There are 42 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 6 min. (Give answers accurate to 3 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?
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(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?
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A) 0.99413

B) 0.00022

Step-by-step explanation:

A) First of all let's find the total grading time from 6:50 P.M to 11:00 P.M.:

Total grading time; X = 11:00 - 6:50 = 4hours 10minutes = 250 minutes

Now since we are given an expected value of 5 minutes, the mean grading time for the whole population would be:

μ = n*μ_s ample = 42 × 5 = 210 minutes

While the standard deviation for the population would be:

σ = √nσ_sample = √(42 × 6) = 15.8745 minutes

To find the z-score, we will use the formula;

z = (x - μ)/σ

Thus;

z = (250 - 210)/15.8745

z = 2.52

From the z-distribution table attached, we have;

P(Z < 2.52) ≈ 0.99413

B) solving this is almost the same as in A above, the only difference is an additional 10 minutes to the time.

Thus, total time is now 250 + 10 = 260 minutes

Similar to the z-formula in A above, we have;

z = (260 - 210)/15.8745

z = 3.15

P(Z > 3.15) = 0.00022

mxb67

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