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leonid [27]
3 years ago
6

For a large data set of a average student grades, the third quartile Q3 is found 78.5 . What does the mean when we say that 78.5

is the third quartile?
Mathematics
1 answer:
xz_007 [3.2K]3 years ago
3 0

Step-by-step explanation:              

We are told that for a large data set of a average student grades, the third quartile Q3 is found 78.5.

Since we know that third quartile represents the number such that 75% of the data is less than this number. Third quartile is the middle value between median and the highest value of a data set. Third quartile is also known as upper quartile. Third quartile splits lower 75% data from highest 25% of data.

Therefore, third quartile represents that 75% of student grades are less than 78.5 and 25% of students grades are greater than 78.5.


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Answer:

Step-by-step explanation:

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2a + 3b = 5<br> b = a - 5<br> What is a and b?
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2a + 3(a-5) = 5, 2a + 3a+15=5, 5a=-10, a=-2, b= (-2) -5, b=-7...a=-2, b=-7
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What is the <br> Quadratic form for (-1,-11) (0,-3) (3,-27)
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The final answer to this question is y=−4x²+4x−3
5 0
3 years ago
A large operator of timeshare complexes requires anyone interested in making a purchase to first visit the site of interest. His
spayn [35]

Answer:

0.087 = 8.7% probability that this person made a day visit.

0.652 = 65.2% probability that this person made a one-night visit.

0.261 = 26.1% probability that this person made a two-night visit.

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

P(B|A) = \frac{P(A \cap B)}{P(A)}

In which

P(B|A) is the probability of event B happening, given that A happened.

P(A \cap B) is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Made a purchase.

Probability of making a purchase:

10% of 20%(day visit)

30% of 50%(one night)

20% of 30%(two night).

So

p = 0.1*0.2 + 0.3*0.5 + 0.2*0.3 = 0.23

How likely is it that this person made a day visit?

Here event B is a day visit.

10% of 20% is the percentage of purchases and day visit. So

P(A \cap B) = 0.1*0.2 = 0.02

So

P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.02}{0.23} = 0.087

0.087 = 8.7% probability that this person made a day visit.

A one-night visit?

Event B is a one night visit.

The percentage of both(one night visit and purchase) is 30% of 50%. So

P(A \cap B) = 0.3*0.5 = 0.15

So

P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.15}{0.23} = 0.652

0.652 = 65.2% probability that this person made a one-night visit.

A two-night visit?

Event B is a two night visit.

The percentage of both(two night visit and purchase) is 20% of 30%. So

P(A \cap B) = 0.2*0.3 = 0.06

Then

P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.06}{0.23} = 0.261

0.261 = 26.1% probability that this person made a two-night visit.

7 0
3 years ago
There are three highways in the county. the number of daily accidents that occur on these highways are poisson random variables
Romashka-Z-Leto [24]
There is a theorem that we can use to find the expected value of a random variable that is a sum of other random variables as follows.

If X=X_1 + X_2 + ...+X_k, then E(X)=E(X_1)+E(X_2)+...+E(X_k).

In this case, let X = the number of accidents that will happen on any of those highways today,
X_1,X_2,X_3 are the numbers of accidents on each highway, respectively.

Then  X = X_1+X_2+X_3.

Since X_1,X_2,X_3 are Poisson variates, their expected values are the parameters given, .3, .5 and .7.

So E(X_1)=.3, E(X_2) = .5, E(X_3) = .7
Thus, E(X)=E(X_1)+E(X_2)+E(X_3) = .3 + .5 + .7 = 1.5

The answer is 1.5.
8 0
2 years ago
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