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
outlier is an observation point that is distant from other observations.
Gidigah
what is its effect on mode?
Usama
Outlier have little effect on the mode of a given set of data.
Gidigah
How can you identify a possible outlier(s) in a data set.
Daniel
The best visualisation method to identify the outlier is box and wisker method or boxplot diagram. The points which are located outside the max edge of wisker(both side) are considered as outlier.
Akash
@Daniel Adunkwah - Usually you can identify an outlier visually. They lie outside the observed pattern of the other data points, thus they're called outliers.
sorry question a bit unclear...do you mean how do you analyze quantitative data? If yes, it depends on the specific question(s) you set in the beginning as well as on the data you collected. So the method of data analysis will be dependent on the data collecter and questions asked.
Bheka
how to solve for degree of freedom
saliou
Quantitative data is the data in numeric form. For eg: Income of persons asked is 10,000. This data is quantitative data on the other hand data collected for either make or female is qualitative data.
Rohan
*male
Rohan
Degree of freedom is the unconditionality. For example if you have total number of observations n, and you have to calculate variance, obviously you will need mean for that. Here mean is a condition, without which you cannot calculate variance. Therefore degree of freedom for variance will be n-1.
Rohan
data that is best presented in categories like haircolor, food taste (good, bad, fair, terrible) constitutes qualitative data
Bheka
vegetation types (grasslands, forests etc) qualitative data
Bheka
I don't understand how you solved it can you teach me
how we make a classes of this(170.3,173.9,171.3,182.3,177.3,178.3,174.175.3)
Sarbaz
6.5
phoenix
11
Shakir
7.5
Ron
why is always lower class bundry used
Caleb
Assume you are in a class where quizzes are 20% of your grade, homework is 20%, exam _1 is 15%,exam _2 is 15%, and the final exam is 20%.Suppose you are in the fifth week and you just found out that you scored a 58/63 on the fist exam. You also know that you received 6/9,8/10,9/9 on the first
You have to plot the class midpoint and the frequency
Wydny
ok so you use those two to draw the histogram right.
Amford
yes
Wydny
ok can i be a friend so you can be teaching me small small
Amford
how do you calculate cost effectiveness?
George
Hi everyone, this is a very good statistical group and am glad to be part of it. I'm just not sure how did I end up here cos this discussion just popes on my screen so if I wanna ask something in the future, how will I find you?
Bheka
To make a histogram, follow these steps:
On the vertical axis, place frequencies. Label this axis "Frequency".
On the horizontal axis, place the lower value of each interval. ...
Draw a bar extending from the lower value of each interval to the lower value of the next interval.
I want to test linear regression data such as maintenance fees vs house size. Can I use R square, F test to test the relationship?
Is the good condition of R square greater than 0.5