Statistics

Question

I DID NOT QUITE UNDERSTAND THIS, IF U CAN EXPLAIN I WOULD GET IT.

 

The scores on a test given to all juniors in a school district are normally distributed with a mean of 74 and a standard deviation of 8.

Find the percent of juniors whose score is more than 99.

Answer 1

Hi Chad,

 

With a normal distribution, 68% of the population is within ±1 standard deviation, 95% of the population is within ±2 standard deviations and 99.7% of the population is within ±3 standard deviations.

 

If the mean is 74 and the standard deviation is 8

then the scores ±1 standard deviation are 66-82

the scores ±2 standard deviations are 58-90

and the scores ±3 standard deviations are 50-98

 

Since 99.7% of the scores are ±3 standard deviations, then 0.15% are below 50 and 0.15% are above 98.

So the percent of scores above 99 is less than 0.15%.

 

Thought:  If perhaps the question was to be find the percent of juniors whose score is more than 90, rather than 99, then we'd recognize that 95% of the scores are ±2 standard deviations, so 2.5% would be greater than +2 standard deviations and the answer would be 2.5%.

 

 

Answer 2

The difference between the mean  score (74) and a score of 99 is  25.   Dividing by the standard deviation (8), gives 3.125 .  This means that a score of 99 is more than three standard deviations above the mean.  Clearly, there will be few scores more than 99.  (However, the assumption of normal distribution is never perfect in the real world.)

 

The numerical answer can be found by using a TI-84 calculator to compute the cumulative normal distribution on the interval 99 to  + ∞, with mean = 74 and standard deviation 8.  On the calculator it is DISTR (2nd VARS).   Select 2: normalcdf.    Then fill in lower = 99,  upper = 1E99 , μ = 74,  σ = 8.  Scroll down to paste , hit enter and then enter again.   The result is 8.89E-4.   This is approximately .089%.    About a tenth of a percent.

 

The function normalcdf computes the area under the normal curve between the limits 99 and  (a very large number which is essentially + ∞) .   The maximum score was probably 100, but the normal distribution does not 'know' about that.   By entering the very large number 1E99,   one is essentially computing the fraction of scores greater than 99 within the assumption of normality.   The TI-84 lets one put in the mean (μ) and standard deviation (σ).   If one were using a tabulated form the the cumulative normal distribution, one would compute the z score ( 99-74)/8 and look up  the fraction above this value.