Researchers published a study in which they considered the incidence among the elderly of various mental health conditions such as dementia, bi-polar disorder, obsessive compulsive disorder, delirium, and Alzheimer's disease. In the U.S., 45% of adults over 65 suffer from one or more of the conditions considered in the study. Calculate the probability that fewer than 320 out of the n

Answer:

The step-by-step procedure to solve this question is given, using the normal approximation to the binomial distribution.

Step-by-step explanation:

We use the normal approximation to the binomial distribution to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

[tex]E(X) = np[/tex]

The standard deviation of the binomial distribution is:

[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]

Normal probability distribution

Problems of normally distributed distributions can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that [tex]\mu = E(X)[/tex], [tex]\sigma = \sqrt{V(X)}[/tex].

In the U.S., 45% of adults over 65 suffer from one or more of the conditions considered in the study.

This means that [tex]p = 0.45[/tex]

Mean and standard deviation for the approximation, for a sample of n:

[tex]\mu = E(X) = np = 0.45n[/tex]

[tex]\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{0.45*0.55n}[/tex]

Calculate the probability that fewer than 320 out of the n

Using continuity correction, this is [tex]P(X < 320 - 0.5) = P(X < 319.5)[/tex], which is the pvalue of Z when X = 319.5. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{319.5 - 0.45n}{\sqrt{0.45*0.55n}}[/tex]

Just have to replace the value of n, find Z, and then find its pvalue.

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