Binomial Distribution: A 2005 Gallup Poll found that 7% of teenagers (ages 13 to 17) suffer from arachnophobia and are extremely afraid of spiders. At a summer camp, 12 teenagers are sleeping in each tent. Assume that these 12 teenagers are independent of each other.
a) Calculate the probability that at least one of them suffers from arachnophobia. (please round to four decimal places)
b) Calculate the probability that exactly 2 of them suffer from arachnophobia. (please round to four decimal places)
c) Calculate the probability that at most 1 of them suffers from arachnophobia. (please round to four decimal places)
d) Calculate the probability that more than 3 teenagers are afraid of spiders. (please round to four decimal places)
e) What is the mean and standard deviation?
Answer:
Given:
The probability of success $(p)=0.07$
The sample size $(n)=12$
(A) The probability that at least one of them suffers from arachnophobia:
$P(x \geq 1) = \sum_{1}^{12} \binom{n}{x} \cdot p^x \cdot (1 – p)^{n – x} = \sum_{1}^{12} \binom{12}{x} \cdot 0.07^x \cdot (1 – 0.07)^{12 – x} = 0.5814$
(B) The probability that exactly 2 of them suffer from arachnophobia:
$P(x = 2) = \binom{n}{x} \cdot p^x \cdot (1 – p)^{n – x} = \binom{12}{2} \cdot 0.07^2 \cdot (1 – 0.07)^{12 – 2} = 0.1565$
(C) The probability that at most 1 of them suffers from arachnophobia:
$P(x \leq 1) = \sum_{0}^{1} \binom{n}{x} \cdot p^x \cdot (1 – p)^{n – x} = \sum_{0}^{3} \binom{12}{x} \cdot 0.07^x \cdot (1 – 0.07)^{12 – x} = 0.7967$
(D) The probability that more than 3 teenagers are afraid of spiders:
$P(x > 3) = \sum_{4}^{12} \binom{n}{x} \cdot p^x \cdot (1 – p)^{n – x} = \sum_{4}^{12} \binom{12}{x} \cdot 0.07^x \cdot (1 – 0.07)^{12 – x} = 0.0075$
(E) The mean $(\mu):$
$\mu = n \cdot p = 12 \cdot 0.07 = 0.84$
(F) The standard deviation $(\sigma):$
$\sigma = \sqrt{n \cdot p(1 – p)} = \sqrt{12 \cdot 0.07(1 – 0.07)} = 0.884$
