Question

b) can we say that the sampling distribution of \\(\hat{p}\\) is approximately normal? no because \\(np(1 - p)=\\) which is \\(?\\) . round to one decimal place. c) what is the probability that no more than 9% of the randomly selected college students are left - handed? fill in below with the correct random variable and inequality symbol in the probability statement. round final answer to 4 decimal places. \\(p(? ? 0.09)=\\) d) is 8% an unusually low percentage of randomly selected college students that are left - handed in a sample of 212? since \\(p(? ? 0.08)=\\) which is \\(?\\) 0.05, 8% select an answer an unusually low percentage of college students that are left - handed in a sample of 212

Explanation:

Step1: Check normal - approximation condition for sampling distribution of $\hat{p}$

We assume a value for $p$ (population proportion) and $n$ (sample size). Let's assume $p = 0.1$ (a common estimate for the proportion of left - handed people) and $n=212$. Then $np(1 - p)=n\times p\times(1 - p)=212\times0.1\times(1 - 0.1)=212\times0.1\times0.9 = 19.1$
Since $np(1 - p)=19.1\geq10$, the sampling distribution of $\hat{p}$ is approximately normal. But since in the problem it says "No", we assume we are using some other values or conditions. Let's continue with the steps for the probability calculations.

Step2: Standardize for part c

The sampling distribution of $\hat{p}$ has mean $\mu_{\hat{p}}=p$ and standard deviation $\sigma_{\hat{p}}=\sqrt{\frac{p(1 - p)}{n}}$. Let $p = 0.1$ and $n = 212$, then $\sigma_{\hat{p}}=\sqrt{\frac{0.1\times(1 - 0.1)}{212}}=\sqrt{\frac{0.09}{212}}\approx0.021$.
We want to find $P(\hat{p}\leq0.09)$. We standardize $\hat{p}$ using $z=\frac{\hat{p}-\mu_{\hat{p}}}{\sigma_{\hat{p}}}=\frac{0.09 - 0.1}{0.021}\approx - 0.48$. Then $P(\hat{p}\leq0.09)=P(Z\leq - 0.48)$. Using the standard normal table, $P(Z\leq - 0.48)=0.3156$.

Step3: Standardize for part d

We want to find $P(\hat{p}\leq0.08)$. First, calculate the $z$ - score: $z=\frac{\hat{p}-\mu_{\hat{p}}}{\sigma_{\hat{p}}}=\frac{0.08 - 0.1}{0.021}\approx - 0.95$. Then $P(\hat{p}\leq0.08)=P(Z\leq - 0.95)$. Using the standard normal table, $P(Z\leq - 0.95)=0.1711$. Since $0.1711>0.05$, 8% is not an unusually low percentage.

Answer:

b) (Assuming values as above) If we assume $p = 0.1$ and $n = 212$, $np(1 - p)=19.1$, which is $\geq10$ (contradicting the "No" in the problem, but for the sake of filling blanks as per problem structure).
c) $P(\hat{p}\leq0.09)=0.3156$
d) Since $P(\hat{p}\leq0.08)=0.1711$, which is $>0.05$, 8% is not an unusually low percentage.