Answer
See below
Work Step by Step
Only 30% of graduating seniors will buy a class ring, hence $p=0.3$
There are 15 people, $n=15$
Mean: $\bar x=np\\=15\times 0.3\\=4.5$
Standard deviation: $\sigma =\sqrt np(1-p)\\=\sqrt 15\times0.3\times(1-0.3)\\=1.77$
We can find the z-score corresponding to an x-value of 4 by applying the formula: $z=\frac{x-\bar x}{\sigma}=\frac{4-4.5}{1.77}\approx -0.3$
We will use the table: $P(x \leq 4)\approx P(z\leq-0.3)\approx0.3821 \gt 0.5$
Hence, we accept the survey’s findings.
