# 5.3 Binomial distribution - university of calgary - base content  (Page 3/30)

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It has been stated that about 41% of adult workers have a high school diploma but do not pursue any further education. If 20 adult workers are randomly selected,

1. find the probability that 12 of them have a high school diploma but do not pursue any further education.
2. find the probability that at most 12 of them have a high school diploma but do not pursue any further education.
3. Find how many adult workers do you expect to have a high school diploma but do not pursue any further education.

Let X = the number of workers who have a high school diploma but do not pursue any further education.

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

1. $P(\mathrm{X=12})=\left(\begin{array}{c}\mathrm{20}\\ \mathrm{12}\end{array}\right)\mathrm{0.41}^{\mathrm{12}}\mathrm{0.59}^{8}=\mathrm{0.04173}$
2. P ( x ≤ 12) = 0.9738. (calculator or computer)

## Using ti-83,83+,84,84+ calculator

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

To calculate ( x = value): binompdf( n , p , number) if "number" is left out, the result is the binomial probability table.
To calculate P ( x ≤ value): binomcdf( n , p , number) if "number" is left out, the result is the cumulative binomial probability table.
For this problem: After you are in 2 nd DISTR, arrow down to binomcdf. Press ENTER. Enter 20,0.41,12). The result is P ( x ≤ 12) = 0.9738.

## Note

If you want to find P ( x = 12), use the pdf (binompdf). If you want to find P ( x >12), use 1 - binomcdf(20,0.41,12).

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

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

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

c. The number of adult workers that you expect to have a high school diploma but not pursue any further education is the mean, μ = np = (20)(0.41) = 8.2.

The formula for the variance is σ 2 = npq . The standard deviation is σ = $\sqrt{npq}$ .
σ = $\sqrt{\left(20\right)\left(0.41\right)\left(0.59\right)}$ = 2.20.

## Try it

About 32% of students participate in a community volunteer program outside of school. If 30 students are selected at random,

1. find the probability that 14 of them participate in a community volunteer program outside of school.
2. find the probability that at most 14 of them participate in a community volunteer program outside of school. Use the calculator or computer to find the answer.

1. $P(\mathrm{X=14})=\left(\begin{array}{c}\mathrm{30}\\ \mathrm{14}\end{array}\right)\mathrm{0.32}^{\mathrm{14}}\mathrm{0.68}^{\mathrm{16}}=\mathrm{0.03588}$
2. 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

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