Normal distribution and Probability

Question

Say we have GRE scores are normally distributed with

mean 500 and standard deviation 100. Find the probability

that a randomly selected GRE score is greater than 620.

 We want to know what’s the probability of

getting a score 620 or beyond.

 p(z > 1.2)

 Result: The probability of randomly

getting a score of 620 is ~.12

 What is the area for scores less than z = -1.5?

 What is the area between z =1 and 1.5?

 What z score cuts off the highest 30% of the

distribution?

 What two z scores enclose the middle 50% of the

distribution?

 If 500 scores are normally distributed with mean = 50

and SD = 10, and an investigator throws out the 20

most extreme scores, what are the highest and lowest scores that are retained

Answer 1

In this problem, we are given a variable X (GRE Scores). We can summarize the information about X as: X~N(500,100) which means X is normally distributed, with a mean of 500 and a standard deviation of 100. We are also given p(X>620) = p(z>1.2) = 0.12. In this example, the z score has been calculated by taking 620-500/100 = 1.2. Then, we go to our z score table. Because we want the "greater than" score, we are going to do 1.0 minus whatever the z score is. So from the table we have a score of 0.88, so that's where the 0.12 comes from.

A diagram such as this is helpful for answering the rest of the questions: https://mathbitsnotebook.com/Algebra2/Statistics/STzScores.html

The area for scores less than z = -1.5 is the sum of areas left of that score, thus 4.4+1.7+0.5+0.1... = ~6.7

The area between scores z=1 and z=1.5 is 9.2.

To find the z score that cuts of the highest 30%, add up from the right until you reach 30: 0.1+0.5+1.7+4.4+9.2+15 =30.9, therefore the z cutoff is 0.5.

The middle 50% is not neatly enclosed by any set of z scores. The pair of -0.5 to +0.5 encloses 38.2% while the pair of -1 to +1 encloses 68.2%. If you add just +1 or just -1 (so going from -1 to +0.5 for example) then it encloses 53.2% but this is not a uniform/symmetrical set.

For the last question, if X~N(50,10), n=500, and 20 outliers are removed, what are the high and low scores? Given this information. we know that 95% of the points are contained within 2 standard deviations of the mean. So 50+10+10 = 70, and 50-10-10=30, therefore 30 and 70 are the points that contain 95% of the scores when all of the points are considered. Throwing out the most extreme scores, assuming they are evenly distributed on both sides, will not affect the mean but will affect the standard deviation, because sd is based on the number of samples. If we use SEM as an intermediary we can back-calculate an estimate of the new SD. The SEM of the original sample is 10/sqrt(500) = 0.447. Changing the n to 480 gives us an SD around 9.8. Now, if we want to do 95% again it is +2SD, or if we want to do 99% it is +3SD.