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alexdok [17]
2 years ago
15

Discuss the validity of the following statement. If the statement is always​ true, explain why. If​ not, give a counterexample.

If the odds for E equal the odds against​ E', then ​P(E)P(F)=P(E∩F)
Mathematics
1 answer:
Zarrin [17]2 years ago
8 0

Correction:

Because F is not present in the statement, instead of working on​P(E)P(F) = P(E∩F), I worked on

P(E∩E') = P(E)P(E').

Answer:

The case is not always true.

Step-by-step explanation:

Given that the odds for E equals the odds against E', then it is correct to say that the E and E' do not intersect.

And for any two mutually exclusive events, E and E',

P(E∩E') = 0

Suppose P(E) is not equal to zero, and P(E') is not equal to zero, then

P(E)P(E') cannot be equal to zero.

So

P(E)P(E') ≠ 0

This makes P(E∩E') different from P(E)P(E')

Therefore,

P(E∩E') ≠ P(E)P(E') in this case.

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Answer:

The change received by the customer is $7.

Step-by-step explanation:

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A local company makes snack-size bags of potato chips. Each day, the company produces batches of 400 snack-size bags using a pro
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Answer:

Probability that a sample of 40 bags has an average weight of at least 2.02 ounces is 0.103.

Step-by-step explanation:

We are given that the company produces batches of 400 snack-size bags using a process designed to fill each bag with an average of 2 ounces of potato chips. Assume the amount placed in each of the 400 bags is normally distributed and has a standard deviation of 0.1 ounce.

Also, sample of 40 bags are selected.

<em>Let </em>\bar X<em> = sample average weight</em>

The z-score probability distribution for sample mean is given by ;

                Z = \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

where, \mu = population mean weight of potato chips = 2 ounces

            \sigma = standard deviation = 0.1 ounces

            n = sample of bags = 40

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

Now, the probability that a sample of 40 bags has an average weight of at least 2.02 ounces is given by = P(\bar X \geq 2.02 ounces)

   P(\bar X \geq 2.02) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } \geq \frac{2.02-2}{\frac{0.1}{\sqrt{40} } } ) = P(Z \geq 1.265) = 1 - P(Z < 1.265)

                                                      = 1 - 0.89707 = 0.103

<em>The above probability is calculated using z table by looking at value of x = 1.265 which will lie between x = 1.26 and x = 1.27 in the z table which have an area of 0.89707.</em>

<em />

Therefore, probability that a sample of 40 bags has an average weight of at least 2.02 ounces is 0.103.

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