# 4.3 Binomial distribution  (Page 3/29)

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## Try it

About 32% of students participate in a community volunteer program outside of school. If 30 students are selected at random, find the probability that at most 14 of them participate in a community volunteer program outside of school. Use the TI-83+ or TI-84 calculator to find the answer.

P ( x ≤ 14) = 0.9695

In the 2013 Jerry’s Artarama art supplies catalog, there are 560 pages. Eight of the pages feature signature artists. Suppose we randomly sample 100 pages. Let X = 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 two pages feature signature artists
2. the probability that at most six pages feature 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, 4, 5, 6, 7, 8
2. X ~ B $\left(100,\frac{8}{560}\right)$
1. P ( x = 2) = binompdf $\left(100,\frac{8}{560},2\right)$ = 0.2466
2. P ( x ≤ 6) = binomcdf $\left(100,\frac{8}{560},6\right)$ = 0.9994
3. P ( x >3) = 1 – P ( x ≤ 3) = 1 – binomcdf $\left(100,\frac{8}{560},3\right)$ = 1 – 0.9443 = 0.0557
1. Mean = np = (100) $\left(\frac{8}{560}\right)$ = $\frac{800}{560}$ ≈ 1.4286
2. Standard Deviation = $\sqrt{npq}$ = $\sqrt{\left(100\right)\left(\frac{8}{560}\right)\left(\frac{552}{560}\right)}$ ≈ 1.1867

## 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

#### Questions & Answers

mean is number that occurs frequently in a giving data
That places the mode and the mean as the same thing. I'd define the mean as the ratio of the total sum of variables to the variable count, and it assigns the variables a similar value across the board.
Samsicker
what is mean
what is normal distribution
What is the uses of sample in real life
pain scales in hospital
Lisa
change of origin and scale
3. If the grades of 40000 students in a course at the Hashemite University are distributed according to N(60,400) Then the number of students with grades less than 75 =*
If a constant value is added to every observation of data, then arithmetic mean is obtained by
sum of AM+Constnt
Fazal
data can be defined as numbers in context. suppose you are given the following set of numbers 18,22,22,20,19,21
what are data
what is mode?
what is statistics
Natasha
statistics is a combination of collect data summraize data analyiz data and interprete data
Ali
what is mode
Natasha
what is statistics
It is the science of analysing numerical data in large quantities, especially for the purpose of inferring proportions in a whole from those in a representative sample.
Bernice
history of statistics
statistics was first used by?
Terseer
if a population has a prevalence of Hypertension 5%, what is the probability of 4 people having hypertension from 8 randomly selected individuals?
Carpet land sales persons average 8000 per weekend sales Steve qantas the firm's vice president proposes a compensation plan with new selling incentives Steve hopes that the results of a trial selling period will enable him to conclude that the compensation plan increases the average sales per sales
Supposed we have Standard deviation 1.56, mean 6.36, sample size 25 and Z-score 1.96 at 95% confidence level, what is the confidence interval?