Respuesta :
Answer:
Approximately [tex]0.038[/tex] (or equivalently [tex]3.8\%[/tex],) 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 [tex]n = 100[/tex] selected residents owns boats as a Bernoulli random variable: for [tex]k = \lbrace 1,\, \dots,\, 100\rbrace[/tex], [tex]X_k \sim \text{Bernoulli}(\underbrace{0.13}_{p})[/tex].
[tex]X_k = 0[/tex] means that the [tex]k[/tex]th resident in this sample does not own boats. On the other hand, [tex]X_k = 1[/tex] means that this resident owns boats. Therefore, the sum [tex](X_1 + \cdots + X_{100})[/tex] would represent the number of residents in this sample that own boats.
Each of these [tex]100[/tex] random variables are all independent from one another. The mean of each [tex]X_k[/tex] would be [tex]\mu = 0.13[/tex], whereas the variance of each [tex]X_k\![/tex] would be [tex]\sigma = p\, (1 - p) = 0.13 \times (1 - 0.13) = 0.1131[/tex].
The sample size of [tex]100[/tex] is a rather large number. Besides, all these samples share the same probability distribution. Apply the Central Limit Theorem. By this theorem, the sum [tex](X_1 + \cdots + X_{100})[/tex] would approximately follow a normal distribution with:
- mean [tex]n\, \mu = 100 \times 0.13 = 13[/tex], and
- variance [tex]\sigma\, \sqrt{n} = p\, (1 - p)\, \sqrt{n} = 0.1131 \times 10 = 1.131[/tex].
[tex]11\%[/tex] of that sample of [tex]100[/tex] residents would correspond to [tex]11\% \times 100 = 11[/tex] residents. Calculate the [tex]z[/tex]-score corresponding to a sum of [tex]11[/tex]:
[tex]\begin{aligned}z &= \frac{11 - 13}{1.131} \approx -1.77 \end{aligned}[/tex].
The question is (equivalently) asking for [tex]P( (X_1 + \cdots + X_{100}) < 11)[/tex]. That is equal to [tex]P(Z < -1.77)[/tex]. However, some [tex]z[/tex]-tables list only probabilities like [tex]P(Z > z)[/tex]. Hence, convert [tex]P(Z < -1.77)\![/tex] to that form:
[tex]\begin{aligned} & P( (X_1 + \cdots + X_{100}) < 11) \\ &= P(Z < -1.77) \\ &= 1 - P(Z >1.77) \end{aligned}[/tex].
Look up the value of [tex]P(Z > 1.77)[/tex] on a [tex]z[/tex]-table:
[tex]P(Z > 1.77) \approx 0.962[/tex].
Therefore:
[tex]\begin{aligned} & P( (X_1 + \cdots + X_{100}) < 11) \\ &= 1 - P(Z >1.77) \\ &\approx 1 - 0.962 = 0.038 \end{aligned}[/tex].
