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Breakdown voltage: normal, variance 2.25

Short answer

Standardize with sigma=1.5. The probabilities are (i) 0.6568, (ii) 0.9962, and (iii) 0.0228.

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) Less than 44 volts iii) More than 43 volts

X ~ N(μ = 40, σ² = 2.25), so σ = √2.25 = 1.5. Use 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(X < 44)

z = (44 − 40)/1.5 = 2.67

P = Φ(2.67) = 0.9962

iii) P(X > 43)

z = (43 − 40)/1.5 = 2.00

P = 1 − Φ(2.00) = 1 − 0.9772 = 0.0228

Answer: (i) 0.6568, (ii) 0.9962, (iii) 0.0228

✓ Answer checked by Claude Sonnet 5.5

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