In the 2013 Jerrys Artarama Art Supplies Catalog There Are 560 Paghes

Discrete Random Variables

Binomial Distribution

OpenStaxCollege

[latexpage]

At that place are three characteristics of a binomial experiment.

There are a stock-still number of trials. Think of trials as repetitions of an experiment. The letter of the alphabet n denotes the number of trials.

There are only two possible outcomes, called "success" and "failure," for each trial. The alphabetic character p denotes the probability of a success on 1 trial, and q denotes the probability of a failure on i trial. p + q = i.

The n trials are independent and are repeated using identical weather condition. Because the n trials are independent, the outcome of one trial does non help in predicting the consequence of some other trial. Another fashion of saying this is that for each private trial, the probability, p, of a success and probability, q, of a failure remain the same. For case, randomly guessing at a true-false statistics question has just two outcomes. If a success is guessing correctly, then a failure is guessing incorrectly. Suppose Joe always guesses correctly on any statistics truthful-fake question with probability p = 0.half-dozen. Then, q = 0.iv. This means that for every true-false statistics question Joe answers, his probability of success (p = 0.vi) and his probability of failure (q = 0.4) remain the same.

The outcomes of a binomial experiment fit a binomial probability distribution. The random variable X = the number of successes obtained in the n contained trials.

The hateful, μ, and variance, σ ⁱⁱ, for the binomial probability distribution are μ = np and σ ² = npq. The standard deviation, σ, is then σ = \(\sqrt{npq}\).

Any experiment that has characteristics two and iii and where n = 1 is called a Bernoulli Trial (named later Jacob Bernoulli who, in the late 1600s, studied them extensively). A binomial experiment takes place when the number of successes is counted in one or more Bernoulli Trials.

At ABC College, the withdrawal rate from an elementary physics grade is 30% for whatever given term. This implies that, for any given term, 70% of the students stay in the class for the entire term. A "success" could exist defined as an individual who withdrew. The random variable X = the number of students who withdraw from the randomly selected elementary physics class.

Attempt It

The state health board is concerned about the amount of fruit available in school lunches. Forty-eight percent of schools in the state offering fruit in their lunches every day. This implies that 52% do not. What would a "success" be in this case?

a schoolhouse that offers fruit in their lunch every twenty-four hour period

Suppose you play a game that you lot can only either win or lose. The probability that you lot win any game is 55%, and the probability that yous lose is 45%. Each game you play is independent. If you play the game 20 times, write the function that describes the probability that y'all win fifteen of the 20 times. Here, if you define X every bit the number of wins, then X takes on the values 0, 1, 2, 3, …, xx. The probability of a success is p = 0.55. The probability of a failure is q = 0.45. The number of trials is n = 20. The probability question can exist stated mathematically every bit P(x = 15).

Endeavour It

A trainer is teaching a dolphin to do tricks. The probability that the dolphin successfully performs the play a trick on is 35%, and the probability that the dolphin does not successfully perform the pull a fast one on is 65%. Out of 20 attempts, you want to notice the probability that the dolphin succeeds 12 times. Land the probability question mathematically.

P(10 = 12)

A fair coin is flipped fifteen times. Each flip is contained. What is the probability of getting more than ten heads? Let X = the number of heads in xv flips of the fair coin. 10 takes on the values 0, 1, 2, 3, …, 15. Since the money is fair, p = 0.5 and q = 0.5. The number of trials is n = 15. State the probability question mathematically.

P(x > 10)

Try It

A fair, six-sided die is rolled ten times. Each ringlet is independent. You desire to find the probability of rolling a 1 more iii times. State the probability question mathematically.

P(x > three)

Approximately seventy% of statistics students do their homework in fourth dimension for it to exist collected and graded. Each educatee does homework independently. In a statistics class of 50 students, what is the probability that at least forty volition exercise their homework on time? Students are selected randomly.

a. This is a binomial problem because there is merely a success or a __________, there are a fixed number of trials, and the probability of a success is 0.70 for each trial.

b. If we are interested in the number of students who do their homework on time, then how practise we ascertain X?

b. X = the number of statistics students who do their homework on time

c. What values does x accept on?

d. What is a "failure," in words?

d. Failure is defined as a pupil who does not complete his or her homework on fourth dimension.

The probability of a success is p = 0.70. The number of trials is n = 50.

e. If p + q = 1, and so what is q?

f. The words "at to the lowest degree" translate as what kind of inequality for the probability question P(x ____ 40).

f. greater than or equal to (≥)

The probability question is P(x ≥ 40).

Try Information technology

Sixty-five percent of people pass the country driver'due south test on the get-go try. A group of 50 individuals who have taken the driver'southward examination is randomly selected. Requite ii reasons why this is a binomial problem.

This is a binomial problem because there is only a success or a failure, and there are a definite number of trials. The probability of a success stays the aforementioned for each trial.

Annotation for the Binomial: B = Binomial Probability Distribution Function

X ~ B(n, p)

Read this as "10 is a random variable with a binomial distribution." The parameters are n and p; due north = number of trials, p = probability of a success on each trial.

It has been stated that about 41% of adult workers have a high school diploma but practice non pursue any further education. If 20 adult workers are randomly selected, detect the probability that at most 12 of them have a high school diploma but do not pursue whatever farther instruction. How many adult workers practice y'all wait to have a high schoolhouse diploma simply do not pursue any further pedagogy?

Let X = the number of workers who have a high schoolhouse diploma merely practice not pursue whatsoever farther education.

X takes on the values 0, 1, 2, …, twenty where n = 20, p = 0.41, and q = 1 – 0.41 = 0.59. Ten ~ B(20, 0.41)

Find P(x ≤ 12). P(ten ≤ 12) = 0.9738. (calculator or estimator)

Go into 2^nd DISTR. The syntax for the instructions are as follows:

To calculate (x = value): binompdf(due north, p, number)
if "number" is left out, the result is the binomial probability tabular array.

To calculate P(x ≤ value): binomcdf(n, p, number) if "number" is left out, the issue is the cumulative binomial probability table.

For this problem: After y'all are in two^nd DISTR, pointer down to binomcdf. Press ENTER. Enter xx,0.41,12). The outcome is P(ten ≤ 12) = 0.9738.

Note

If you want to find P(x = 12), apply the pdf (binompdf). If y'all want to find P(ten > 12), use 1 – binomcdf(20,0.41,12).

The probability that at most 12 workers have a high school diploma just do not pursue any further pedagogy is 0.9738.

The graph of X ~ B(20, 0.41) is as follows:

The y-axis contains the probability of x, where Ten = the number of workers who take only a high school diploma.

The number of adult workers that you expect to have a high school diploma merely not pursue any further education is the mean, μ = np = (xx)(0.41) = viii.2.

The formula for the variance is σ^two = npq. The standard divergence is σ = \(\sqrt{npq}\).

σ = \(\sqrt{\left(xx\right)\left(0.41\correct)\left(0.59\right)}\) = ii.20.

Endeavour It

About 32% of students participate in a community volunteer plan outside of school. If xxx students are selected at random, find the probability that at most 14 of them participate in a community volunteer program outside of school. Utilise the TI-83+ or TI-84 calculator to find the answer.

P(x ≤ xiv) = 0.9695

In the 2013 Jerry's Artarama art supplies itemize, there are 560 pages. Viii of the pages feature signature artists. Suppose we randomly sample 100 pages. Permit 10 = the number of pages that feature signature artists.

1. What values does x take on?
2. What is the probability distribution? Find the following probabilities:
   1. the probability that 2 pages feature signature artists
   2. the probability that at most half-dozen pages characteristic signature artists
   3. the probability that more than three pages feature signature artists.
3. Using the formulas, calculate the (i) mean and (ii) standard deviation.

1. x = 0, 1, 2, 3, iv, 5, half dozen, 7, 8
2. 10 ~ B\(\left(100,\frac{eight}{560}\right)\)
   1. P(x = 2) = binompdf\(\left(100,\frac{eight}{560},2\right)\) = 0.2466
   2. P(10 ≤ 6) = binomcdf\(\left(100,\frac{8}{560},6\right)\) = 0.9994
   3. P(10 > iii) = 1 – P(x ≤ 3) = ane – binomcdf\(\left(100,\frac{eight}{560},iii\correct)\) = 1 – 0.9443 = 0.0557
3. 1. Mean = np = (100)\(\left(\frac{8}{560}\right)\) = \(\frac{800}{560}\) ≈ 1.4286
   2. Standard Divergence = \(\sqrt{npq}\) = \(\sqrt{\left(100\right)\left(\frac{8}{560}\right)\left(\frac{552}{560}\right)}\) ≈ 1.1867

Try It

Co-ordinate to a Gallup poll, 60% of American adults adopt saving over spending. Let X = the number of American adults out of a random sample of l who prefer saving to spending.

1. What is the probability distribution for X?
2. Utilise your calculator to observe the following probabilities:
   1. the probability that 25 adults in the sample prefer saving over spending
   2. the probability that at virtually twenty adults prefer saving
   3. the probability that more than 30 adults adopt saving
3. Using the formulas, summate the (i) mean and (ii) standard difference of X.

1. X ∼ B(50, 0.6)
2. Using the TI-83, 83+, 84 figurer with instructions equally provided in [link]:
   1. P(x = 25) = binompdf(50, 0.6, 25) = 0.0405
   2. P(x ≤ 20) = binomcdf(50, 0.six, 20) = 0.0034
   3. P(x > xxx) = one – binomcdf(fifty, 0.6, 30) = 1 – 0.5535 = 0.4465
3. 1. Mean = np = 50(0.half dozen) = 30
   2. Standard Deviation = \(\sqrt{npq}\) = \(\sqrt{50\left(0.6\right)\left(0.4\right)}\) ≈ iii.4641

The lifetime risk of developing pancreatic cancer is nigh i in 78 (1.28%). Suppose we randomly sample 200 people. Let X = the number of people who will develop pancreatic cancer.

1. What is the probability distribution for X?
2. Using the formulas, calculate the (i) mean and (ii) standard deviation of X.
3. Utilise your calculator to notice the probability that at most eight people develop pancreatic cancer
4. Is it more likely that five or six people will develop pancreatic cancer? Justify your answer numerically.

1. X ∼ B(200, 0.0128)
2. 1. Mean = np = 200(0.0128) = 2.56
   2. Standard Deviation = \(\sqrt{npq}\text{ =  }\sqrt{\text{(200)(0}\text{.0128)(0.9872)}}\approx ane.\text{5897}\)
3. Using the TI-83, 83+, 84 computer with instructions as provided in [link]:

   P(x ≤ 8) = binomcdf(200, 0.0128, 8) = 0.9988

4. P(10 = 5) = binompdf(200, 0.0128, v) = 0.0707

   P(x = half dozen) = binompdf(200, 0.0128, 6) = 0.0298

   And then P(x = v) > P(x = 6); information technology is more than likely that 5 people will develop cancer than six.

Try It

During the 2013 regular NBA season, DeAndre Hashemite kingdom of jordan of the Los Angeles Clippers had the highest field goal completion rate in the league. DeAndre scored with 61.3% of his shots. Suppose you cull a random sample of eighty shots made by DeAndre during the 2013 flavour. Let Ten = the number of shots that scored points.

1. What is the probability distribution for Ten?
2. Using the formulas, calculate the (i) mean and (ii) standard divergence of 10.
3. Utilize your estimator to find the probability that DeAndre scored with lx of these shots.
4. Find the probability that DeAndre scored with more 50 of these shots.

1. X ~ B(fourscore, 0.613)
2. 1. Mean = np = lxxx(0.613) = 49.04
   2. Standard Deviation = \(\sqrt{npq}=\sqrt{lxxx\left(0.613\right)\left(0.387\right)}\approx iv.3564\)
3. Using the TI-83, 83+, 84 calculator with instructions equally provided in [link]:

   P(x = sixty) = binompdf(lxxx, 0.613, 60) = 0.0036

4. P(ten > 50) = 1 – P(x ≤ 50) = 1 – binomcdf(fourscore, 0.613, 50) = 1 – 0.6282 = 0.3718

The following example illustrates a problem that is not binomial. It violates the condition of independence. ABC Higher has a educatee advisory commission fabricated up of ten staff members and vi students. The committee wishes to cull a chairperson and a recorder. What is the probability that the chairperson and recorder are both students? The names of all committee members are put into a box, and two names are drawn without replacement. The first name drawn determines the chairperson and the second name the recorder. There are two trials. However, the trials are not contained because the consequence of the first trial affects the outcome of the second trial. The probability of a student on the first draw is \(\frac{6}{16}\). The probability of a student on the 2nd draw is \(\frac{five}{xv}\), when the first draw selects a student. The probability is \(\frac{6}{xv}\), when the outset draw selects a staff member. The probability of drawing a student's name changes for each of the trials and, therefore, violates the condition of independence.

Endeavor It

A lacrosse squad is selecting a captain. The names of all the seniors are put into a hat, and the first 3 that are drawn will be the captains. The names are not replaced one time they are fatigued (one person cannot be ii captains). Yous want to see if the captains all play the same position. State whether this is binomial or not and state why.

This is non binomial because the names are non replaced, which means the probability changes for each time a name is fatigued. This violates the condition of independence.

References

"Admission to electricity (% of population)," The Globe Bank, 2013. Available online at http://information.worldbank.org/indicator/EG.ELC.ACCS.ZS?order=wbapi_data_value_2009%20wbapi_data_value%20wbapi_data_value-first&sort=asc (accessed May fifteen, 2015).

"Distance Education." Wikipedia. Available online at http://en.wikipedia.org/wiki/Distance_education (accessed May 15, 2013).

"NBA Statistics – 2013," ESPN NBA, 2013. Available online at http://espn.go.com/nba/statistics/_/seasontype/ii (accessed May 15, 2013).

Newport, Frank. "Americans Still Enjoy Saving Rather than Spending: Few demographic differences seen in these views other than by income," GALLUP® Economy, 2013. Available online at http://www.gallup.com/poll/162368/americans-enjoy-saving-rather-spending.aspx (accessed May xv, 2013).

Pryor, John H., Linda DeAngelo, Laura Palucki Blake, Sylvia Hurtado, Serge Tran. The American Freshman: National Norms Autumn 2011. Los Angeles: Cooperative Institutional Research Program at the Higher Education Inquiry Establish at UCLA, 2011. Also available online at http://heri.ucla.edu/PDFs/pubs/TFS/Norms/Monographs/TheAmericanFreshman2011.pdf (accessed May xv, 2013).

"The World FactBook," Central Intelligence Agency. Available online at https://www.cia.gov/library/publications/the-earth-factbook/geos/af.html (accessed May 15, 2013).

"What are the fundamental statistics about pancreatic cancer?" American Cancer Society, 2013. Available online at http://www.cancer.org/cancer/pancreaticcancer/detailedguide/pancreatic-cancer-key-statistics (accessed May 15, 2013).

Chapter Review

A statistical experiment can be classified every bit a binomial experiment if the following atmospheric condition are met:

1. In that location are a stock-still number of trials, n.
2. There are only ii possible outcomes, chosen "success" and, "failure" for each trial. The letter p denotes the probability of a success on one trial and q denotes the probability of a failure on one trial.
3. The north trials are contained and are repeated using identical weather condition.

The outcomes of a binomial experiment fit a binomial probability distribution. The random variable X = the number of successes obtained in the n independent trials. The mean of X can exist calculated using the formula μ = np, and the standard deviation is given by the formula σ = \(\text{ }\sqrt{npq}\).

Formula Review

Ten ~ B(n, p) means that the discrete random variable X has a binomial probability distribution with n trials and probability of success p.

X = the number of successes in n independent trials

northward = the number of independent trials

X takes on the values 10 = 0, i, 2, 3, …, due north

p = the probability of a success for any trial

q = the probability of a failure for whatsoever trial

p + q = 1

q = 1 – p

The hateful of X is μ = np. The standard difference of X is σ = \(\sqrt{npq}\).

Use the following information to answer the next eight exercises: The Higher Education Research Institute at UCLA collected data from 203,967 incoming first-time, full-fourth dimension freshmen from 270 4-year colleges and universities in the U.Southward. 71.iii% of those students replied that, yes, they believe that same-sex activity couples should have the correct to legal marital condition. Suppose that you randomly pick eight starting time-time, total-time freshmen from the survey. You are interested in the number that believes that same sex-couples should have the right to legal marital status.

In words, ascertain the random variable Ten.

Ten = the number that reply "yes"

X ~ _____(_____,_____)

<!– <solution id="id7617599″>
B(viii,0.713)
–>

What values does the random variable Ten take on?

0, one, ii, three, 4, five, 6, 7, 8

Construct the probability distribution function (PDF).

x	P(x)

<!– <solution id="fs-idm163794480″>

–>

On average (μ), how many would you expect to answer yes?

5.7

What is the standard deviation (σ)?

<!– <solution id="id8562245″>
1.2795
–>

What is the probability that at about five of the freshmen reply "yes"?

0.4151

What is the probability that at to the lowest degree ii of the freshmen reply "yes"?

<!– <solution id="id8562331″>
0.9990 –>

HOMEWORK

According to a recent commodity the average number of babies born with meaning hearing loss (deafness) is approximately two per 1,000 babies in a salubrious baby nursery. The number climbs to an average of thirty per 1,000 babies in an intensive intendance nursery.

Suppose that one,000 babies from healthy infant nurseries were randomly surveyed. Discover the probability that exactly two babies were built-in deaf.

<!– <solution id="eip-idm92090528″>
0.2709
–>

Employ the following information to respond the adjacent four exercises. Recently, a nurse commented that when a patient calls the medical advice line claiming to take the flu, the chance that he or she truly has the influenza (and non just a nasty cold) is simply near 4%. Of the adjacent 25 patients calling in challenge to accept the flu, we are interested in how many actually have the flu.

Ascertain the random variable and list its possible values.

X = the number of patients calling in claiming to have the flu, who really take the influenza.

X = 0, 1, 2, …25

Country the distribution of X.

<!– <solution id="eip-idm5232336″>
B(25,0.04)
–>

Find the probability that at least 4 of the 25 patients actually have the flu.

0.0165

On average, for every 25 patients calling in, how many practise you wait to have the flu?

<!– <solution id="eip-idm142934512″>
one
–>

People visiting video rental stores often hire more than i DVD at a time. The probability distribution for DVD rentals per customer at Video To Go is given [link]. There is five-video limit per customer at this store, so nobody e'er rents more than 5 DVDs.

x	P(x)
0	0.03
ane	0.50
ii	0.24
3
4	0.07
v	0.04

1. Describe the random variable 10 in words.
2. Find the probability that a customer rents iii DVDs.
3. Detect the probability that a client rents at least four DVDs.
4. Find the probability that a customer rents at most two DVDs.

1. X = the number of DVDs a Video to Go client rents
2. 0.12
3. 0.eleven
4. 0.77

A school newspaper reporter decides to randomly survey 12 students to see if they will attend Tet (Vietnamese New Yr) festivities this year. Based on by years, she knows that xviii% of students attend Tet festivities. We are interested in the number of students who will attend the festivities.

1. In words, ascertain the random variable X.
2. List the values that X may take on.
3. Give the distribution of X. X ~ _____(_____,_____)
4. How many of the 12 students exercise we expect to attend the festivities?
5. Find the probability that at most four students will attend.
6. Find the probability that more than than ii students will attend.

<!– <solution id="id17988422″>
X = the number of students who will nourish Tet.
0, 1, 2, iii, 4, 5, 6, seven, 8, 9, ten, 11, 12
X ~ B(12,0.xviii)
2.16
0.9511
0.3702 –>

Apply the following information to answer the next two exercises: The probability that the San Jose Sharks will win any given game is 0.3694 based on a 13-year win history of 382 wins out of 1,034 games played (as of a certain appointment). An upcoming monthly schedule contains 12 games.

The expected number of wins for that upcoming calendar month is:

1. 1.67
2. 12
3. \(\frac{382}{1043}\)
4. 4.43

d. 4.43

Let X = the number of games won in that upcoming month.

What is the probability that the San Jose Sharks win half dozen games in that upcoming month?

1. 0.1476
2. 0.2336
3. 0.7664
4. 0.8903

<!– <solution id="id17994552″>
a
–>

What is the probability that the San Jose Sharks win at to the lowest degree five games in that upcoming calendar month

1. 0.3694
2. 0.5266
3. 0.4734
4. 0.2305

c

A student takes a x-question true-fake quiz, but did not study and randomly guesses each answer. Find the probability that the student passes the quiz with a grade of at least 70% of the questions correct.

<!– <solution id="eip-id1171731523478″>
X = number of questions answered correctly
X ~ B(ten, 0.v)
We are interested in AT LEAST 70% of ten questions right. 70% of ten is seven. We want to find the probability that X is greater than or equal to seven. The consequence "at to the lowest degree 7" is the complement of "less than or equal to six".
Using your calculator's distribution menu: 1 – binomcdf(ten, .5, half-dozen) gives 0.171875
The probability of getting at least 70% of the ten questions right when randomly guessing is approximately 0.172. –>

A pupil takes a 32-question multiple-choice exam, but did not study and randomly guesses each respond. Each question has three possible choices for the answer. Find the probability that the educatee guesses more 75% of the questions correctly.

• Ten = number of questions answered correctly
• X ~ B\(\left(\text{32, }\frac{\text{1}}{\text{3}}\correct)\)
• We are interested in More than THAN 75% of 32 questions correct. 75% of 32 is 24. We want to find P(ten > 24). The upshot "more than 24" is the complement of "less than or equal to 24."
• Using your calculator's distribution menu: 1 – binomcdf\(\left(\text{32, }\frac{\text{1}}{\text{3}},\text{ 24}\right)\)
• P(x > 24) = 0
• The probability of getting more than 75% of the 32 questions right when randomly guessing is very minor and practically cypher.

Half-dozen different colored dice are rolled. Of involvement is the number of dice that show a one.

1. In words, define the random variable Ten.
2. List the values that X may accept on.
3. Give the distribution of X. X ~ _____(_____,_____)
4. On average, how many dice would you look to show a one?
5. Observe the probability that all six dice bear witness a ane.
6. Is it more probable that three or that four dice volition bear witness a one? Utilise numbers to justify your reply numerically.

<!– <solution id="id17531763″>
X = the number of dice that show a ane
0, 1, 2, iii, iv, five, vi
X ~ B

(

six,
1
six

)

1
0.00002
three die –>

More than 96 pct of the very largest colleges and universities (more than than 15,000 total enrollments) have some online offerings. Suppose y'all randomly pick xiii such institutions. Nosotros are interested in the number that offer distance learning courses.

1. In words, define the random variable 10.
2. List the values that X may accept on.
3. Give the distribution of X. 10 ~ _____(_____,_____)
4. On average, how many schools would you wait to offering such courses?
5. Find the probability that at about ten offer such courses.
6. Is information technology more than likely that 12 or that xiii will offer such courses? Use numbers to justify your answer numerically and answer in a complete sentence.

1. X = the number of college and universities that offering online offerings.
2. 0, 1, two, …, thirteen
3. 10 ~ B(13, 0.96)
4. 12.48
5. 0.0135
6. P(ten = 12) = 0.3186 P(x = 13) = 0.5882 More probable to get 13.

Suppose that near 85% of graduating students nourish their graduation. A group of 22 graduating students is randomly chosen.

1. In words, define the random variable Ten.
2. List the values that 10 may accept on.
3. Give the distribution of 10. X ~ _____(_____,_____)
4. How many are expected to attend their graduation?
5. Notice the probability that 17 or 18 attend.
6. Based on numerical values, would you be surprised if all 22 attended graduation? Justify your answer numerically.

<!– <solution id="fs-idm62699008″>
X = the number of students who attend their graduation
0, one, 2, …, 22
X ~ B(22, 0.85)
18.7
0.3249
P(ten = 22) = 0.0280 (less than 3%) which is unusual
–>

At The Fencing Center, 60% of the fencers use the foil every bit their main weapon. Nosotros randomly survey 25 fencers at The Fencing Heart. We are interested in the number of fencers who do not use the foil equally their main weapon.

1. In words, define the random variable Ten.
2. List the values that X may take on.
3. Give the distribution of X. X ~ _____(_____,_____)
4. How many are expected to not to use the foil every bit their master weapon?
5. Find the probability that six do not use the foil equally their main weapon.
6. Based on numerical values, would yous be surprised if all 25 did not use foil equally their primary weapon? Justify your answer numerically.

1. X = the number of fencers who practise non use the foil as their master weapon
2. 0, 1, two, 3,… 25
3. Ten ~ B(25,0.40)
4. ten
5. 0.0442
6. The probability that all 25 not utilize the foil is most zero. Therefore, it would be very surprising.

Approximately viii% of students at a local high school participate in later on-school sports all four years of high school. A group of 60 seniors is randomly called. Of interest is the number who participated in subsequently-school sports all four years of loftier school.

1. In words, define the random variable 10.
2. Listing the values that X may have on.
3. Give the distribution of 10. X ~ _____(_____,_____)
4. How many seniors are expected to have participated in after-school sports all four years of loftier schoolhouse?
5. Based on numerical values, would y'all exist surprised if none of the seniors participated in after-school sports all four years of loftier schoolhouse? Justify your reply numerically.
6. Based upon numerical values, is information technology more than likely that four or that five of the seniors participated in later-schoolhouse sports all four years of high schoolhouse? Justify your respond numerically.

<!– <solution id="fs-idm58938512″>
10 = the number of high school students who participate in afterward school sports all four years of high school.
0, 1, 2, …, sixty
10 ~ B(sixty, 0.08)
4.8
Aye, P(x = 0) = 0.0067, which is a pocket-size probability
P(x = four) = 0.1873, P(x = 5) = 0.1824. More likely to get four.
–>

The chance of an IRS audit for a tax return with over 💲25,000 in income is almost two% per year. We are interested in the expected number of audits a person with that income has in a 20-year period. Assume each twelvemonth is independent.

1. In words, ascertain the random variable Ten.
2. List the values that X may take on.
3. Requite the distribution of Ten. X ~ _____(_____,_____)
4. How many audits are expected in a 20-yr period?
5. Detect the probability that a person is not audited at all.
6. Find the probability that a person is audited more than twice.

1. 10 = the number of audits in a xx-year period
2. 0, ane, ii, …, twenty
3. X ~ B(twenty, 0.02)
4. 0.four
5. 0.6676
6. 0.0071

Information technology has been estimated that merely nigh 30% of California residents have adequate convulsion supplies. Suppose you randomly survey 11 California residents. We are interested in the number who have acceptable earthquake supplies.

1. In words, define the random variable 10.
2. List the values that X may take on.
3. Give the distribution of X. 10 ~ _____(_____,_____)
4. What is the probability that at to the lowest degree eight have adequate convulsion supplies?
5. Is information technology more likely that none or that all of the residents surveyed volition have adequate convulsion supplies? Why?
6. How many residents do you await will take acceptable earthquake supplies?

<!– <solution id="id16028424″>

Ten = the number of California residents who do have adequate earthquake supplies.
0, 1, 2, 3, 4, five, 6, 7, eight, 9, x, xi
B(eleven, 0.30)
0.0043
P(x = 0) = 0.0198. P(x = eleven) = 0 or none
iii.3 –>

There are two similar games played for Chinese New year's day and Vietnamese New year's day. In the Chinese version, off-white dice with numbers 1, ii, 3, iv, 5, and six are used, along with a board with those numbers. In the Vietnamese version, off-white dice with pictures of a gourd, fish, rooster, crab, crayfish, and deer are used. The board has those six objects on it, too. We will play with bets being 💲1. The player places a bet on a number or object. The "house" rolls three die. If none of the die testify the number or object that was bet, the house keeps the 💲1 bet. If one of the dice shows the number or object bet (and the other two do non show it), the player gets dorsum his or her 💲1 bet, plus 💲1 profit. If two of the dice show the number or object bet (and the third die does not bear witness information technology), the thespian gets dorsum his or her 💲one bet, plus 💲2 profit. If all three dice show the number or object bet, the player gets back his or her 💲1 bet, plus 💲3 profit. Let X = number of matches and Y = profit per game.

1. In words, define the random variable X.
2. List the values that X may accept on.
3. Give the distribution of Ten. X ~ _____(_____,_____)
4. List the values that Y may have on. So, construct one PDF table that includes both X and Y and their probabilities.
5. Calculate the average expected matches over the long run of playing this game for the player.
6. Summate the average expected earnings over the long run of playing this game for the player.
7. Determine who has the advantage, the player or the house.

1. Ten = the number of matches
2. 0, i, ii, iii
3. X ~ B\(\left(3,\frac{1}{half dozen}\right)\)
4. In dollars: −ane, 1, 2, three
5. \(\frac{one}{two}\)
6. Multiply each Y value by the corresponding X probability from the PDF table. The answer is −0.0787. You lot lose virtually eight cents, on boilerplate, per game.
7. The house has the advantage.

According to The World Bank, just ix% of the population of Republic of uganda had admission to electricity equally of 2009. Suppose we randomly sample 150 people in Republic of uganda. Permit Ten = the number of people who accept admission to electricity.

1. What is the probability distribution for X?
2. Using the formulas, calculate the hateful and standard difference of X.
3. Employ your figurer to discover the probability that fifteen people in the sample have access to electricity.
4. Find the probability that at almost x people in the sample have access to electricity.
5. Find the probability that more than 25 people in the sample have admission to electricity.

<!– <solution id="fs-idm37905376″>
X ~ B(150,0.09)

Mean = np = 150(0.09) = 13.5
Standard Divergence =

npq

=

150(0.09)(0.91)

≈ three.5050

P(x = xv) = binompdf(150, 0.09, 15) = 0.0988
P(x ≤ x) = binomcdf(150, 0.09, 10) = 0.1987
P(ten > 25) = i – P(10 ≤ 25) = 1 – binomcdf(150, 0.09, 25) = i – 0.9991 = 0.0009
–>

The literacy rate for a nation measures the proportion of people age 15 and over that can read and write. The literacy charge per unit in Afghanistan is 28.ane%. Suppose you choose 15 people in Afghanistan at random. Let 10 = the number of people who are literate.

1. Sketch a graph of the probability distribution of X.
2. Using the formulas, calculate the (i) mean and (ii) standard deviation of X.
3. Find the probability that more than five people in the sample are literate. Is information technology is more likely that three people or four people are literate.

1. Ten ~ B(xv, 0.281)

   

2. 1. Mean = μ = np = 15(0.281) = 4.215
   2. Standard Deviation = σ = \(\sqrt{npq}\) = \(\sqrt{15\left(0.281\right)\left(0.719\right)}\) = 1.7409
3. P(x > 5) = ane – P(x ≤ 5) = 1 – binomcdf(15, 0.281, 5) = one – 0.7754 = 0.2246

   P(10 = 3) = binompdf(15, 0.281, iii) = 0.1927

   P(x = 4) = binompdf(15, 0.281, four) = 0.2259

   It is more likely that 4 people are literate that three people are.

Glossary

Binomial Experiment

a statistical experiment that satisfies the following three conditions:

1. At that place are a fixed number of trials, n.
2. At that place are only two possible outcomes, called "success" and, "failure," for each trial. The letter p denotes the probability of a success on ane trial, and q denotes the probability of a failure on ane trial.
3. The n trials are independent and are repeated using identical conditions.

Bernoulli Trials

an experiment with the following characteristics:

1. There are only two possible outcomes called "success" and "failure" for each trial.
2. The probability p of a success is the aforementioned for whatever trial (so the probability q = ane − p of a failure is the aforementioned for whatsoever trial).

Binomial Probability Distribution

a discrete random variable (RV) that arises from Bernoulli trials; at that place are a stock-still number, n, of independent trials. "Independent" means that the upshot of any trial (for case, trial i) does not touch the results of the following trials, and all trials are conducted under the aforementioned weather condition. Under these circumstances the binomial RV X is defined equally the number of successes in n trials. The notation is: 10 ~ B(north, p). The mean is μ = np and the standard departure is σ = \(\sqrt{npq}\). The probability of exactly 10 successes in n trials is

P(10 = x) = \(\left(\begin{array}{l}n\\ x\end{assortment}\correct)\)p ^x q ^{n − x}.

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Source: http://pressbooks-dev.oer.hawaii.edu/introductorystatistics/chapter/binomial-distribution/

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