This is a conditional probability question.
The conditional probability is given as follows: The probability of an event A given B is given by
[tex]P(A|B)= \frac{P(A\cap B)}{P(B)} [/tex]
Thus, given that the
fracture toughness of a particular steel alloy is known to be
normally distributed with a mean of 28.3 and a standard deviation of
0.77.
The probability that the fracture toughness will be between 29 and 40.3 given that the fracture toughness is at least 27 is given by
[tex]P(29 \leq X \leq 40.3 \, | \, X \geq 27)= \frac{P(29 \leq X \leq 40.3)\cap P(X \geq 27)}{P(X \geq 27)} \\ \\ =\frac{P(29 \leq X \leq 40.3)}{P(X \geq 27)}[/tex]
[tex]P(29 \leq X \leq 40.3)=P\left(z \ \textless \ \frac{40.3-28.3}{0.77}\right)- P\left(z \ \textless \ \frac{29-28.3}{0.77}\right) \\ \\ =P(z\ \textless \ 15.58)-P(z\ \textless \ 0.91)=1-0.81835=0.18165[/tex]
[tex]P(X \geq 27)=1-P(x\ \textless \ 27) \\ \\ =1-P\left(z\ \textless \ \frac{27-28.3}{0.77}\right)=1-P(z\ \textless \ -1.688) \\ \\ =1-0.04568=0.95432[/tex]
Thus, the probability that the fracture toughness will be between 29 and 40.3 given that the fracture toughness is at least 27 is given by
[tex]\frac{P(29 \leq X \leq 40.3)}{P(X \geq 27)}= \frac{0.18165}{0.95432} =\bold{0.1903}[/tex]
