Question by Stefanie Mathers: Statistics – IQR? Expected Cost? Experts possibly?
Hello,
I was wondering if you guys could help with me out with a few problems!
A researcher is researching the age at which patients are initially diagnosed with Alzheimer’s. Suppose the age at which patients develop it is normally distributed with a mean of 70 and a standard deviation of 4.6 years.
1) What is the probability a randomly selected patient was over age 70 upon initial diagnosis?
I think it is 50% because the mean is 70, and on a standard distribution the mean divides the graph into two halves, correct?
2) What is the IQR for the ages of these patients?
I know It is Q3 – Q1 which is .75 – .33 but there has to be more. Do I get the z score of the two values and just subtract them?
3) Suppose the cost is based upon the age of initial diagnosis. Suppose that the yearly cost is $ 2000 for patients that are 2.53 standard deviations below the mean in age,
$ 3000 for patients that are within 2.53 standard deviations below the mean or 2.53 standard deviations above the mean,
and $ 5000 for patients that are 2.53 standard deviations above the mean in age.
What is the expected cost?
Is it just (2000)(2.53) + (3000)(2.53) + (5000)(2.53)? I am completely lost on this one.
I did attempt the problems so any help would be appreciated.
Most detailed answer:
Answer by CPlusPlus Guru
1) Correct.
2) The IQR is the interquartile range of your distribution, or Q3 – Q1, the difference between the third and first quartile. So far, so good. But it isn’t the difference between 75% and 33%, or between 75% and 25%, but between the levels of your variable that lie above 75% and above 25% of outcomes – that’s what those two quartiles are.
For the standard normal distribution, Q3 = 0.6745, because P(Z<0.6745) = 0.75 (both numbers within 4 decimal places), and Q1 = -Q3 = -0.6745, because this (and every) normal distribution is symmetric about its mean. For *your* normal distribution, Q3 = mean + sigma * 0.6745 and Q1 = mean – sigma * 0.6745, so IQR = Q3 – Q1 = mean + sigma * 0.6745 – (mean – sigma * 0.6745) = 2 * sigma * 0.6745 = 2 * 4.6 * 0.6745 = 6.2054 years.
3) How *do* you calculate the expected cost? For each possible outcome (cost level), you multiply it by the probability it will occur, and add all such products. We’ll need the probability that a randomly chosen patient was (or is? or has developed the disease? I assume not) more than 2.53 standard deviations below the mean. That is P(Z< -2.53) = 0.0057. By symmetry, P(Z>2.53) = 0.0057. And the middle probability is P(-2.532.53)) = 1 – 2 * 0.0057 = 0.9886.
So the expected cost is $ 2000 * 0.0057 + $ 3000 * 0.9886 + $ 5000 * 0.0057 = $ 3005.70. (Notice that the probability distribution is symmetric, and the cost distribution is not, so the less costly younger patients do not offset the more costly older ones; still, the overall expected cost is not much different from that of the middle group, because that’s where almost 99% of the patients are.)
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