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andrey2020 [161]
2 years ago
9

Two factories — Factory A and Factory B — design batteries to be used in mobile phones. Factory A produces 60% of all batteries,

and Factory B produces the other 40%. 2% of Factory A's batteries have defects, and 4% of Factory B's batteries have defects. What is the probability that a battery is both made by Factory A and defective? *
Mathematics
1 answer:
elena-s [515]2 years ago
4 0

Answer:

Probability that a battery is both made by Factory A and defective is 0.012 or 1.2%.

Step-by-step explanation:

We are given that Two factories — Factory A and Factory B — design batteries to be used in mobile phones. Factory A produces 60% of all batteries, and Factory B produces the other 40%.

2% of Factory A's batteries have defects, and 4% of Factory B's batteries have defects.

Let the Probability that factory A produces batteries = P(A) = 0.60

Probability that factory B produces batteries = P(B) = 0.40

<em>Also, let D = event that battery is defective</em>

Probability that batteries have defects given that it is produced by Factory A = P(D/A) = 0.02

Probability that batteries have defects given that it is produced by Factory B = P(D/B) = 0.04

Now, <u>probability that a battery is both made by Factory A and defective</u> = Probability that factory A produces batteries \times Probability that batteries have defects given that it is produced by Factory A

                       =  P(A) \times P(D/A)

                       =  0.60 \times 0.02 = 0.012  or  1.2%

Therefore, the probability that a battery is both made by Factory A and defective is 1.2%.

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Percentage Rate=6%

Step-by-step explanation:

Total borrowed=$2,100

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X=$2,478 - $2,100

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Answer:

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And we can find this probability using the complement rule and the normal standard distribution and we got:

P(z>6.302)=1-P(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".  

Solution to the problem

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

X \sim N(510,14.28)  

Where \mu=510 and \sigma=14.28

We are interested on this probability

P(X>600)

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(X>600)=P(\frac{X-\mu}{\sigma}>\frac{600-\mu}{\sigma})=P(Z>\frac{600-510}{14.28})=P(z>6.302)

And we can find this probability using the complement rule and the normal standard distribution and we got:

P(z>6.302)=1-P(z

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