Question: It is claimed that the expected of time some computer part may work before requiring a reboot is 26 days. In order to examine this claim 80 identical parts are set to work. Assume that the distribution of the length of time the part can work (in days) is Exponential. The central region that contains 70% of the distribution of the average is of the form $E(X)\pm c$, where $E(X)$ is the expectation of the sample average. The value of c is (You are asked to apply the Normal approximate to the distribution of the average of the 80 parts that are examined. The answer may be rounded up to 3 decimal places of the actual value.):
Its answer: 3.012796
My thought
70% of the distribution of the average is: $\mu \pm$ qnorm(0.75) $\cdot \sigma$
$\mu = E(X) = 26$
$\sigma = \frac {\sqrt{E(X)^2}} {\sqrt{80}} = 2.90688837075$
With given $E(X) \pm c$ and since $E(X)$ is $\mu$,
$c=$ qnorm(0.75) $\cdot \sigma$
$c=0.6744898 \cdot 2.90688837075 = 1.960666$
How come 3.012796 is the correct answer?
Your error is the same you did here.
The central region that contains 70% of the distribution is the one that excludes 15% in the left tail and another 15% in the right one.
Thus the quantile you are interested in is
qnorm(0.7+0.15)=qnorm(0.85)1.036433Concluding
$$1.036433\times 2.906888=3.012796$$
as claimed