Question

The fracture toughness (in ) of a particular steel alloy is known to be normally distributed with a mean of 28.3 and a standard deviation of 0.77. We select one sample of this alloy at random and measure its fracture toughness. Given that the fracture toughness is at least 27, what is the probability that the fracture toughness will be between 29 and 40.3? is it just P(X <= 40.3) - P(X <= 29)? How do I use the 27?

Answer 1

This is a conditional probability question.

The conditional probability is given as follows: The probability of an event A given B is given by

$P(A|B)= \frac{P(A\cap B)}{P(B)}$

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

$P(29 \leq X \leq 40.3 \, | \, X \geq 27)= \frac{P(29 \leq X \leq 40.3)\cap P(X \geq 27)}{P(X \geq 27)} \\ \\ =\frac{P(29 \leq X \leq 40.3)}{P(X \geq 27)}$

$P(29 \leq X \leq 40.3)=P\left(z \ \textless \ \frac{40.3-28.3}{0.77}\right)- P\left(z \ \textless \ \frac{29-28.3}{0.77}\right) \\ \\ =P(z\ \textless \ 15.58)-P(z\ \textless \ 0.91)=1-0.81835=0.18165$

$P(X \geq 27)=1-P(x\ \textless \ 27) \\ \\ =1-P\left(z\ \textless \ \frac{27-28.3}{0.77}\right)=1-P(z\ \textless \ -1.688) \\ \\ =1-0.04568=0.95432$

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

$\frac{P(29 \leq X \leq 40.3)}{P(X \geq 27)}= \frac{0.18165}{0.95432} =\bold{0.1903}$