Average truancy per student is 8 days per semester with a standard deviation 2.2 and a population of 250 students.
a. What percent of the population is truant more than 9 days?
b. What percent of the population is truant less than 3 days?
c. What percent of the population is truant between 5 and 10 days?
d. What percent of the population is truant between 8.5 and 10 days?
e. What raw score would produce a truancy percent of the top 15% of the students who missed the least amount of school?
To solve these questions, we will use the concept of the normal distribution and z-scores. The formula to calculate a z-score is:
z = (x - μ) / σ
Where:
- x is the raw score
- μ is the mean of the distribution
- σ is the standard deviation of the distribution
a. To find the percent of the population truant more than 9 days, we need to find the z-score for 9 and then calculate the area under the curve to the right of that z-score.
First, calculate the z-score:
z = (9 - 8) / 2.2
z = 0.45
To find the area to the right of the z-score, we look up the z-score in a standard normal distribution table or use a calculator.
Using the table, we find that the area to the right of z = 0.45 is approximately 0.3259. So, approximately 32.59% of the population is truant more than 9 days.
b. To find the percent of the population truant less than 3 days, we need to find the z-score for 3 and calculate the area under the curve to the left of that z-score.
z = (3 - 8) / 2.2
z = -2.27
Using the table, we find that the area to the left of z = -2.27 is approximately 0.0111. So, approximately 1.11% of the population is truant less than 3 days.
c. To find the percent of the population truant between 5 and 10 days, we need to find the z-scores for both values and calculate the area between those z-scores.
z1 = (5 - 8) / 2.2
z1 = -1.36
z2 = (10 - 8) / 2.2
z2 = 0.91
Using the table, we find that the area to the left of z = -1.36 is approximately 0.0869, and the area to the left of z = 0.91 is approximately 0.8192.
The area between these two z-scores is 0.8192 - 0.0869 = 0.7323. So, approximately 73.23% of the population is truant between 5 and 10 days.
d. To find the percent of the population truant between 8.5 and 10 days, we need to find the z-scores for both values and calculate the area between those z-scores.
z1 = (8.5 - 8) / 2.2
z1 = 0.23
z2 = (10 - 8) / 2.2
z2 = 0.91
Using the table, we find that the area to the left of z = 0.23 is approximately 0.5910, and the area to the left of z = 0.91 is approximately 0.8189.
The area between these two z-scores is 0.8189 - 0.5910 = 0.2279. So, approximately 22.79% of the population is truant between 8.5 and 10 days.
e. To find the raw score that would produce a truancy percent of the top 15% of the students who missed the least amount of school, we need to find the z-score corresponding to the top 15% and then solve for the raw score.
Using the table, we find that the z-score corresponding to the top 15% is approximately 1.04.
Now, rearranging the z-score formula, we solve for x:
1.04 = (x - 8) / 2.2
Multiply both sides by 2.2:
2.288 = x - 8
Add 8 to both sides:
x = 10.288
So, the raw score that would produce a truancy percent of the top 15% is approximately 10.288 days.