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.
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.
What values does
x take on?
What is the probability distribution? Find the following probabilities:
the probability that two pages feature signature artists
the probability that at most six pages feature signature artists
the probability that more than three pages feature signature artists.
Using the formulas, calculate the (i) mean and (ii) standard deviation.
x = 0, 1, 2, 3, 4, 5, 6, 7, 8
X ~
B$\left(100,\frac{8}{560}\right)$
P (
x = 2) = binompdf
$\left(100,\frac{8}{560},2\right)$ = 0.2466
P (
x ≤ 6) = binomcdf
$\left(100,\frac{8}{560},6\right)$ = 0.9994
P (
x >3) = 1 –
P (
x ≤ 3) = 1 – binomcdf
$\left(100,\frac{8}{560},3\right)$ = 1 – 0.9443 = 0.0557
Mean =
np = (100)
$\left(\frac{8}{560}\right)$ =
$\frac{800}{560}$ ≈ 1.4286
Standard Deviation =
$\sqrt{npq}$ =
$\sqrt{(100)\left(\frac{8}{560}\right)\left(\frac{552}{560}\right)}$ ≈ 1.1867
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.
What is the probability distribution for
X ?
Use your calculator to find the following probabilities:
the probability that 25 adults in the sample prefer saving over spending
the probability that at most 20 adults prefer saving
the probability that more than 30 adults prefer saving
Using the formulas, calculate the (i) mean and (ii) standard deviation of
X .
X ∼
B (50, 0.6)
Using the TI-83, 83+, 84 calculator with instructions as provided in
[link] :
P (
x = 25) = binompdf(50, 0.6, 25) = 0.0405
P (
x ≤ 20) = binomcdf(50, 0.6, 20) = 0.0034
P (
x >30) = 1 - binomcdf(50, 0.6, 30) = 1 – 0.5535 = 0.4465
Mean =
np = 50(0.6) = 30
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.
What is the probability distribution for
X ?
Using the formulas, calculate the (i) mean and (ii) standard deviation of
X .
Use your calculator to find the probability that at most eight people develop pancreatic cancer
Is it more likely that five or six people will develop pancreatic cancer? Justify your answer numerically.
X ∼
B (200, 0.0128)
Mean =
np = 200(0.0128) = 2.56
Standard Deviation =
$\sqrt{npq}\text{=}\sqrt{\text{(200)(0}\text{.0128)(0.9872)}}\approx 1.\text{5897}$
Using the TI-83, 83+, 84 calculator with instructions as provided in
[link] :
P (
x ≤ 8) = binomcdf(200, 0.0128, 8) = 0.9988
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.
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.
What is the probability distribution for
X ?
Using the formulas, calculate the (i) mean and (ii) standard deviation of
X .
Use your calculator to find the probability that DeAndre scored with 60 of these shots.
Find the probability that DeAndre scored with more than 50 of these shots.
X ~
B (80, 0.613)
Mean =
np = 80(0.613) = 49.04
Standard Deviation =
$\sqrt{npq}=\sqrt{80(0.613)(0.387)}\approx 4.3564$
Using the TI-83, 83+, 84 calculator with instructions as provided in
[link] :
P (
x = 60) = binompdf(80, 0.613, 60) = 0.0036
P (
x >50) = 1 –
P (
x ≤ 50) = 1 – binomcdf(80, 0.613, 50) = 1 – 0.6282 = 0.3718
Sixty-four third year high school students were given a standardized reading comprehension test. The mean and standard deviation obtained were 52.27 and 8.24, respectively. Is the mean significantly different from the population mean of 50? Use the 5% level of significance.
Simple Bar Chart is a Diagram which shows the data values in form of horizontal bars. It shows categories along y-axis and values along x-axis. The x-axis displays above the bars and y-axis displays on left of the bars with the bars extending to the right side according to their values.
add all the data and divide by the number of data sets. For example, if test scores were 70, 60, 70, 80 the total is 280 and the total data sets referred to as N is 4. Therfore the mean or arthritmatic average is 70. I hope this helps.
Jim
*Tan A - Tan B = sin(A-B)/CosA CosB ...
*2sinQ/Cos 3Q = tan 3Q - tan Q