The final exam in a one-term statistics course is taken in the december exam period. students who are sick or have other legitimate reasons for missing the exam are allowed to write a deferred exam scheduled for the first week in january. a statistics professor has observed that only 2% of all students legitimately miss the december final exam. suppose that the professor has 40 students registered this term. a) how many students can the professor expect to miss the december exam? b) what is the probability that the professor will not have to create a deferred exam?

a) 1 student

b) probability = .4457

Step-by-step explanation:

let x = number of students who legitimately missed the exam

a) given that the percentage of students who legitimately missed the exam = 2%

and total number of students registered =40

the number of students expected to miss the exam would be = ?\frac{2%}{100%}*80 = 0.8

rounding off to the nearest student, only 1 student would miss the exam

b) the probability that the professor will not have to create a deferred exam will be;

since 2% is observed to miss the exam, the percentage that would not miss the exam will be 100%-2% =98%

the probabiity that the professor will not have to create a deferred exam is P[E(x)=0] = \frac{98}{100}^40 = .4457

n/b it is raised to 40 because there are 40 students

L = 9.91 decibels.

Step-by-step explanation:

The loudness, L is inversely proportional to the square of the distance, d, from the source of the sound.

i.e L

L =

Where k is the constant of proportionality.

When d = 14 feet and L = 85 dB, then;

k = L x

  = 85 x

 = 16660

k = 16660

L =

Thus, when d = 41 feet, then;

L =

  =

  = 9.91077

L = 9.91 dB

The loudness is 9.91 decibels.