Let’s walk through the given question step-by-step to determine which description best fits the sampling distribution of [tex]\(\hat{\rho}\)[/tex], which represents the proportion of students who live within five miles of the school.
1. Determine the values for [tex]\( n \)[/tex] and [tex]\( p \)[/tex]:
- The proportion of students living within five miles of the school is [tex]\( p = 0.19 \)[/tex].
- The sample size of students taken by the principal is [tex]\( n = 17 \)[/tex].
2. Calculate [tex]\( n(1 - p) \)[/tex]:
[tex]\[
n(1 - p) = 17 \times (1 - 0.19) = 17 \times 0.81 = 13.77
\][/tex]
Since [tex]\( n(1 - p) = 13.77 \)[/tex], it satisfies the condition [tex]\( n(1 - p) > 10 \)[/tex]. This means for the calculation of [tex]\( n(1 - p) \)[/tex], we have a sufficiently large sample size such that this part would suggest closer to a Normal distribution if it were the only condition.
3. Calculate [tex]\( n \times p \)[/tex]:
[tex]\[
n \times p = 17 \times 0.19 = 3.23
\][/tex]
Since [tex]\( n \times p = 3.23 \)[/tex], it does not satisfy the condition [tex]\( n \times p > 10 \)[/tex]. This indicates that the sample size is not large enough for the binomial distribution to be approximated by a Normal distribution, which results in the sampling distribution of [tex]\(\hat{p}\)[/tex] not being approximately Normal.
4. Analyze the skewness:
Since [tex]\( p = 0.19 \)[/tex] is closer to 0 than to 1, this means that the distribution is skewed towards the right because there are more samples of students who do not live within five miles.
Based on these calculations and conditions, the best description for the sampling distribution of [tex]\(\hat{\rho}\)[/tex] (proportion of students within five miles) is:
Because [tex]\( np = 3.23 < 10 \)[/tex], the sampling distribution of [tex]\(\hat{p}\)[/tex] is not approximately Normal. Because [tex]\( p = 0.19 \)[/tex] is closer to 0 than 1, the sampling distribution of [tex]\(\widehat{\rho}\)[/tex] is skewed to the right.
