Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Baby elk Biologists estimate that a randomly selected baby elk has a 44 % chance of surviving to adulthood. Assume this estimate is correct. Suppose researchers choose 7 baby elk at random to monitor. Let X= the number that survive to adulthood.

Short Answer

Expert verified

The given statement is correct because According to biologists, a baby elk chosen at random has a 44 percent chance of living to adulthood.

Step by step solution

01

Given Information

The Trials is conducted n=7

Success of probability P=44%=0.44

02

According to the given question

The random variable X satisfies the following requirements:

a. Probability of success, defined as the likelihood of surviving to adulthoodpwhich corresponds corresponding0.44is now fixed.

b. The number of baby elk chosen is set.

c. Baby elks are completely self-sufficient.

d. There are two possible outcomes: either the newborn elk survives to adulthood or it does not.

Here, all of the binomial criteria are met. As a result, it is possible to conclude that X the binomial distribution is followed.

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 Vaia!

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

Skee BallAna is a dedicated Skee Ball player (see photo in Exercise 4) who always rolls for the 50-point slot. The probability distribution of Anaโ€™s score Xon a randomly selected roll of the ball is shown here. From Exercise 8, ฮผX=23.8.

(a) Find the median of X.

(b) Compare the mean and median. Explain why this relationship makes sense based on the probability distribution.

Easy-start mower? A company has developed an "easy-start" mower that cranks the engine with the push of a button. The company claims that the probability the mower will start on any push of the button is 0.9. Assume for now that this claim is true. On the next 30 uses of the mower, let T=the number of times it starts on the first push of the button. Here is a histogram of the probability distribution of T :

a. What probability distribution does T have? Justify your answer.

b. Describe the shape of the probability distribution.

Large Counts condition To use a Normal distribution to approximate binomial probabilities, why do we require that both npand n(1-p) be a t least 10?

Life insurance If four 21-year-old men are insured, the insurerโ€™s average income is

V=X1+X2+X3+X44=0.25X1+0.25X2+0.25X3+0.25X4

where Xiis the income from insuring one man. Assuming that the amount of income earned on individual policies is independent, find the mean and standard deviation of V. (If you compare with the results of Exercise 57, you should see that averaging over more insured individuals reduces risk.)

Loser buys the pizza leona and Fred are friendly competitors in high school. Both are about to take the ACT college entrance examination, They agree that if one of them scores 5ar more points better than the other, the loser will buy the winner a pizza. Suppose that in fact Fred and Leona have equal ability, so that each score varies Normally with mean 24and standard deviation data-custom-editor="chemistry" 2. (The variation is due to luck in guessing and the accident of the specific questions being familiar to the student.) The two scores are independent. What is the probability that the scores differ by 5or more points in either direction? Follow the four-step process.

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.

Sign-up for free