Question

assume 11% of the population is left - handed. assume this percentage is also true for all college students. a random sample of 212 college students from a campus with 5300 students is taken and whether or not they are left - handed is recorded.
list the given information with the correct symbols:
? = 0.11
? = 212
? = 5300
a) identify the individual, variable, random variable x and the random variable \\(\widehat{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 \\(\widehat{p}\\) is as follows:
\\(\widehat{p}\\) = select an answer
b) can we say that the sampling distribution of \\(\widehat{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: Identify symbols

Let $p = 0.11$ (population - proportion of left - handed people), $n=212$ (sample size), $N = 5300$ (population size).

Step2: Define individual, variable, random variables (a)

• The individual is a college student.
• The variable information collected from each individual is whether the student is left - handed or not.
• $X$ is the number of left - handed students in the sample of 212 college students.
• $\hat{p}=\frac{X}{n}$ is the proportion of left - handed students in the sample of 212 college students.

Step3: Check normality of sampling distribution (b)

We check the condition $np(1 - p)$. Substitute $n = 212$ and $p=0.11$ into $np(1 - p)$:

$$np(1 - p)=212\times0.11\times(1 - 0.11)=212\times0.11\times0.89 = 212\times0.0979 = 20.7548\approx20.8$$

Since $np(1 - p)=20.8\geq10$, the sampling distribution of $\hat{p}$ is approximately normal.

Step4: Calculate probability (c)

The mean of the sampling distribution of $\hat{p}$ is $\mu_{\hat{p}}=p = 0.11$ and the standard deviation is $\sigma_{\hat{p}}=\sqrt{\frac{p(1 - p)}{n}}=\sqrt{\frac{0.11\times(1 - 0.11)}{212}}=\sqrt{\frac{0.11\times0.89}{212}}=\sqrt{\frac{0.0979}{212}}\approx\sqrt{0.0004618}\approx0.0215$.
We want to find $P(\hat{p}\leq0.09)$. First, calculate the z - score: $z=\frac{\hat{p}-\mu_{\hat{p}}}{\sigma_{\hat{p}}}=\frac{0.09 - 0.11}{0.0215}=\frac{- 0.02}{0.0215}\approx - 0.93$.
Using the standard normal table, $P(\hat{p}\leq0.09)=P(Z\leq - 0.93)=0.1762$.

Step5: Determine if 8% is unusual (d)

First, find the z - score for $\hat{p}=0.08$. $z=\frac{0.08 - 0.11}{0.0215}=\frac{-0.03}{0.0215}\approx - 1.40$.
Since $|z| = 1.40<2$, 8% is not an unusually low percentage.

Answer:

a)

• The individual: A college student
• The variable: Whether the student is left - handed or not
• $X$: The number of left - handed students in the sample of 212 college students
• $\hat{p}$: The proportion of left - handed students in the sample of 212 college students

b) Yes, because $np(1 - p)=20.8\geq10$
c) $P(\hat{p}\leq0.09)=0.1762$
d) No