For the population whose distribution is $E\left(\frac{1}{6}\right)$, random sample of size $n=36$ are repeatedly taken.
Find the following. Round to four decimals if needed. Answers of 0 and 1 are possible due to rounding.
a. $P(\bar{x}< 5.5)$ :
b. $P(\bar{x}> 6.9)$
c. The value of $a$ such that $P(\bar{x}> a)=0.08$ :

Step 1 ：The problem is asking for probabilities related to the sample mean of a population with an exponential distribution. The exponential distribution has a mean of \(\frac{1}{\lambda}\), so in this case, the population mean is \(\frac{1}{\frac{1}{6}} = 6\).

Step 2 ：The Central Limit Theorem tells us that for large sample sizes, the distribution of the sample mean will be approximately normal with mean equal to the population mean and standard deviation equal to the population standard deviation divided by the square root of the sample size.

Step 3 ：The standard deviation of an exponential distribution is also \(\frac{1}{\lambda}\), so the standard deviation of the sample mean will be \(\frac{6}{\sqrt{36}} = 1\).

Step 4 ：We can use these parameters to calculate the z-scores for the given values and then use the standard normal distribution to find the probabilities.

Step 5 ：For part a, the z-score is -0.5, which corresponds to a probability of 0.3085. So, \(P(\bar{x}<5.5) = \boxed{0.3085}\).

Step 6 ：For part b, the z-score is 0.9, which corresponds to a probability of 0.1841. So, \(P(\bar{x}>6.9) = \boxed{0.1841}\).

Step 7 ：For part c, we need to find the value of a such that the probability of the sample mean being greater than a is 0.08. The z-score corresponding to a probability of 0.08 is approximately 1.4051. So, the value of a is \(6 + 1.4051 = \boxed{7.4051}\).