

Given that a machine produces parts that are either defect free (90%), slightly defective (3%), or obviously defective (7%).

Let A, B, and C be the events of defect-free, slightly defective, and the defective parts produced by the machine.

So, from the given data:

P(A)=0.90, P(B)=0.03, and P(C)=0.07.

Let E be the event that the part is disregarded by the inspection machine.

As a part is incorrectly identified as defective and discarded 2% of the time that a defect free part is input.

So, $P\left(\frac{E}{A}\right)=0.02$

Now, from the conditional probability,

$P\left(\frac{E}{A}\right)=\frac{P(E\cap A)}{P(A)}$

$\Rightarrow P(E\cap A)=P\left(\frac{E}{A}\right)\times P(A)$

$\Rightarrow P(E\cap A)=0.02\times 0.90=0.018\cdots(i)$

This is the probability of disregarding the defect-free parts by inspection machine.

Similarly,

$P\left(\frac{E}{A}\right)=0.40$

and $\Rightarrow P(E\cap B)=0.40\times 0.03=0.012\cdots(ii)$

This is the probability of disregarding the partially defective parts by inspection machine.

$P\left(\frac{E}{A}\right)=0.98$

and $\Rightarrow P(E\cap C)=0.98\times 0.07=0.0686\cdots(iii)$

This is the probability of disregarding the defective parts by inspection machine.

(a) The total probability that a part is marked as defective and discarded by the automatic inspection machine

$=P(E\cap C)$

$= 0.0686$ [from equation (iii)]

(b) The total probability that the parts produced get disregarded by the inspection machine,

$P(E)=P(E\cap A)+P(E\cap B)+P(E\cap C)$

$\Rightarrow P(E)=0.018+0.012+0.0686$

$\Rightarrow P(E)=0.0986$

So, the total probability that the part produced get shipped

$=1-P(E)=1-0.0986=0.9014$

The probability that the part is good (either defect free or slightly defective)

$=\left(P(A)-P(E\cap A)\right)+\left(P(B)-P(E\cap B)\right)$

$=(0.9-0.018)+(0.03-0.012)$

$=0.9$

So, the probability that a part is 'good' (either defect free or slightly defective) given that it makes it through the inspection machine and gets shipped

$=\frac{\text{Probabilily that shipped part is 'good'}}{\text{Probability of total shipped parts}}$

$=\frac{0.9}{0.9014}$

$=0.9984$

(c) The probability that the 'bad' (defective} parts get shipped

=1- the probability that the 'good' parts get shipped

=1-0.9984

=0.0016

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Answer:

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b) (38.02, 61.98) is the 95% as it covers more values than the other.

Step-by-step explanation:

a) Upper and lower values for the confidence intervals are at the same distance from the mean, as they represent an upper and lower limit for the value, and the most probable value is the one in the middle, which, in a normal distribution is the mean.

As they derive from the same data, both pairs of values should give us the same mean:

$\mbox {mean}=\frac{\mbox{upper value+lower value}}{\mbox{2} }\\\mbox {mean from the first set}=\frac{38.02+61.98}{2}=50 \\\mbox {mean from the first set}=\frac{39.95+60.05}{2}=50$

b) The percentage of the confidence interval tells us which is the area below the bell curve (of the normal distribution) between the limits given.

One could give a value with a 100% CI, but limits should be minus and plus infinity, and as the interval gets smaller, less values are covered (less area), and less certain we are about our value being inside the interval.

Then, we can say that the widest interval must be the 95% CI, i.e. (38.02, 61.98).

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