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KatRina [158]
3 years ago
6

Research results showed that experimental Drug R had a minimal to modest effect in reducing the symptoms of generalized anxiety

disorder (GAD). The researchers decide that their earlier hypothesis, that 200 mg of Drug R will reduce GAD symptoms by 25 percent as evidenced by the Hamilton Anxiety Scale, needs to be revised. They decide to increase the dosage to 300 mg in their next study. The variable that may change in response to this increase is the:
Mathematics
1 answer:
Angelina_Jolie [31]3 years ago
3 0

Answer:

The variable that may change in response to the increase of the drug is the GAD symptoms by a 37,5%.

Step-by-step explanation:

According to the results of the first experiment with a mass of 200 mg of Drug R, they obtain a reduced of the GAD symptoms by a 25 percent evidenced by the Hamilton Anxiety Scale.

If they decided to increase the mass of Drug R to 300 mg the results expected are a increase of the porcentange of the reduced symptoms of generalized anxiety disorder, according to the tendence of the first hypothesis and the Hamilton Anxiety Scale.

We can express this increase by using the three simple rule. Where if 200 mg of Drug R reduced the 25% of the GAD symptoms, if we increase to 300 mg of Drug R how much porcentage this amount will be reduced.

Doing the maths 300mg × 25%=7500mg%,

    ⇒ 7500mg% ÷ 200mg = 37,5%.

<u>In conclusion</u> if they increased the mas of Drug R to 300 mg they will be reduced the generalized anxiety disorder (GAD) to a 37,5%.

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Step-by-step explanation:

The complete question is:

Of the entering class at a college, 71% attended public high school, 21% attended private high school, and 8% were home schooled. Of those who attended public high school, 16% made the Dean's list, 19% of those who attended private high school made the Dean's list, and 15% of those who were home schooled made the Dean's list.

a) Find the probability that the student made the Dean's list.

b) Find the probability that the student came from a private high school, given that the student made the Dean's list.

c) Find the probability that the student was not home schooled, given that the student did not make the Dean's list.

Solution:

Denote the events as follows:

<em>A</em> = a student attended public high school

<em>B</em> = a student attended private high school

<em>C</em> = a student was home schooled

<em>D</em> = a student made the Dean's list

The provided information is as follows:

P (A) = 0.71

P (B) = 0.21

P (C) = 0.08

P (D|A) = 0.16

P (D|B) = 0.19

P (D|C) = 0.15

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The law of total probability states that:

P(X)=\sum\limits_{i} P(X|Y_{i})\cdot P(Y_{i})

Compute the probability that the student made the​ Dean's list as follows:

P(D)=P(D|A)P(A)+P(D|B)P(B)+P(D|C)P(C)

         =(0.16\times 0.71)+(0.19\times 0.21)+(0.15\times 0.08)\\=0.1136+0.0399+0.012\\=0.1655

Thus, the probability that the student made the​ Dean's list is 0.1655.

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Compute the probability that the student came from a private high school, given that the student made the Dean's list as follows:

P(B|D)=\frac{P(D|B)P(B)}{P(D)}

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Compute the probability that the student was not home schooled, given that the student did not make the Dean's list as follows:

P(C^{c}|D^{c})=1-P(C|D^{c})

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Thus, the probability that the student was not home schooled, given that the student did not make the Dean's list is 0.9185.

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