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Fiesta28 [93]
2 years ago
11

A health fitness research group wishes to estimate the mean amount of time (in hours) that members of a fitness center spend eac

h week exercising at the center. They want to estimate the mean within a margin of error of 0.4 hours with a 95% level of confidence. Previous data suggest that the population standard deviation (sigma) is 2.6 hours. What is the smallest sample size that meets these criteria
Mathematics
1 answer:
Arlecino [84]2 years ago
6 0

Answer:

the smallest sample size is 163

Step-by-step explanation:

The computation of the smallest sample size that meets these criteria is shown below:

n = (Z a/2 ×  Standard deviation  ÷ Margin of error ) ^2

Zα/2 at 0.05% LOS is = 1.96 ( From Standard Normal Table )

Standard Deviation ( S.D) = 2.6

Margin of error i.e. ME =0.4

n = ( 1.96 × 2.6 ÷ 0.4) ^2

= 163

Hence, the smallest sample size is 163

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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
madreJ [45]

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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Decimal Answer: 5.3928571
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Which expression represents h(x)?
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Answer:

(g-f)x is the answers for the question

Step-by-step explanation:

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Which of the following is a radical equation?
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It is A because it can’t be B and it can’t be C and it can’t be D so my answer is A
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Answer:

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Step-by-step explanation:

We know that rational numbers are numbers that are fractions and decimals.

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