In: Statistics and Probability
An expensive watch is powered by a 3-volt lithium battery expected to last three years. Suppose the life of the battery has a standard deviation of 0.3 year and is normally distributed.a. Determine the probability that the watch's battery will last longer than 3.8 years.b. Calculate the probability that the watch's battery will last more than 2.15 years.c. Compute the length-of-life value for which 15% of the watch's batteries last longer.
a. The probability that the battery will last longer than 3.8 years is
b. The probability that the battery will last more than 2.15 years is
c. The length-of-life value for which 15% of the batteries last longer is years.
Answer:
Given that,
An expensive watch is powered by a 3-volt lithium battery expected to last three years.
Suppose the life of the battery has a standard deviation of 0.3 year and is normally distributed.
i.e,
The mean= 3 years
The standard deviation= 0.3 years
(a).
Determine the probability that the watch's battery will last longer than 3.8 years:
P(X>3.8) = 1 - P(X<3.8)
= 1 - P(Z < (X - mean)/standard deviation)
= 1 - P(Z < (3.8 - 3)/0.3)
= 1 - P(Z < 2.67)
= 1 - 0.9962
= 0.0038
Therefore, the probability that the watch's battery will last longer than 3.8 years = 0.0038.
(b).
Calculate the probability that the watch's battery will last more than 2.15 years:
P(X > 2.15) = 1 - P(X<2.15)
= 1 - P(Z < (2.15-3)/0.3)
= 1 - P(Z < -2.83)
= 1 - 0.0023
= 0.9977
Therefore, the probability that the watch's battery will last longer than 2.15 years = 0.9977.
(c).
Compute the length-of-life value for which 15% of the watch's batteries last longer:
Let the value for which 15% of the watch's batteries last longer be A
P(X > A) = 0.15
P(X < A) = 1-0.15 = 0.85
P(Z < (A - mean)/standard deviation) = 0.85
From standard normal distribution table,
(A - 3)/0.3 = 1.0364
A-3=1.0364 (0.3)
A-3=0.3109
A=0.3109+3
A = 3.3109