Question

There are 46 students in an elementary statistics class. On the basis of years of experience, the instructor knows that the time needed to grade a randomly chosen first examination paper is a random variable with an expected value of 5 min and a standard deviation of 4 min. (Round your answers to four decimal places.)(a) 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?(b) 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?

Answer 1

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