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Chapter 5: Q7 (page 220)

In Exercises 6鈥�10, use the following: Five American Airlines flights are randomly selected, and the table in the margin lists the probabilities for the number that arrive on time (based on data from the Department of Transportation). Assume that five flights are randomly selected.

Find the mean of the number of flights among five that arrive on time.

x

P(x)

0

0+

1

0.006

2

0.051

3

0.205

4

0.409

5

0.328

Short Answer

Expert verified

The mean number of flights that arrive on time is 4.

Step by step solution

01

Given information

The probability distribution for the five American airlines flights.

02

Calculate the mean

Let x represents the number of flights that arrive on time.

The mean for random variable x is computed as,

渭=鈭�x脳Px=0脳0.0+1脳0.006+2脳0.051+...+5脳0.328=3.999鈮�4

Thus, the mean number of flights that arrive on time is 4.0.

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Most popular questions from this chapter

In Exercises 6鈥�10, use the following: Five American Airlines flights are randomly selected, and the table in the margin lists the probabilities for the number that arrive on time (based on data from the Department of Transportation). Assume that five flights are randomly selected.

What does the probability of 0+ indicate? Does it indicate that among five randomly selectedflights, it is impossible for none of them to arrive on time?

x

P(x)

0

0+

1

0.006

2

0.051

3

0.205

4

0.409

5

0.328

In Exercises 7鈥�14, determine whether a probability

distribution is given. If a probability distribution is given, find its mean and standard deviation. If a probability distribution is not given, identify the requirements that are not satisfied.

Five males with an X-linked genetic disorder have one child each. The random variable xis the number of children among the five who inherit the X-linked genetic disorder.

x

P(x)

0

0.031

1

0.156

2

0.313

3

0.313

4

0.156

5

0.031

Hypergeometric Distribution If we sample from a small finite population without replacement, the binomial distribution should not be used because the events are not independent. If sampling is done without replacement and the outcomes belong to one of two types, we can use the hypergeometric distribution. If a population has A objects of one type (such as lottery numbers you selected), while the remaining B objects are of the other type (such as lottery numbers you didn鈥檛 select), and if n objects are sampled without replacement (such as six drawn lottery numbers), then the probability of getting x objects of type A and n - x objects of type B is

\(P\left( x \right) = \frac{{A!}}{{\left( {A - x} \right)!x!}} \times \frac{{B!}}{{\left( {B - n + x} \right)!\left( {n - x} \right)!}} \div \frac{{\left( {A + B} \right)!}}{{\left( {A + B - n} \right)!n!}}\)

In New Jersey鈥檚 Pick 6 lottery game, a bettor selects six numbers from 1 to 49 (without repetition), and a winning six-number combination is later randomly selected. Find the probabilities of getting exactly two winning numbers with one ticket. (Hint: Use A = 6, B = 43, n = 6, and x = 2.)

In Exercises 15鈥�20, refer to the accompanying table, which describes results from groups of 8 births from 8 different sets of parents. The random variable x represents the number of girls among 8 children.

Using Probabilities for Significant Events

a. Find the probability of getting exactly 7 girls in 8 births.

b. Find the probability of getting 7 or more girls in 8 births.

c. Which probability is relevant for determining whether 7 is a significantly high number ofgirls in 10 births: the result from part (a) or part (b)?

d. Is 7 a significantly high number of girls in 8 births? Why or why not?

Number of girls x

P(x)

0

0.004

1

0.031

2

0.109

3

0.219

4

0.273

5

0.219

6

0.109

7

0.031

8

0.004

For 100 births, P(exactly 56 girls) = 0.0390 and P(56 or more girls) = 0.136. Is 56 girls in 100 births a significantly high number of girls? Which probability is relevant to answering that question?
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