Question 926245: The professor of a introductory calculus class has stated that, historically, the distribution of final exam grades in the course resemble a Normal distribution with a mean final exam mark of μ=61 % and a standard deviation of σ=9 %.
c) The top 15% of students writing the final exam will receive a letter grade of at least an A in the course. To two decimal places, find the minimum final exam mark needed on the calculus final to earn a letter grade of at least an A in the course.
Suppose this professor randomly picked 26 final exams, observing the earned mark on each. What is the probability that 4 of these have a final exam grade of less than 50%? Use four decimals in your answer.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! the mean is 61
the standard deviation is 9.
the top 15% of the students will get an A.
you want to know the minimum final exam grade to earn an A.
in the normal distribution curve, getting a score in the top 15% of the class will require a z-score of 1.03643338
15% being greater than a certain score means that 85% are less than that score.
a z-score required for 85% of the scores to be less than that z-score is equal to 1.03643338.
you can find this using a z-score calculator or from the tables.
the calculator will give you much greater accuracy.
i use the ti-84 plus.
the formula for z-score is z = (x-m) / s
x is the raw score, m is the mean, and s is the standard deviation.
you know z and m and s and you want to solve for x.
from z = (x-m) / s, you can derive x = s * z + m
this becomes:
x = 9 * 1.03643338 + 61 which results in:
x = 70.32790042
students needs a score greater than or equal to 70.32790042 in order to get an A.
this can be visually shown as:
next question.
you sample 26 exams.
you want to know the probability that 4 of them have a raw score of 50% or less.
the probability of getting a raw score of 50% or less is as follows:
z = (x-m) / s
x = 50, m = 61, s = 9
calculate z-score to get z = (50-61)/9 which is equal to -11/9.
use your z-score calculator to find that the probability of getting a z-score of -11/9 or less is equal to .1108118649.
your results might vary depending on the calculator you use, but they'll be pretty close to this.
a picture of what this looks like is shown below:
the probability of getting a z-score greater than -11/9 is equal to 1 minus the probability of getting a z-score less than -11/9 which makes the probability of getting a z-score greater than -11/9 equal to .8891881351.
you can use the binomial distribution to find the probability that exactly 4 out of the 26 exams will be 50% or less.
the binomial probability formula is p(x) = c(n,x) * p^x * q^(n-x)
x = 4
n = 26
n-x = 22
p = .1108118649
q = .8891881351
c(26,4) = 14950
p(4) = c(26,4) * .1108118649^4 * .8891881351^22 which becomes:
p(4) = .1701557961.
this equals .1702 rounded to 4 decimal places.
a picture of the binomial distribution calculations is shown below:
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