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Breakdown voltage: normal probabilities (1.5 V)

Short answer

Standardize with Z=(X-40)/1.5 and use the normal table. The probabilities are (a) 0.6568, (b) 0.9772, and (c) 0.7486.

The breakdown voltage X of a randomly chosen diode of a particular type is known to be normally distributed with mean 40 and standard deviation 1.5 volts. What is the probability that the breakdown voltage will be (a) between 39 and 42 volts; (b) at most 43 volts; (c) at least 39 volts.

X ~ N(μ = 40, σ = 1.5). Standardise with Z = (X − μ)/σ and use the standard normal table Φ(z) = P(Z ≤ z).

(a) P(39 < X < 42)

z₁ = (39 − 40)/1.5 = −0.67, z₂ = (42 − 40)/1.5 = 1.33

P = Φ(1.33) − Φ(−0.67) = 0.9082 − 0.2514 = 0.6568

(b) P(X ≤ 43)

z = (43 − 40)/1.5 = 2.00

P = Φ(2.00) = 0.9772

(c) P(X ≥ 39)

z = (39 − 40)/1.5 = −0.67

P = 1 − Φ(−0.67) = 1 − 0.2514 = 0.7486

(Values are from standard normal tables with z rounded to 2 decimals; exact computation gives 0.6563, 0.9772, 0.7475.)

Answer: (a) 0.6568, (b) 0.9772, (c) 0.7486

✓ Answer checked by Claude Sonnet 5.5

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