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 $[20000 - 3 σ, 20000 + 3 σ]$ 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 $σ = 6.000$ , 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. $Z = \frac{X - μ}{σ} \Rightarrow X = σ \cdot Z + μ$ $P (X \geq 25.000 hours) = P (σ \cdot Z + μ \geq 25.000) = P (Z \geq 0, 833)$ Looking at the table it is possible to see that: $p (X < 25.000) = 0, 7967$ $p (X \geq 25.000) = 1 - 0, 7967 = 0, 2033$
Observe that: $p (10.000 \leq X \leq 15.000) = p (X \leq 15.000) - p (X \leq 10.000) =$ $= p (Z \leq \frac{15.000 - 20.000}{6.000}) - p (Z \leq \frac{10.000 - 20.000}{6.000}) =$ $= p (Z \leq - 0, 83) - p (Z \leq - 1, 67)$ For symmetry, it is possible to say: $p (10.000 \leq X \leq 15.000) = p (Z \leq 1, 67) - p (Z \leq 0, 83) =$ $= 0, 9525 - 0, 7967 = 0, 1558$

Solution:

$σ = 6.000$
$p (X < 25.000) = 0, 7967$ ; $p (X \geq 25.000) = 1 - 0, 7967 = 0, 2033$
$p (10.000 \leq X \leq 15.000) = 0, 1558$

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