Answer
See below
Work Step by Step
85% are generally happy, hence $p=0.85$
There are 26 people, $n=26$
Mean: $\bar x=np\\=26\times 0.85\\=22.1$
Standard deviation: $\sigma =\sqrt np(1-p)\\=\sqrt 26\times0.85\times(1-0.85)\\=1.82$
We can find the z-score by applying the formula: $z=\frac{x-\bar x}{\sigma}=\frac{19-22.1}{1.81}\approx-1.7$
We will use the table to find $P(x \leq 19)\approx P(z\leq-1.7)\approx0.0446$
Hence, we reject the survey’s findings.
