Hi Angel,
The problem seems to be incomplete, please post more information about the variable IQ score and tell us where you're stuck.
Hey this is my first time doing this but I am makeing a b in the elementary statistics class that I am in I need help finding out how to find the probability such as the question what is the probability that an adult individual selected randomly will have an IQ score of greater than 125. please help. Thanks, angel
Hi Angel,
The problem seems to be incomplete, please post more information about the variable IQ score and tell us where you're stuck.
Hello Angel,
It sounds like the missing elements in your problem might be the mean and standard deviation of IQ scores in the population.
If you knew the mean and standard deviation and if you assumed that IQ scores are normally distributed in the population, you could easily find the cumulative probability that an IQ score is less than 125. Call this PCum. The probability that the IQ score is greater than 125 would be 1 - PCum.
On the internet, there are free, online tutorials that explain how to solve this type of problem (i.e., how to find PCum). A lesson that addresses exactly the type of problem you are facing can be found at http://stattrek.com/Lesson2/Normal.aspx . The lesson describes the normal distribution, references a helpful calculator (at http://stattrek.com/Tables/Normal.aspx ), and even works through a couple of sample problems that are very similar to the problem you face.
As I said, however, to solve the problem, you need to know the mean and standard deviation of IQ scores in the population.
I hope this helps.
Harvey
If this is the one that IQ scores are normall distribued with a mean of 100 and a std dev of 16. I believe it goes like this....
P(x>125)
Z1= 90-100/16=-10/16 = -.63
Z2=120-100/16=1.25
therefore:
P(Z<-.625) P(z<1.25)
1-.26435= .089435
.73565
so 0.89435-0.73565=0.1587
I hope this helps!
Does anyone know about What is the cutoff IQ score for the top 1%?
If IQs are normally distributed with mean=100 and s.d.=16, then:
z = (125 - 100)/16 = 25/16 = 1.563
The area under the standard normal curve above z=1.563 is 0.059, so there is a 5.9% chance that a randomly selected individual will have an IQ above 125.
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For the IQ score that cuts off the top 1%, find the z score that has 99% of the area below it in the standard normal distribution, then plug it into:
z = (x - 100)/16
and solve for x.
I accidently gave you the answer to the worng problem. I guess I looked at the answer at the one below that. The work should all be right, just put the wrong answer. (Should be the 0.05938) Sorry about that!
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