Question

Merta reports that 74% of its trains are on time. A check of 60 randomly selected trains shows that 38 of them arrived on time. Find the probability that among the 60 trains, 38 or fewer arrive on time. Based on the result, does it seem plausible that the "on-time" rate of 74% could be correct?

Answer 1

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 = $\frac{38}{60} = 0.634$

Therefore, the probability is given by:

$P(p\leq 0.634) = [\frac{p - P'}{\sqrt{\frac{P'q'}{n}}}]\leq [\frac{0.634 - 0.74}{\sqrt{\frac{0.74\times 0.26}{60}}}]$

P$(p\leq 0.634) = P[z\leq -1.87188]$

P$(p\leq 0.634) = P[z\leq -1.87] = 0.0298$

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.