There are 43 students in an elementary statistics class. On the basis of years o
ID: 3131851 • Letter: T
Question
There are 43 students in an elementary statistics class. On the basis of years of experience, the instructor know 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. If grading times are independent and the instructor beings 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? If the sports report begins at 11:10, what is the probability that he misses part of the report if the wants until grading is done before turning on the TV?Explanation / Answer
a)
From 6:50 to 11:00 is 250 minutes. Hence, it is asking the probability of a sample mean less than 250/43 = 5.813953488.
We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as
x = critical value = 5.813953488
u = mean = 5
n = sample size = 43
s = standard deviation = 4
Thus,
z = (x - u) * sqrt(n) / s = 1.33436249
Thus, using a table/technology, the left tailed area of this is
P(z < 1.33436249 ) = 0.908957457 [ANSWER]
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b)
From 6:50 to 11:10 is 260 minutes. Hence, it is asking the probability of a sample mean greater than 260/43.
We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as
x = critical value = 6.046511628
u = mean = 5
n = sample size = 43
s = standard deviation = 4
Thus,
z = (x - u) * sqrt(n) / s = 1.715608916
Thus, using a table/technology, the right tailed area of this is
P(z > 1.715608916 ) = 0.043116823 [ANSWER]
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