A worn poorly set up machine is observed to produce components whose length x follows a normal distribution with mean 20 cm and variance 2.56 cm. Calculate:
a) the probability that a component is is at least 24 cm long
b) the probability that the length of a coponent lies between 19 cm and 21 cm
a) $Z=\frac{24-20}{\sqrt{2.56}}=2.5$
This corresponds to a value of .9938 on the z table. Then do I take $1-.9938=.0062$?
b)
$\frac{19-20}{\sqrt{2.56}} \leq Z \leq \frac{21-20}{\sqrt{2.56}}$
$-.625 \leq Z \leq .625$-Is this work correct? Where do I go from here?
How is it that $\Phi(.625) -\Phi(-.625)=.46.....$
when $\Phi(.625)$ corresponds to an area of .73405?
a) Yes, you are correct. We have that $$ P(X>24)=P\biggl(\frac{X-20}{\sqrt{2.56}}>2.5\biggr)=1-\Phi(2.5)\approx0.00621, $$ where $\Phi$ is the cumulative distribution function of the standard normal distribution.
b) You work is correct. You only need one final step \begin{align*} P(19<X<21)&=P\biggl(\frac{19-20}{\sqrt{2.56}}<\frac{X-20}{\sqrt{2.56}}<\frac{21-20}{\sqrt{2.56}}\biggr)\\ &=P(-0.625<Z<0.625)\\ &=\Phi(0.625)-\Phi(-0.625)\\ &\approx0.73401-0.26599\\ &=0.46802 \end{align*} since $$ P(x<\xi\le y)=P(\xi\le y)-P(\xi\le x) $$ for any random varialbe $\xi$ and any $x<y$.