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lana66690 [7]
2 years ago
11

The lengths of a professor's classes has a continuous uniform distribution between 50.0 min and 52.0 min. If one such class is r

andomly selected, find the probability that the class length is between 51.5 and 51.7 min. P(51.5 < X < 51.7)
Mathematics
1 answer:
svp [43]2 years ago
3 0

Answer:

0.1 = 10% probability that the class length is between 51.5 and 51.7 min, that is, P(51.5 < X < 51.7) = 0.1.

Step-by-step explanation:

A distribution is called uniform if each outcome has the same probability of happening.

The uniform distributon has two bounds, a and b, and the probability of finding a value between c and d is given by:

P(c \leq X \leq d) = \frac{d - c}{b - a}

The lengths of a professor's classes has a continuous uniform distribution between 50.0 min and 52.0 min.

This means that a = 50, b = 52

If one such class is randomly selected, find the probability that the class length is between 51.5 and 51.7 min.

P(51.5 \leq X \leq 51.7) = \frac{51.7 - 51.5}{52 - 50} = \frac{0.2}{2} = 0.1

0.1 = 10% probability that the class length is between 51.5 and 51.7 min, that is, P(51.5 < X < 51.7) = 0.1.

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A, D, E and F are all true statements.



When looking for congruence in these type of problems, you have to look for the order of the letters in the original listing. Since they are listed LMN and PQR, this means we have to focus on the placement of those letters. Since L is the first listed in the sequence, it is congruent with P, which is also listed first. M is congruent with Q since they are both in the middle, and N and R are congruent since they are both at the end.



Now to find true statements, you need to make sure everything matches up in the statement.



Let's take A for instance. It states that MN = QR. Now we know M is in the middle and N is at the end. Since Q is in the middle and R is at the end, they match and are therefore true.


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2 years ago
student randomly receive 1 of 4 versions(A, B, C, D) of a math test. What is the probability that at least 3 of the 5 student te
alexdok [17]

Answer:

1.2%

Step-by-step explanation:

We are given that the students receive different versions of the math namely A, B, C and D.

So, the probability that a student receives version A = \frac{1}{4}.

Thus, the probability that the student does not receive version A = 1-\frac{1}{4} = \frac{3}{4}.

So, the possibilities that at-least 3 out of 5 students receive version A are,

1) 3 receives version A and 2 does not receive version A

2) 4 receives version A and 1 does not receive version A

3) All 5 students receive version A

Then the probability that at-least 3 out of 5 students receive version A is given by,

\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}\times \frac{3}{4}+\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}+\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}

= (\frac{1}{4})^3\times (\frac{3}{4})^2+(\frac{1}{4})^4\times (\frac{3}{4})+(\frac{1}{4})^5

= (\frac{1}{4})^3\times (\frac{3}{4})[\frac{3}{4}+\frac{1}{4}+(\frac{1}{4})^2]

= (\frac{3}{4^4})[1+\frac{1}{16}]

= (\frac{3}{256})[\frac{17}{16}]

= 0.01171875 × 1.0625

= 0.01245

Thus, the probability that at least 3 out of 5 students receive version A is 0.0124

So, in percent the probability is 0.0124 × 100 = 1.24%

To the nearest tenth, the required probability is 1.2%.

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3 years ago
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