/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q7BSC a.聽The probability of 7 hurrica... [FREE SOLUTION] | 91影视

91影视

a.The probability of 7 hurricanes in a year is equal to 0.140.

b. Thus, the expected number of years to have 7 hurricanes in a 55-year period is equal to 7.7 years.

c. The expected number of years that have 7 hurricanes is approximately equal to the actual number of years that have 7 hurricanes in a 55-year period.Since the expected and the actual values are approximately equal, the Poisson distribution works well here.

Short Answer

Expert verified

a.The probability of 7 hurricanes in a year is equal to 0.140.

b. Thus, the expected number of years to have 7 hurricanes in a 55-year period is equal to 7.7 years.

c. The expected number of years that have 7 hurricanes is approximately equal to the actual number of years that have 7 hurricanes in a 55-year period.Since the expected and the actual values are approximately equal, the Poisson distribution works well here.

Step by step solution

01

Given information

The mean number of Atlantic hurricanes in the United States is given to be equal to 6.1 per year.

02

Poisson probability

a.

Let X be the number of Atlantic hurricanes in one year. Here, X follow a Poisson distribution with mean equal to\({\kern 1pt} \mu = 6.1\).

The probability of 7 hurricanes in a year is computed below:

\[\begin{aligned}{c}P\left( x \right) = \frac{{{\mu ^x}{e^{ - \mu }}}}{{x!}}\\P\left( 7 \right) = \frac{{{{\left( {6.1} \right)}^7}{{\left( {2.71828} \right)}^{ - 6.1}}}}{{7!}}\\ = 0.139856\\ \approx 0.140\end{aligned}\]

Therefore, the probability of 7 hurricanes in a year is equal to 0.140.

03

Expected number of hurricanes

b.

The expected number of years to have 7 hurricanes in a 55-year period is computed below:

\(\begin{aligned}{c}55 \times P\left( 7 \right) = 55 \times 0.140\\ \approx 7.7\end{aligned}\)

Thus, the expected number of years to have 7 hurricanes in a 55-year period is equal to 7.7 years.

04

Comparison of actual and expected values

c.

It is given that the actual number of years that had 7 hurricanes in the recent 55-year period is equal to 7.

The expected number of years that have 7 hurricanes in a 55-year period is equal to 7.7.

Thus, the expected number of years is equal to the actual number of years.

Since the expected and the actual values are approximately equal, the Poisson distribution works well here.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

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?

Bone Density Test A bone mineral density test is used to identify a bone disease. The result of a bone density test is commonly measured as a z score, and the population of z scores is normally distributed with a mean of 0 and a standard deviation of 1.

a. For a randomly selected subject, find the probability of a bone density test score less than 1.54.

b. For a randomly selected subject, find the probability of a bone density test score greater than -1.54.

c. For a randomly selected subject, find the probability of a bone density test score between -1.33 and 2.33.

d. Find \({Q_1}\), the bone density test score separating the bottom 25% from the top 75%.

e. If the mean bone density test score is found for 9 randomly selected subjects, find the probability that the mean is greater than 0.50.

Groups of people aged 15鈥65 are randomly selected and arranged in groups of six. The random variable xis the number in the group who say that their family and / or partner contribute most to their happiness (based on a Coca-Cola survey). The accompanying table lists

the values of xalong with their corresponding probabilities. Does the table describe a probability distribution? If so, find the mean and standard deviation.

x

P(x)

0

0+

1

0.003

2

0.025

3

0.111

4

0.279

5

0.373

6

0.208

Identifying Binomial Distributions. In Exercises 5鈥12, determine whether the given procedure results in a binomial distribution (or a distribution that can be treated as binomial). For those that are not binomial, identify at least one requirement that is not satisfied.

Investigating Dates In a survey sponsored by TGI Friday鈥檚, 1000 different adult respondents were randomly selected without replacement, and each was asked if they investigate dates on social media before meeting them. Responses consist of 鈥測es鈥 or 鈥渘o.鈥

Identifying Binomial Distributions. In Exercises 5鈥12, determine whether the given procedure results in a binomial distribution (or a distribution that can be treated as binomial). For those that are not binomial, identify at least one requirement that is not satisfied.

Surveying Senators The Senate members of the 113th Congress include 80 males and 20 females. Forty different senators are randomly selected without replacement, and the gender of each selected senator is recorded.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.