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.
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.
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
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.
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.
Source:
OpenStax, Introduction to statistics i - stat 213 - university of calgary - ver2015revb. OpenStax CNX. Oct 21, 2015 Download for free at http://legacy.cnx.org/content/col11874/1.3
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