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harkovskaia [24]
4 years ago
12

The mathematics club will select a president, a vice president, and a treasurer for the club. if there are 15 members in the clu

b, how many different selections of a president, a vice president, and a treasurer are possible if each club member can be selected to only one position? here are the choices i forgot to include them 42 455 2730 3375
Mathematics
1 answer:
Masteriza [31]4 years ago
4 0
This is a problem of variations because the order is important, it can not be selected all of members, and can not repeat.
 15V3 = 15!/(15-3)! = 2730
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Answer:

0.2333 = 23.33% probability this student's score will be at least 2100.

Step-by-step explanation:

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Normal Probability Distribution:

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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

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

P(A) is the probability of A happening.

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This means that \mu = 1490, \sigma = 295

In this question:

Event A: Student was recognized.

Event B: Student scored at least 2100.

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Probability of scoring at least 1900, which is 1 subtracted by the pvalue of Z when X = 1900. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{1900 - 1490}{295}

Z = 1.39

Z = 1.39 has a pvalue of 0.9177

1 - 0.9177 = 0.0823

This means that P(A) = 0.0823

Probability of a student being recognized and scoring at least 2100:

Intersection between at least 1900 and at least 2100 is at least 2100, so this is 1 subtracted by the pvalue of Z when X = 2100.

Z = \frac{X - \mu}{\sigma}

Z = \frac{2100 - 1490}{295}

Z = 2.07

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P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.0192}{0.0823} = 0.2333

0.2333 = 23.33% probability this student's score will be at least 2100.

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