Question 160595: Please help! Homework question.
A person must score in the upper 2% of the population on an IQ test to qualify for membership in Mensa, the international high-IQ society. If IQ scores are normally distributed with a mean of 100 and a standard deviation of 15, what score must a person have to qualify for Mensa?
Found 2 solutions by nabla, MathLover1:
Answer by nabla(475)   (Show Source):
You can put this solution on YOUR website! First we standardize the score with
z=(x-u)/o
where z is the distance in standard deviations from the mean, x is the score to be standardized, u is the mean of the population, and o is the standard deviation of the population [or sample].
Now, we can suspect from a percentile rank chart that the upper 2% is 2 standard deviations from the mean. In fact, however there is approximately only 100-2.27=97.73 within two standard deviations above the mean. Chances are, however, that your instructor only wants 2 standard deviations above the mean:
z=(x-100)/15
15z+100=x
This gives a score, x, as a function of its distance from the mean[z]. So if we want 2 standard deviations above the mean, 15(2)+100=130. You will need an IQ of 130 to get into MENSA accordingly. Strictly speaking, the score would be a little higher than 130 (in order for to be ahead of 98% of the population) due to the reasoning above.
Answer by MathLover1(20849)  (Show Source):
|
|
|
