Homework SourseThe lifetime of a certain type of battery is normally distri
The lifetime of a certain type of battery is normally distributed with mean value 10 hours and standard deviation 1 hour. There are four batteries in a package. What is the probability that the total lifetime of these batteries is
(a) at most 42 hours?
(b) between 37 and 41 hours?
(c) at least 38 hours?
Solution
The mean sum here is
u(sum) = n*u = 4*10 = 40
Note that the standard deviation of the sum is
s(sum) = s*sqrt(n) = 1*sqrt(4) = 2
a)
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 42
u = mean = 40
s = standard deviation = 2
Thus,
z = (x - u) / s = 1
Thus, using a table/technology, the left tailed area of this is
P(z < 1 ) = 0.841344746 [ANSWER]
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B)
We first get the z score for the two values. As z = (x - u) / s, then as
x1 = lower bound = 37
x2 = upper bound = 41
u = mean = 40
s = standard deviation = 2
Thus, the two z scores are
z1 = lower z score = (x1 - u)/s = -1.5
z2 = upper z score = (x2 - u) / s = 0.5
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0.066807201
P(z < z2) = 0.691462461
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.62465526 [ANSWER]
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c)
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 38
u = mean = 40
s = standard deviation = 2
Thus,
z = (x - u) / s = -1
Thus, using a table/technology, the right tailed area of this is
P(z > -1 ) = 0.841344746 [ANSWER]
