4.The College Boards, which are administered each year to many thousands of high
school students, are scored so as to yield a mean of 500 and standard deviation of 100.
These scores are close to be normally distributed.
a. What percentage of the scores can be expected ...
... to satisfy the condition of being between 450 and 600.
(450-500)/100 =-.5 (600-500)/100 =1
.3085 .8413
.8413-.3085 = .5328
Right, but to get percentage, you need to multiply by 100.
To find the percentage of scores that can be expected to be between 450 and 600, we can use the standard normal distribution (also known as the Z-score).
First, we need to convert the values 450 and 600 into Z-scores. The formula for calculating the Z-score is:
Z = (X - μ) / σ
Where:
- X is the value we want to convert to a Z-score (450 or 600)
- μ is the mean of the distribution (500)
- σ is the standard deviation of the distribution (100)
For 450:
Z = (450 - 500) / 100 = -0.5
For 600:
Z = (600 - 500) / 100 = 1
Now that we have the Z-scores, we can look up the percentage of scores within these Z-scores using a standard normal distribution table (also known as a Z-table).
The Z-table provides the area under the curve for different Z-scores. We need to find the area between -0.5 and 1.
Looking up the Z-score -0.5 in the Z-table, we find that the area to the left of -0.5 is 0.3085.
Looking up the Z-score 1 in the Z-table, we find that the area to the left of 1 is 0.8413.
To find the percentage of scores between -0.5 and 1, we subtract the area to the left of -0.5 from the area to the left of 1:
0.8413 - 0.3085 = 0.5328
So, approximately 53.28% of scores can be expected to be between 450 and 600.