Population = 135 students
Mean score = 72.3
Standard deviation of the scores = 6.5
Part (a): Students from 2SD and 3SD above the mean
2SD below and above the mean includes 95% of the population while 3SD includes 99.7% of the population.
95% of population = 0.95*135 ≈ 129 students
99.7% of population = 0.997*135 ≈ 135 students
Therefore, number of students from 2SD to 3SD above and below the bean = 135 - 129 = 6 students.
In this regard, Students between 2SD and 3SD above the mean = 6/2 = 3 students
Part (b): Students who scored between 65.8 and 72.3
The first step is to calculate Z values
That is,
Z = (mean-X)/SD
Z at 65.8 = (72.3-65.8)/6.5 = 1
Z at 72.3 = (72.3-72.3)/6.5 = 0
Second step is to find the percentages at the Z values from Z table.
That is,
Percentage of population at Z(65.8) = 0.8413 = 84.13%
Percentage of population at (Z(72.3) = 0.5 = 50%
Third step is to calculate number of students at each percentage.
That is,
At 84.13%, number of students = 0.8413*135 ≈ 114
At 50%, number of students = 0.5*135 ≈ 68
Therefore, students who scored between 65.8 and 72.3 = 114-68 = 46 students
Answer:
Anna will need to deposit 
Step-by-step explanation:
we know that
The compound interest formula is equal to
where
A is the Final Investment Value
P is the Principal amount of money to be invested
r is the rate of interest in decimal
t is Number of Time Periods
n is the number of times interest is compounded per year
in this problem we have
substitute in the formula above and solve for P
It isn't easy really but you can always divide the number out. For example 50% of 200 is 100 by dividing 200 by 2 or 25% of 100 is 25 by dividing 100 by 4 since 25% is equivalent to 1/4
Answer:
20%
Step-by-step explanation:
Probability=5/(5+12+8)*100=5/25*100=20%
