#### Answer

The observed rate seems to be very unlikely.

#### Work Step by Step

p=0.61
$q=1-p=1-0.61=0.39$
$n⋅p=1002⋅0.61=611.22≥5.$
$n⋅q=1002⋅0.39=390.78≥5.$
Hence, the requirements are satisfied.
mean: $\mu=n\cdotp=1002\cdot0.61=611.22.$
standard deviation: $\sigma=\sqrt{n\cdot p\cdot q}=\sqrt{1002\cdot0.61\cdot0.39}=15.44.$
700.5 is the first value lower than 701, hence:
$z=\frac{value-mean}{standard \ deviation}=\frac{700.5-611.22}{15.44}=5.78.$
By using the table, the probability belonging to z=5.78: 0.9999, hence the probability: 1-0.9999=0.0001. This probability is really close to 0, therefore the observed rate seems to be very unlikely.