An investigator is looking at the relationship between periodontal disease and t

Question

An investigator is looking at the relationship between periodontal disease and the onset of hypertension. Suppose the investigator decides to look at height as a variable that may confound the relationship between exposure and case status in his study. After collecting information about the height of each participant, he assembles a database that approximates height with the normal distribution. a) Assuming the mean of the distribution is 168 cm with a standard deviation of 14 cm, what are the highest and lowest 1% of heights in the population? (1 point) Highest (Example answer ###.##) Lowest (Example answer ###.##) b) Assume the investigator is now looking at the population of Dutch men in his study, and that the mean height there is 190 cm with a standard deviation of 13 cm, answer the previous question. (1 point) Highest (Example answer ###.##) Lowest (Example answer ###.##)

Explanation / Answer

Assuming the sample consists of a large no. Of observations, the heights, H can be assumed to be distributed Normally, with mean = 168cm and standard deviation = 14cm.

Therefore, H ~ N(168, 14)

Now, let the highest 1% height be h1 => P(H >= h1) = 0.01

Now, if H follows a normal distribution with mean mu, and standard deviation sigma, then Z = (H - my)/sigma follows a standard normal distribution.

Therefore, P(H >= h1) = 0.01 can be written as,

P((H-mu)/sigma >= (h1-mu)/sigma) = 0.01

=> P(Z >= (h1-168)/14) = 0.01

=> 1 - P(Z < (h1-168)/14) = 0.01

=> P(Z <= (h1-168)/14) = 0.99

=> (h1-168)/14 = phi_inverse(0.99),

Now, phi_inverse(0.99) can be looked up from a standard normal table, phi_inverse(0.99) = 2.327

=> h1 = 2.327*14 + 168 = 200.58cm

Similarly, let lowest 1% height be h2, => P(H <= h2) = 0.01

=> P(Z <= (h2-168)/14) = 0.01

=> (h2-168)/14 = phi_inverse(0.01)

phi_inverse(0.01) = -2.327 (from standard normal table)

=> h2 = -2.327*14 + 168 = 135.42cm

b) let the heights of Dutch men be D, we can assume that D follows a normal distribution for large no. Of observations.

D ~ N(190, 13)

Let highest height be d1 => P(D >= d1) = 0.01

Similar to a)

=> P((D-190)/13 >= (d1-190)/13) = 0.01

=> P(Z >= (d1-190)/13) = 0.01

=> 1 - P(Z < (d1-190)/13) = 0.01

=> (d1-190)/13 = phi_inverse(0.99)

=> d1 = 13*2.327 + 190 = 220.25cm

Now, let lowest height be d2, so similar to above,

P(D <= d2) = 0.01

=> d2 = -2.327*13+190 = 159.75cm

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