Part a: The probability that a given battery will last between 2.3 and 3.6 years is 80.41%.
Part b: The probability that a given battery will last longer than 24 months is 97.72%.
For the given question,
The data for the storage battery are-
Part a: The probability that a given battery will last between 2.3 and 3.6 years.
Find z score for x = 2.3 years.
z(2.3) = (x - μ )/σ
Put the values,
z(2.3) = (2.3 - 3.0)/0.5
z(2.3) = -1.4
Use negative z score table;
P(z = -1.4) = 0.0808
Similarly, for x = 3.6
z = (3.6 - 3)/0.5
z = 1.2
Use positive z table;
z = 0.8849
P(2.3<x<3.6) = 0.8849 - 0.0808
P(2.3<x<3.6) = 0.8041
Thus, the probability that a given battery will last between 2.3 and 3.6 years is 80.41%.
Part b: The probability that a given battery will last longer than 24 months = 2 years.
x = 2 years
z (2) = ( 2 - 3.0 )/ 0.5
z = 2
P(x > 2) = 0.9772
Thus, the probability that a given battery will last longer than 24 months is 97.72%.
To know more about the normal distribution, here
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