## Elementary Statistics (12th Edition)

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.