Question

According to New Jersey Transit, the 8:00 A.M. weekday train from Princeton to New York City has a 90% chance of arriving on time on a randomly selected day. Suppose this claim is true. Choose 6 days at random. Let W

Answer 1

Answer:

See explanation below.

Step-by-step explanation:

Assuming this problem: "According to New Jersey Transit, the 8:00 A.M. weekday train from Princeton to New York City has a 90% chance of arriving on time on a randomly selected day. Suppose this claim is true. Choose 6 days at random. Let W = the number of days on which the train arrives late. "

For this case we can check if the binomial model can be used checking conditions:

1) We satisfy that we have independent trials and is assumed for this case

2) We have a fixed value for the trials n =6 and for the probability p = 0.9 on each trial

3) We have bernoulli experiments on each trial since we have success or failure for each case.

So then since all the conditions are satisfied we can assume that the binomial model can be used here.

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Let W the random variable of interest "the number of days on which the train arrives late", on this case we now that:

$W \sim Binom(n=6, p=0.9)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

We can find all the possible probabilities for all the possible values of X like this:

$P(X=0)=(6C0)(0.9)^0 (1-0.9)^{6-0}=1x10^{-6}$

$P(X=1)=(6C1)(0.9)^1 (1-0.9)^{6-1}=5.4x10^{-5}$

$P(X=2)=(6C2)(0.9)^2 (1-0.9)^{6-2}=0.001215$

$P(X=3)=(6C3)(0.9)^3 (1-0.9)^{6-3}=0.01458$

$P(X=4)=(6C4)(0.9)^4 (1-0.9)^{6-4}=0.0984$

$P(X=5)=(6C5)(0.9)^5 (1-0.9)^{6-5}=0.354$

$P(X=5)=(6C5)(0.9)^6 (1-0.9)^{6-6}=0.531$