Answer to Question #191136 in Statistics and Probability for Hasnain Ali

Question

A and B manufacture two types of cables, having mean breaking strengths

of 4000 and 4500 lb and standard deviations of 300 and 200 lb,

respectively. If 100 cables of brand A and 50 cables of brand B are tested,

what is the probability that the mean breaking strength of B will be (a) at

least 600 lb more than A, (b) at least 450 lb more than A?

Answer 1

We have given that,

$x_A = 4000,x_B = 4500$

$n_A = 100, n_B = 50$

$\sigma_A = 300, \sigma_B = 200$

Mean of $A -B = x_A-x_B = 4000-4500 = -500$

Also, standard deviation of the difference

$\sigma_{A-B} = \sqrt{(\dfrac{300^2}{100}+\dfrac{200^2}{50})} = 41.23$

a.) The probability that the mean breaking strength of B will be at least 600 lb more than A

$= P(A-B<-600)$

$= P(Z<\dfrac{-600+500}{41.23})=P(Z<-2.4254) = 0.0076$

b.) The probability that the mean breaking strength of B at least 450 lb more than A.

$P(A-B<-450) = P(Z<\dfrac{-450+500}{41.23})=P(Z<1.2127) = 0.8874$