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.
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.
-
1. Your answer of 50% is correct.
2. Q3 and Q1 is .75 and .25 (not .33). There is a short cut though. Since the normal dbn is symmetrical around its mean, you can simply check the z score for .75 and multiply your reading by 2! Thus the answer is approximately .67x2 = 1.34xSigma = 1.34x4.6=...
3. Check the Z score for 2.53, which is .4943. The probability weight for the $2,000 and $5,000 are both just equal to .0047.
Thus the expected cost is ($2,000)(.0047)+($3,000)(2x.4943)+($5,00… The answer should be pretty close to $3,000, slightly above it!
2. Q3 and Q1 is .75 and .25 (not .33). There is a short cut though. Since the normal dbn is symmetrical around its mean, you can simply check the z score for .75 and multiply your reading by 2! Thus the answer is approximately .67x2 = 1.34xSigma = 1.34x4.6=...
3. Check the Z score for 2.53, which is .4943. The probability weight for the $2,000 and $5,000 are both just equal to .0047.
Thus the expected cost is ($2,000)(.0047)+($3,000)(2x.4943)+($5,00… The answer should be pretty close to $3,000, slightly above it!