Answer to Question #225804 in Statistics and Probability for Prathyush

Question #225804
The life length of a certain brand A of an electronic device is normally distributed with mean
40 and standard deviation 3, while that of another brand B is normally distributed with mean
45 and standard deviation 6. Which of the two brands should be preferred for a continuous use
of
(a) 45 hours ?
(b) 55 hours ?
1
Expert's answer
2021-08-15T18:11:42-0400

Let X=X= the life length of a certain brand A of an electronic device: XN(μ1,σ12).X\sim N(\mu_1, \sigma_1^2).

Let Y=Y= the life length of a certain brand B of an electronic device: YN(μ2,σ22).Y\sim N(\mu_2, \sigma_2^2).

Given


μ1=40 h,σ1=3 h,\mu_1=40\ h, \sigma_1=3\ h,

μ2=45 h,σ2=6 h\mu_2=45\ h, \sigma_2=6\ h

(a)


P(X45)=1P(X<45)P(X\geq45)=1-P(X<45)

=1P(Z<45403)1P(Z<1.6667)=1-P(Z<\dfrac{45-40}{3})\approx1-P(Z<1.6667)

0.0478\approx0.0478


P(Y45)=1P(Y<45)P(Y\geq45)=1-P(Y<45)

=1P(Z<45456)=1P(Z<0)=1-P(Z<\dfrac{45-45}{6})=1-P(Z<0)

=0.5=0.5

Brands B should be preferred for a continuous use of 45 hours.


(b)


P(X55)=1P(X<55)P(X\geq55)=1-P(X<55)

=1P(Z<55403)1P(Z<5)=1-P(Z<\dfrac{55-40}{3})\approx1-P(Z<5)

0.0000003\approx0.0000003




P(Y55)=1P(Y<55)P(Y\geq55)=1-P(Y<55)

=1P(Z<55456)=1P(Z<1.6667)=1-P(Z<\dfrac{55-45}{6})=1-P(Z<1.6667)

0.0478\approx0.0478

Brands B should be preferred for a continuous use of 55 hours.



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