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I am trying to solve this problem:

A factory produces batteries whose duration in hours for a particular has a normal distribution with $\mu_0=53$ and ${\sigma_0}^2=25$. Now suppose that there is a change in the quality of the batteries such that of the total produced, $70$ % of them has the correct duration but the remaining $30$ % have a duration in hours with normal distribution of new parameters $\mu_1$ and $\sigma_1^2$. Let $D$ be the duration in hours of a battery chosen at random from the lot of production.

1) Calculate $\mu_1$ and ${\sigma_1}^2$ knowing that $P(D \geq 47)=0,82688$ and $P(D \geq 60)=0,05746$.

2) Calculate the density function of the duration in hours for a battery chosen at random from the lot.

I am a bit lost with the exercise, I'll write what I could do:

We have two normal distributions $X_0 \sim N(53,25)$ and $X_1 \sim N(\mu_1,\sigma_1)$, I'll call now $Z_0=\dfrac{X_0-53}{5}$ and $Z_1=\dfrac{X_1-\mu_1}{\sigma_1}$ to the standarized normal distributions, notice that $Z_0=Z_1$. I define the event $A=\{\text{a battery is not damaged}\}$. We have the following $$(*) P(D \geq 47)=P(D \geq 47 | A)P(A)+P(D \geq 47 |A^c)P(A^c)$$$$=P(X_0 \geq 47)\dfrac{7}{10}+P(X_1 \geq 47)\dfrac{3}{10}$$$$=\dfrac{7}{10}P(Z_0 \geq \dfrac{-6}{5})+\dfrac{3}{10}P(Z_1 \geq \dfrac{47-\mu_1}{\sigma_1})$$ $$=\dfrac{7}{10}F_{Z_0}(\dfrac{6}{5})+\dfrac{3}{10}P(Z_1 \geq \dfrac{47-\mu_1}{\sigma_1})$$

If I didn't mess up with the calculations, we get $$P(Z_1 \geq \dfrac{47-\mu_1}{\sigma_1})=0,6915$$

In a similar manner, we arrive to $$P(Z_1 \geq \dfrac{60-\mu_1}{\sigma_1})=0,003$$

I don't know what to do from here in order to solve for $\mu_1$ and $\sigma_1$. I would appreciate suggestions to complete the solution and to find the density function.

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Actually, you are very close to the answer, what you have now are $$ \begin{align} Pr (Z_1 \ge \frac{47-\mu_1}{\sigma_1}) = 0.6915 &, Pr (Z_1 \ge \frac{60-\mu_1}{\sigma_1}) = 0.003 \end{align} $$

You just need to find $a$ and $b$ such that $Pr(Z \ge a)=0.6915$ and $Pr(Z \ge b)=0.003$ where $Z \sim N(0,1)$

Then you will have $$ \begin{align} a= \frac{47-\mu_1}{\sigma_1} & , b= \frac{60-\mu_1}{\sigma_1} \end{align} $$ which will lead you to $\mu_1$ and $\sigma_1$

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