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2 years ago

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An oil exploration company currently has two active proj- ects, one in Asia and the other in Europe. Let A be the event that the

Asian project is successful and B be the event that the European project is successful. Suppose that A and B are independent events with and . a. If the Asian project is not successful, what is the proba- bility that the European project is also not successful? Explain your reasoning. b. What is the probability that at least one of the two proj- ects will be successful? c. Given that at least one of the two projects is successful, what is the probability that only the Asian project is successful? P(A) 5 .4 P(B) 5 .7

1 answer:

sukhopar [10]2 years ago

4 0

Answer:

a) P(B'|A') = P(B') = 1 - 0.7 = 0.3.

The reasoning is that: since, events A and B are independent, then, events A' and B' are also independent and for independent events, P(A|B) = P(A).

b) P(A u B) = 0.82

c) P[(A n B')|(A u B)] = 0.146

Step-by-step explanation:

P(A) = 0.4

P(B) = 0.7

A and B are independent events.

P(A') = 1 - 0.4 = 0.6

P(B') = 1 - 0.7 = 0.3

a) If the Asian project is not successful, what is the proba- bility that the European project is also not successful?

P(B'|A') = P(B') = 1 - 0.7 = 0.3

The reasoning is that: since, events A and B are independent, then, events A' and B' are also independent and for independent events, P(A|B) = P(A).

b) The probability that at least one of the two projects will be successful = P(A u B)

P(A u B) = P(A) + P(B) - P(A n B) = 0.4 + 0.7 - (0.4)(0.7) = 0.82

OR

P(A u B) = P(A n B') + P(A' n B) + P(A n B) = (0.4)(0.3) + (0.6)(0.7) + (0.4)(0.7) = 0.82

c) Given that at least one of the two projects is successful, what is the probability that only the Asian project is successful?

P[(A n B')|(A u B)] = }P(A n B') n P(A u B)]/P(A u B) = P(A n B')/P(A u B) = (0.4)(0.3)/(0.82)

P[(A n B')|(A u B)] = 0.146

Hope this Helps!!!

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Answer:

99% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is [$(-31.82) , $12.02].

Step-by-step explanation:

We are given that a random sample of 16 sales receipts for mail-order sales results in a mean sale amount of $74.50 with a standard deviation of $17.25.

A random sample of 9 sales receipts for internet sales results in a mean sale amount of $84.40 with a standard deviation of $21.25.

The pivotal quantity that will be used for constructing 99% confidence interval for true mean difference is given by;

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where, $\bar X_1$ = sample mean for mail-order sales = $74.50

$\bar X_2$ = sample mean for internet sales = $84.40

$s_1$ = sample standard deviation for mail-order purchases = $17.25

$s_2$ = sample standard deviation for internet purchases = $21.25

$n_1$ = sample of sales receipts for mail-order purchases = 16

$n_2$ = sample of sales receipts for internet purchases = 9

Also, $s_p =\sqrt{\frac{(n_1-1)\times s_1^{2}+(n_2-1)\times s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(16-1)\times 17.25^{2}+(9-1)\times 21.25^{2} }{16+9-2} }$ = 18.74

The true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is represented by ( $\mu_1-\mu_2$ ).

Now, 99% confidence interval for ( $\mu_1-\mu_2$ ) is given by;

= $(\bar X_1-\bar X_2) \pm t_(_\frac{\alpha}{2}_) \times s_p \times \sqrt{\frac{1}{n_1} +\frac{1}{n_2}}$

Here, the critical value of t at 0.5% level of significance and 23 degrees of freedom is given as 2.807.

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Now we need to find the y-coordinate of the center of mass.

$\bar{y}=\frac{1}{M}\int\int y*\rho dydx$

$\bar{y}=\frac{1}{M}\int^{9}_{1}\int^{4}_{1}y*ky^{2} dydx$

$\bar{y}=\frac{k}{168k}\int^{9}_{1}\int^{4}_{1}y^{3} dydx$

$\bar{y}=\frac{1}{168}\int^{9}_{1}\frac{y^{4}}{4}|^{4}_{1}dx$

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$\bar{y}=\frac{255}{672}\int^{9}_{1}dx$

$\bar{y}=\frac{255}{672}8=\frac{2040}{672}$

$\bar{y}=\frac{85}{28}$

Therefore the center of mass is:

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