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Breakdown voltage: three ranges

Short answer

Standardize with sigma=1.5. The probabilities are (i) 0.6568, (ii) 0.4772, and (iii) 0.9962.

The breakdown voltage X of a randomly chosen diode of a particular type is known to be normally distributed with mean 40 volts and variance 2.25 volts. What is the probability that the breakdown voltage will be: i) Between 39 and 42 volts ii) Between 40 and 43 volts iii) Less than 44 volts

X ~ N(μ = 40, σ² = 2.25), so σ = 1.5, and Z = (X − 40)/1.5.

i) 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

ii) P(40 < X < 43)

z₁ = 0, z₂ = (43 − 40)/1.5 = 2.00

P = Φ(2.00) − Φ(0) = 0.9772 − 0.5000 = 0.4772

iii) P(X < 44)

z = (44 − 40)/1.5 = 2.67

P = Φ(2.67) = 0.9962

Answer: (i) 0.6568, (ii) 0.4772, (iii) 0.9962

✓ Answer checked by Claude Sonnet 5.5

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