Question

Statistics:

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

its on a normal distribution

Answer 2

This is a problem involving conditional probability. You want to find
\[P(29\le X\le40.3~|~X\ge27)\]
This means the probability that \(X<27\) is zero, but \(P(29\le X\le40.3)\) alone assumes it's not zero.

Answer 3

\[\begin{align*}P(29\le X\le40.3~|~X\ge27)&=\frac{P(29\le X\le40.3)\cap P(X\ge27)}{P(X\ge27)}\\\\
&=\frac{P(29\le X\le40.3)}{P(X\ge27)}\\\\
&=\frac{P\left(\dfrac{29-28.3}{0.77}\le Z\le\dfrac{40.3-28.3}{0.77}\right)}{P\left(Z\ge\dfrac{27-28.3}{0.77}\right)}\\\\
&=\frac{P\left(0.91\le Z\le15.58\right)}{P\left(Z\ge-1.69\right)}\\\\
\end{align*}\]

Answer 4

Thank you! your awesome.