What Is the Probability of Age-Related Onset for a Disease?

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SUMMARY

The discussion focuses on calculating probabilities related to the age of onset for a disease, which follows a normal distribution with a mean of 11 years and a standard deviation of 2 years. The specific probabilities calculated include the likelihood of a child being between the ages of 9.5 and 15.5 years, which involves finding z-scores of -0.75 and 2.25, and consulting a standard normal distribution table. Additionally, it confirms that 50% of the population is over 11 years of age, as this is the mean, and calculates the probability of being under 10 years by finding the z-score of -0.5.

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  • Understanding of normal distribution concepts
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  • Familiarity with standard normal distribution tables
  • Basic statistics knowledge, including mean and standard deviation
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hello i really have a problem in my statistic class. please help me
i'm begging you, i don't want to fail. please.. please
the problem is...

suppose the ages at time of onset of a certain disease are approximately normally distributed with a mean of 11 years and a standard daviation of 2 years. A child has just come down with the isease. What is the probability that the child is:
a) between the ages of 9.5 and 15.5 years?
b) Over 11 years of age?
c) Under 10?

it's just one problem please solve it for me
please..
shahi
freshmen student
 
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How is our solving this one problem going to help you? It would be much better for you to go to your teacher. The last thing in the world you want to do is to trick your teacher into thinking you know the subject better than you do!
(By the way, this should be posted in the homework section and you should show us what you have tried.)

In this problem, as in any "normal distribution" problem, the first thing you do is find the "standard variable": z= (x- mean)/standard deviation. In this problem you are told that the mean is 11 and the standard deviation is 2:

a) between the ages of 9.5 and 15.6 years"
translates into z= (9.5-11)/2= -0.75 and z= (15.5- 11)/2= 2.25.
Consult a table of the standard normal distribution (There should be one in your text. If not, there is a nice one at http://people.hofstra.edu/faculty/Stefan_Waner/RealWorld/normaltable.html ) to find the probability that z is between -0.75 and 2.25.

b) Over 11 years of age?
That's easy! 11 is the mean. What percentage of any normal population is above the mean? (Of course, that translates to z= 0.)

c) Under 10?
This translates to z= (10-11)/2= -0.5. Find the probability that z is less than -0.5.
(Most tables of the standard normal distribution have only positive z, saving space, since the distribution is symmetric. The probability that z< -0.5 is exactly the same as the probability that z> 0.5. To find the probability -0.75< z< 2.25, find the probability that 0< z< 0.75, the probability that 0<z< 2.25 and add.)
 
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