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Statistics / Statistics: Informed Decisions Using Data 4 / Chapter 6.2 / Problem 43E

Statistics: Informed Decisions Using Data | 4th Edition | ISBN: 9780321757272 | Authors: Michael Sullivan, III

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Solved: On-Time Flights According to flightstats.com,

Chapter 6, Problem 43E

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QUESTION:

Problem 43E

On-Time Flights According to flightstats.com, American Airlines flights from Dallas to Chicago are on time 80% of the time. Suppose 100 flights are randomly selected.

(a) Compute the mean and standard deviation of the random variable X, the number of on-time flights in 100 trials of the probability experiment.

(b) Interpret the mean.

(c) Would it be unusual to observe 75 on-time flights in a random sample of 100 flights from Dallas to Chicago? Why?

Questions & Answers

QUESTION:

Problem 43E

On-Time Flights According to flightstats.com, American Airlines flights from Dallas to Chicago are on time 80% of the time. Suppose 100 flights are randomly selected.

(a) Compute the mean and standard deviation of the random variable X, the number of on-time flights in 100 trials of the probability experiment.

(b) Interpret the mean.

(c) Would it be unusual to observe 75 on-time flights in a random sample of 100 flights from Dallas to Chicago? Why?

ANSWER:

Answer

Step 1 of 3

(a)

American Airlines flights from Dallas to Chicago are on time 80% of the time.

Suppose 100 flights are randomly selected.

We are asked to compute the mean and standard deviation of the random variable $X,$ the number of on-time flights in 100 trials of the probability experiment.

Random variable $X$ represents the number of on-time flights in 100 trials.

Since trials are independent and each trial has the same success probability hence $X$ follows the binomial distribution.

Given value $n\ =100,\ p=80\%=0.8\ and\ q\ =20\%\ =0.2.$

We can write the formula for mean and standard deviation,

$mean\ or\ expectation\ \mu{}\ =n\times{}p\ and\ standard\ deviation\ \sigma{}=\sqrt{npq}$

$\mu{}\ =100\times{}0.8=80\ flights$

$\sigma{}=\sqrt{100\times{}0.8\times{}0.2}=4\ flights$

Hence the mean and standard deviation of the random variable $X$ is $80\ flights\ and\ 4\ flights$ respectively.

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