A firm producing batteries for mobile phones say that the average duration until they get damaged is $$25.000$$ hours. Many customers went to the consumers' association to complain and a study was done. The result is that batteries follow a normal distribution with mean life equal to $$20.000$$ hours.

Bearing in mind that the battery lives outside the interval $$[200003\sigma, 20000+3\sigma]$$ cannot happen, define a reasonable value of sigma.

What is the probability that one battery lasts the promised $$25.000$$ hours or more? And the probability of lasting less time?
 What is the probability for the battery to live between $$10.000$$ and $$15.000$$ hours?
Development:

It is possible to suppose that $$\sigma=6.000$$, so that the batteries never last less than $$2.000$$ hours, and no more than $$38.000$$.

It will be necessary to transform the variable $$X$$ ($$N(20.000, 6.000)$$) into the variable $$Z$$ ($$N(0,1)$$) to be able to use the table. $$$Z=\dfrac{X\mu}{\sigma} \Rightarrow X=\sigma\cdot Z+\mu$$$ $$$P(X\geq25.000 \mbox{ hours } )=P(\sigma\cdot Z+\mu \geq 25.000)=P(Z\geq 0,833)$$$ Looking at the table it is possible to see that: $$$p(X < 25.000)=0,7967$$$ $$$p(X\geq25.000)=10,7967=0,2033$$$
 Observe that: $$$p(10.000 \leq X \leq 15.000)=p(X\leq15.000)p(X\leq10.000)=$$$ $$$=p(Z\leq\dfrac{15.00020.000}{6.000})p(Z\leq\dfrac{10.00020.000}{6.000})=$$$ $$$=p(Z\leq0,83)p(Z\leq1,67)$$$ For symmetry, it is possible to say: $$$p(10.000\leq X \leq 15.000)=p(Z\leq1,67)p(Z\leq0,83)=$$$ $$$=0,95250,7967=0,1558$$$
Solution:
 $$\sigma=6.000$$
 $$p(X < 25.000)=0,7967$$; $$p(X\geq25.000)=10,7967=0,2033$$
 $$p(10.000\leq X \leq 15.000)=0,1558$$