Question

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 = 750 adults over 65 in the study suffer from one or more of the conditions under consideration. Give your answer accurate to three decimal places in decimal form.

Answer 1

Answer:

Probability that fewer than 320 out of the n = 750 adults over 65 in the study suffer from one or more of the conditions under consideration is 0.0934 .

Step-by-step explanation:

We are given that In the U.S., 45% of adults over 65 suffer from one or more of the conditions considered in the study. And we have to find the probability that fewer than 320 out of the n = 750 adults over 65 in the study suffer from one or more of the conditions under consideration.

Firstly, the above situation can be represented through binomial distribution, i.e.;

$P(X=r) = \binom{n}{r} p^{r} (1-p)^{2} ;x=0,1,2,3,....$

where, n = number of samples taken = 750

            r = number of success

            p = probability of success, i.e. 45%

Now, we can't calculate the required probability using binomial distribution because here n is very large, so we will convert this distribution into normal distribution using continuity correction.

So, Let X = No. of adults suffering from one or more of the conditions under consideration

Mean of X, $\mu$ = $n \times p$ = $750 \times 0.45$ = 337.5

Standard deviation of X, $\sigma$ = $\sqrt{np(1-p)}$ = $\sqrt{750 \times 0.45 \times (1-0.45)}$ = 13.62

So, X ~ N($\mu = 337.5, \sigma^{2} = 13.62^{2}$)

Now, the z score probability distribution is given by;

         Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

So, probability that fewer than 320 out of the n = 750 adults over 65 in the study suffer from one or more of the conditions under consideration is given by = P(X < 319.5) ---- using continuity correction

P(X < 319.5) = P( $\frac{X-\mu}{\sigma}$ < $\frac{319.5-337.5}{13.62}$ ) = P(Z < -1.32) = 1 - P(Z $\leq$ 1.32)

                                                         = 1 - 0.90658 = 0.0934

Therefore, required probability is 0.0934.