Answer to Question #225804 in Statistics and Probability for Prathyush

Question

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 ?

Answer 1

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

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

Given

$$\mu_1=40\ h, \sigma_1=3\ h,$$

$$\mu_2=45\ h, \sigma_2=6\ h$$

(a)

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

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

$$\approx0.0478$$

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

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

$$=0.5$$

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

(b)

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

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

$$\approx0.0000003$$

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

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

$$\approx0.0478$$

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