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FromTheMoon [43]

3 years ago

12

Mark is a nurse anesthetist. He administers medication that puts people to sleep for surgeries and other invasive medical proced

ures. For healthy adults, the manufacturer recommends using a dose of 0.6mcg/kg, which means that Mark must give his patients 0.6mcg of medication for each kg they weight. If Mark's patient weights 60kg, how much medication should he administer?

1 answer:

natka813 [3]3 years ago

6 0

Answer:

Quantity of medicine provide to patient = 36 mcg

Step-by-step explanation:

Given:

Quantity recommend = 0.6 mcg / kg

Weight of patient = 60 kg

Find:

Quantity of medicine provide to patient

Computation:

Quantity of medicine provide to patient = Weight of patient × Quantity recommend

Quantity of medicine provide to patient = 60 × 0.6

Quantity of medicine provide to patient = 36 mcg

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a) 91.33% probability that at most 6 will come to a complete stop

b) 10.91% probability that exactly 6 will come to a complete stop.

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Step-by-step explanation:

For each driver, there are only two possible outcomes. Either they will come to a complete stop, or they will not. The probability of a driver coming to a complete stop is independent of other drivers. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

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20% of all drivers come to a complete stop at an intersection having flashing red lights in all directions when no other cars are visible.

This means that $p = 0.2$

20 drivers

This means that $n = 20$

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$P(X \leq 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

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$P(X = 2) = C_{20,2}.(0.2)^{2}.(0.8)^{18} = 0.1369$

$P(X = 3) = C_{20,3}.(0.2)^{3}.(0.8)^{17} = 0.2054$

$P(X = 4) = C_{20,4}.(0.2)^{4}.(0.8)^{16} = 0.2182$

$P(X = 5) = C_{20,5}.(0.2)^{5}.(0.8)^{15} = 0.1746$

$P(X = 6) = C_{20,6}.(0.2)^{6}.(0.8)^{14} = 0.1091$

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91.33% probability that at most 6 will come to a complete stop

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$P(X < 6) + P(X \geq 6) = 1$

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$P(X \geq 6) = 1 - P(X < 6)$

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