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3 years ago

BEST ANSWER GETS BRAINLY (EMERGENCY)

Mathematics

2 answers:

Nostrana [21]3 years ago

5 0

Answer:

Part A: 5

Part B: 2

Step-by-step explanation:

to find the mean all you have to do is add all the numbers together and then divide it by how many numbers there are in the data set

3+4+4+4+5+6+6+8=40

40÷8=5

so the mean is 5

next to find the mean of absolute deviations you need to subtract each number in the data set from the mean then divide it by how many numbers are in the data set (if you get a negative like 6-5=-1 just take away the negative sign to make it positive)

5-3=2

5-4=1

5-5=0

6-5=-1

8-5=-3

2+1+1+1+0+1+1+3=10

10÷5=2

so the mean absolute deviation is 2

MrRissso [65]3 years ago

4 0

Answer:

Part A: 5

Part B: 2

Step-by-step explanation:

to find the mean all you have to do is add all the numbers together and then divide it by how many numbers there are in the data set

3+4+4+4+5+6+6+8=40

40÷8=5

so the mean is 5

5-3=2

5-4=1

5-5=0

6-5=-1

8-5=-3

2+1+1+1+0+1+1+3=10

10÷5=2

so the mean absolute deviation is 2

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An alarming number of U.S. adults are either overweight or obese. The distinction between overweight and obese is made on the ba

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Answer:

(A) The probability that a randomly selected adult is either overweight or obese is 0.688.

(B) The probability that a randomly selected adult is neither overweight nor obese is 0.312.

(D) The events "overweight" and "obese" mutually exclusive.

Step-by-step explanation:

Denote the events as follows:

X = a person is overweight

Y = a person is obese.

The information provided is:

A person is overweight if they have BMI 25 or more but below 30.

A person is obese if they have BMI 30 or more.

P (X) = 0.331

P (Y) = 0.357

(A)

The events of a person being overweight or obese cannot occur together.

Since if a person is overweight they have (25 ≤ BMI < 30) and if they are obese they have BMI ≥ 30.

So, P (X ∩ Y) = 0.

Compute the probability that a randomly selected adult is either overweight or obese as follows:

$P(X\cup Y)=P(X)+P(Y)-P(X\cap Y)\\=0.331+0.357-0\\=0.688$

Thus, the probability that a randomly selected adult is either overweight or obese is 0.688.

(B)

Commute the probability that a randomly selected adult is neither overweight nor obese as follows:

$P(X^{c}\cup Y^{c})=1-P(X\cup Y)\\=1-0.688\\=0.312$

Thus, the probability that a randomly selected adult is neither overweight nor obese is 0.312.

(C)

If two events cannot occur together, but they form a sample space when combined are known as exhaustive events.

For example, flip of coin. On a flip of a coin, the flip turns as either Heads or Tails but never both. But together the event of getting a Heads and Tails form a sample space of a single flip of a coin.

In this case also, together the event of a person being overweight or obese forms a sample space of people who are heavier in general.

Thus, the events "overweight" and "obese" exhaustive.

(D)

Mutually exclusive events are those events that cannot occur at the same time.

The events of a person being overweight and obese are mutually exclusive.

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Answer:

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Step-by-step explanation:

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