Answer :
Answer:
a) P ( T < 250 mins ) = 0.7695
b) P ( T > 260 mins ) = 0.1344
Step-by-step explanation:
- The RV from a sample has the following parameters that are mean = 5 mins, and standard deviation s = 4 mins.
- The entire population has n = 46 students.
- We will first compute the population mean u and population standard deviation σ as follows:
u = n*mean
u = 46*5 = 230 mins
σ = sqt ( n ) * s
σ = sqt ( 46 ) * 4
σ = 27.129 mins
- Approximating that the time taken T to grade the population of entire class follows a normal distribution with parameters u and σ as follows:
T~ N ( 230 , 27.129 )
Find:
- If grading times are independent and the instructor begins grading at 6:50 P.M. and grades continuously, what is the (approximate) probability that he is through grading before the 11:00 P.M. TV news begins?
- The total time till 6:50 PM to 11:00 PM is ( 4 hr and 10 mins ) = 250 mins.
- We will compute the Z-value as follows:
Z = ( 250 - 230 ) / 27.129
Z = 0.7372
- Then use the Z-Tables and determine the probability:
P ( T < 250 mins ) = P ( Z < 0.7372 )
P ( T < 250 mins ) = 0.7695
Find:
- If the sports report begins at 11:10, what is the probability that he misses part of the report if he waits until grading is done before turning on the TV?
- For the teacher to miss the sports report he must take more time than 6:50 PM to 11:10 P.M.
- The total time till 6:50 PM to 11:10 PM is ( 4 hr and 20 mins ) = 260 mins.
- We will compute the Z-value as follows:
Z = ( 260 - 230 ) / 27.129
Z = 1.10582
- Then use the Z-Tables and determine the probability:
P ( T > 260 mins ) = P ( Z > 1.10582 )
P ( T > 260 mins ) = 0.1344