Respuesta :
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 = [tex]\frac{38}{60} = 0.634[/tex]
Therefore, the probability is given by:
[tex]P(p\leq 0.634) = [\frac{p - P'}{\sqrt{\frac{P'q'}{n}}}]\leq [\frac{0.634 - 0.74}{\sqrt{\frac{0.74\times 0.26}{60}}}][/tex]
P[tex](p\leq 0.634) = P[z\leq -1.87188][/tex]
P[tex](p\leq 0.634) = P[z\leq -1.87] = 0.0298[/tex]
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.
