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WINSTONCH [101]
2 years ago
6

Sara has intense problems with leaving her house. She often refuses to do so, and in cases when she will go out, she simply must

be by her mother's side at all times. This goes as far as requiring her mother to be in physical contact with Sara (holding hands, clinging to her mother's clothing) whenever they go shopping. Sara also refuses to go to a peer's house for a play date unless her mother stays the whole time and is never more than one room away. Sara may be diagnosed with:
a. generalized anxiety disorder (GAD)b. oppositional-defiant disorderc. specific phobiad. separation anxiety disorder
Health
1 answer:
lapo4ka [179]2 years ago
5 0

Answer:separation anxiety disorder

Explanation:

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Answer:

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Explanation:

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You are performing a cohort study to look at risk factors for breast cancer in postmenopausal women. If the probability of devel
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Answer:

E(X) = np = 100 *0.05 = 5

So we expect about 5 people with the disease in the random sample.

Explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

The probability mass function for the Binomial distribution is given as:  

P(X)=(nCx)(p)^x (1-p)^{n-x}  

Where (nCx) means combinatory and it's given by this formula:  

nCx=\frac{n!}{(n-x)! x!}  

Solution to the problem

For this case we can assume that our random variable who represent the number of students that develop the disease is distributed on this way X \sim Bin(n=100, p =0.05)

And from definition the expected value is given by:

E(X) = np = 100 *0.05 = 5

And the variance is given by:

Var (X) = np(1-p) = 100*0.05 *(1-0.05) =4.75

And the deviation is given by:

Sd(X) = \sqrt{4.75}= 2.179

So we expect about 5 people with the disease in the random sample.

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