Question

1. the los angeles times (dec. 13, 1992) reported that 80% of airline passengers prefer to sleep on long flights rather than watch movies, read, etc. consider randomly selecting 25 passengers from a particular long flight. define a random variable x and answer the following questions. use correct notation.

a. what is the probability that exactly 11 of those selected passengers prefer sleeping?
$p(x=11)=_{25}c_{11}(0.8)^{11}(1-0.8)^{25-11}=0.000003$
$0.00031$
b. what is the probability that at least 19 passengers prefer sleeping on the long flight?
$p(x\geq19) = binomcdf(25,0.8,19,25)=0.78$
$78?$
c. what is the average number of passengers that prefer sleeping on the long flight?
e. calculate and interpret the standard deviation of x.

Explanation:

Step1: Define binomial variables

Let $X$ = number of passengers who prefer sleeping. $X \sim Binomial(n=25, p=0.8)$

Step2: Solve part (a): Exact 11 successes

Use binomial probability formula: $P(X=k) = \binom{n}{k}p^k(1-p)^{n-k}$
$$P(X=11) = \binom{25}{11}(0.8)^{11}(0.2)^{14}$$
Calculate $\binom{25}{11}=4457400$, $(0.8)^{11}\approx0.085899$, $(0.2)^{14}\approx1.6384\times10^{-10}$
$$P(X=11)=4457400\times0.085899\times1.6384\times10^{-10}\approx0.000063$$

Step3: Solve part (b): At least 19 successes

Use complement rule: $P(X\geq19)=1-P(X\leq18)$
Using binomial CDF:
$$P(X\geq19)=1-\sum_{k=0}^{18}\binom{25}{k}(0.8)^k(0.2)^{25-k}\approx0.780$$

Step4: Solve part (c): Mean of binomial variable

Use formula $\mu = np$
$$\mu = 25\times0.8=20$$

Step5: Solve part (e): Standard deviation of X

Use formula $\sigma = \sqrt{np(1-p)}$
$$\sigma = \sqrt{25\times0.8\times0.2}=\sqrt{4}=2$$
Interpretation: The number of passengers who prefer sleeping typically varies by about 2 from the mean of 20.

Answer:

a. $\boldsymbol{P(X=11)\approx0.000063}$
b. $\boldsymbol{P(X\geq19)\approx0.780}$
c. $\boldsymbol{20}$
e. Standard deviation: $\boldsymbol{2}$; On average, the number of passengers who prefer sleeping differs from the mean (20) by about 2 passengers.