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:
12
Step-by-step explanation:
2+5=7
7+5=12
Pattern of +5
Answer: 0.00102%
Step-by-step explanation:
Given : Human body temperatures are normally distributed with a mean of 
and a standard deviation of 
A hospital uses 
as the lowest temperature considered to be a fever.
Let x be the random variable that represents the human body temperatures.
For x= 100.6, 
Using normal distribution table for z-values for right-tailed area ,
P(x>100.6)=
Hence, the required probability = 0.00102%
Answer:
B
Step-by-step explanation:
.
Answer:
Area= 32 (formula is: length x width)
Perimeter= 8 + 4 + 8 + 4= 24
Step-by-step explanation:
