Suppose the scores of students on a Statistics course are Normally distributed with a mean of 563 and a standard deviation of 37
1 answer:
Answer:
47.72% of students scored between 563 and 637 on the exam .
Step-by-step explanation:
The percentage of the students scored between 563 and 637 on the exam
= The percentage of the students scored lower than 637 on the exam -
the percentage of the students scored lower than 563 on the exam.
Since 563 is the mean score of students on the Statistics course, 50% of students scored lower than 563. that is P(x<563)=0.5
P(x<637)=P(z<z*) where z* is the z-statistic of the score 637.
z score can be calculated using the formula
z*= where
- X =637
- M is the mean score (563)
- s is the standard deviation of the score distribution (37)
Then z*= =2
P(z<2)=0.9772, which means that 97.72% of students scored lower than 637 on the exam.
As a Result, 97.72%-50%=47.72% of students scored between 563 and 637 on the exam
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Answer:
The picture that represent her drawing is uploaded below. The height is 8 ft and the base is 6 ft.
Step-by-step explanation:
Renee garden is triangular in shape . The area of the triangular garden is given as 24 ft² . The area of a triangle can be represented below.
Area of a triangle = 1/2bh
where
b = base
h = height
The drawing that represent Renee drawing is given below. The height is 8 ft and the base is 6 ft .
Using the formula
Area of a triangle = 1/2bh
Area of a triangle = 1/2 × 6 × 8
Area of a triangle = 48/2
Area of a triangle = 24 ft²
