Answer with explantion please 6 Many veterans who bravely se

Answer with explantion please

6. Many veterans who bravely served in Iraq during Operation Enduring Freedom or Operation Iraqi Freedom have developed symptoms of post-traumatic stress disorder (PTSD) or similar ailments. A large study of returning veterans showed that 52.7% of the soldiers suffered from a PTSD or similar ailment after serving

in Iraq. A random sample of 200 veterans from Iraq were selected for additional research. The number suffering from PTSD or a similar condition was recorded.
(Source: Frayne SM, et al. (2010) Medical care needs of returning veterans with
PTSD: their other burden. Journal of General Internal Medicine 26(1).)

a. Explain why this is a binomial experiment.

b. What is the mean of X, the number of people suffering from PTSD in a sample of 200 veterans?

c. In the sample of 200, the expected number of veterans with PTSD is __________.

d. Compute the standard deviation.

e. Would it be unusual to find 110 individuals suffering from PTSD in a group of 200 veterans?

Solution

6. Many veterans who bravely served in Iraq during Operation Enduring Freedom or Operation Iraqi Freedom have developed symptoms of post-traumatic stress disorder (PTSD) or similar ailments. A large study of returning veterans showed that 52.7% of the soldiers suffered from a PTSD or similar ailment after serving

in Iraq. A random sample of 200 veterans from Iraq were selected for additional research. The number suffering from PTSD or a similar condition was recorded.
(Source: Frayne SM, et al. (2010) Medical care needs of returning veterans with
PTSD: their other burden. Journal of General Internal Medicine 26(1).)

The binomial distribution describes the behavior of a count variable X if the following conditions apply:

1: The number of observations n is fixed.

2: Each observation is independent.

3: Each observation represents one of two outcomes (\"success\" or \"failure\").

4: The probability of \"success\" p is the same for each outcome.

In our case

1: The number of observations n is fixed. 200

2: Each observation is independent.

3: Each observation represents one of two outcomes, suffered from a PTSD or similar ailment after serving

4: The probability of \"success\" p is the same for each outcome. 0.527

n=200

p=0.527

mean of x = np=200*0.527=105.4

Expectation = np =200*0.527= 105.4

Variance = np(1 - p) = 200*0.527*0.473= 49.8542

Standard deviation = sqrt( variance) = 7.0608

e. Would it be unusual to find 110 individuals suffering from PTSD in a group of 200 veterans?

P(X=x) = (_nC_x) p^x (1-p)^n-x   

P(X=110) = (₂₀₀C₁₁₀) 0.527¹¹⁰ (1-0.527)^200-110

P(X = 110) = 0.04578

P(X = 110) = 0.04578   which is < 0.05.

So it is unusual to find 110 individuals suffering from PTSD in a group of 200 veterans.