Wright et al. [A-2] used the 1999-2000 National Health and Nutrition Examination Survey(NHANES) to estimate dietary intake of 10 key nutrients. One of those nutrients was calcium (mg). They found in all adults 60 years or older a mean daily calcium intake of 721 mg with a standard deviation of 454. Using these values for the mean and standard deviation for the U.S. population, find and interpret the probability that a random sample of size 50 will a mean:

Answers

Answer 1

Complete question :

Wright et al. [A-2] used the 1999-2000 National Health and Nutrition Examination Survey NHANES) to estimate dietary intake of 10 key nutrients. One of those nutrients was calcium in all adults 60 years or older a mean daily calcium intake of 721 mg with a standard deviation of 454. Usin these values for the mean and standard deviation for the U.S. population, find the probability that a randonm sample of size 50 will have a mean: (mg). They found a) Greater than 800 mg b) Less than 700 mg. c) Between 700 and 850 mg.

Answer:

0.10935

0.3718

0.9778

0.606

Step-by-step explanation:

μ = 721 ; σ = 454 ; n = 50

P(x > 800)

Zscore = (x - μ) / σ/sqrt(n)

P(x > 800) = (800 - 721) ÷ 454/sqrt(50)

P(x > 800) = 79 / 64.205295

P(x > 800) = 1.23

P(Z > 1.23) = 0.10935

2.)

Less than 700

P(x < 700) = (700 - 721) ÷ 454/sqrt(50)

P(x < 700) = - 21/ 64.205295

P(x < 700) = - 0.327

P(Z < - 0.327) = 0.3718

Between 700 and 850

P(x < 850) = (850 - 721) ÷ 454/sqrt(50)

P(x < 850) = 129/ 64.205295

P(x < 700) = 2.01

P(Z < 2.01) = 0.9778

P(x < 850) - P(x < 700) =

P(Z < 2.01) - P(Z < - 0.327)

0.9778 - 0.3718

= 0.606


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The Gable Construction Corporation has placed bids for installing electronic message boards on two different highways.
Because of competition from other companies, Gable Construction Corporation officials estimate that the probability that
they will be awarded the Macon Expressway contract is 0.65 and the probability that they will be awarded the Seaview
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Answers

Answer:

0.64 = 64% probability that they will be awarded the contract for either the Macon Expressway or the Seaview Expressway

Step-by-step explanation:

We solve this question using these events as Venn sets.

I am going to say that:

Event A: Awarded the Macon Expressway.

Event B: Awarded the Seaview Expressway.

Probability that they will be awarded the Macon Expressway contract is 0.65

This means that [tex]P(A) = 0.65[/tex]

The probability that they will be awarded the Seaview Expressway contract is 0.43.

This means that [tex]P(B) = 0.43[/tex]

Furthermore, the probability that they will be awarded both contracts is 0.22.

This means that [tex]P(A \cap B) = 0.22[/tex]

What is the

Probability that they will be awarded the contract for either the Macon Expressway or the Seaview Expressway?

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[tex]P(A) - P(A \cap B) = 0.65 - 0.22 = 0.43[/tex]

Only B:

[tex]P(B) - P(A \cap B) = 0.43 - 0.22 = 0.21[/tex]

Either

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0.64 = 64% probability that they will be awarded the contract for either the Macon Expressway or the Seaview Expressway

ill so you the picture of it​

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Answer:

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Given

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In 10 to 99, there is only 1 32.

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[tex]P(32) = \frac{1}{90}[/tex]

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[tex]n(Odd) = 45[/tex]

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[tex]P(Odd) = \frac{n(Odd)}{n}[/tex]

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Answers

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Answers

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