Equations of significance probabilities

Consider a population of independent light bulbs with an exponential lifetime distribution with mean $\mu $. It is claimed that their expected lifetime is 1000 hours. A definition of a 100(1−$\alpha $)% confidence interval obtained from an observation to is the set of all ${\mu}_{0}$ which are not rejected in a test of a null hypothesis ${\mu}_{0}$ against an alternative hypothesis $\ne $.

One particular light bulb fails after 622 hours. Solve the equations of the two significance probabilities Pr(T "$$" 622 |${\mu}_{0}$) = 0.05 (for a test of ${\mu}_{0}$ versus ${\mu}_{0}$) and Pr(T "$$" 622 |${\mu}_{0}$ = 0.05 (for a test of ${\mu}_{0}$ versus ${\mu}_{0}$) for $\mu $. Determine the range of values of $\mu $ such that both of the probabilities Pr(T "$$" 622 | $\mu $) and Pr(T "$$" 622 |$\mu $) are at least 0.05. (This range gives an equi-tailed 90% confidence interval for $\mu $.)

I don't seem to understand what they mean by 'solve the equations'. Do I have to find a specific value for T or compute Pr(T $\ge $ 622 |${\mu}_{0}$), Pr(T $\le $ 622 |${\mu}_{0}$) and compare with 0.05? I believe I will get the second part after I understand this bit.

Consider a population of independent light bulbs with an exponential lifetime distribution with mean $\mu $. It is claimed that their expected lifetime is 1000 hours. A definition of a 100(1−$\alpha $)% confidence interval obtained from an observation to is the set of all ${\mu}_{0}$ which are not rejected in a test of a null hypothesis ${\mu}_{0}$ against an alternative hypothesis $\ne $.

One particular light bulb fails after 622 hours. Solve the equations of the two significance probabilities Pr(T "$$" 622 |${\mu}_{0}$) = 0.05 (for a test of ${\mu}_{0}$ versus ${\mu}_{0}$) and Pr(T "$$" 622 |${\mu}_{0}$ = 0.05 (for a test of ${\mu}_{0}$ versus ${\mu}_{0}$) for $\mu $. Determine the range of values of $\mu $ such that both of the probabilities Pr(T "$$" 622 | $\mu $) and Pr(T "$$" 622 |$\mu $) are at least 0.05. (This range gives an equi-tailed 90% confidence interval for $\mu $.)

I don't seem to understand what they mean by 'solve the equations'. Do I have to find a specific value for T or compute Pr(T $\ge $ 622 |${\mu}_{0}$), Pr(T $\le $ 622 |${\mu}_{0}$) and compare with 0.05? I believe I will get the second part after I understand this bit.