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stealth61 [152]
3 years ago
12

A computer consulting firm presently has bids out on three projects. Let Ai = {awarded project i}, for i = 1, 2, 3, and suppose

that P(A1) = 0.23, P(A2) = 0.25, P(A3) = 0.29, P(A1 ∩ A2) = 0.09, P(A1 ∩ A3) = 0.11, P(A2 ∩ A3) = 0.07, P(A1 ∩ A2 ∩ A3) = 0.02. Use the probabilities given above to compute the following probabilities, and explain in words the meaning of each one. (Round your answers to four decimal places.)
a. P(A2 | A1)
b. P(A2 ∩ A3 | A1)
c. P(A2 ∪ A3 | A1)
d. P(A1 ∩ A2 ∩ A3 | A1 ∪ A2 ∪ A3)
Mathematics
1 answer:
Karo-lina-s [1.5K]3 years ago
6 0

Step-by-step explanation:

The data below is what was provided in the question and it is what I solved the question with

P(A1) = 0.23

P(A2) = 0.25

P(A3) = 0.29

P(A1 n A2 ) = 0.09

P(A1 n A3) = 0.11

P(A2 n A3) = 0.07

P(A1 n A2 n A3) = 0.02

a

P(A2|A1) = P(A1 n A2)/P(A1)

= 0.09/0.23

= 0.3913

We have 39.13% confidence that event A2 will occur given that event A1 already occured

b.)

P(A3 n A3|A1) = P(A2 n A3)n A1)/P(A1)

= 0.02/0.23

= 0.08695

We have about 8.7% chance of events A2 and A3 occuring given that A1 already occured.

C.

P(A2 u A3|A1)

= P(A1 n A2)u(A1 n A3)/P(A1)

= P( A1 n A2) + P(A1 n A3) - P(A1 n A2 n A3) / P(A1)

= (0.09+0.11-0.02)/0.23

= 0.18/0.23

= 0.7826

We have 78.26% chance of A2 or A3 happening given that A1 has already occured.

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For the hypothesis test H0: μ = 10 against H1: μ >10 and variance known, calculate the Pvalue for each of the following test
Basile [38]

Answer:

a) p_v =P(Z>2.05)=1-P(z

b) p_v =P(Z>-1.84)=1-P(z

c) p_v =P(Z>0.4)=1-P(z

Step-by-step explanation:

Some previous concepts

The p-value is the probability of obtaining the observed results of a test, assuming that the null hypothesis is correct.

A z-test for one mean "is a hypothesis test that attempts to make a claim about the population mean(μ)".

The null hypothesis attempts "to show that no variation exists between variables or that a single variable is no different than its mean"

The alternative hypothesis "is the hypothesis used in hypothesis testing that is contrary to the null hypothesis"

Hypothesis

Null hypothesis: \mu=10

Alternative hypothesis: \mu >10

If the random variable is distributed like this: X \sim N(\mu,\sigma)

We assume that the variance is known so the correct test to apply here is the z test to compare means, the statistic is given by the following formula:

z_o=\frac{\bar X -\mu}{\sigma}

Since we have the values for the statistic already calculated we can calculate the p value using the following formulas:

Part a

p_v =P(Z>2.05)=1-P(z

And in order to find the answer using excel we can use the following code:

"=1-NORM.DIST(2.05,0,1,TRUE)"

Part b

p_v =P(Z>-1.84)=1-P(z

And in order to find the answer using excel we can use the following code:

"=1-NORM.DIST(-1.84,0,1,TRUE)"

Part c

p_v =P(Z>0.4)=1-P(z

And in order to find the answer using excel we can use the following code:

"=1-NORM.DIST(0.4,0,1,TRUE)"

Conclusions

If we use a reference value for the significance, let's say \alpha=0.05. For part a the p_v so then we can reject the null hypothesis at this significance level.

For part b the p_v>\alpha so then we FAIL to reject the null hypothesis at this significance level.

For part c the p_v>\alpha so again we FAIL to reject the null hypothesis at this significance level.

3 0
3 years ago
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