# 4.3 Binomial distribution -- rrc math 1020 v2015rb  (Page 4/30)

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## Try it

According to a Gallup poll, 60% of American adults prefer saving over spending. Let X = the number of American adults out of a random sample of 50 who prefer saving to spending.

1. What is the probability distribution for X ?
2. Use your calculator to find the following probabilities:
1. the probability that 25 adults in the sample prefer saving over spending
2. the probability that at most 20 adults prefer saving
3. the probability that more than 30 adults prefer saving
3. Using the formulas, calculate the (i) mean and (ii) standard deviation of X .
1. X B (50, 0.6)
2. Using the TI-83, 83+, 84 calculator with instructions as provided in [link] :
1. P ( x = 25) = binompdf(50, 0.6, 25) = 0.0405
2. P ( x ≤ 20) = binomcdf(50, 0.6, 20) = 0.0034
3. P ( x >30) = 1 - binomcdf(50, 0.6, 30) = 1 – 0.5535 = 0.4465
1. Mean = np = 50(0.6) = 30
2. Standard Deviation = $\sqrt{npq}$ = $\sqrt{50\left(0.6\right)\left(0.4\right)}$ ≈ 3.4641

The lifetime risk of developing pancreatic cancer is about one 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. Use your calculator to find 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)
1. Mean = np = 200(0.0128) = 2.56
2. Standard Deviation =
2. Using the TI-83, 83+, 84 calculator with instructions as provided in [link] :
P ( x ≤ 8) = binomcdf(200, 0.0128, 8) = 0.9988
3. P ( x = 5) = binompdf(200, 0.0128, 5) = 0.0707
P ( x = 6) = binompdf(200, 0.0128, 6) = 0.0298
So P ( x = 5)> P ( x = 6); it is more likely that five people will develop cancer than six.

## Try it

During the 2013 regular NBA season, DeAndre 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 choose a random sample of 80 shots made by DeAndre during the 2013 season. Let X = the number of shots that scored points.

1. What is the probability distribution for X ?
2. Using the formulas, calculate the (i) mean and (ii) standard deviation of X .
3. Use your calculator to find the probability that DeAndre scored with 60 of these shots.
4. Find the probability that DeAndre scored with more than 50 of these shots.
1. X ~ B (80, 0.613)
1. Mean = np = 80(0.613) = 49.04
2. Standard Deviation = $\sqrt{npq}=\sqrt{80\left(0.613\right)\left(0.387\right)}\approx 4.3564$
2. Using the TI-83, 83+, 84 calculator with instructions as provided in [link] :
P ( x = 60) = binompdf(80, 0.613, 60) = 0.0036
3. P ( x >50) = 1 – P ( x ≤ 50) = 1 – binomcdf(80, 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 College has a student advisory committee made up of ten staff members and six students. The committee wishes to choose 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 independent because the outcome 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 second draw is $\frac{5}{15}$ , when the first draw selects a student. The probability is $\frac{6}{15}$ , when the first 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.

#### Questions & Answers

find the 15th term of the geometric sequince whose first is 18 and last term of 387
I know this work
salma
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
hmm well what is the answer
Abhi
how do they get the third part x = (32)5/4
can someone help me with some logarithmic and exponential equations.
sure. what is your question?
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
hmm
Abhi
is it a question of log
Abhi
🤔.
Abhi
I rally confuse this number And equations too I need exactly help
salma
But this is not salma it's Faiza live in lousvile Ky I garbage this so I am going collage with JCTC that the of the collage thank you my friends
salma
Commplementary angles
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Sherica
im all ears I need to learn
Sherica
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Tamia
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Uday
hi
salma
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a perfect square v²+2v+_
kkk nice
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Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
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China
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many many of nanotubes
Porter
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Cesar
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In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
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anybody can imagine what will be happen after 100 years from now in nano tech world
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silver nanoparticles could handle the job?
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not now but maybe in future only AgNP maybe any other nanomaterials
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Hello
Uday
I'm interested in Nanotube
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this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
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can nanotechnology change the direction of the face of the world
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
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