Normal Distribution and Z scores

Question

The SAT Mathematics are know to follow an approximate normal distribution with a mean of 500 and a standard devation of 100. The ACT Mathematics are known to follow an approximate normal distribution with a mean of 21 and a standard deviation of 6. The student scored a 720 on the SAT math test and a 33 on the ACT Math test.

A)What is the probability that a randomly selected student scored lower than this student on the SAT mathematics?

B)What is the probability that a randomly selected student score higher than this student on the SAT mathematics

C)What is the probability that a randomly selected student scored lower than this student on the ACT mathematics.

D) What is the probabability that a randomly selected student score higher than this student on the ACT mathematics

Answer 1

For a question like this, the first thing you want to do is calculate the z-score. A z-score is just a measure of how many standard deviations away from the mean you are. For example, if the value of your SAT score is 600, your z-score is +1, because you are 100 points above the mean of 500, which is 1 standard deviation higher. It is calculated as z = (value - mean)/standard deviation. For the student in the problem, the z-scores are:

SAT: z = (720 - 500)/100 = 2.20

ACT: z = (33 - 21)/6 = 2.00

So, the student scored 2.20 standard deviations above the mean on the SAT and 2.00 standard deviations above the mean on the ACT. Now we can answer the other questions by looking at a z-score table (for example, http://www.z-table.com/).

A) You want the table value that is in the row labelled 2.2 and the column labelled .00 (you will have to scroll down a bit). This corresponds to the percent of observations (in this case, SAT scores) that are less than 2.20 standard deviations above the mean. In this case, the answer is 0.9861 (AKA 98.61%).

B) Because there is a 0.9861 probability of a randomly selected student having a lower score, the probability of a higher score is just the opposite, or 1 - 0.9861. This is equal to 0.0139 (or 1.39%).

C) Following the same process as part A) for the ACT z-score of 2.00, the answer is 0.9772 (97.72%)

D) Following the same reasoning as part B), the answer is 1 - 0.9772 = 0.0228 (or 2.28%)

Answer 2

Normal Distribution and Z scores.

The SAT Mathematics are know to follow an approximate normal distribution with a mean of 500 and a standard deviation of 100. The ACT Mathematics are known to follow an approximate normal distribution with a mean of 21 and a standard deviation of 6. The student scored a 720 on the SAT math test and a 33 on the ACT Math test.

A)What is the probability that a randomly selected student scored lower than this student on the SAT mathematics?

B)What is the probability that a randomly selected student score higher than this student on the SAT mathematics

C)What is the probability that a randomly selected student scored lower than this student on the ACT mathematics.

D) What is the probability that a randomly selected student score higher than this student on the ACT mathematics

Answer.

Let’s deal with the SAT first.

Mean = 500

Standard Deviation = 100

Student Score = 720

Z = (x – u) / st.dev = (720 – 500) / 100 = 220 / 100 = 2.20

Without the ability to draw a picture, imagine that the right tail contains the scores higher than 720 and the remainder of the curve to the left contains the scores below 720.

Using the Z-table, we see that 0.0139 is in the tail and the area to the left of the tail is 1 – 0.0139 = 0.9861.

Thus, 1.39 percent of students scored higher than this student on the SAT and 98.61 percent of all students scored below this student on the SAT.

Now, let’s deal with the ACT exam.

Mean 21

Standard Deviation = 6

Student Score = 33

Z = (x – u) / st.dev = (33 – 21) / 6 = 9 / 6 = 1.50

Without the ability to draw a picture, imagine that the right tail contains the scores higher than 33 and the remainder of the curve to the left contains the scores below 33.

Using the Z-table, we see that 0.0668 is in the tail and the area to the left of the tail is 1 – 0.0668 = 0.9332.

Thus, 6.68 percent of students scored higher than this student on the ACT and 93.32 percent of all students scored below this student on the ACT.

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