Question

an electrical power supply is used to charge battery of cell phone this power supply is capable to maintain the mean output 2.6 A with it standard deviation 0.2 an average cell phone consume a charging current of 100 mA THE Cellphone charging current show standard deviation 30A how many cell phone would have to be connected to the power supply for the probability of structural damage to exceed 1 percent?

Answer 1

Solution:

Let X is random variable of charging cellphone in A, $X \sim N\left(100,30^2\right)$ and $Y_{i}$ is the random variable for charing battery such that $Y \sim N(2.6,{0.2}^2)$ , then by CLT, $\sum_{i} Y_{i} / n \sim N\left(2.6, \frac{{0.2}^2}{n}\right)$ , i.e. $\sum_{i} Y_{i} \sim N\left(2.6 n, {0.2}^2n\right)$ and we want to calculate n such that $P\left(\sum_{i} Y_{i} \geq X\right)=.01$ .

Let $U=\sum_{i} Y_{i}-X$ . Note that $U \sim N\left(2.6 n-100,900+0.04n\right)$ .

$\begin{aligned} P\left(\sum_{i} Y_{i} \geq X\right) &=P(U \geq 0)=P\left(\frac{U-E[U]}{\sqrt{900+0.04n}} \geq \frac{2.6n-100}{\sqrt{900+0.04n}}\right) \\ &=P\left(Z \geq \frac{2.6n-100}{\sqrt{900+0.04n}}\right)=.01 \end{aligned}$

From the normal table

$\dfrac{2.6n-100}{\sqrt{900+0.04n}}=2.326$

$\Rightarrow 2.6n-100=2.326\sqrt{900+0.04n} \\\Rightarrow 6.76n^2-520n+10000=4869.2484+0.21641104n \\\Rightarrow n=\frac{520.21641\dots +\sqrt{131889.59105\dots }}{13.52},\:n=\frac{520.21641\dots -\sqrt{131889.59105\dots }}{13.52}$

$\Rightarrow n=65.33894\dots \approx65$