# 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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