-Final grades in Professor Albert\'s large calculus class are approximately norm

Question

-Final grades in Professor Albert's large calculus class are approximately normally distributed with a mean of 76 (%) and standard deviation of 8 (%).

A- Draw a sketch of this normal distribution and label at least three points on the horizontal axis.

B- In Professor Albert's course, students who earn less than a 60% in the class are assigned a failing grade

C- What proportion of the students earned F's? Report your answer with three decimal places.

D- In Professor Albert's course, students who earn above a 94% are assigned an "A". What proportion of students earned A's? Report your answer with three decimal places.

F- What proportion of students earn between an 82% and 88% in this class? Report your answer with three decimal places.

What is the 25th percentile in this course? Report your answer with one decimal place.

G-The top 30% of students earned scores above what value? Report your answer with one decimal place.

H- The top 30% of students earned scores above what value? Report your answer with one decimal place.

G-The top 30% of students earned scores above what value? Report your answer with one decimal place.

H- The top 30% of students earned scores above what value? Report your answer with one decimal place.

Explanation / Answer

Normal Distribution
Mean ( u ) =0.76
Standard Deviation ( sd )=0.08
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
B.
P(X < 0.6) = (0.6-0.76)/0.08
= -0.16/0.08= -2
= P ( Z <-2) From Standard Normal Table
= 0.0228                  

D.
P(X > 0.94) = (0.94-0.76)/0.08
= 0.18/0.08 = 2.25
= P ( Z >2.25) From Standard Normal Table
= 0.0122 ~ 1.22% are earned about A
F.
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 0.82) = (0.82-0.76)/0.08
= 0.06/0.08 = 0.75
= P ( Z <0.75) From Standard Normal Table
= 0.77337
P(X < 0.88) = (0.88-0.76)/0.08
= 0.12/0.08 = 1.5
= P ( Z <1.5) From Standard Normal Table
= 0.93319
P(0.82 < X < 0.88) = 0.93319-0.77337 = 0.1598~ 15.98% are earned b/w this

G.
P ( Z > x ) = 0.3
Value of z to the cumulative probability of 0.3 from normal table is 0.52
P( x-u/ (s.d) > x - 0.76/0.08) = 0.3
That is, ( x - 0.76/0.08) = 0.52
--> x = 0.52 * 0.08+0.76 = 0.8019

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