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jonny [76]
2 years ago
7

In Europe, 53% of the flowers of the Rewardless Orchid, Dactylorhiza sambucina, are yellow, whereas the remaining flowers are pu

rple. For this problem, only use the normal approximation where it is appropriate. Use the binomial distribution where the normal approximation is inappropriate.
Required:
a. If we took a random sample of a single individual from this population, what is the probability that it would be purple?
b. If we took a random sample of five individuals, what is the probability that at least three are yellow?
c. If we took many samples of n=5 individuals, what is the expected standard deviation of the sampling distribution for the proportion of yellow flowers?
d. If we took a random sample of 263 individuals, what is the probability that no more than 150 are yellow?
Mathematics
1 answer:
stich3 [128]2 years ago
7 0

Answer:

a. The probability of taking out a purple flower is 0.47.

b. The probability that out of a sample of 5 flowers at least 3 flowers are yellow 0.56.

c. The standard deviation of the distribution for proportion of yellow flowers is 0.22.

d. The probability that out of the 263 samples, at most 150 of them are yellow is 0.90.

Step-by-step explanation:

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Answer:

5

Step-by-step explanation:

1. Add -8 to -12. This equals -20.

2. Now you have -20/(-4)

3. Solve this by dividing -20 and -4.

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2 years ago
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1 year ago
Suppose that you take 120 mg of an antibiotic every 4 hr. The​ half-life of the drug is 4 hr​ (the time it takes for half of the
vodomira [7]

Answer:

The steady state amount of antibiotic in the bloodstream when t --> ∞ is 240 mg.

Step-by-step explanation:

Let the amount of antibiotic in one's bloodstream be given as Aₙ (where n = the number of half lives since the start of usage)

Let's follow the time line of events.

At t = 0 hr, the drug is taken

A₀ = 120 mg

At t = 4 hrs, n = 1, the drug is taken again

A₁ = (0.5×A₀) + 120

A₁ = (0.5×120) + 120 = 180 mg

At t = 8 hrs, n = 2, the drug is taken again,

A₂ = (0.5×A₁) + 120

A₂ = (0.5×180) + 120 = 210 mg

At t = 12 hrs, n = 3, the drug is taken again

A₃ = (0.5×A₂) + 120

A₃ = (0.5×210) + 120 = 225 mg

At this point, it becomes evident that at t = 4n hrs, n = n i.e. n half lives later, the general formula for the amount of the antibiotic in the bloodstream is

Aₙ = 0.5Aₙ₋₁ + 120

where Aₙ₋₁ = The amount of antibiotic in the bloodstream at the time t = 4(n-1) and (n-1) half lives later.

For infinite series, that are increasing in this order, as the value of n --> ∞,

Aₙ = Aₙ₋₁ = K

And our general formula becomes

K = 0.5K + 120

0.5K = 120

K = (120/0.5)

K = 240 mg

Hence, the steady state amount of antibiotic in the bloodstream when t --> ∞ is 240 mg.

Hope this Helps!!!

5 0
3 years ago
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