Question

In a beach town, 13% of the residents own boats. A random sample of 100 residents was selected. What is the probability that less than 11% of the residents in the sample own boats?
Find the z-table here.

Answer 1

Answer:

0.276

Step-by-step explanation:

got it right on edge

Answer 2

Answer:

Approximately $0.038$ (or equivalently $3.8\%$,) assuming that whether each resident owns boats is independent from one another.

Step-by-step explanation:

Assume that whether each resident of this town owns boats is independent from one another. It would be possible to model whether each of the $n = 100$ selected residents owns boats as a Bernoulli random variable: for $k = \lbrace 1,\, \dots,\, 100\rbrace$, $X_k \sim \text{Bernoulli}(\underbrace{0.13}_{p})$.

$X_k = 0$ means that the $k$th resident in this sample does not own boats. On the other hand, $X_k = 1$ means that this resident owns boats. Therefore, the sum $(X_1 + \cdots + X_{100})$ would represent the number of residents in this sample that own boats.

Each of these $100$ random variables are all independent from one another. The mean of each $X_k$ would be $\mu = 0.13$, whereas the variance of each $X_k\!$ would be $\sigma = p\, (1 - p) = 0.13 \times (1 - 0.13) = 0.1131$.

The sample size of $100$ is a rather large number. Besides, all these samples share the same probability distribution. Apply the Central Limit Theorem. By this theorem, the sum $(X_1 + \cdots + X_{100})$ would approximately follow a normal distribution with:

• mean $n\, \mu = 100 \times 0.13 = 13$, and
• variance $\sigma\, \sqrt{n} = p\, (1 - p)\, \sqrt{n} = 0.1131 \times 10 = 1.131$.

$11\%$ of that sample of $100$ residents would correspond to $11\% \times 100 = 11$ residents. Calculate the $z$-score corresponding to a sum of $11$:

$\begin{aligned}z &= \frac{11 - 13}{1.131} \approx -1.77 \end{aligned}$.

The question is (equivalently) asking for $P( (X_1 + \cdots + X_{100}) < 11)$. That is equal to $P(Z < -1.77)$. However, some $z$-tables list only probabilities like $P(Z > z)$. Hence, convert $P(Z < -1.77)\!$ to that form:

$\begin{aligned} & P( (X_1 + \cdots + X_{100}) < 11) \\ &= P(Z < -1.77) \\ &= 1 - P(Z >1.77) \end{aligned}$.

Look up the value of $P(Z > 1.77)$ on a $z$-table:

$P(Z > 1.77) \approx 0.962$.

Therefore:

$\begin{aligned} & P( (X_1 + \cdots + X_{100}) < 11) \\ &= 1 - P(Z >1.77) \\ &\approx 1 - 0.962 = 0.038 \end{aligned}$.