#### Answer

0.2358, the observed rate could be possible.

#### Work Step by Step

p=0.00034
$q=1-p=1-0.00034=0.99966$
$n⋅p=420095⋅0.00034=143≥5.$
$n⋅q=420095⋅0.99966=419952≥5.$
Hence, the requirements are satisfied.
mean: $\mu=n\cdotp=420095\cdot0.00034=143.$
standard deviation: $\sigma=\sqrt{n\cdot p\cdot q}=\sqrt{420095\cdot0.00034\cdot0.99966}=11.95.$
135.5 is the first value more than 135, hence:
$z=\frac{value-mean}{standard \ deviation}=\frac{135.5-143}{11.95}=-0.72.$
By using the table, the probability belonging to z=-0.72: 0.2358, hence the probability: 0.2358. This probability is not too close to 0, therefore the observed rate could be possible.