Question

a) identify the individual, variable, random variable x and the random variable \\(\hat{p}\\) in the context of this problem.
the individual is select an answer
the variable information collected from each individual is select an answer
the wording in context for the random variable x is as follows:
x = select an answer
the wording in context for the random variable \\(\hat{p}\\) is as follows:
\\(\hat{p}\\) = select an answer
b) can we say that the sampling distribution of \\(\hat{p}\\) is approximately normal?
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?

Explanation:

Step1: Define terms

The individual is a single college - student. The variable is whether a college - student is left - handed or not (a categorical variable). The random variable $X$ is the number of left - handed college students in the sample. The random variable $\hat{p}$ is the sample proportion of left - handed college students, i.e., $\hat{p}=\frac{X}{n}$, where $n$ is the sample size.

Step2: Check normality condition

To check if the sampling distribution of $\hat{p}$ is approximately normal, we use the condition $np(1 - p)\geq10$. Let's assume we know the population proportion $p$. If we assume $p$ is some value (say we know from prior studies that the proportion of left - handed people is around 0.1), and $n = 212$. Then $np(1 - p)=212\times0.1\times(1 - 0.1)=212\times0.1\times0.9 = 19.08\geq10$.

Step3: Standardize for probability calculation

If $\hat{p}$ is approximately normal with mean $\mu_{\hat{p}}=p$ and standard deviation $\sigma_{\hat{p}}=\sqrt{\frac{p(1 - p)}{n}}$, to find $P(\hat{p}\leq0.09)$, we first standardize $\hat{p}$ using $z=\frac{\hat{p}-p}{\sqrt{\frac{p(1 - p)}{n}}}$. Let $p = 0.1$ and $n = 212$, then $\sigma_{\hat{p}}=\sqrt{\frac{0.1\times(1 - 0.1)}{212}}=\sqrt{\frac{0.09}{212}}\approx0.0207$. And $z=\frac{0.09 - 0.1}{0.0207}\approx - 0.48$. Then $P(\hat{p}\leq0.09)=P(Z\leq - 0.48)\approx0.3156$.

Step4: Determine unusualness

We can use the range - rule of - thumb for significant values. The mean of $\hat{p}$ is $p$ and the standard deviation is $\sigma_{\hat{p}}=\sqrt{\frac{p(1 - p)}{n}}$. For $p = 0.1$ and $n = 212$, $\sigma_{\hat{p}}\approx0.0207$. The significant low values are $\mu_{\hat{p}}-2\sigma_{\hat{p}}=0.1-2\times0.0207 = 0.0586$ and significant high values are $\mu_{\hat{p}}+2\sigma_{\hat{p}}=0.1 + 2\times0.0207=0.1414$. Since $0.08>0.0586$, 8% is not an unusually low percentage.

Answer:

a)

• The individual: A single college student
• The variable: Whether a college student is left - handed or not
• $X$: The number of left - handed college students in the sample
• $\hat{p}$: The sample proportion of left - handed college students ($\hat{p}=\frac{X}{n}$, where $n$ is the sample size)

b) Yes, because if we assume $p = 0.1$ and $n = 212$, $np(1 - p)=19.1\geq10$
c) $P(\hat{p}\leq0.09)\approx0.3156$
d) No