statistics probability

Question

IQ scores are normally distributed with a mean of 100 and a standard deviation of 15.

A) A person is considered intellectually very superior if his or her IQ score exceeds 132, what percentage of people falls into that category?

B) if we redefined the category of intellectually very superior to be the IQ scores in the top 3% under the normal curve find the new minimum score falls into that category

C) what percentage of IQ scores fall within 1.5 standard deviation of the mean?

D) find the highest IQ score in the bottom 30% under the normal curve

E) find the two IQ scores separate in the bottom 65% from the top 11%

Answer 1

I am going to answer these questions using different methods to show that there are multiple ways of approaching things in inferential statistics.

A. Using the Z score and a Z table. with a mu of 100, sigma of 15 we calculate the Z score for an IQ of 132 as follows. Z = (132 - 100) / 15 which is 2.13

Using a standard table (ref https://www.z-table.com/) we find the left tail (equal to or less than the Z score) is 0.9834.

The answer for the right side of the curve (greater than or equal to) is 1 - 0.9834 or 0.0166

B. Using the Z table again, in revers, we look for the value inside the table which represents (1-.03) , where .03 is 3%. The Z value for the number closest to .9700 is 1.88 (actual p value is .9699)

With a critical Z value of 1.88 we calculate the x value which would represent the IQ level for very superior. using the standard formula for Z and some algebra we get x = (Z * sigma) + mu.

Our answer is 1.88 * 15 + 100 or 128.2

C. Using a TI calculator or similar scientific calculator which has a normal distribution function of the general format. nmcdf(lowerbound,upperbound[,mu,sigma]). Where mu and sigma are NOT given, they are assumed to be 0 and 1, which is a Z distribution. We can get our answer directly using the full form

nmcdf-1.5,1.5) = .8664. This is the cumulative distribution between -1.5 Z and 1.5 Z scores

D. The higest IQ score of the lower 30% is where the cumulative distribution (left tail) is = 30% We can get this with an inverse normal distribution function . These are available on line , or on TI calculators.

format is invnm(p, mu, sigma) where p is the left tail cumulative percent.

invnm(.3,100,15) = 92.134 which is to say a person with an IQ of 92.134 would have an IQ at the lower 30% level.

E. This is a two part question. First find the lower bound for 65%. and the upper bound for 11% (which is lower bound for 89%)

We can get this using an on line calculater at https://onlinestatbook.com/2/calculators/inverse_normal_dist.html

lower IQ = 101.95

upper IQ = 106.13