Answer:
No, the on-time rate of 74% is not correct.
Solution:
As per the question:
Sample size, n = 60
The proportion of the population, P' = 74% = 0.74
q' = 1 - 0.74 = 0.26
We need to find the probability that out of 60 trains, 38 or lesser trains arrive on time.
Now,
The proportion of the given sample, p =

Therefore, the probability is given by:
![P(p\leq 0.634) = [\frac{p - P'}{\sqrt{(P'q')/(n)}}]\leq [\frac{0.634 - 0.74}{\sqrt{(0.74* 0.26)/(60)}}]](https://img.qammunity.org/2020/formulas/mathematics/high-school/ecc6oqcbxvk87mfltgtc67y41htk353f0l.png)
P
![(p\leq 0.634) = P[z\leq -1.87188]](https://img.qammunity.org/2020/formulas/mathematics/high-school/3tajsv3nsu79mly95z4vmn0py35hk8uwug.png)
P
![(p\leq 0.634) = P[z\leq -1.87] = 0.0298](https://img.qammunity.org/2020/formulas/mathematics/high-school/pqprh03yyjhkelh11iztzh4tvq7996mjf5.png)
Therefore, Probability of the 38 or lesser trains out of 60 trains to be on time is 0.0298 or 2.98 %
Thus the on-time rate of 74% is incorrect.