Answer to Question #281280 in Statistics and Probability for Phathuynh

Question

The breakdown voltage of a randomly chosen diode of a certain type is known to be normally distributed with mean value 39 V and standard

deviation 1.1 V.

a) What is the probability that the voltage of a single diode is between 37.98 and 41.24?

b) What value c is such that only 1% of all diodes have voltages exceeding that value?

c) If 6 diodes are independently selected, what is the probability that at least 2 has a voltage exceeding 41.24?

Answer 1

Sol) Given,

Mean =39, S. D=1.1

a)

$\begin{aligned} & P(37.98<x<41.24) \\ & P(x<42)-P(x<39) \\ z=& \frac{x-\text { mean }}{\text { S.D }} \\=& P\left(z<\frac{42-40}{1.5}\right)-P\left(z<\frac{39-40}{1.5}\right) \\=& P(z<1.33333)-P(z<-0.66666) \\=& 0.9088-0.2525=0.6563 \end{aligned}$

 

b) 1% of diodes

$z=\frac{x-\text { mean }}{S \cdot D}\Rightarrow \quad 0.1=\frac{x-\text { mean }}{S \cdot D}$

z-value corresponding to 0.15 is 1.04

$\begin{aligned} 1.04 &=\frac{x-40}{15} \\ (1.04)(15) & \neq 40=x \\ x &=41.56 \end{aligned}$

 

c)

$\begin{aligned} P \text { (atleast } 41.24 \text { ) } &=1-P(\text { less then } 41.24) \\ &=1-P(x<41.24) \\ &=1-P\left(z<\frac{41.24-40}{1.5}\right) \\ &=1-P(z<.826) \\ &=1-0.5000 \\ &=0.5 \\ \text{As 6 diode are selected}\\ &=1-(0.5)^{6} \\ &=1-0.015625 \\ &=0.9843 \end{aligned}$