A.) A recent midterm was given out in a large mathematics course of 400 students
ID: 2922006 • Letter: A
Question
A.) A recent midterm was given out in a large mathematics course of 400 students at a major university. The scores were normally distributed, and had an average score of 83 points with a standard deviation of 6 points. If Amy earned 72 points on her midterm, what would be her z-score? Bold one.
1.07 -1.83 9 1.83 None of these
B.) What is the distribution that Amy’s z-score (from above) will follow? Bold one
N(0, 1) N(72, 6) N(83, 6) B(83, 6) None of these
C.) Joe scored in the 70th percentile of their section. Approximately how many points did Joe earn on his midterm? Bold one.
76 points 86 points 75 points None of these
D.) Approximately what percent of students would you estimate earned between 80 and 92 points on the midterm? [Provide numerical support].
Explanation / Answer
Q1.
NORMAL DISTRIBUTION
the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd ~ N(0,1)
mean ( u ) = 83
standard Deviation ( sd )= 6
P(X < 72) = (72-83)/6
= -11/6= -1.8333
Q2.
N(83, 6)
Q3.
P ( Z < x ) = 0.7
Value of z to the cumulative probability of 0.7 from normal table is 0.5244
P( x-u/s.d < x - 83/6 ) = 0.7
That is, ( x - 83/6 ) = 0.5244
--> x = 0.5244 * 6 + 83 = 86.1464 ~ 86 points
Q4.
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 80) = (80-83)/6
= -3/6 = -0.5
= P ( Z <-0.5) From Standard Normal Table
= 0.3085
P(X < 92) = (92-83)/6
= 9/6 = 1.5
= P ( Z <1.5) From Standard Normal Table
= 0.9332
P(80 < X < 92) = 0.9332-0.3085 = 0.6247
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