The Food Marketing Institute shows that $17\%$ of households spend more than $$100$ per week on groceries. Assume the population proportion is

$p=0\mathrm{.}17$ and a simple random sample of $800$ households will be selected from the population. Use the $z-$table.

a$.$ Show the sampling distribution of $\frac{?}{\text{b}\text{a}\text{r}}(p),$ the sample proportion of households spending more than $$100$ per week on groceries.

$E(\frac{?}{\text{b}\text{a}\text{r}}(p))=(to2$ decimals$)$

${}_{\frac{?}{\text{b}\text{a}\text{r}}(p)}=(to4$ decimals$)$

b$.$ What is the probability that the sample proportion will be within $\backslash$pm $0\mathrm{.}02$ of the population proportion $($to $4$ decimals$)?$

c$.$ Answer part $($b$)$ for a sample of $1600$ households $($to $4$ decimals$).$