It is known from past experience that the life of traffic li
Solution
5.
Note that
n = z(alpha/2)^2 s^2 / E^2
where
alpha/2 = (1 - confidence level)/2 = 0.05
Using a table/technology,
z(alpha/2) = 1.644853627
Also,
s = sample standard deviation = 270
E = margin of error = 115
Thus,
n = 14.91373292
Rounding up,
n = 15 [ANSWER, D]
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6)
We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as
x = critical value = 105/20 = 5.25
u = mean = 5
n = sample size = 20
s = standard deviation = 1.1
Thus,
z = (x - u) * sqrt(n) / s = 1.016394535
Thus, using a table/technology, the left tailed area of this is
P(z < 1.016394535 ) = 0.845279227 = 0.85 [ANSWER, D]
