Answer :
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.;
[tex]P(X=r) = \binom{n}{r} p^{r} (1-p)^{2} ;x=0,1,2,3,....[/tex]
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, [tex]\mu[/tex] = [tex]n \times p[/tex] = [tex]750 \times 0.45[/tex] = 337.5
Standard deviation of X, [tex]\sigma[/tex] = [tex]\sqrt{np(1-p)}[/tex] = [tex]\sqrt{750 \times 0.45 \times (1-0.45)}[/tex] = 13.62
So, X ~ N([tex]\mu = 337.5, \sigma^{2} = 13.62^{2}[/tex])
Now, the z score probability distribution is given by;
Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ 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( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{319.5-337.5}{13.62}[/tex] ) = P(Z < -1.32) = 1 - P(Z [tex]\leq[/tex] 1.32)
= 1 - 0.90658 = 0.0934
Therefore, required probability is 0.0934.