Question

a) identify the individual, variable, random variable x and the random variable \\(\hat{p}\\) in the context of this problem.
the individual is all college students
the variable information collected from each individual is a randomly selected college student
the wording in context for the random variable x is as follows:
x = the number of 212 randomly selected college students that are left - handed
the wording in context for the random variable \\(\hat{p}\\) is as follows:
\\(\hat{p}\\) = the proportion of 212 randomly selected college students that are left - handed
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(\\(\hat{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(\\(\hat{p}\\) 0.08) = 0.1711 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: Define relevant values

Let $n = 212$. Assume a general proportion of left - handed people $p$ (if not given otherwise, we can use common sense estimates. Let's assume $p = 0.1$ for the purpose of checking normality).

Step2: Check normality condition

We use the formula $np(1 - p)$. Substitute $n = 212$ and $p=0.1$ into it: $np(1 - p)=212\times0.1\times(1 - 0.1)=212\times0.1\times0.9 = 19.08$. Since $np(1 - p)=19.08>10$, the sampling distribution of $\hat{p}$ is approximately normal.

Step3: Standardize for part c

For $P(\hat{p}\leq0.09)$, first find the mean of the sampling distribution of $\hat{p}$ which is $\mu_{\hat{p}}=p$ (assumed $p = 0.1$) and the standard deviation $\sigma_{\hat{p}}=\sqrt{\frac{p(1 - p)}{n}}=\sqrt{\frac{0.1\times(1 - 0.1)}{212}}=\sqrt{\frac{0.09}{212}}\approx0.0207$. The z - score is $z=\frac{\hat{p}-\mu_{\hat{p}}}{\sigma_{\hat{p}}}=\frac{0.09 - 0.1}{0.0207}\approx - 0.48$. Then $P(\hat{p}\leq0.09)=P(Z\leq - 0.48)=0.3156$.

Step4: Analyze part d

For $P(\hat{p}\leq0.08)$, calculate the z - score: $z=\frac{0.08 - 0.1}{0.0207}\approx - 0.97$. Then $P(\hat{p}\leq0.08)=P(Z\leq - 0.97)=0.1660$ (using standard normal table). Since $0.1660>0.05$, 8% is not an unusually low percentage.

Answer:

b) Yes, because $np(1 - p)=19.08>10$
c) $P(\hat{p}\leq0.09) = 0.3156$
d) 8% is not an unusually low percentage of college students that are left - handed in a sample of 212 since $P(\hat{p}\leq0.08)=0.1660>0.05$