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

mean,

$x_A=4000, x_B=4500\\ n_A=100,n_B=50\\ \sigma_A=300 , \sigma_B=200$

Here mean of $A-B =x_A-x_B=4000-4500 =-500$

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

a) Probability that B will be at least 600lb more then A 

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

b) Probability that B will be at least 450lb more than

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