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jarptica [38.1K]
2 years ago
6

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

sis of body mass index (BMI), expressed as weight/height2. An adult is considered overweight if the BMI is 25 or more but less than 30. An obese adult will have a BMI of 30 or greater. According to a January 2012 article in the Journal of the American Medical Association, 33.1% of the adult population in the United States is overweight and 35.7% is obese. Use this information to answer the following questions.A. What is the probability that a randomly selected adult is either overweight or obese? (Round your answer to 3 decimal places.)B. What is the probability that a randomly selected adult is neither overweight nor obese? (Round your answer to 3 decimal places.)C. Are the events "overweight" and "obese" exhaustive?D. Are the events "overweight" and "obese" mutually exclusive?
Mathematics
1 answer:
madreJ [45]2 years ago
5 0

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.

(C) The events "overweight" and "obese" exhaustive.

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

Step-by-step explanation:

Denote the events as follows:

<em>X</em> = a person is overweight

<em>Y</em> = 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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