Question

because np(1 - p) = 19.08 which is < 10. 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(̂p ≤ 0.09) = 0.3156 d) is 8% an unusually low percentage of randomly selected college students that are left - handed in a sample of 212? since p(̂p ≤ 0.08) = 0.1660 which is > 0.05, 8% is not an unusually low percentage of college students that are left - handed in a sample of 212

Explanation:

Step1: Identify relevant concepts

We are dealing with sampling - distribution of sample proportion. Let $n = 212$ be the sample size and assume the population proportion of left - handed students is $p$ (not given in the problem - setup but we use the sample - proportion $\hat{p}$ for calculations). The mean of the sampling distribution of the sample proportion is $\mu_{\hat{p}}=p$ and the standard deviation is $\sigma_{\hat{p}}=\sqrt{\frac{p(1 - p)}{n}}$.

Step2: Calculate for part c

First, we need to standardize the sample proportion $\hat{p}=0.09$. The $z$ - score is calculated as $z=\frac{\hat{p}-\mu_{\hat{p}}}{\sigma_{\hat{p}}}=\frac{\hat{p}-p}{\sqrt{\frac{p(1 - p)}{n}}}$. If we assume a common value for the proportion of left - handed people in the general population $p = 0.12$ (a common estimate). Then $\sigma_{\hat{p}}=\sqrt{\frac{0.12\times(1 - 0.12)}{212}}=\sqrt{\frac{0.12\times0.88}{212}}\approx\sqrt{\frac{0.1056}{212}}\approx0.0223$. The $z$ - score for $\hat{p}=0.09$ is $z=\frac{0.09 - 0.12}{0.0223}=\frac{- 0.03}{0.0223}\approx - 1.35$. Then $P(\hat{p}\leq0.09)=P(Z\leq - 1.35)$. Looking up in the standard normal table, $P(Z\leq - 1.35)=0.0885$.

Step3: Calculate for part d

For $\hat{p}=0.08$, the $z$ - score is $z=\frac{0.08 - 0.12}{0.0223}=\frac{-0.04}{0.0223}\approx - 1.79$. Then $P(\hat{p}\leq0.08)=P(Z\leq - 1.79)$. Looking up in the standard normal table, $P(Z\leq - 1.79) = 0.0367$. Since $0.0367<0.05$, 8% is an unusually low percentage.

Answer:

c) $P(\hat{p}\leq0.09)=0.0885$
d) Since $P(\hat{p}\leq0.08)=0.0367$ which is $<0.05$, 8% is an unusually low percentage of college students that are left - handed in a sample of 212.