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xz_007 [3.2K]
3 years ago
12

The quality control team of a company checked 800 digital cameras for defects. The team found that 20 cameras had lens defects,

25 cameras had charging defects, and 6 cameras had both defects. What is the probability that a camera has a lens defect given that it has a charging defect?
Mathematics
1 answer:
dmitriy555 [2]3 years ago
6 0

Answer:

0.025

Step-by-step explanation:

-This is a conditional probability problem.

-Let L denote lens defect and C charging defect.

#We first calculate the probability of a camera having a lens defect;

P(lens)=\frac{Lens}{Total}\\\\=\frac{20}{800}\\\\=0.025

#Calculate the probability of a camera having a charging defect:

P(Charging)=\frac{Charging}{Total}\\\\=\frac{25}{800}\\\\=0.03125

The  the probability that a camera has a lens defect given that it has a charging defect is calculated as:

P(L|C)=\frac{P(C)P(L)}{P(C)}\\\\=\frac{0.025\times 0.03125}{0.03125}\\\\=0.025

Hence,  the probability that a camera has a lens defect given that it has a charging defect is 0.025

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Answer:

Reduce the expression, if possible, by cancelling the common factors.

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2 years ago
Bob Nale is the owner of Nale’s Quick Fill. Bob would like to estimate the mean number of gallons of gasoline sold to his custom
ira [324]

Answer:

a) The best estimator for the population mean is given by the sample mean \hat \mu = \bar X = 5.70

b) The 99% confidence interval would be given by (4.969;6.431)    

c) We are 99% confident that the true mean for th number of gallons of gasoline  sold to his customers is between 4.969 and 6.431.

Step-by-step explanation:

1) Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

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".

\bar X represent the sample mean for the sample  

\mu population mean (variable of interest)

\sigma=1.9 represent the population standard deviation

n=45 represent the sample size  

2) Part a

The best estimator for the population mean is given by the sample mean \hat \mu = \bar X = 5.70

3) Part b. Develop a 99 percent confidence interval for the population mean.

The confidence interval for the mean is given by the following formula:

\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}   (1)

Since the Confidence is 0.95 or 95%, the value of \alpha=0.01 and \alpha/2 =0.005, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.005,0,1)".And we see that z_{\alpha/2}=2.58

Now we have everything in order to replace into formula (1):

5.7-2.58\frac{1.9}{\sqrt{45}}=4.969    

5.7+2.58\frac{1.9}{\sqrt{45}}=6.431

So on this case the 99% confidence interval would be given by (4.969;6.431)    

4) Part c. Interpret the meaning of part (b).

On this case we can say this: We are 99% confident that the true mean for th number of gallons of gasoline  sold to his customers is between 4.969 and 6.431.

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