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OleMash [197]
3 years ago
5

Conditional Distribution, Marginal Distribution, Joint Distribution. What’s the difference?

Mathematics
1 answer:
lutik1710 [3]3 years ago
3 0

Explanation:

Marginal distribution: This distribution gives the probability for each possible value of the Random variable ignoring other random variables. Basically, the values of other variables is not considered in the marginal distribution, they can be any value possible. For example, if you have two variables X and Y, the probability of X being equal to a value, lets say, 4, contemplates every possible scenario where X is equal to 4, independently of the value Y has taken. If you want the probability of a dice being a multiple of 3, you are interested that the dice is either 3 or 6, but you dont care if the dice is even or odd.

Conditional distribution: This distribution contrasts from the previous one in the sense that we are restricting the universe of events to specific condition for other variable, making a modification of our marginal results. If we know that throwing a dice will give us a result higher than 2, then to in order to calculate the probability of the dice being a multiple of 3 using that condition, we have two favourable cases (3 and 6) from 4 total possible results (3,4,5 and 6) discarding the impossible values (1 and 2) from this universe since they dont match the condition given (note that the restrictions given can also reduce the total of favourable cases).

The joint distribution calculates the probabilities for two different events (related to two different random variables) occuring simultaneously. If we want to calculate the joint probability of a dice being multiple of 3 and greater than 2 at the same time, our possible cases in this case are 3 and 6 from 6 possible results. We are not discarding 1 or 2 as possible results because we are not assuming, that the dice is greater than 2, that is another condition that we should met in the combination of events.

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On Monday a Farmer sold 25196 pounds of potatoes on tuesday he sold 18023pounds on Wednesday he sold some more potatoes in all h
arlik [135]

Answer:

<em>estimated sales on Wednesday is 19000 pounds.</em>

<em></em>

Step-by-step explanation:

On Monday, he sold 25196 pounds. Estimated to the nearest thousand that is 25000 pounds.

On Tuesday, he sold 18023 pounds. Estimated to the nearest thousand, that is 18000 pounds

Wednesday's sales is unknown. We designate as x

All in all he sold 62409. Estimated to the nearest thousand, that is 62000

The sales on Monday, plus sales on  Tuesday, plus sales on Wednesday, must all sum up to the total sales.

25000 + 18000 + x = 62000

43000 + x = 62000

x = 62000 - 43000 = 19000

therefore <em>estimated sales on Wednesday is 19000 pounds.</em>

3 0
3 years ago
Which answer choice shows 4 + 0.3 + 0.09 written in standard form?A.43.9B.4.39C.0.439D.0.0439
Oliga [24]

Place values:

As each is in a different place value, when we add them neither will affect the other, meaning:

4+0.3+0.09=4.39

Answer: B 4.39

5 0
1 year ago
drivers have no prior driving record, an insurance company considers each driver to be randomly selected from the pool. This mon
Kisachek [45]

Complete question is:

A large pool of adults earning their first drivers license includes 50% low- risk drivers, 30% moderate-risk drivers, and 20% high-risk drivers. Because these drivers have no prior driving record, an insurance company considers each driver to be randomly selected from the pool. This month, the insurance company writes 4 new policies for adults earning their first drivers license. What is the probability that these 4 will contain at least two more high-risk drivers than low-risk drivers?

Answer:

probability that these 4 will contain at least two more high-risk drivers than low-risk drivers = 0.0488

Step-by-step explanation:

Let H represent High risk

M represent moderate risk

L represent Low risk.

The following combinations will satisfy the condition that there are at least two more high-risk drivers than low-risk drivers: HHHH, HHHL, HHHM, HHMM

The HHHH case has probability 0.2 ⁴ = 0.0016

The HHHL case has probability 4 × 0.2³ × 0.3 = 0.0096 (This is because L can be in four different places)

Similarly, the HHHM case has probability 4 × 0.2 ³ × 0.5 = 0.016

Lastly, the HHMM case has probability 6 × 0.2 ² × 0.3 ² = 0.0216 (This is because the number of ways to choose places for two M letters in this way is 6)

Summing all these probabilities, we have;

0.0016 + 0.0096 + 0.016 + 0.0216 = 0.0488

5 0
3 years ago
Find the margin of error for a 90% confidence interval when the standard deviation is LaTeX: \sigma= 50????=50 and LaTeX: n = 25
Murrr4er [49]

Answer:

The margin of error  for a 90% confidence interval is 16.4

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 25

Standard deviation = 50

z_{critical}\text{ at}~\alpha_{0.10} = \pm 1.64

Margin of error =

z_{critical}\times \dfrac{\sigma}{\sqrt{n}}

Putting the values, we get,

1.64\times \dfrac{50}{\sqrt{25}} = 16.4

Thus, the margin of error  for a 90% confidence interval is 16.4

8 0
3 years ago
Plz help me. Plz calculate 4(x−2)−3
Alisiya [41]

Answer:

4x-11

Step-by-step explanation:

open parentheses: 4x-8-3

solve: 4x-11

7 0
3 years ago
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