Answer
See below
Work Step by Step
Only 80% of graduating seniors will buy a class ring, hence $p=0.8$
There are 20 people, $n=20$
Mean: $\bar x=np\\=20\times 0.8\\=16$
Standard deviation: $\sigma =\sqrt np(1-p)\\=\sqrt 20\times0.8\times(1-0.8)\\=1.79$
We can find the z-score corresponding to an x-value of 4 by applying the formula: $z=\frac{x-\bar x}{\sigma}=\frac{12-16}{1.79}\approx -2.2$
We will use the table: $P(x \leq 12)\approx P(z\leq-2.2)\approx0.0139 \lt 0.5$
Hence, we reject the survey’s findings.
