Question

The Los Angeles Times (Dec. 13, 1992) reported that 75% of airline passengers prefer to sleep on long flights than watch movies, read, etc. Consider randomly selecting 25 passengers from a particularly long flight and you are willing to consider this as a random sample of the population of all passengers on long flights.

(1) What is the probability that exactly 18 passengers slept on the flight?

(Show what you typed into the calculator, define n, p, and r, and round your answer to 3 decimal places.)

(2) What is the probability that at least 20 people slept on the plane?

(Show what you typed into the calculator, define n, p, and r, and round your answer to 3 decimal places.)

(3) What are the mean and standard deviation for the random variable X in this situation?

(Round to 3 decimal places and clearly label the mean and standard deviation.)

(4) Show that the requirements for a binomial setting are satisfied.

Answer 1

Let X denotes the number of airline passengers prefer to sleep on long flights than watch movies, read, etc out of randomly selected 25 passengers.

X ~ Binomial( n = 25, p = 0.75)

The probability mass function of X is

1) The probability that exactly 18 passengers slept on the flight

$= \binom{25}{18}*0.75^{18}*(1-0.75)^{25-18}$

2) The probability that at least 20 people slept on the plane

$= P(X\geq 20)$

$= 1 - P(X\leq 19)$

$=1 - \sum_{x=0}^{19} \binom{25}{x}*0.75^{x}*(1-0.75)^{25-x}$

3) Mean of X = E(X) = n*p = 25*0.75 = 18.75

Standard deviation of X = sd(X) = $\sqrt{25*0.75*(1-0.75)} = 2.16506$

4) Here

i) number of customers in the sample is fixed( = 25)

ii) The probability that a randomly selected customer sleeps or not does not depend on other customer.

iii) The probability that a randomly selected customer sleeps or not is same for all customers ( = 0.75)