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Normal: storage battery life

Short answer

For X~N(3,0.5²): (i) z=(2.3−3)/0.5=−1.4, P(X<2.3)=0.0808. (ii) z=(2.5−3)/0.5=−1.0, P(X≤2.5)=0.1587.

A certain type of storage battery lasts, on average, 3.0 years with a standard deviation of 0.5 year. Assuming that the battery lives are normally distributed, find the probability that a given battery will (i) last less than 2.3 years, and (ii) will last at most 2.5 years.

Battery life X ~ N(μ = 3.0 years, σ = 0.5 year).

(i) Less than 2.3 years

z = (2.3 − 3.0)/0.5 = -1.40

P(X < 2.3) = P(Z < −1.40) = 0.5 − P(0 < Z < 1.40) = 0.5 − 0.4192 = 0.0808

(ii) At most 2.5 years

z = (2.5 − 3.0)/0.5 = -1.00

P(X ≤ 2.5) = P(Z ≤ −1.00) = 0.5 − 0.3413 = 0.1587

Answer: (i) 0.0808 (8.08%) (ii) 0.1587 (15.87%)

✓ Answer checked by Claude Sonnet 5.5

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