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

The probability that fewer than 320 out of 750 adults over 65 in the study suffer from one or more of the conditions under consideration is approximately 0.109 or 10.9%.

The calculation of the probability that fewer than 320 out of n = 750 adults over 65 in the study suffer from one or more of the conditions under consideration:

Given:

p = 0.45 (proportion of adults over 65 suffering from one or more conditions)

n = 750 (sample size)

x = 320 (number of adults suffering from one or more conditions)

Using the normal approximation to the binomial distribution:

Calculate the mean (µ) and standard deviation (σ) of the binomial distribution:

µ = np = 750 * 0.45 = 337.5

σ = √(npq) = √(750 * 0.45 * 0.55) = 14.26

Calculate the z-score (Z) for x:

Z = (x - µ) / σ = (320 - 337.5) / 14.26 = -1.23

Find the area to the left of Z under the standard normal curve using a z-table or calculator. This represents the probability that fewer than 320 adults suffer from one or more conditions.

Probability = P(Z < -1.23) ≈ 0.109

Therefore, the probability that fewer than 320 out of 750 adults over 65 in the study suffer from one or more of the conditions under consideration is approximately 0.109 or 10.9%.

Answer 2

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 .

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 * p` =
`750 * 0.45` = 337.5

Standard deviation of X,
`\sigma` =
`√(np(1-p))` =
`√(750 * 0.45 * (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 =
`(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(
`(X-\mu)/(\sigma)` <
`(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.