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tekilochka [14]

rational irrational irrational

7 0

3 years ago

(10 points)Assume IQs of adults in a certain country are normally distributed with mean 100 and SD 15. Suppose a president, vice

vesna_86 [32]

Answer:

0.0139 = 1.39% probability that the president will have an IQ of at least 107.5 and that at least one of the other two leaders (vice president and/or secretary of state) will have an IQ of at least 130.

Step-by-step explanation:

To solve this question, we need to use the binomial and the normal probability distributions.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Probability the president will have an IQ of at least 107.5

IQs of adults in a certain country are normally distributed with mean 100 and SD 15, which means that $\mu = 100, \sigma = 15$

This probability is 1 subtracted by the p-value of Z when X = 107.5. So

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

$Z = \frac{107.5 - 100}{15}$

$Z = 0.5$

$Z = 0.5$ has a p-value of 0.6915.

1 - 0.6915 = 0.3085

0.3085 probability that the president will have an IQ of at least 107.5.

Probability that at least one of the other two leaders (vice president and/or secretary of state) will have an IQ of at least 130.

First, we find the probability of a single person having an IQ of at least 130, which is 1 subtracted by the p-value of Z when X = 130. So

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

$Z = \frac{130 - 100}{15}$

$Z = 2$

$Z = 2$ has a p-value of 0.9772.

1 - 0.9772 = 0.0228.

Now, we find the probability of at least one person, from a set of 2, having an IQ of at least 130, which is found using the binomial distribution, with p = 0.0228 and n = 2, and we want:

$P(X \geq 1) = 1 - P(X = 0)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{2,0}.(0.9772)^{2}.(0.0228)^{0} = 0.9549$

$P(X \geq 1) = 1 - P(X = 0) = 0.0451$

0.0451 probability that at least one of the other two leaders (vice president and/or secretary of state) will have an IQ of at least 130.

What is the probability that the president will have an IQ of at least 107.5 and that at least one of the other two leaders (vice president and/or secretary of state) will have an IQ of at least 130?

0.3085 probability that the president will have an IQ of at least 107.5.

0.0451 probability that at least one of the other two leaders (vice president and/or secretary of state) will have an IQ of at least 130.

Independent events, so we multiply the probabilities.

0.3082*0.0451 = 0.0139

0.0139 = 1.39% probability that the president will have an IQ of at least 107.5 and that at least one of the other two leaders (vice president and/or secretary of state) will have an IQ of at least 130.

8 0

3 years ago

Determine the number of ways of placing the numbers $1, 2, 3, \dots, 9$ in a circle, so that the sum of any three numbers in con

solmaris [256]

Answer:

144 ways

Step-by-step explanation:

The arrangements to have the sum of three numbers being divisible by three can be approached by modular arithmetic such that the difference between the third integer and the sum of the first two integers is divisible by 3 making them congruent modulus 3

We have;

With the first number as 1

The second number can be 2, 5 or 8

The third number can then be 3, 6, 9

The fourth number can then be 4 or 7

The sixth number can be 3, 6, 9

The seventh number can be 4 or 7

The eight number can be 2, 5, 8

and the ninth number can be 3, 6, 9

Therefore, we have

the first number as 1

The second, fifth, and eight number can be 2, 5 or 8

The third, sixth and ninth number can then be 3, 6, 9

The fourth, and seventh number can then be 4 or 7 Which gives,

The number of possible arrangements for the numbers 2, 5, and 8 = 3! = 6

The number of possible arrangements for the numbers 3, 6, and 9 = 3! = 6

The number of possible arrangements for the numbers 4, and 7 = 2! = 2

he total number of possible arrangements = 2×6×6 =72

The arrangement can be reversed in the counter clockwise directions given the total number of ways of placing the numbers 1, 2, 3,..., 9 in a circle, so that the sum of any three numbers in consecutive positions is divisible by 3 = 72 × 2 = 144 ways.

7 0

3 years ago

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What is an example of dilation in the real world?

IgorC [24]

In eye doctors offices they dilate you’re eyes mailing your pupils bigger. Dilation is used in eye exams so the eye doctor can see your eye better

6 0

3 years ago

For all real numbers, x^2 ≥x.

BlackZzzverrR [31]

1/2 because it is a fraction

8 0

3 years ago