# Problems from The normal (or Gaussian) distribution

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 $$[20000-3\sigma, 20000+3\sigma]$$ 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?
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### 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: $$p(X < 25.000)=0,7967$$$$$p(X\geq25.000)=1-0,7967=0,2033$$$

• Observe that: $$p(10.000 \leq X \leq 15.000)=p(X\leq15.000)-p(X\leq10.000)=$$$$$=p(Z\leq\dfrac{15.000-20.000}{6.000})-p(Z\leq\dfrac{10.000-20.000}{6.000})=$$$ $$=p(Z\leq-0,83)-p(Z\leq-1,67)$$$For symmetry, it is possible to say: $$p(10.000\leq X \leq 15.000)=p(Z\leq1,67)-p(Z\leq0,83)=$$$ $$=0,9525-0,7967=0,1558$$\$

### Solution:

• $$\sigma=6.000$$
• $$p(X < 25.000)=0,7967$$; $$p(X\geq25.000)=1-0,7967=0,2033$$
• $$p(10.000\leq X \leq 15.000)=0,1558$$
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