Answer:
Explanation below.
Step-by-step explanation:
For this case we have the following dataset:
46, 16, 41, 26, 22 ,33, 30, 22 ,36, 34,
63, 21, 26, 18, 27, 24, 31, 38, 26, 55,
31, 47, 27, 43 ,35, 22 ,64,40, 58, 20,
49, 37, 53, 25, 29, 32, 23, 49, 39, 40,
24, 56, 30, 51, 21, 45, 27, 34, 47, 35
So we have 50 values. The first step on this case would be order the dataset on increasing way and we got:
16, 18, 20, 21, 21, 22, 22, 22, 23, 24,
24, 25, 26, 26, 26, 27, 27, 27, 29, 30
30, 31, 31, 32, 33, 34, 34, 35, 35, 36,
37, 38, 39, 40, 40, 41, 43, 45, 46, 47,
47, 49, 49, 51, 53, 55, 56, 58, 63, 64
We can find the range for this dataset like this:
[tex] Range = Max-Min = 64-16 =48[/tex]
Then since we need 7 classes we can find the length for each class doing this:
[tex] W = \frac{48}{7}=6.86[/tex]
And now we can define the classes like this and counting how many observations lies on each interval we got the frequency:
Class Frequency Midpoint RF CF
________________________________________________
[16-22.86) 8 19.43 (8/50)=0.16 0.16
[22.86-29.71) 11 26.29 (11/50)=0.22 0.38
[29.71-36.57) 11 33.14 (11/50)=0.22 0.6
[36.57-43.43) 7 40.0 (7/50)=0.14 0.74
[43.43-50.29) 6 46.86 (6/50)=0.12 0.86
[50.29-57.14) 4 53.72 (4/50)=0.08 0.94
[57.14-64] 3 60.57 (3/50)=0.06 1.0
________________________________________________
Total 50 1.00
RF= Relative frequency. CF= Cumulative frequency
The relative frequency was calculated as the individual frequency for a class divided by the total of observations (50)
The mid point is the average between the limits of the class.
And the cumulative frequency is calculated adding the relative frequencies for each class.
Suppose two dice, one orange and one blue, are rolled. Define the following events: A: The product of the two numbers that show is 12 B: The number on the orange die is strictly larger than the number on the blue die. C: The sum of the numbers is divisible by four. D: The number on the orange die is either 1 or 3
Answer:
A = {2,6; 6,2; 3,4; 4,3}
B = {6,5; 6,4; 6,3; 6,2; 6,1; 5,4; 5,3; 5,2; 5,1; 4,3; 4,2; 4,1; 3,2; 3,1; 2,1}
C = {1,3; 3,1; 2,2; 3,5; 5,3; 4,4; 6,6}
D = {1,1; 1,2; 1,3; 1,4; 1,5; 1,6; 3,1; 3,2; 3,3; 3,4; 3,5; 3,6}
Step-by-step explanation:
Let the pair (O,B) be the number rolled on the orange and blue dice respectively.
For event A: The product of the two numbers that show is 12, the sample space is:
A = {2,6; 6,2; 3,4; 4,3}
For event B: The number on the orange die is strictly larger than the number on the blue die, the sample space is:
B = {6,5; 6,4; 6,3; 6,2; 6,1; 5,4; 5,3; 5,2; 5,1; 4,3; 4,2; 4,1; 3,2; 3,1; 2,1}
For event C: The sum of the numbers is divisible by four, the sum must be 4, 8 or 12 and the sample space is:
C = {1,3; 3,1; 2,2; 3,5; 5,3; 4,4; 6,6}
For event D: The number on the orange die is either 1 or 3, the sample space is:
D = {1,1; 1,2; 1,3; 1,4; 1,5; 1,6; 3,1; 3,2; 3,3; 3,4; 3,5; 3,6}
A missile protection system consists of n radar sets operating independently, each with a probability of .9 of detecting a missile entering a zone that is covered by all of the units.
a If n = 5 and a missile enters the zone, what is the probability that exactly four sets detect the missile? At least one set?
b How large must n be if we require that the probability of detecting a missile that enters the zone be .999?
Answer:
a. probability that exactly four sets detect the missile is 0.06561
probability that at least 1 set detect the missile is 0.99999
b. n = 3
Step-by-step explanation:
a. The probability that exactly 4 sets with probability of detection being 0.9 and 1 set fail with probability of 1 - 0.9 = 0.1 is
0.9*0.9*0.9*0.9*0.1 = 0.06561
The probability of having at least 1 set detect the missile is the inverse of the probability of having none of the set detecting the missile, which means all of the set fail to detect the missile, which is
0.1*0.1*0.1*0.1*0.1 = 0.00001
So the probability that at least 1 set detect the missile is
1 - 0.00001 = 0.99999
b. For the system to have a success rate of 0.999, this means at least 1 radar could detect the missile with probability of 0.999, which means all of them can fail with probability of 0.001. For this to happen:
[tex]0.1^n = 0.001[/tex]
[tex](10^{-1})^n = 10^{-3}[/tex]
[tex]10^{-1n} = 10^{-3}[/tex]
[tex]-n = -3[/tex]
[tex]n = 3[/tex]
You need 3 radars
The probability that exactly four out of five radar sets detect a missile is about 0.33, while the probability that at least one set detects it is almost 1 (0.99999). In order to achieve a detection probability of .999 or higher, there should be at least 11 radar sets.
Explanation:The subject matter of this question is in the realm of probability and statistics, specifically binomial distributions. Probability is the measure of the likelihood that an event will occur in a random experiment.
a) To find the probability that exactly four sets detect the missile, we use the formula for a binomial probability, that is P(X=k) = C(n, k) * (p^k) * ((1-p)^(n-k)), where n is the number of trials, k is the number of successes, p is the probability of success, and C(n, k) is the number of combinations of n items taken k at a time. So, the probability is C(5, 4) * (0.9^4) * ((1-0.9)^(5-4)) = 0.32805. The probability that at least one set detects the missile equals to 1 minus the probability that none of the sets detect the missile, which is 1-(0.1^5) = 0.99999.
b) In order to achieve a probability of at least .999 of detecting a missile, we'd need to solve the inequality 1-(1-p)^n >= .999 for n. This yields n as greater than or equal to log(.001)/log(.1), which rounded up to the nearest whole number is 11.
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For each initial value problem, determine whether Picard's Theorem can be used to show the existence of a unique solution in an open interval containing t = 0. Justify your answer.
(a) y' = ty4/3, y(0) = 0
(b) y' = tył/3, y(0) = 0
(c) y' = tył/3, y(0) = 1
Answer:
Part a: [tex]f , \, f_y[/tex] is continuous at the initial value (0,0) so due to Picardi theorem there exists an interval such that the IVP has a unique solution.
Part b: [tex]f_y[/tex] is not continuous at the initial value (0,0) so due to Picardi theorem there does not exist an interval such that the IVP has a unique solution.
part c: [tex]f , \, f_y[/tex] is continuous at the initial value (0,1) so due to Picardi theorem there exists an interval such that the IVP has a unique solution.
Step-by-step explanation:
Part a
as [tex]y^{' }=ty^{4/3}[/tex]
Let
[tex]f(t,y)=ty^{4/3}[/tex]
Now derivative wrt y is given as
[tex]f_y=\frac{4}{3}ty^{1/3}[/tex]
Finding continuity via the initial value
[tex]f[/tex] is continuous on [tex]R^2[/tex] also [tex]f_y[/tex] is also continuous on [tex]R^2[/tex]
Also
[tex]f , \, f_y[/tex] is continuous at the initial value (0,0) so due to Picardi theorem there exists an interval such that the IVP has a unique solution.
Part b
as [tex]y^{' }=ty^{1/3}[/tex]
Let
[tex]f(t,y)=ty^{1/3}[/tex]
Now derivative wrt y is given as
[tex]f_y=\frac{1}{3}ty^{-2/3}[/tex]
Finding continuity via the initial value
[tex]f[/tex] is continuous on [tex]R^2[/tex] also [tex]f_y[/tex] is also continuous on [tex]R^2[/tex]
Also
[tex]f_y[/tex] is not continuous at the initial value (0,0) so due to Picardi theorem there does not exist an interval such that the IVP has a unique solution.
Part c
as [tex]y^{' }=ty^{1/3}[/tex]
Let
[tex]f(t,y)=ty^{1/3}[/tex]
Now derivative wrt y is given as
[tex]f_y=\frac{1}{3}ty^{-2/3}[/tex]
Finding continuity via the initial value
[tex]f[/tex] is continuous on [tex]R^2[/tex] also [tex]f_y[/tex] is also continuous on [tex]R^2[/tex] when [tex]y\neq 0[/tex]
Also
[tex]f , \, f_y[/tex] is continuous at the initial value (0,1) so due to Picardi theorem there exists an interval such that the IVP has a unique solution.
5. Two vertices of a triangle lie at (4, 0) and (8, 0). The perimeter of the triangle is 12 units. What are all the possible locations of the third vertex? How do you know you have found them all? Can you determine which of these vertices will produce a triangle with the largest area?
Final answer:
The possible locations of the third vertex of the triangle with a perimeter of 12 units, given two vertices at (4, 0) and (8, 0), are (12, 0) and (0, 0). The vertex at (0, 0) will produce a triangle with the largest area equals to 8 square units.
Explanation:
Possible Locations of Third Vertex:
To find the possible locations of the third vertex of the triangle, we need to consider the perimeter of the triangle. Given that the perimeter is 12 units and two vertices are located at (4, 0) and (8, 0), we can deduce that the third vertex must lie on the line segment between these two points.
Calculating the Third Vertex:
Since the distance between the two given vertices is 8 - 4 = 4 units, the remaining distance to achieve a perimeter of 12 units would be 12 - 4 = 8 units.
Therefore, the possible locations of the third vertex are (4 + 8, 0) = (12, 0) or (8 - 8, 0) = (0, 0).
Determining the Largest Area:
To determine which vertex would produce a triangle with the largest area, we need to consider the base and the height of the triangle. Since both the possible vertices lie on the x-axis, the base would be the same for both triangles, which is 4 units.
The height of the triangle can be determined by finding the distance between the base and the y-coordinate of the third vertex. If the third vertex is (12, 0), the height would be 0 units. If the third vertex is (0, 0), the height would be 4 units.
Therefore, the triangle with the third vertex at (0, 0) would produce the largest area, which is equal to 1/2 * base * height = 1/2 * 4 * 4 = 8 square units.
The possible locations of the third vertex of the triangle form a line segment parallel to the x-axis and symmetric about the y-axis, extending from (4, 3) to (8, -3) or from (4, -3) to (8, 3). The triangle with the largest area will be the one where the third vertex is directly above or below the midpoint of the line segment joining the first two vertices, at either (6, 3) or (6, -3).
Given two vertices of a triangle at (4, 0) and (8, 0), the distance between them is the length of the line segment joining them, which is 8 - 4 = 4 units. Since the perimeter of the triangle is 12 units, the sum of the lengths of the other two sides must be 12 - 4 = 8 units.
Let's denote the third vertex as (x, y). The distance from (4, 0) to (x, y) is one side of the triangle, and the distance from (8, 0) to (x, y) is the other side. Using the distance formula, we have:
[tex]\[ \sqrt{(x - 4)^2 + y^2} + \sqrt{(x - 8)^2 + y^2} = 8 \][/tex]
To simplify the problem, we can look for points where the sum of the distances to the two fixed points is 8 units. Since the two given vertices are on the x-axis, the third vertex must be such that its x-coordinate is between 4 and 8 (inclusive) to form a triangle.
For the y-coordinate, we know that the sum of the distances from the third vertex to the two fixed points is 8 units. If we consider the case where y = 0, the third vertex would be on the x-axis, which would not form a triangle. Therefore, y must be non-zero.
Let's consider the two sides of the triangle formed by the third vertex and the two fixed points. The lengths of these sides can be thought of as the radii of two circles, one centered at (4, 0) and the other at (8, 0), both with a radius of 4 units (since each circle's radius is half the sum of the lengths of the two sides not on the x-axis).
The third vertex must lie on the intersection of these two circles. Since the circles are of equal radius and their centers are 4 units apart, they will intersect in two points, which will be equidistant from the line segment joining the first two vertices.
To find these points, we can set up the following system of equations based on the distance formula:
[tex]\[ (x - 4)^2 + y^2 = 16 \][/tex]
[tex]\[ (x - 8)^2 + y^2 = 16 \][/tex]
The line segment joining these two points, (6, 3) and (6, -3), represents all possible locations of the third vertex. This line segment is parallel to the x-axis and symmetric about the y-axis, and it extends from (4, 3) to (8, -3) or from (4, -3) to (8, 3), depending on which side of the x-axis the third vertex is located.
To find the vertex that produces the largest area, we note that for a given base length (which is fixed at 4 units in this case), the area of a triangle is maximized when the height is maximized. The height is maximized when the third vertex is directly above or below the midpoint of the base, which is at (6, 0). Therefore, the third vertex should be at either (6, 3) or (6, -3) to produce the triangle with the largest area. The area of this triangle is 1/2 * base * height = 1/2 * 4 * 3 = 6 square units.
We have found all possible locations of the third vertex because any point outside the line segment joining (6, 3) and (6, -3) would result in a perimeter greater than 12 units, and any point inside would result in a perimeter less than 12 units. Thus, the line segment from (4, 3) to (8, -3) or from (4, -3) to (8, 3) represents the complete set of solutions for the third vertex that satisfy the given perimeter constraint.
in a random sample of 72 adults in santa clarita, CA each person was asked if they support the death penalty. 31 adults in the sample said they dp support the death penalty. What was the sample proportion of adults in Santa CLarita that support the death penalty?Now calculate a 95% confience interval population estimate of people in Santa Clarita that support the death penalty.
Answer:
a) [tex] \hat p = \frac{X}{n}= \frac{31}{72}= 0.431[/tex]
b) [tex]0.431 - 1.96 \sqrt{\frac{0.431(1-0.431)}{72}}=0.317[/tex]
[tex]0.431 + 1.96 \sqrt{\frac{0.431(1-0.431)}{72}}=0.545[/tex]
And the 95% confidence interval would be given (0.317;0.545).
Step-by-step explanation:
Part a
The best estimator for the population proportion is the sample proportion given by:
[tex] \hat p = \frac{X}{n}= \frac{31}{72}= 0.431[/tex]
Where X represent the adults in the sample that support the death penalty and n the sample size selected
Part b
The confidence interval would be given by this formula
[tex]\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]
For the 95% confidence interval the value of [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2=0.025[/tex], with that value we can find the quantile required for the interval in the normal standard distribution.
[tex]z_{\alpha/2}=1.96[/tex]
And replacing into the confidence interval formula we got:
[tex]0.431 - 1.96 \sqrt{\frac{0.431(1-0.431)}{72}}=0.317[/tex]
[tex]0.431 + 1.96 \sqrt{\frac{0.431(1-0.431)}{72}}=0.545[/tex]
And the 95% confidence interval would be given (0.317;0.545).
The sample proportion is approximately 0.431, and the 95% confidence interval for the population proportion is (0.317, 0.545), providing a range for the likely proportion of adults in Santa Clarita who support the death penalty based on the sample data.
In a random sample of 72 adults in Santa Clarita, California, the sample proportion ([tex]\hat{p}[/tex]) of adults who support the death penalty is calculated using the formula [tex]\hat{p}[/tex] = X/n, where X is the number of adults in the sample supporting the death penalty (31), and n is the sample size (72).
In this case, [tex]\hat{p}[/tex] = 31/72 ≈ 0.431, representing the estimated proportion of adults in Santa Clarita supporting the death penalty.
To construct a 95% confidence interval for the population proportion, the formula for the confidence interval is employed:
Confidence Interval = [tex]\left( \hat{p} - Z \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}, \hat{p} + Z \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \right)[/tex]
Here, Z is the z-score associated with a 95% confidence interval, which is approximately 1.96. Substituting the given values into the formula, the confidence interval is calculated as (0.317, 0.545), indicating that we can be 95% confident that the true proportion of adults in Santa Clarita supporting the death penalty lies within this range.
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Remington needs at least 3,000 to buy used car .He already has 1,800 . If he saves $50 per week , write and solve and inequality to find out how many weeks he must save to buy the car . Interpret the solution
Answer: $100
good luck!!
Find a parametrization of the line in which the planes x + y + z = -6 and y + z = -8 intersect.
Find the parametrization of the line. Use a point with z = 0 on the line to determine the parametrization.
Answer:
L(x,y) = (2,-8,0) + (0,-1,1)*t
Step-by-step explanation:
for the planes
x + y + z = -6 and y + z = -8
the intersection can be found subtracting the equation of the planes
x + y + z - ( y + z ) = -6 - (-8)
x= 2
therefore
x=2
z=z
y= -8 - z
using z as parameter t and the point (2,-8,0) as reference point , then
x= 2
y= -8 - t
z= 0 + t
another way of writing it is
L(x,y) = (2,-8,0) + (0,-1,1)*t
Final answer:
The parametrization of the line where the planes x + y + z = -6 and y + z = -8 intersect is found by solving the equations together and using a point with z = 0. This leads to parametric equations x(t) = 2, y(t) = -8 - t, and z(t) = t.
Explanation:
To find a parametrization of the line in which the planes x + y + z = -6 and y + z = -8 intersect, we first solve these two equations together to find the relationship between x, y, and z. Since both equations involve y and z, we can set them equal to isolate x.
1. Subtract the second equation from the first to isolate x: x = 2.
2. Using the second equation y + z = -8, we express y in terms of z: y = -8 - z.
Now, to use a point with z = 0 to determine the parametrization, we plug z = 0 into our equations. This gives us x = 2 and y = -8 for the point (2, -8, 0).
With z as our parameter t, the parametrization of the line can be given as x = 2, y = -8 - t, and z = t. Therefore, the parametric equations describing the intersection line are x(t) = 2, y(t) = -8 - t, and z(t) = t.
Compact and oversized tires are produced on two different machines. The table shows the number of each type of tire produced, y, depending on the number of hours, x, the machines work.
Answer:
OPTION B: 57x - 3
Step-by-step explanation:
Total tires produced is the sum of number of over-sized tires and the number of compact tires.
So, we have when x = 1, tires, y = 23 + 31 = 54
When x = 2, total tires y = 48 + 63 = 111
When x = 3, total tires y = 73 + 95 = 168
When x = 4, total tires y = 98 + 127 = 225
We can substitute the options to check which one will be the correct representation.
Option A: 55x - 1
When x = 1, y = 55(1) - 1 = 54
When x = 2, y = 55(2) - 1 = 109 [tex]$ \ne $[/tex] 111
So, this option is eliminated.
Option B: 57x - 1
When x = 1, y = 57(1) - 3 = 54
When x = 2, y = 57(2) - 3 = 111
When x = 3, y = 57(3) - 3 = 168
When x = 4, y = 57(4) - 3 = 225
These values exactly match with the values from the table. So, Option B is the right answer.
Option C: 110x - 2
When x = 1, y = 110(1) - 2 = 108 [tex]$ \ne $[/tex] 54
Hence, this option is incorrect.
Option D: 114x - 6
When x = 1, y = 114(1) - 6 [tex]$ \ne $[/tex] 54
Hence, this option is also eliminated.
Option B is the required answer.
Time value of money calculations can be solved using a mathematical equation, a financial calculator, or a spreadsheet. Which of the following equations can be used to solve for the present value of a perpetuity? PMT x {1 – [1 / (1+r)n1+rn ]} PV x (1+r)n1+rn FV / (1+r)n1+rn PMTr
The formula for the present value (PV) of a perpetuity is \[PV = \frac{FV}{(1 + r)^n}\]. Here option C is correct.
The formula for calculating the present value (PV) of a perpetuity is given by:
\[PV = \frac{FV}{(1 + r)^n}\]
Where:
PV (Present Value) is what we want to find.
FV (Future Value) is the fixed payment that will continue indefinitely.
r (Discount Rate) represents the interest rate or required rate of return.
n represents the number of time periods (infinite in the case of a perpetuity).
This formula takes into account the infinite nature of the perpetuity and discounts future cash flows to their equivalent value in today's dollars, considering the time value of money. The discount factor \(\frac{1}{(1 + r)^n}\) ensures that the cash flows in the future are worth less in present terms. Therefore, option C is correct.
Complete question:
Which of the following equations can be used to solve for the present value (PV) of a perpetuity?
A) \(PV = PMT \cdot \left(1 - \frac{1}{{(1 + r)^{n(1+r)}}}\right)\)
B) \(PV = PV_0 \cdot (1 + r)^n\)
C) \(PV = \frac{FV}{{(1 + r)^n}}\)
D) \(PV = PMT \cdot \frac{1 - \left(\frac{1}{{(1 + r)^{n(1+r)}}}\right)}{r}\)
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Astronomers treat the number of stars in a given volume of space as a Poisson random variable. The density in the Milky Way Galaxy in the vicinity of our solar system is one star per 16 cubic light years.
How many cubic light years of space must be studied so that the probability of one or more stars exceeds 0.94?[Round your answer to the nearest integer.]
Answer:
t = 45 cubic light years to find a star with this certainty.
Step-by-step explanation:
The Poisson random probability equation is given by:
[tex]P(k events in interval t)=\frac{(\lambda t)^{k}e^{-\lambda t}}{k!}[/tex]
λ is the density (1/16 star/cubic light years)t is the parameter in cubic light yearsWe can use the next equation to quantify how many cubic light years of space must be studied so that the probability of one or more stars exceeds 0.94.
[tex]P(k\ge 1) \ge 0.94[/tex]
[tex]1-f(0)=1-\frac{(\frac{1}{16}*t)^{0}e^{-\frac{1}{16}*t}}{0!}=1-e^{-\frac{1}{16}*t}} \ge 0.94[/tex]
So, here we just need to solve it for t:
[tex]1-e^{-\frac{1}{16}*t}} \ge 0.94[/tex]
[tex]e^{-\frac{1}{16}*t}} \ge 0.06[/tex]
[tex]ln(e^{-\frac{1}{16}*t}}) \ge ln(0.06)[/tex]
[tex]-\frac{1}{16}*t \ge -2.8[/tex]
[tex]t \ge 44.8[/tex]
Therefore t = 45 cubic light years to find a star with this certainty.
I hope it helps you!
To find the volume of space where the probability of encountering one or more stars exceeds 0.94 in the Milky Way, use the inverse of the cumulative distribution function of the Poisson distribution, with the given star density of one per 16 cubic light years, and solve for the volume V such that the probability of zero stars is below 0.06.
In order to solve the problem of how many cubic light years of space must be studied so that the probability of one or more stars exceeds 0.94, we need to use the properties of the Poisson distribution. The Poisson distribution is often used for modeling the number of events in fixed intervals of time or space under certain conditions. If we let λ denote the average number of stars in a given volume, then for our Milky Way Galaxy in the vicinity of our solar system where the density is one star per 16 cubic light years, λ is 1/16 per cubic light year. The probability of finding no stars in a volume V is given by e^{-λV}. To find when the probability of one or more stars exceeds 0.94, we need to find V such that the probability of finding no stars is less than 0.06 (which is 1 - 0.94).
Let's calculate the volume V:
We first solve the inequality e^{-λV} < 0.06 for V.
Taking the natural logarithm of both sides gives us -λV < ln(0.06).
We then solve for V: V > ln(0.06) / -λ.
Substituting λ (1/16) gives us V > ln(0.06) / -(1/16).
Finally, calculate the numeric value and round to the nearest integer to find the minimum volume V that meets the requirement.
By performing these calculations, you can find how many cubic light years of space must be studied so that the probability of one or more stars exceeds 0.94.
Find a system of two equations in two variables, x1 and x2, that has the solution set given by the parametric representation x1 = t and x2 = 5t − 6, where t is any real number. (Enter your answer as a comma-separated list of equations.)
Answer:
The required system of equations to the given parametric equations are:
5x1 - x2 = 6
x1 + x2 = -6
Step-by-step explanation:
Given the parametric equations:
x1 = t
x2 = -6 + 5t
Eliminating the parameter t, we obtain one of the equations of a system in two variables, x1 and x2 that has the solution set given by the parametric equations.
Doing that, we have:
5x1 - x2 = 6
Again a second equation can be a linear combination of x1 and x2
x1 + x2 = -6 + 6t
x1 + x2 = -6 (putting t=0)
And they are the required equations.
Use the null hypothesis H0 : μ = 98.6, alternative hypothesis Ha: μ < 98.6, and level of significance α = 0.05. 98 99.6 97.8 97.6 98.7 98.4 98.9 97.1 99.2 97.4 99.1 96.9 98.8 99.9 96.8 97 98.7 97.6 98.7 98.2 whats the t-score?
Answer:
The t-score is -1.8432
Step-by-step explanation:
We are given the following in the question:
98, 99.6, 97.8, 97.6, 98.7, 98.4, 98.9, 97.1, 99.2, 97.4, 99.1, 96.9, 98.8, 99.9, 96.8, 97, 98.7, 97.6, 98.7, 98.2
Formula:
[tex]\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}[/tex]
where [tex]x_i[/tex] are data points, [tex]\bar{x}[/tex] is the mean and n is the number of observations.
[tex]Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}[/tex]
[tex]Mean =\displaystyle\frac{1964.4}{20} = 98.22[/tex]
Sum of squares of differences = 16.152
[tex]s = \sqrt{\dfrac{16.152}{49}} = 0.922[/tex]
Population mean, μ = 98.6
Sample mean, [tex]\bar{x}[/tex] = 98.22
Sample size, n = 20
Sample standard deviation, s = 0.922
First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu = 98.6\\H_A: \mu < 98.6[/tex]
Formula:
[tex]t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }[/tex]
Putting all the values, we have
[tex]t_{stat} = \displaystyle\frac{98.22 - 98.6}{\frac{0.922}{\sqrt{20}} } = -1.8432[/tex]
The t-score is -1.8432
Final answer:
The t-score is calculated using the sample mean, the population mean under the null hypothesis, the sample's standard deviation, and the sample size. For the provided data, these values lead to a computed t-score of approximately -0.4594.
Explanation:
To calculate the t-score for a set of data when testing a hypothesis, you can use the following formula:
t = (X - \/u0) / (s/√n)
Where X is the sample mean, \/u0 is the population mean according to the null hypothesis, s is the sample standard deviation, and n is the sample size. To find the t-score, you need to know the sample mean (X), the standard deviation (s), and the sample size (n), all of which are usually provided in the problem or can be calculated from the data. In this case:
X (sample mean) = 98.59µ0 (population mean under H0) = 98.6s (standard deviation) = 0.0973n (sample size) = 20 (number of data points provided)Plugging these values into the t-score formula gives us:
t = (98.59 - 98.6) / (0.0973/√20)
Perform the calculations:
t = -0.01 / (0.0973/4.4721)
t = -0.01 / 0.02176
t ≈ -0.4594
Thus, the calculated t-score is approximately -0.4594.
Would you use a sample or a census to measure each of the following? (a) The number of cans of Campbell’s soup on your local supermarket's shelf today at 6:00 p.m. (b) The proportion of soup sales last week in Boston that was sold under the Campbell's brand. (c) The proportion of Campbell’s brand soup cans in your family's pantry.
Answer:
b
Step-by-step explanation:
a) Census. It would be easy enough to count all of them.
b) Sample. It would be too costly to track each can.
c) Census. You can count them all quickly and cheaply.
What is sample space?The sample space for a given set of events is the set of all possible values the events may assume.
A) The number of cans of Campbell’s soup on your local supermarket's shelf today at 6:00 p.m.
Census. It would be easy enough to count all of them.
B) The proportion of soup sales last week in Boston that was sold under the Campbell's brand.
Sample. It would be too costly to track each can.
C) The proportion of Campbell’s brand soup cans in your family's pantry.
Census. You can count them all quickly and cheaply.
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Whats an explicit rule for this? 1, -4, -9, -14, etc. Write an explicit formula for the nth term an.
Answer:
-5n + 6
Step-by-step explanation:
It goes up in the - 5 times tables and you have to add 6 to get the real answer.
Find the difference between numbers to find the n. (-5n bit)
Calculate the data value that corresponds to each of the following z-scores.
a. Final exam scores: Allison’s z-score = 2.30, μ = 74, σ = 7.
b. Weekly grocery bill: James’ z-score = –1.45, μ = $53, σ = $12.
c. Daily video game play time: Eric’s z-score = –0.79, μ = 4.00 hours, σ = 1.15 hours.
Answer:
a) 90.1
b) $35.6
c) 3.0915 hours
Step-by-step explanation:
The z-score measures how many standard deviations a score X is above or below the mean.
It is given by the following formula:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
In which [tex]\mu[/tex] is the mean and [tex]\sigma[/tex] is the standard deviaition.
In all three cases, we have to find X
a. Final exam scores: Allison’s z-score = 2.30, μ = 74, σ = 7.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]2.30 = \frac{X - 74}{7}[/tex]
[tex]X - 74 = 7*2.3[/tex]
[tex]X = 90.1[/tex]
b. Weekly grocery bill: James’ z-score = –1.45, μ = $53, σ = $12.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-1.45 = \frac{X - 53}{12}[/tex]
[tex]X - 53 = -1.45*12[/tex]
[tex]X = 35.6[/tex]
Mean and standard deviation in dollars, so the answer also in dollars.
c. Daily video game play time: Eric’s z-score = –0.79, μ = 4.00 hours, σ = 1.15 hours.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-0.79 = \frac{X - 4}{1.15}[/tex]
[tex]X - 4 = -0.79*1.15[/tex]
[tex]X = 3.0915[/tex]
Mean and standard deviation in hours, answer in hours.
Suppose the manager of a gas station monitors how many bags of ice he sells daily along with recording the highest temperature each day during the summer. The data is plotted with temperature, in degrees Fahrenheit (degree), on the horizontal axis and the number of ice bags sold on the vertical axis. One of the plotted points on the graph is (82.67). The least squares regression line for this data is y =-192.20 + 3.13x. Determine the predicted number of bags of ice sold, y, when the temperature is 82 degree F. Round the predicted value to the nearest whole number. y = ice bags Compute the residual at this temperature. Round the value to the nearest whole number. residual = ice bags
Answer:
64bags of ice were sold when the temperature is 82 degree F
Step-by-step explanation:
From the equation ; Y =-192.20 + 3.13x
where Y = predicted number of bags of ice sold
x = temperature, in degrees Fahrenheit (degree)
To find Y when x = 82
substitute the value of x in the equation given
= Y = -192.20 + 3.13(82)
Y = 64.46 and approximately, Y = 64
To Compute the residual at this temperature;
residual = 67 - 64.46= 2.53 which is approximately 3
A survey was conducted to study the relationship between the annual income of a family and the amount of money the family spends on entertainment. Data were collected from a random sample of 280 families from a certain metropolitan area. A meaningful graphical display of these data would be:
(a) side-by-side boxplots
(b) a pie chart
(c) a stemplot
(d) a scatterplot
(e) a contingency table
Answer:
The correct option is d i.e. scatter plot
Step-by-step explanation:
The correct option is d i.e. scatter plot
scatter plot will be the best option to display the variation of expenditure with respect to annual income.
on one axis annual income is used and on the other expenditure of the family. hence for a particular change in annual income, an impact on expenditure will easily be predicted.
A scatterplot would be a meaningful graphical display for studying the relationship between annual income and entertainment spending.
Explanation:A meaningful graphical display of the relationship between the annual income of a family and the amount of money the family spends on entertainment would be a scatterplot.
A scatterplot shows the relationship between two variables by plotting each data point as a dot on a graph. In this case, the x-axis would represent the annual income and the y-axis would represent the amount spent on entertainment. Each dot on the scatterplot would represent a family and its corresponding values for income and entertainment spending.
By examining the scatterplot, it can be determined whether there is a correlation between income and entertainment spending. For example, if most dots are clustered around a certain line or pattern, it suggests a relationship between the two variables.
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A research study estimated that under a certain condition, the probability a subject would be referred for heart catheterization was 0.906 for whites and 0.847 for blacks. a. A press release about the study stated that the odds of referral for cardiac catheterization for blacks are 60% of the odds for whites. Explain how they obtained 60% (more accurately, 57%). b. An Associated Press story21 that described the study stated "Doctors were only 60% as likely to order cardiac catheterization for blacks as for whites." What is wrong with this interpretation? Give the correct percentage for this interpretation. (In stating results to the general public, it is better to use the relative risk than the odds ratio. It is simpler to understand and less likely to be misinterpreted.)
Answer and Step-by-step explanation:
a) The press release uses some weird relative risk method to arrive at this value.
P(B) = 0.847, Probability that a black is safe from being referred for cardiac carthetirization, P(B') = 1 - 0.847 = 0.153
P(W) = 0.907, P(W') = 1 - 0.907 = 0.094
The press release's relative risk = (0.847/0.153)/(0.907/0.094) = 0.574 = 57.4%
b) This is interpretation for relative risk, not the odds ratio. The actual relative risk is
(0.847/0.906) = 0.935: i.e., 60% should have been 93.5%.
Hope this helps!
An apartment building is planning on replacing refrigerators in 37 of its units. If the refrigerators cost $565 each, estimate the total cost by rounding both numbers to the nearest 10.
Researchers studied 208 infants whose brains were temporarily deprived of oxygen due to complications at birth. When researchers detected oxygen deprivation, they randomly assigned babies to either usual care or to a whole-body cooling group. The goal was to see whether reducing body temperature for three days after birth increased the rate of survival without brain damage.
Which of the following is used in the design of this experiment? Check all that apply.
a. Random assignment
b. No answer text provided
c. Double blinding
d. Control group
Answer:
Correct option: c. Double blind.
Step-by-step explanation:
A double blind experiment is an experiment where the participants are divided into two groups: one is the experimental group and the other is a control group. The participants in the experimental group are provided with a treatment and those in the control group are not provided with the treatment but are given a placebo.
In this experiment neither the researcher nor the participants know to which group a participant is placed.
After the experiment the results for the two groups are compared and the conclusion is hence drawn.
Here the researcher randomly assigned babies to either usual care or to a whole-body cooling group. The experimental group is the cooling group and the control group is the group provided with usual treatment.
Thus, the researchers are conducting a double blind experiment to determine whether reducing body temperature for three days after birth increased the rate of survival without brain damage or not.
Thus, the correct option is (c).
could someone help me understand this?
Answer:
8 < x < 40
Step-by-step explanation:
x − 8 must be more than 0, but it can't be greater than 32.
0 < x − 8 < 32
8 < x < 40
A more precise answer would require law of cosines and calculus.
An apartment has two fire alarms. In the event of a fire, the probability that alarm A will fail is 0.004, and the probability that alarm B will fail is 0.01. Assume the two failures are independent, what is the probability that at least one alarm fails in the event of a fire
Answer: 0.0136
Step-by-step explanation:
Let events are:
A = alarm A will fail
B= alarm B will fail
We have given ,
P(A) = 0.004 , P(B)=0.01
Since both events are independent , so
P(A and B) = P(A) x P(B)
= 0.04 x 0.01 =0.0004
i.e. P(A and B) =0.0004
Now , P(A or B) = P(A)+P(B)-P(A and B)
= 0.004+ 0.01-0.0004=0.0136
Hence, the probability that at least one alarm fails in the event of a fire is 0.0136 .
The probability that at least one alarm fails in the event of a fire is approximately 0.014 or 1.4%.
The probability that at least one alarm fails in the event of a fire is given by:
[tex]\[ P(\text{at least one fails}) = 1 - P(\text{both work}) \][/tex]
Since the failures are independent, the probability that both alarms work is the product of their individual probabilities of working:
[tex]\[ P(\text{both work}) = P(\text{A works}) \times P(\text{B works}) \][/tex]
The probability that alarm A works is the complement of the probability that it fails, which is [tex]\( 1 - 0.004 \)[/tex]. Similarly, the probability that alarm B works is 1 - 0.01 .
Thus, we have:
[tex]\[ P(\text{A works}) = 1 - 0.004 = 0.996 \] \[ P(\text{B works}) = 1 - 0.01 = 0.99 \][/tex]
Now we can calculate [tex]\( P(\text{both work}) \)[/tex]:
[tex]\[ P(\text{both work}) = 0.996 \times 0.99 \] \[ P(\text{both work}) = 0.98604 \][/tex]
Finally, we find the probability that at least one alarm fails:
[tex]\[ P(\text{at least one fails}) = 1 - 0.98604 \] \[ P(\text{at least one fails}) = 0.01396 \][/tex]
List the Octal and Hexadecimal numbers from 16 to 32. Using A and B as the last two digits (A representing a value of 10 and B representing a value of 11), list the numbers from 8 to 28 in base 12.
Answer:
see attached
Step-by-step explanation:
The attached list shows base-10 numbers in the left column, followed by their equivalents in base 8, base 16, and base 12.
Counting is done in the usual way: when you come to the last digit of the particular base, you increment the next digit to the left, and start over.
The Octal numbers from 16 to 32 and the Hexadecimal numbers from 16 to 32 are listed. Additionally, numbers from 8 to 28 are provided in base 12 system, where A and B are two last digits representing 10 and 11 respectively.
Explanation:The Octal numbers from 16 to 32 are as follows: 20, 21, 22, 23, 24, 25, 26, 27, 30, 31, 32, 33, 34, 35, 36, 37, 40.
The Hexadecimal numbers from 16 to 32 are: 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 1A, 1B, 1C, 1D, 1E, 1F, 20.
For base 12 numbers from 8 to 28, where A stands for 10 and B for 11, they are: 8, 9, A, B, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 1A, 1B, 20, 21, 22, 23.
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The vertices of a triangle are given. Determine whether the triangle is an acute triangle, an obtuse triangle, or a right triangle. (-4, 0, 0), (0, 0, 0), (7, 2, 6)
Answer:
Obtuse triangle
Step-by-step explanation:
Given are the vertices of a triangle
Let A (-4, 0, 0),B (0, 0, 0),C(7, 2, 6)
Let us find angles between AB, BC and CA
AB = (4, 0,0): BC = (7,2,6) : CA = (11, 2,6)
Cos B = [tex]\frac{AB.BC}{|AB||BC|} \\[/tex]
B = arc cos [tex]\frac{AB.BC}{|AB||BC|} \\[/tex]=137 deg 54 min 7 sec
Similarly
A=29 deg 53 min 53 seconds
C = 12 deg 12 min 3 sec
Obtuse triangle since one angle > 90 degrees
To determine the type of triangle, find the lengths of the sides using the distance formula. Then compare the sum of squares of the two shortest sides with the square of the longest side.
Explanation:To determine whether a triangle is acute, obtuse, or right, we need to find the lengths of its three sides and then use the Pythagorean theorem. The distance formula can be used to find the lengths of the sides by finding the distances between the given vertices. After finding the lengths, we can compare the sum of the squares of the two shortest sides with the square of the longest side to determine the type of triangle.
Using the distance formula, we find that the lengths of the sides are 4,7, and 9. The shortest side is 4, so we calculate the sum of the squares of 4 and 7, which equals 65. The square of the longest side (9) is 81. Since 65 < 81, the triangle is an acute triangle.
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In a standard Normal distribution, if the area to the left of a z-score is about 0.3500, what is the approximate z-score? Draw a sketch of the Normal curve, showing the area and z-score.
Answer:
z-score=0.385
(See attached picture)
Step-by-step explanation:
The procedure to find the z-score will depend on the resources we have available. I have a table with the area between the mean and the value we wish to normalize, so the very first thing we need to do is precisely find this area we need to analyze.
Everything to the left of thte mean will represent 50% of the data, so we start by subtracting:
50%-35%=15%
so we need to look in the table for the value 0.15.
In my table I can see that for an area of 0.15, the z-score will be between 0.38 (z-score of 0.1480) and 0.39 (z-score of 0.1517).
By doing some interpolation, you can determine a more accurate value of the z-score to be 0.385.
The z-score corresponding to an area of 0.3500 in a standard normal distribution is approximately -0.39. This z-value indicates that the data point is 0.39 standard deviations below the mean.
Explanation:In a standard Normal distribution, the z-score is equivalent to the number of standard deviations a given data point is from the mean. If you know the area to the left of the z-score (which in this case is 0.3500), you can use a z-score table (also known as a standard normal table) to find the corresponding z-score.
Normally, the z-score table gives the area to the left of the score. However, in this case, the value (0.3500) does not appear in the body of the z-score table because it corresponds to a negative z-score (since 0.3500 < 0.5). Thus, we will first find the equivalent positive area (1- 0.3500 = 0.6500) and look up that value in the z-score table. The value 0.6500 corresponds approximately to a z-score of 0.39. Since our original question gives an area less than 0.5 (indicating a z-score below the mean) the z-score is -0.39.
Please note that for illustrating the Normal curve, any standard statistics textbook or online resource will have a diagram illustrating the curve, with a vertical line indicating the z-score (in this case -0.39) and shading demonstrating the area to the left of the z-score. These images typically aren't included in text-based tutoring platforms.
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Consider purchasing a system of audio components consisting of a receiver, a pair of speakers, and a CD player. Let A1 be the event that the receiver functions properly throughout the warranty period. Let A2 be the event that the speakers function properly throughout the warranty period. Let A3 be the event that the CD player functions properly throughout the warranty period. Suppose that these events are (mutually) independent with P(A1) = 0.91, P(A2) = 0.85, and P(A3) = 0.77.(a) What is the probability that at least one component needs service during the warranty period?(b) What is the probability that exactly one of the components needs service during the warranty period?
Answer:
(a) Probability that at least one component needs service during the warranty period = 0.4044.
(b) Probability that exactly one of the components needs service during the warranty period = 0.3419.
Step-by-step explanation:
Given A1 be the event that the receiver functions properly throughout the warranty period.
A2 be the event that the speakers function properly throughout the warranty period.
A3 be the event that the CD player functions properly throughout the warranty period.
Also P(A1) = 0.91, P(A2) = 0.85, and P(A3) = 0.77.
Now P(A1)' means that the receiver need service during the warranty period which is 1 - P(A1) = 1 - 0.91 = 0.09.
Similarly,P(A2)' =1 - P(A2) =1 - 0.85 =0.15 and P(A3)' =1 - P(A3)=1 - 0.77 = 0.23
Note: The ' sign on the P(A1) represent compliment of A1 or not A1.
(a) The probability that at least one component needs service during the warranty period = 1 - none of the component needs service during the warranty period
And none of the component needs service during the warranty period means that all the three components functions properly during the warranty period .
So, Probability that at least one component needs service during the warranty period = 1 - P(A1) x P(A2) x P(A3) = 1 - (0.91 x 0.85 x 0.77) = 0.4044.
(b) Now to find the Probability that exactly one of the components needs service during the warranty period, there would be three cases for this:
Receiver needs service and other two does not need during the warranty period.Speaker needs service and other two does not need during the warranty period.CD player needs service and other two does not need during the warranty period.And we have to add these three cases to calculate above probability.
Probability that exactly one of the components needs service during the warranty period = P(A1)' x P(A2) x P(A3) + P(A1) x P(A2)' x P(A3) + P(A1) x P(A2) x P(A3)'
= 0.09 x 0.85 x 0.77 + 0.91 x 0.15 x 0.77 + 0.91 x 0.85 x 0.23
= 0.3419.
Answer:
(a) P (At least one component needs service) = 0.404
(b) P (Either component A₁ or A₂ or A₃) = 0.997
Step-by-step explanation:
Given:
[tex]A_{1}=[/tex] Event that the receiver functions properly throughout the warranty period.
[tex]A_{2}=[/tex] Event that the speakers function properly throughout the warranty period.
[tex]A_{3}=[/tex] Event that the CD player functions properly throughout the warranty period.
[tex]P(A_{1})=0.91,\ P(A_{2})=0.85\ and\ P(A_{3})=0.77[/tex]
(a)
Compute the probability that at least one component needs service during the warranty period:
P (At least one component needs service) = 1 - P (None of the one component needs service)
= 1 - {P ([tex]A_{1}[/tex]) × P ([tex]A_{2}[/tex]) × P ([tex]A_{3}[/tex])}
[tex]=1 - (0.91\times0.85\times0.77)\\=1-0.595595\\=0.404405\\\approx0.404[/tex]
Thus, the probability that at least one component needs service during the warranty period is 0.404.
(b)
Compute the probability that exactly one of the components needs service during the warranty period, i.e. P (Either A₁ or A₂ or A₃):
[tex]P(A_{1}\cup A_{2}\cup A_{3})=P(A_{1})+P(A_{2})+P(A_{3})-P(A_{1}\cap A_{2})-P(A_{2}\cap A_{3})-P(A_{3}\cap A_{1})+P(A_{1}\cap A_{2}\cap A_{3})\\=P(A_{1})+P(A_{2})+P(A_{3})-[P(A_{1})\tmes P(A_{2})]-[P(A_{2})\tmes P(A_{3})]-[P(A_{3})\tmes P(A_{1})] +[P(A_{1})\tmes P(A_{2})\times P(A_{3})]\\=0.91+0.85+0.77-(0.91\times0.85)-(0.85\times0.77)-(0.77\times0.91)+(0.91\times0.85\times0.77)\\=0.996895\\\approx0.997[/tex]
Thus, the probability that exactly one of the components needs service during the warranty period is 0.997
A construction firm bids on two different contracts. Let E1 be the event that the bid on the first contract is successful, and define E2 analogously for the second contract. Suppose that P(E1) = 0.7 and P(E2) = 0.8 and that E1 and E2 are independent events.
(a) Calculate the probability that both bids are successful (the probability of the event E1and E2).
(b) Calculate the probability that neither bid is successful (the probability of the event (not E1) and (not E2)).
(c) What is the probability that the firm is successful in at least one of the two bids?
Answer:
(a) 0.56
(b) 0.06
(c) 0.94
Step-by-step explanation:
P(E1) = 0.7 and P(E2) = 0.8
(a) The probability that both bids are successful is given by the product of the probability of success of each bid:
[tex]P(E1\ and\ E2) = 0.7*0.8=0.56[/tex]
(b) The probability that neither bid is successful is given by the product of the probability of failure of each bid:
[tex]P(not\ E1\ and\ not\ E2)= (1-P(E1))*(1-P(E2))\\P(not\ E1\ and\ not\ E2)=0.3*0.2=0.06[/tex]
(c) The probability that the firm is successful in at least one of the two bids is given by the sum of the probability of success of each bid subtracted by the probability that both bids are successful:
[tex]P(E1\ or\ E2)=P(E1)+P(E2) - P(E1\ and\ E2)\\P(E1\ or\ E2)=0.7+0.8-0.56\\P(E1\ or\ E2)=0.94[/tex]
Using probability concepts, it is found that there is a:
a) 0.56 = 56% probability that both bids are successful.
b) 0.06 = 6% probability that neither bid is successful.
c) 0.94 = 94% probability that the firm is successful in at least one of the two bids.
Item a:
These two events are independent, hence, the probability of both is the multiplication of the probabilities of each, thus:
[tex]p = 0.7(0.8) = 0.56[/tex]
0.56 = 56% probability that both bids are successful.
Item b:
E1 has a 1 - 0.7 = 0.3 probability of being unsuccessful, while E2 has a 0.2 probability, hence:
[tex]p = 0.3(0.2) = 0.06[/tex]
0.06 = 6% probability that neither bid is successful.
Item c:
1 - 0.06 = 0.94
0.94 = 94% probability that the firm is successful in at least one of the two bids.
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For 14 baseball teams , the correlation with number of wins in the regular season is 0.51 for shutouts, 0.61 for hits made, -0.70 for runs allowed and -0.56 for homeruns allowed.
1. Which variable has the strongest linear association with number of wins?
O shutouts, runs allowed, homeruns allowed, or hits made.
Answer:
For this case the strongest linear association is given by the greatest correlation coeffcient in absolute value from the list provided. We have:
[tex] |r_3|>|r_2| > |r_4| > |r_1|[/tex]
So on this case we can conclude that the strongest linear association with number of wins is for runs allowed.
Step-by-step explanation:
Previous concepts
The correlation coefficient is a "statistical measure that calculates the strength of the relationship between the relative movements of two variables". It's denoted by r and its always between -1 and 1.
And in order to calculate the correlation coefficient we can use this formula:
[tex]r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^2 -(\sum x)^2][n\sum y^2 -(\sum y)^2]}}[/tex]
Solution to the problem
For this case we have a list of correlation coefficients given:
[tex] r_1 = 0.51[/tex] represent the correlation between number of wins and shutouts
[tex] r_2 = 0.61[/tex] represent the correlation between number of wins and hits made
[tex] r_3 = -0.7[/tex] represent the correlation between number of wins and runs allowed
[tex] r_4 = -0.56[/tex] represent the correlation between number of wins and homeruns allowed
When we analyze linear association we are interested just in the absolute value for r since if r is near to +1 we have positive linear association but on the case that r is near to -1 we have an strong linear association but inversely proportional.
For this case the strongest linear association is given by the greatest correlation coeffcient in absolute value from the list provided. We have:
[tex] |r_3|>|r_2| > |r_4| > |r_1|[/tex]
So on this case we can conclude that the strongest linear association with number of wins is for runs allowed.
In relation to the number of wins for 14 baseball teams, the variable 'runs allowed' holds the strongest linear association. This is represented by its correlation coefficient of -0.70, indicating a strong inverse relationship. As 'runs allowed' increase, the 'number of wins' decrease.
Explanation:In context of these 14 baseball teams, correlations are being determined with the number of wins in the regular season and four variables: shutouts, hits made, runs allowed, and homeruns allowed. The correlation coefficient represents the strength and direction of a linear relationship between two variables. Coefficients close to +1 or -1 indicate a strong linear association, while those near 0 suggest a weak association. The sign of the correlation indicates the direction of the relationship, either positive or negative.
The variable with the strongest linear association with the number of wins is 'runs allowed', which bears a correlation of -0.70. This implies a strong inverse relationship where as 'runs allowed' increase, the 'number of wins' tends to decrease.
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The soccer league in 1 community has 8 teams. You are required to predict, in order, the top 3 teams at the end of the season. Ignoring the possibility of ties, calculate the number of different predictions you could make. What is the probability of making the correct prediction by chance?
Answer:
336 different predictions.
1/336 probability of making the correct prediction by chance
Step-by-step explanation:
The order is important.
For example, Team A, B and C is a different outcome than team B, A, C.
So we use the permutations formula to solve this problem:
Permutations formula:
The number of possible permutations of x elements from a set of n elements is given by the following formula:
[tex]P_{(n,x)} = \frac{n!}{(n-x)!)}[/tex]
In this problem, we have that:
Permutations of 3 from a set of 8. So
[tex]P_{(8,3)} = \frac{8!}{(8-3)!} = 336[/tex]
What is the probability of making the correct prediction by chance?
There are 336 possible outcomes.
By chance, you predict 1.
So there is a 1/336 probability of making the correct prediction by chance
Final answer:
You can make 336 different predictions for the top 3 soccer teams out of 8, and the chance of making the correct prediction by chance is approximately 0.298%.
Explanation:
To calculate the number of different predictions you could make for the top 3 teams out of 8, without considering ties, you use permutations since the order matters. The formula for permutations is P(n, r) = n! / (n-r)!, where n is the total number of teams and r is the number of positions to fill.
In this case, n = 8 teams and r = 3 positions. Therefore, the calculation is P(8, 3) = 8! / (8-3)! = 8 x 7 x 6 = 336 different predictions.
To find the probability of making the correct prediction by chance, since there is only one correct prediction out of all possible predictions, the probability is 1 / 336. Thus, the probability is approximately 0.00298, or 0.298%.
Telephone interviews of 1, 502 adults 18 years of age or older found that only 69% could identify the current vice-president.
Is the value a parameter or a statistic?
A. The value is a parameter because the 1, 502 adults 18 years of age or older are a sample.
B. The value is a parameter because the 1, 502 adults 18 years of age or older are a population.
C. The value is a statistic because the 1, 502 adults 18 years of age or older are a population.
D. The value is a statistic because the 1, 502 adults 18 years of age or older are a sample.
Answer:
D
Step-by-step explanation:
The population consists of all characteristics of interest and sample is portion or a subset of population. Here, 1502 adults 18 years or older are select ted from a population of all adults 18 years or older, so, 1502 adults are the sample. The measurement taken from sample is termed as statistic. The given value 69% is computed from a sample and thus it is a sample statistic.