Applied research in political science

Question

1. The quantitative part of a national exam is scaled so that the mean score is 500 and the standard deviation is 100. If the distribution of scores is normally distributed (a) what proportion of the students scored between 500 and 682? (b) What proportion scored between 340 and 682?



Answer 1

To solve this problem, we'll use the properties of the standard normal distribution and the z-score formula.

(a) To find the proportion of students who scored between 500 and 682, we first need to find the z-scores corresponding to these scores using the formula:

�=�−��z=σX−μ​

Where:

1. �X is the score
2. �μ is the mean score (500 in this case)
3. �σ is the standard deviation (100 in this case)

For �=500X=500: �500=500−500100=0z500​=100500−500​=0

For �=682X=682: �682=682−500100=1.82z682​=100682−500​=1.82

Next, we'll use a standard normal distribution table or a calculator to find the proportion corresponding to �=1.82z=1.82. From the table or calculator, we find that the proportion corresponding to �=1.82z=1.82 is approximately 0.9656.

Therefore, the proportion of students who scored between 500 and 682 is approximately 0.9656−0.5=0.46560.9656−0.5=0.4656.

(b) To find the proportion of students who scored between 340 and 682, we'll repeat the above steps for the score of 340:

For �=340X=340: �340=340−500100=−1.6z340​=100340−500​=−1.6

Again, we'll use a standard normal distribution table or a calculator to find the proportion corresponding to �=−1.6z=−1.6. From the table or calculator, we find that the proportion corresponding to �=−1.6z=−1.6 is approximately 0.0548.

Therefore, the proportion of students who scored between 340 and 682 is approximately 0.9656−0.0548=0.91080.9656−0.0548=0.9108.