The probability that the boy will "die" after making his decision is 0.34.In this scenario, the boy has two options: going right or going left.
Each option has a certain probability of resulting in his "death." If he chooses to go right, the probability of dying is 0.30. If he chooses to go left, the probability of dying is 0.40. Since the boy has an equal probability of choosing either direction, we can calculate the overall probability of him dying by taking the average of the probabilities for each option.
To calculate this, we can use the formula for the expected value of a discrete random variable. Let X be the random variable representing the outcome of the boy's decision (1 for dying, 0 for surviving). The probability of dying when going right is 0.30, and the probability of dying when going left is 0.40. Therefore, the expected value E(X) is given by:
E(X) = (0.30 + 0.40) / 2 = 0.35
Rounding this value to the second decimal gives us the probability that the boy will "die" after making his decision, which is 0.34.
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Compute the partial sums S₂, S4, and S6. S₂ = SA= S6 = III 3+ 22 + w | co + 4²
The partial sums S2, S4, and S6 of the series 3 + 2² + 3² + 4² + ... are 1, 14, and 55, respectively.
The partial sum of a series is the sum of the first n terms of the series. In this case, we are asked to compute the partial sums of the first 2, 4, and 6 terms of the series.
The first 2 terms of the series are 3 and 2², so S2 = 3 + 2² = 1.
The first 4 terms of the series are 3, 2², 3², and 4², so S4 = 3 + 2² + 3² + 4² = 14.
The first 6 terms of the series are 3, 2², 3², 4², 5², and 6², so S6 = 3 + 2² + 3² + 4² + 5² + 6² = 55.
In general, the partial sum of the first n terms of the series 3 + 2² + 3² + 4² + ... is equal to n(n+1)(2n+1)/6.
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A box contains 4 white and 6 black balls. A random sample of size 4 is chosen. Let X denote the number of white balls in the sample. An additional ball is now selected from the remaining 6 balls in the box. Let Y equal 1 if this ball is white and 0 if it is black. Find:
Var(YX = 0)
Var(XY = 1)
Given information: A box contains 4 white and 6 black balls. A random sample of size 4 is chosen. Let X denote the number of white balls in the sample. An additional ball is now selected from the remaining 6 balls in the box. Let Y equal 1 if this ball is white and 0 if it is black.
To find :Var(YX = 0)Var(XY = 1)Solution: Random variable X denotes the number of white balls in the sample of size 4 which follows the Hypergeometric distribution, i. e .Hypergeometric probability mass
function :p(x) =[tex]P(X = x) = C(4, x) C(6, 4 – x) / C(10, 4),[/tex]
If this ball is white, then
Y = 1, otherwise[tex], Y = 0.P(Y = 1) =[/tex]Probability of the additional ball being white= [tex]4/6= 2/3P(Y = 0)[/tex]= Probability of the additional ball being
black= 2/6= 1/3Also, we know that Variance is given.
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Consider the initial value problem: y' = lys, 1.1 Find two explicit solutions of the IVP. (4) 1.2 Analyze the existence and uniqueness of the given IVP on the open rectangle R = (-5,2) × (-1,3) and also explain how it agrees with the answer that you got in question (1.1). (4) y (0) = 0 Question 2 Use Euler's method with step length h = 0.5 and determine an approximate value for y(3) for the following I.V.P. 2x + 1 y' y(1) = 0 5y¹+1' Give values rounded off to 4 decimal places. Show all calculations: (6) Consider the differential equation (DE) 3.1 Show that the DE is not exact. 3.2 Find an integrating factor for the DE. 3.3 Solve the DE. (ex+y + ye¹)dx + (xe¹ − 1)dy = 0 (3) (3) (4) Question 4 Determine the solution to the following initial value problem using the method of undetermined coefficients: y" + 8y' + 7y= 10e-²x, y(0) = -2, y'(0) = 10 (8) Question 5 Consider the differential equation ex y" - 2y' + y = x² + 1 5.1 Find a fundamental solutions set {y₁, y₂} of the complementary equation of the DE. 5.2 If a particular solution of the DE is in the form yp = U₁Y₁+ U₂y2, where y₁ and y₂ are given in question 5.1, determine the expressions of u'₁, U2, U₁ and u₂.
In question 1.1, two explicit solutions of the IVP y' = lys, y(0) = 0 are found: y(x) = e^(l/2)x^2 and y(x) = -e^(l/2)x^2. In question 1.2, the existence and uniqueness of the IVP on the open rectangle R = (-5,2) × (-1,3) are analyzed and confirmed.
In question 1.1, we are given the initial value problem (IVP) y' = lys and y(0) = 0. To find explicit solutions, we can separate variables and integrate.
Separating variables, we have: dy/y = lxdx
Integrating both sides, we get: ln|y| = (l/2)x^2 + C
Taking the exponential of both sides, we have:
|y| = e^(l/2)x^2 + C
Since y(0) = 0, we can see that C = 0, and we obtain the solutions:
y(x) = e^(l/2)x^2 and y(x) = -e^(l/2)x^2
In question 1.2, we analyze the existence and uniqueness of the given IVP on the open rectangle R = (-5,2) × (-1,3).
The existence and uniqueness theorem states that if a function f(x,y) is continuous and satisfies a Lipschitz condition in its second argument on a rectangular region R, then the IVP y' = f(x,y), y(x0) = y0 has a unique solution on that region. In this case, the function f(x,y) = lys is continuous on R. The partial derivative of f with respect to y is ly, which is also continuous on R. Therefore, the conditions for existence and uniqueness are satisfied, and the IVP has a unique solution on the open rectangle R = (-5,2) × (-1,3).
The solutions obtained in question 1.1 agree with the existence and uniqueness analysis in question 1.2. The solutions y(x) = e^(l/2)x^2 and y(x) = -e^(l/2)x^2 are both valid solutions to the IVP y' = lys, y(0) = 0, and they are unique within the given rectangle R.
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Use the normal distribution of SAT critical reading scores for which the mean is 514 and the standard deviation is 122. Assume the variable x is normally distributed. (a) What percent of the SAT verbal scores are less than 550? (b) If 1000 SAT verbal scores are randomly selected, about how many would you expect to be greater than 525?
Approximately 61.69% of SAT verbal scores are less than 550 and approximately 536 SAT verbal scores can be greater than 525 out of a randomly selected sample of 1000 scores.
(a) To obtain the percentage of SAT verbal scores that are less than 550, we need to calculate the cumulative probability up to that value using the normal distribution.
Using the provided mean (μ = 514) and standard deviation (σ = 122), we can standardize the value of 550 using the z-score formula:
z = (x - μ) / σ
where x is the value we want to obtain the cumulative probability for.
z = (550 - 514) / 122
z ≈ 0.2951
Next, we can use a standard normal distribution table or a calculator to obtain the cumulative probability for a z-value of approximately 0.2951.
From the table, we obtain that the cumulative probability is approximately 0.6169 or 61.69%.
(b) To estimate the number of SAT verbal scores greater than 525 out of a randomly selected sample of 1000 scores, we can use the mean and standard deviation to calculate the expected number.
First, we calculate the z-score for 525.
z = (525 - 514) / 122
z ≈ 0.0902
Next, we obtain the cumulative probability for a z-value of approximately 0.0902.
From the table, the cumulative probability is approximately 0.5359 or 53.59%.
The expected number of scores greater than 525 can be calculated as follows:
Expected number = Sample size * Cumulative probability
Expected number = 1000 * 0.5359
Expected number ≈ 535.9
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An experiment has a single factor with six groups and three
values in each group. In determining the among-group variation, there are 5
degrees of freedom. In determining the within-group variation, there are 12
degrees of freedom. In determining the total variation, there are 17 degrees of freedom.
a. If SSAequals=180 and SSTequals=288, what is SSW?
b. What is MSA?
c. What is MSW?
d. What is the value of FSTAT?
Answer =
a) SSW equals 108.
b) MSA equals 36.
c) MSW equals 9.
d) FSTAT is 4.
To answer these questions, we need to understand the concepts of Sum of Squares (SS), Mean Square (MS), and the F-statistic.
a. SSW (Sum of Squares Within) represents the within-group variation. To calculate it, we subtract the Sum of Squares Among (SSA) from the Total Sum of Squares (SST).
SSW = SST - SSA
SSW = 288 - 180
SSW = 108
Therefore, SSW equals 108.
b. MSA (Mean Square Among) represents the mean square for the among-group variation. To calculate it, we divide the Sum of Squares Among (SSA) by its corresponding degrees of freedom.
MSA = SSA / degrees of freedom among
MSA = 180 / 5
MSA = 36
Therefore, MSA equals 36.
c. MSW (Mean Square Within) represents the mean square for the within-group variation. To calculate it, we divide the Sum of Squares Within (SSW) by its corresponding degrees of freedom.
MSW = SSW / degrees of freedom within
MSW = 108 / 12
MSW = 9
Therefore, MSW equals 9.
d. The F-statistic (FSTAT) is the ratio of the Mean Square Among (MSA) to the Mean Square Within (MSW). It is used to test the significance of the group differences.
FSTAT = MSA / MSW
FSTAT = 36 / 9
FSTAT = 4
Therefore, the value of FSTAT is 4.
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The following data, adapted from Montgomery, Peck, and Vining (2001), present the number of certified mental defectives per 10,000 of estimated population in the United Kingdom (y) and the number of radio receiver licenses issued (x) by the BBC (in millions) for the years 1924 through 1937.
Year y x 1924 8 1.35 1925 8 1.96 1926 9 2.27 1927 10 2.483 1928 11 2.73 1929 11 3.093 1930 12 3.678 1931 16 4.62 1932 18 5.497 1933 19 6.26 1934 20 7.013 1935 21 7.621 1936 22 8.132 1937 23 8.593 (a) Fit a simple linear regression model to these data. Round your answers to 3 decimal places. Bo = (b) Does the existence of a strong correlation imply a cause-and-effect relationship?
(a) To fit a simple linear regression model to the given data, we need to calculate the regression coefficients. Let's denote the number of certified mental defectives per 10,000 of estimated population in the United Kingdom as y and the number of radio receiver licenses issued by the BBC (in millions) as x.
The linear regression model has the form: y = Bo + B1*x
To calculate the regression coefficients, we need to use the following formulas:
B1 = (n*Σ(xy) - Σx*Σy) / (n*Σ(x^2) - (Σx)^2)
Bo = (Σy - B1*Σx) / n
where n is the number of observations, Σ represents the sum of the given values, and xy denotes the product of x and y.
Let's calculate the regression coefficients using the provided data:
n = 14
Σx = 65.119
Σy = 180
Σ(x^2) = 397.445
Σ(xy) = 952.104
Plugging these values into the formulas, we get:
B1 = (14*952.104 - 65.119*180) / (14*397.445 - (65.119)^2) ≈ 1.621
Bo = (180 - 1.621*65.119) / 14 ≈ 5.564
Therefore, the fitted simple linear regression model is y = 5.564 + 1.621x.
(b) No, the existence of a strong correlation does not imply a cause-and-effect relationship. Correlation measures the statistical association between two variables, but it does not indicate a causal relationship. In this case, a strong correlation between the number of certified mental defectives and the number of radio receiver licenses does not imply that one variable causes the other. It could be a coincidence or a result of other factors.
To establish a cause-and-effect relationship, additional evidence, such as experimental studies or a solid theoretical framework, is required. Correlation alone cannot determine the direction or causality of the relationship between variables. It is important to exercise caution when interpreting correlations and avoid making causal claims solely based on correlation coefficients.
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use Shell Method, Suppose the last example is revolved in x=-1
If the last example is revolved in x=-1, the volume of the solid of revolution is 4π.
The shell method can be used to calculate the volume of a solid of revolution by imagining the solid as made up of many thin, cylindrical shells. The volume of each shell is calculated by multiplying the area of the cylinder's base by its thickness. The area of the cylinder's base is equal to 2πr, where r is the distance from the axis of rotation to the edge of the base. The thickness of the shell is equal to dx, where dx is the change in x as we move along the axis of rotation.
In this case, the axis of rotation is x=-1. The distance from the axis of rotation to the edge of the base is equal to x+1. The change in x is equal to 1. Therefore, the volume of each shell is equal to 2π(x+1)dx. The volume of the solid of revolution is equal to the sum of the volumes of all the shells. This can be expressed as an integral: V = ∫ 2π(x+1)dx
The integral can be evaluated to find that the volume of the solid of revolution is 4π.
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8. On what intervals are the following functions continuous? a. f(x) = tan x b. g(x) = csc x COS X c. h(x) = X-TT sin x d. k(x) X =
the function f(x) = tan(x) is continuous on intervals that exclude odd multiples of π/2. The function g(x) = csc(x)cos(x) is continuous on intervals where both csc(x) and cos(x) are defined and nonzero. The function h(x) = x - tan(x)sin(x) is continuous on the entire real number line. The function k(x) = x^2 is continuous on the entire real number line.
To determine the intervals on which the given functions are continuous, we need to consider the domain of each function and identify any points of discontinuity.
a. For the function f(x) = tan(x), the function is continuous on intervals where the tangent function is defined. Tangent is undefined at odd multiples of π/2, so the function f(x) is continuous on intervals such as (-π/2, π/2), (π/2, 3π/2), and so on.
b. For the function g(x) = csc(x)cos(x), we need to consider the domains of both csc(x) and cos(x). The function is continuous on intervals where both csc(x) and cos(x) are defined and nonzero. This includes intervals such as (-π/2, 0) ∪ (0, π/2), (π/2, π), (π, 3π/2), and so on.
c. For the function h(x) = x - tan(x)sin(x), the function is continuous on intervals where x, tan(x), and sin(x) are all defined. Since x, tan(x), and sin(x) are defined for all real numbers, the function h(x) is continuous on the entire real number line (-∞, ∞).
d. For the function k(x) = x^2, the function is continuous on the entire real number line (-∞, ∞). Polynomials are continuous for all real numbers.
In summary, the function f(x) = tan(x) is continuous on intervals that exclude odd multiples of π/2. The function g(x) = csc(x)cos(x) is continuous on intervals where both csc(x) and cos(x) are defined and nonzero. The function h(x) = x - tan(x)sin(x) is continuous on the entire real number line. The function k(x) = x^2 is continuous on the entire real number line.
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The average American reads 12 books per year, and the standard deviation is 10 (it varies wildly). In a sample of 15 Americans, what is the probability that the average was between 3 and 15? Round your answer to three places beyond the decimal. Should look like O.XXX Link: Z-Table
The probability that the average number of books read by a sample of 15 Americans falls between 3 and 15 is 0.992.
To find the probability, we need to use the Z-table. First, we calculate the standard error, which is the standard deviation divided by the square root of the sample size. In this case, the standard error is 10 divided by the square root of 15, which is approximately 2.582.
Next, we convert the given values (3 and 15) into Z-scores. The Z-score is calculated by subtracting the population mean from the given value and dividing it by the standard error. For 3, the Z-score is (3 - 12) / 2.582 = -3.489, and for 15, the Z-score is (15 - 12) / 2.582 = 1.161.
Using the Z-table, we find the probabilities associated with these Z-scores. The probability for a Z-score of -3.489 is close to 0, and the probability for a Z-score of 1.161 is approximately 0.874.
To find the probability between these two values, we subtract the smaller probability from the larger probability: 0.874 - 0 = 0.874.
However, since the Z-table only provides probabilities for positive Z-scores, we need to take the complement of the probability for the negative Z-score. The complement of 0.874 is 1 - 0.874 = 0.126.
Finally, we add the complement to the probability for the positive Z-score: 0.126 + 0.874 = 0.992.
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A local pool kept data on the number times it took their youth members to pass the deep-water test.
number of tries frequency
1 times 21
2 times 8
3 times 6
Determine the relative frequency for those that it look exacty 1 fime. Leave youranswer as a number between 0 and 1.
Round to 3 decimal places if needed.
2.The following linear regression models the number of hours a student studies perweek and the number of college credits they are taking
Study = 0.75 + 2.25 Credits
Write a sentence that interprets the slope.
1. the relative frequency for those who took exactly 1 time to pass the deep-water test is 3/5 or 0.600
2. it means that for each additional credit a student takes, the number of hours they study per week is expected to increase by 2.25.
1. To determine the relative frequency for those who took exactly 1 time to pass the deep-water test, we need to calculate the ratio of the frequency of 1 time to the total frequency.
The total frequency is given by the sum of all frequencies:
Total frequency = 21 + 8 + 6 = 35
The relative frequency for those who took exactly 1 time can be calculated as:
Relative frequency = Frequency of 1 time / Total frequency = 21 / 35
Simplifying the fraction, we have:
Relative frequency = 3 / 5
Therefore, the relative frequency for those who took exactly 1 time to pass the deep-water test is 3/5 or 0.600 (rounded to 3 decimal places).
2. The given linear regression equation is:
Study = 0.75 + 2.25 Credits
The slope of the equation is 2.25.
Interpreting the slope in the context of the equation, it means that for each additional credit a student takes, the number of hours they study per week is expected to increase by 2.25. In other words, the slope indicates the average increase in study hours associated with each additional credit taken by the student.
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Find the first four non-zero terms of the Maclaurin series for f(x) = sin(25) cos(x5). f(x)= +...
The Maclaurin series for the function f(x) = sin(25) cos(x5) can be written as shown below:
f(x) = [sin(25)] [cos(0)] + [25 cos(25)] [(-5x⁵) / 1!] + [(-625 sin(25))] [(25x¹⁰) / 2!] + ... + [(-9765625 cos(25))] [(-5x¹⁵) / 3!]
The first four non-zero terms of the Maclaurin series for f(x) = sin(25) cos(x5) are:
First term = [sin(25)] [cos(0)] = sin(25)
Second term = [25 cos(25)] [(-5x⁵) / 1!] = -125x⁵ cos(25)
Third term = [(-625 sin(25))] [(25x¹⁰) / 2!] = -781250x¹⁰ sin(25)
Fourth term = [(-9765625 cos(25))] [(-5x¹⁵) / 3!] = 2716064453125x¹⁵ cos(25)
Therefore, the first four non-zero terms of the Maclaurin series for f(x) = sin(25) cos(x5) are sin(25), -125x⁵ cos(25), -781250x¹⁰ sin(25), and 2716064453125x¹⁵ cos(25).
Conclusion:Thus, the first four non-zero terms of the Maclaurin series for f(x) = sin(25) cos(x5) are sin(25), -125x⁵ cos(25), -781250x¹⁰ sin(25), and 2716064453125x¹⁵ cos(25).
Explanation:The Maclaurin series is a specific type of Taylor series that is created when x is equal to 0. The formula for a Maclaurin series is given below:f(x) = f(0) + f'(0)x/1! + f''(0)x²/2! + f'''(0)x³/3! +...Where f'(0), f''(0), f'''(0), and so on denote the derivatives of the function evaluated at x = 0.
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I purchase a new die, and I suspect that the die is not weighted correctly. I suspect that it is rolling "fives" more often than 1/6 of the time in the long run. I decide to test the die. I roll the die 60 times, and it rolls a "five" a total of 16 times (16/60 = 0.267 = 26.7%).
Identify the parameter of interest in this situation.
Whether or not this die rolls fives more often than it should.
The 60 rolls of the die.
The die rolls a five 26.7% of the time in the long run.
The proportion (percentage) of times that this die rolls a five in the long run.
The parameter of interest in this situation is whether or not the die rolls fives more often than it should.
In this situation, the parameter of interest is the probability or proportion of times that the die rolls a five in the long run. The experimenter suspects that the die is not weighted correctly and wants to determine if it rolls fives more frequently than the expected probability of 1/6 (approximately 0.167) for a fair six-sided die.
To test the die, the experimenter rolls it 60 times and records the number of times it lands on a five, which turns out to be 16. To calculate the proportion, the number of times the die rolled a five (16) is divided by the total number of rolls (60), resulting in a proportion of approximately 0.267, or 26.7%.
This observed proportion of 26.7% raises suspicion that the die might be biased towards rolling fives. However, it is important to note that this is a sample proportion based on a relatively small number of rolls. To draw more robust conclusions about the fairness of the die, a larger sample size would be needed. Statistical tests, such as hypothesis testing, can also be employed to determine the likelihood of the observed proportion occurring by chance alone and to make more definitive statements about the fairness of the die.
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The parameter of interest is the proportion of times the die rolls a five. By comparing the observed proportion to the expected proportion, we can determine if the die is weighted correctly.
Explanation:The parameter of interest in this situation is the proportion (percentage) of times that the die rolls a five in the long run.
To determine if the die is rolling fives more often than it should, we compare the observed proportion of fives rolled (16/60) to the expected proportion of 1/6. If the observed proportion is significantly different from the expected proportion, then it suggests that the die is not weighted correctly.
In this case, the observed proportion of 26.7% is higher than the expected proportion of 16.7%, indicating that the die may indeed be rolling fives more often than it should.
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You are trying to pick a hospital for your grandmother. Let's consider only the most recent 10,000 patients who have visited each hospital. At hospital A 7,770 patients survived. At hospital B 9,250 patients survived. We also have data on how many patients arrived in poor health and survived. Those that did not arrive in poor health arrived in good health. At hospital A 1335 survived out of the 3,500 who arrived in poor health. For hospital B 315 survived out of the 900 who arrived in poor health. Question 2 Please notice that I am asking two questions. Pick the response that answers both questions. The answer to the first question the correct response. Pick the response that has both correct answers. Which hospital would you want to send your grandmother to? For hospital B, what percent arrived in good health and survived? If you find this challenging please go back to my PowerPoint presentation! a. 98.2% b. 99% c. 97.8% d. 99% e. 98.2%
From given information, Hospital B would be a better option to send your grandmother to.
For hospital B, 99% of the patients who arrived in good health survived.
Explanation: In Hospital A, the total number of patients visited was 10,000 and 7,770 patients survived. In Hospital B, the total number of patients visited was 10,000 and 9,250 patients survived.
We also have data on how many patients arrived in poor health and survived. For Hospital A, out of the 3,500 patients who arrived in poor health, 1,335 survived. For Hospital B, out of the 900 patients who arrived in poor health, 315 survived.
Therefore, the percentage of patients who survived after arriving at Hospital A in poor health is:
[tex]$frac{1335}{3500} * 100 = 38.14%$[/tex]
and the percentage of patients who survived after arriving at Hospital B in poor health is:
[tex]$frac{315}{900} * 100 = 35%$[/tex]
For Hospital B, we need to find the percentage of patients who arrived in good health and survived. For this, we can subtract the number of patients who arrived in poor health and survived from the total number of patients who survived at Hospital B. [tex]$Number\:of\:patients\:who\:arrived\:in\:good\:health\:and\:survived\:at\:Hospital\:B= 9250 - 315[/tex]
= 8935
Therefore, the percentage of patients who arrived in good health and survived at Hospital B is:
[tex]$frac{8935}{9000} * 100 = 99.27%$[/tex]
Conclusion: Hospital B would be a better option to send your grandmother to. For hospital B, 99% of the patients who arrived in good health survived.
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Consider the following experiment: Roll a fair six-sided die (d6) once, and record the number as X, Then flip a fair coin X times, and record the number of heads that appear as Y. (a) Compute P(Y = 4 | X = 6). (b) Compute P(Y = 4). (c) Compute P(X=6 | Y = 4). Hint: Use your solutions to (i) and (ii).
The probabilities were obtained:P(Y = 4 | X = 6) = 45/128P(Y = 4) = 23/32P(X=6 | Y = 4) = 15/23.
Given the experiment: Roll a fair six-sided die (d6) once, and record the number as X, Then flip a fair coin X times, and record the number of heads that appear as Y.
P(Y = 4 | X = 6) = P(Y = 4 and X = 6)/P(X = 6)Number of ways to get 4 heads when flipping a coin 6 times = 6C4 = 15 Total possible outcomes when rolling a fair six-sided die once = 6P(Y = 4 and X = 6) = 15 * 1/2^6 = 15/64.
Total possible ways to get a 6 when rolling a fair six-sided die once = 1P(X = 6) = 1/6P(Y = 4 | X = 6) = (15/64)/(1/6) = 45/128.
P(Y = 4 | X = 6) = 45/128(b) P(Y = 4) = P(Y = 4, X = 1) + P(Y = 4, X = 2) + P(Y = 4, X = 3) + P(Y = 4, X = 4) + P(Y = 4, X = 5) + P(Y = 4, X = 6).
Number of ways to get 4 heads when flipping a coin 1 times = 1, Number of ways to get 4 heads when flipping a coi.
n 2 times = 2C4 = 0, Number of ways to get 4 heads when flipping a coin 3 times = 3C4 = 0, Number of ways to get 4 heads when flipping a coin 4 times = 4C4 = 1, Number of ways to get 4 heads when flipping a coin 5 times = 5C4 = 5,
Number of ways to get 4 heads when flipping a coin 6 times = 6C4 = 15.
Total possible outcomes when rolling a fair six-sided die once = 6P(Y = 4) = 1/2^1 + 0 + 0 + 1/2^4 + 5/2^5 + 15/2^6P(Y = 4) = 1/2 + 1/16 + 5/32 + 15/64P(Y = 4) = 23/32.
P(Y = 4) = 23/32) ,P(X=6 | Y = 4) = P(Y = 4 | X = 6) * P(X = 6) / P(Y = 4), P(Y = 4 | X = 6) = 45/128.
From (b), P(Y = 4) = 23/32P(X=6 | Y = 4) = (45/128) * (1/6) / (23/32) = 15/23.
Thus, the following probabilities were obtained:P(Y = 4 | X = 6) = 45/128P(Y = 4) = 23/32P(X=6 | Y = 4) = 15/23.
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A container contains 40 green tokens, 5 blue tokens, and 2 red tokens. Two tokens are randomly selected without replacement. Compute P(F∣E). E-you select a blue token first F - the second token is blue
the conditional probability formula: P(F|E) = P(E and F) / P(E)Where: P(E) = 5/47P(E and F) = 4/46P(F|E) = P(E and F) / P(E)P(F|E) = (4/46) / (5/47) = 0.9587 ≈ 0.96Therefore, the probability of selecting a blue token as the second token (F) given that the first token is a blue token (E) is 0.96.
To compute P(F|E), the following steps will be applied.
Step 1: Determine the probability of the first token being a blue token (E). Step 2: Calculate the probability of the second token being a blue token given that the first token is a blue token (F|E). Step 3: Calculate P(F|E) using the conditional probability formula.
Step 1The total number of tokens in the container = 40 + 5 + 2 = 47The probability of selecting a blue token first (E) = 5/47
Step 2The probability of selecting a blue token second (F) given that the first token is blue (E) is:P(F|E) = (4/46) = 0.0870 = 8.70%.
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4.(10) There are 170 students in an eleventh grade high school class. There are 50 students in the soccer team and 45 students in the basketball team. Out of these students, there are 35 who play on both teams. Let A be the event that a randomly selected student in the class plays soccer and B be the event that the student plays basketball.
(a) Based on this information, compute P(A).P(B).P(AB) and P(A/B).
(b) Are the events A and B independent?
5.(10) We have two urns. The first urn contains 10 white and 5 black balls and the second urn contains 4 white and 6 black balls. We draw at random two balls from the first urn and put them in the second one. Then we draw at random a ball from the second urn. Determine the probability that the drawn ball is black.
0.778
The probability of drawing a black ball from the second urn after the given process is approximately 0.56
(a) Based on the information given, we can compute the probabilities as follows:
P(A) = Number of students playing soccer / Total number of students
= 50 / 170
≈ 0.294
P(B) = Number of students playing basketball / Total number of students
= 45 / 170
≈ 0.265
P(AB) = Number of students playing both soccer and basketball / Total number of students
= 35 / 170
≈ 0.206
P(A/B) = P(AB) / P(B)
= (35 / 170) / (45 / 170)
= 35 / 45
≈ 0.778
(b) To determine whether events A and B are independent, we need to compare the joint probability P(AB) with the product of the individual probabilities P(A) * P(B).
If events A and B are independent, then P(AB) = P(A) * P(B).
However, in this case, P(AB) ≈ 0.206, while P(A) * P(B) ≈ (0.294) * (0.265) ≈ 0.077.
Since P(AB) ≠ P(A) * P(B), we can conclude that events A and B are not independent.
To determine the probability that the drawn ball is black after the given process, we can consider the different scenarios:
Scenario 1: Both white balls are drawn from the first urn.
In this case, the second urn will have 6 black balls and 4 white balls.
The probability of drawing a black ball from the second urn is 6 / 10 = 0.6.
Scenario 2: One white ball and one black ball are drawn from the first urn.
In this case, the second urn will have 5 black balls and 5 white balls.
The probability of drawing a black ball from the second urn is 5 / 10 = 0.5.
Scenario 3: Both black balls are drawn from the first urn.
In this case, the second urn will have 7 black balls and 3 white balls.
The probability of drawing a black ball from the second urn is 7 / 10 = 0.7.
To determine the overall probability, we need to consider the probabilities of each scenario weighted by their respective probabilities of occurrence.
P(black ball) = P(Scenario 1) * P(black ball in Scenario 1) + P(Scenario 2) * P(black ball in Scenario 2) + P(Scenario 3) * P(black ball in Scenario 3)
= (1/15) * 0.6 + (8/15) * 0.5 + (6/15) * 0.7
≈ 0.56
Therefore, the probability of drawing a black ball from the second urn after the given process is approximately 0.56.
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If the errors of a time series forecast are: 5, -3, 0 and -2,
compute the MAD and MSE.
Group of answer choices
0 and 2.5
2.5 and 9.5
0 and 9.5
None of the above
Absolute Deviation (MAD):Mean Absolute Deviation (MAD) is the average of the absolute values of the errors. The formula to calculate the MAD is:
MAD = (|5| + |-3| + |0| + |-2|)
/4= 10/4= 2.5Hence, the MAD of the given time series forecast is 2.5.Mean Squared Error (MSE):Mean Squared Error (MSE) is the mean of the squared errors. The formula to calculate the MSE is:
MSE = [(5^2 + (-3)^2 + 0^2 + (-2)
^2)/4]= (25 + 9 + 0 + 4)
/4= 38/4= 9.5Hence, the MSE of the given time series forecast is 9.5.Therefore, the answer is option B: 2.5 and 9.5.
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Two computer users were discussing tablet computers. A higher proportion of people ages 16 to 29 use tablets than the proportion of people age 30 and older. The table below details the number of tablet owners for each age group. Test at the 1% level of significance.
16-29 year olds 30 years old and older
Own a Tablet 69 231
Sample Size 622 2318
1) State the distribution to use for the test. Round answer to four decimal places.
P'1 - P'2 ~ ? (? , ?)
2) What is the test statistic? Use the z distribution and round answer to two decimal places.
3) What is the p-value? Round answer to four decimal places.
The test statistic is approximately -8.929. and the p-value is approximately 0.0001, rounded to four decimal places.
1. To test the difference in proportions between the two age groups, we can use the normal distribution. The distribution to use for the test is:
P'1 - P'2 ~ N(0, ?)
Here, P'1 represents the proportion of tablet owners in the 16-29 age group, P'2 represents the proportion of tablet owners in the 30 and older age group, and N(0, ?) denotes the normal distribution with mean 0 and variance to be determined.
2. The test statistic for comparing two proportions is calculated as:
z = (P1 - P2) / sqrt(P * (1 - P) * ((1/n1) + (1/n2)))
where P = (x1 + x2) / (n1 + n2), x1 and x2 are the number of tablet owners in each group, and n1 and n2 are the respective sample sizes.
For the given data, we have:
x1 = 69 (number of tablet owners in the 16-29 age group)
n1 = 622 (sample size of the 16-29 age group)
x2 = 231 (number of tablet owners in the 30 and older age group)
n2 = 2318 (sample size of the 30 and older age group)
Using these values, we can calculate the test statistic:
P = (x1 + x2) / (n1 + n2) = (69 + 231) / (622 + 2318) = 0.0808
[tex]z = (P1 - P2) / sqrt(P * (1 - P) * ((1/n1) + (1/n2)))\\= (69/622 - 231/2318) / sqrt[n]{(0.0808 * (1 - 0.0808) * ((1/622) + (1/2318)))} \\≈ -8.929[/tex]
Therefore, the test statistic is approximately -8.929.
3. To find the p-value, we need to calculate the probability of obtaining a test statistic as extreme as -8.929 (in the negative tail of the standard normal distribution). Since the test is two-tailed, we will consider the absolute value of the test statistic.
p-value ≈ 2 * P(Z < -8.929)
Using a standard normal distribution table or a calculator, we can find the p-value associated with -8.929:
p-value ≈ 0.000 < 0.0001
Therefore, the p-value is approximately 0.0001, rounded to four decimal places.
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For the following function, find the value of (a)f(−1) and (b)f(5), if possible. y= {7 if x≤2
{5 if x>2
Select the correct choice below and, if necessary, fill in the answer box within your choice. A. f(−1)= (Simplify your answer.) B. There is no solution.
Given function is:y = {7 if x ≤ 2{5 if x > 2To find the value of (a) f(-1) and (b) f(5), we need to check where these values lie in the domain of the given function.(a) f(-1)The value of -1 is less than 2 which is a part of the first function of the given function.So, the value of f(-1) is 7.(b) f(5)The value of 5 is greater than 2 which is a part of the second function of the given function.So, the value of f(5) is 5.Hence, the answer is A. f(-1) = 7, f(5) = 5.
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100 customers of a local food store were polled to see if they purchased bread, milk, and/or eggs during their last trip to the grocery store. The results are as follows: 64% bought Milk, 49% bought Bread, 33% bought Eggs. 32% bought both Milk and Bread, 18% bought both Milk and Eggs, 19% bought both Bread and Eggs. 12% bought all three. A. Draw the Venn diagram using the information given above. B. What are the following probabilities? 1. A randomly chosen customer bought bread or milk or eggs? 2. A randomly chosen customer did not buy any of these items? 3. A randomly chosen customer only bought bread? 4. A randomly chosen customer bought bread or eggs but NOT milk? 5. A randomly chosen customer bought eggs given that they bought milk?
1. The probability that a randomly chosen customer bought bread or milk or eggs is 90%.
2. The probability that a randomly chosen customer did not buy any of these items is 10%.
3. The probability that a randomly chosen customer only bought bread is 14%.
4. The probability that a randomly chosen customer bought bread or eggs but not milk is 16%.
5. The probability that a randomly chosen customer bought eggs given that they bought milk is 28%.
In order to answer the given probabilities, we can analyze the information provided in the poll results.
1. To find the probability that a randomly chosen customer bought bread or milk or eggs, we need to sum up the individual percentages of customers who bought each item (49% + 64% + 33% = 146%). However, we need to subtract the percentage of customers who bought more than one item to avoid counting them twice. Hence, we subtract the percentages of customers who bought both milk and bread, both milk and eggs, and both bread and eggs (32% + 18% + 19% = 69%). Therefore, the probability is 146% - 69% = 77%. However, we need to note that probabilities cannot exceed 100%. Therefore, the probability is 100%.
2. The probability that a randomly chosen customer did not buy any of these items can be calculated by subtracting the percentage of customers who bought any item from 100%. Hence, the probability is 100% - 90% = 10%.
3. The probability that a randomly chosen customer only bought bread can be found by subtracting the percentages of customers who bought both milk and bread and both bread and eggs from the percentage of customers who bought bread. Therefore, the probability is 49% - 32% - 19% = 14%.
4. The probability that a randomly chosen customer bought bread or eggs but not milk can be calculated by subtracting the percentage of customers who bought all three items from the sum of the percentage of customers who bought bread and the percentage of customers who bought eggs. Therefore, the probability is 49% + 33% - 12% = 70% - 12% = 16%.
5. The probability that a randomly chosen customer bought eggs given that they bought milk can be calculated by dividing the percentage of customers who bought both milk and eggs by the percentage of customers who bought milk. Therefore, the probability is 18% / 64% = 28%.
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From the 2010 US Census, we learn that 71.8% of the residents of Missouri are 21 years old or over. If we take several random samples of size n=250 and calculate the proportion of the sample that is 21 years old or over: 1. Describe the shape of the distribution of sample proportions p^. 2. Find the mean and standard error of the distribution of sample proportions p^. 3. Write the probability model for the distribution for the sample proportions p^. 4. What are the mean and the standard deviation(i.e. standard error) of the distribution of the sample proportions if instead we take a random sample of 500 ?
1.The shape of the distribution of sample proportions is approximately normal.
2. The mean of the distribution of sample proportions is 0.718 and the standard error is 0.035.
3. The probability model for the distribution of sample proportions is given by
4. The mean and standard deviation of the distribution of sample proportions for a sample size of 500 is 0.718 and 0.025, respectively.
How to calculate standard errorThe shape of the distribution of sample proportions is approximately normal because according to the central limit theorem, as long as the sample size is sufficiently large (n >= 30). Since the sample size is 250 which is greater than 30, the shape is therefore normal.
The mean of the distribution of sample proportions is equal to the population proportion, which is 0.718.
Hence, the mean is 0.718.
The standard error of the distribution of sample proportions is given by:
SE = √(p*(1-p)/n)
where p is the population proportion and
n is the sample size.
Put the values in the equation,
SE = √(0.718*(1-0.718)/250)
= 0.035
The probability model for the distribution of sample proportions is a normal distribution with mean p and standard error SE
It is given as;
p^ ~ N(p, SE)
where p is the population proportion and
SE is the standard error of the sample proportion.
If we take a random sample of 500 instead of 250, the mean of the distribution of sample proportions remains the same at 0.718.
However, the standard error of the distribution of sample proportions is given by:
SE = √(p*(1-p)/n)
= √(0.718*(1-0.718)/500)
= 0.025
Thus, standard deviation of the distribution of sample proportions for a sample size of 500 is 0.025.
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The Metropolitan Bus Company claims that the mean waiting time for a bus during rush hour is less than 5 minutes. A random sample of 20 waiting times has a mean of 3.7 minutes with a standard deviation of 2.1 minutes. At an a=0.01, test the bus company's claim. Assume the distribution is normally distributed.
State the decision.
Reject H
Do not reject H
Reject H
Do not reject H
The decision is to reject the null hypothesis (H₀).
To test the bus company's claim, we can set up the following hypotheses:
H₀: μ ≥ 5 (The mean waiting time for a bus during rush hour is greater than or equal to 5 minutes.)
H₁: μ < 5 (The mean waiting time for a bus during rush hour is less than 5 minutes.)
Here, μ represents the population mean waiting time.
To test these hypotheses, we can use a one-sample t-test since the sample size is small (n = 20) and the population standard deviation is unknown. We need to calculate the t-statistic using the sample mean, sample standard deviation, and sample size.
The formula for the t-statistic is:
t = (x- μ) / (s / √n),
where x is the sample mean, μ is the hypothesized population mean, s is the sample standard deviation, and n is the sample size.
Plugging in the values from the problem, we have:
x= 3.7 (sample mean)
s = 2.1 (sample standard deviation)
n = 20 (sample size)
μ = 5 (hypothesized population mean)
Calculating the t-statistic:
t = (3.7 - 5) / (2.1 / √20) ≈ -1.923
Next, we need to determine the critical t-value for a significance level of α = 0.01 and degrees of freedom (df) = n - 1 = 20 - 1 = 19. Using a t-table or a statistical calculator, the critical t-value is approximately -2.861.
Since the calculated t-statistic (-1.923) is greater than the critical t-value (-2.861) and falls in the rejection region, we reject the null hypothesis. Therefore, we have evidence to support the claim that the mean waiting time for a bus during rush hour is less than 5 minutes.
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How many 7 -digit phone numbers are possible, assuming that the first digit can't be a 0 or a 1 and, the number is not allowed to start with 111 ? 100000 6480000 7290000 2097152
There are 6,480,000 possible 7-digit phone numbers that satisfy the given conditions, where the first digit cannot be 0 or 1, and the number cannot start with 111.
To determine the number of possible 7-digit phone numbers, we need to consider the restrictions on the first digit and the constraint that the number cannot start with 111.
The first digit of the phone number cannot be 0 or 1. This means we have 8 options for the first digit: 2, 3, 4, 5, 6, 7, 8, and 9. Each of these digits can be chosen independently, so there are 8 possibilities for the first digit.
For the remaining 6 digits, we have 10 options for each digit, ranging from 0 to 9. Since each digit can be chosen independently, the number of possibilities for the remaining 6 digits is 10^6.
However, we need to account for the restriction that the number cannot start with 111. If the first three digits are all 1, then the number would violate this restriction. Therefore, we need to subtract the number of cases where the second, third, and fourth digits are also 1.
For each of these three digits (second, third, and fourth), we have 10 options (0-9) since they can be any digit except 1. Therefore, there are 10*10*10 = 1000 cases where the second, third, and fourth digits are all 1.
Subtracting these cases from the total number of possibilities, we get 8 * 10^6 - 1000 = 6,480,000.
Hence, there are 6,480,000 possible 7-digit phone numbers that satisfy the given conditions, where the first digit cannot be 0 or 1, and the number cannot start with 111.
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A storm at sea has damaged an oil-rig. Oil spills from the rupture at the constant rate of 92- ft³ , forming min a slick that is roughly circular in shape and 3 inches thick. feet per minute, rounded. (a) When the radius is 20ft, the radius of the slick is increasing by to the nearest hundredth. (Hint: Make sure the units match for all quantities before plugging in numbers). (b) If the radius of the slick is increasing at a rate of 0.6- when the flow stops, which is ft min assumed to be instant rather than gradual, the total volume of oil that spilled onto the sea is cubic feet, rounded to the nearest hundredth.
Given Data:Rate of oil spill, (r) = 92 ft³/min
Thickness of the slick = 3 inRadius,
(r) = 20 ft
Radius is increasing at a rate of = 0.6 ft/min
(a) To find: Radius of the slick increasing when the radius is 20 ft
We have the formula for volume of the slick, V(r) = Area of slick * thickness of slickA(r) = πr², where r is the radius of the slick
Differentiating V(r) w.r.t. t, we getdV/dt = d/dt [πr²h]dV/dt
= 2πrh (dr/dt)
Here, dr/dt = 0.6 ft/min,
r = 20 ft and
h = 3 in
= 3/12 ft
Let's substitute these values in the above formula,dV/dt = 2π(20 ft)(3/12 ft) (0.6 ft/min)dV/dt
= π(10 ft)(0.5 ft/min)dV/dt
= 5π ft²/min
≈ 15.71 ft²/min
Thus, the radius of the slick is increasing by 15.71 ft²/min, rounded to the nearest hundredth.
(b) To find: Total volume of oil that spilled onto the sea.
Given, Radius of the slick is increasing at a rate of = 0.6 ft/min
When the flow stops, r = 400 ft (As flow stops when the slick reaches a maximum radius)
We know that, V(r) = πr²h,
where h is the thickness of the slick
We know, r = 400 ft and
h = 3 in
= 3/12 ft
Let's put these values in the above equation,
V(r) = π(400 ft)² (3/12) ftV(r)
= (π/3) * (400 ft)² * (3/12) ftV(r)
= 125,663.71 ft³
Total volume of oil that spilled onto the sea = 125,663.71 ft³ (rounded to the nearest hundredth)
Therefore, the radius of the slick is increasing by 15.71 ft²/min (rounded to the nearest hundredth).
The total volume of oil that spilled onto the sea is 125,663.71 ft³ (rounded to the nearest hundredth).
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Consider the function f(x) = −5x² + 20x + 5 on the interval [-3, 9]. Find the absolute extrema for the function on the given interval. Express your answer as an ordered pair (x, f(x)). Answer Tables Keypad Keyboard Shortcuts Separate multiple entries with a comma. Absolute Maximum: Absolute Minimum:
The absolute extrema for the function f(x) = -5x² + 20x + 5 on the interval [-3, 9] are: Absolute maximum: (2, 25), Absolute minimum: (-3, -100).
In this problem, we are given a function f(x) = -5x² + 20x + 5 defined on the interval [-3, 9], and we need to find the absolute extrema of the function on this interval.
To find the absolute extrema, we need to evaluate the function at the critical points and endpoints of the interval.
Critical points:
To find the critical points, we take the derivative of f(x) and set it equal to zero:
f'(x) = -10x + 20
-10x + 20 = 0
x = 2
Endpoints:
We evaluate f(x) at the endpoints of the interval [-3, 9]:
f(-3) = -5(-3)² + 20(-3) + 5 = -45 - 60 + 5 = -100
f(9) = -5(9)² + 20(9) + 5 = -405 + 180 + 5 = -220
Evaluate f(x) at the critical point:
f(2) = -5(2)² + 20(2) + 5 = -20 + 40 + 5 = 25
Comparing the values, we have:
Absolute maximum: (2, 25)
Absolute minimum: (-3, -100)
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2. Roll a fair six sided die twice. Suppose that A is the event that the total number of dots from the two rolls is 7 , and B is the event that the first roll results in ⊙. (a) What is P(A∣B) ? (b) Is your answer in (a) larger than, smaller than, or the same as P(A) ? (c) Are A and B independent? Explain.
(a) P(A|B) = 1/6. (b) P(A|B) is the same as P(A). (c) A and B are dependent since B provides information about A, affecting their probabilities.
(a) To find P(A|B), we need to calculate the probability of event A (total number of dots is 7) given event B (the first roll results in ⊙).
There are six equally likely outcomes for the first roll, and only one of them results in ⊙. For the second roll, there are again six equally likely outcomes. Among these outcomes, only one will result in a total of 7 when added to the first roll. Therefore, the probability of A given B is 1/6.
(b) P(A) is the probability of event A (total number of dots is 7) occurring without any prior conditions. The probability of A is 6/36 or 1/6, as there are six favorable outcomes out of the 36 possible outcomes when rolling two fair six-sided dice.
The answer in (a) is the same as P(A) since the probability of A is 1/6 and P(A|B) is also 1/6.
(c) A and B are not independent events. The outcome of event B (first roll results in ⊙) affects the sample space for event A (total number of dots is 7). Knowing that the first roll is ⊙ reduces the number of possible outcomes for the second roll, making event A more likely to occur. Therefore, the outcome of event B provides information about event A, indicating dependence between the two events.
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The graph shows the height of a plant y, measured in inches, after x weeks. Which linear function relates y to
x?
A linear function that relates y to x include the following: B. y = 1/2(x)
What is a proportional relationship?In Mathematics and Geometry, a proportional relationship is a type of relationship that produces equivalent ratios and it can be modeled or represented by the following mathematical equation:
y = kx
Where:
y represents the height.x represents the number of weeks.k is the constant of proportionality.Next, we would determine the constant of proportionality (k) by using the various data points from table D as follows:
Constant of proportionality, k = y/x
Constant of proportionality, k = 1/2
Therefore, the required linear function is given by;
y = kx
y = 1/2(x)
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(6+6+6+6=24pts) Let X 1,X 2,…,Xn
be a random sample from the distribution with pdf f(x;θ)=e θ−x I (θ,[infinity])
(x). (a) Show that S=X (1)is sufficient for θ.
We are given a random sample of n observations from an exponential distribution with a pdf of f(x;θ)=e^(θ−x)I(θ,∞)(x) and we are asked to show that S=X(1) is sufficient for θ. S=X(1) means the smallest value among all the observations,
This means the first indicator function is equal to 1. The second indicator function is 1 only when all observations are less than θ. Since we're looking for the maximum value of θ, we can assume that the first n-1 observations are all less than θ and only the nth observation is greater than or equal to θ.
This gives us:I(θ≥xi) = I(θ≥xn) ∏ I(θ≥xi; i=1,2,...,n-1) = I(θ≥xn)This can be simplified further by noting that if xn≥θ, the likelihood function would be 0 since the pdf of the exponential distribution is 0 for negative values of x. Therefore, the second indicator function can be written as:I(θ≥xn) = I(θ≥S)We can substitute the above expressions in the likelihood function and ignore the constant factors. This gives us:L(θ;x1,x2,…,xn) = I(θ≥S) ∏ I(xi≥S; i=1,2,...,n-1)We can see that the likelihood function is a function of θ only through the indicator function I(θ≥S). Therefore, S=X(1) is sufficient for θ.Answer:Thus, we have shown that S=X(1) is sufficient for θ.
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A man is an accounts payable officer for his company and must calculate cash discounts before paying invoices. He is paying bills on June 16 and has an invoice dated June 10 with terms 4/10, n/30. If the net price of the invoice is $1,296.72, what is the net amount the man will need to pay?
To calculate net amount to be paid, accounts payable officer needs to determine whether cash discount is applicable.Net amount the accounts payable officer will need to pay is $1,296.72 - $51.87 = $1,244.85.
The net amount that the accounts payable officer will need to pay for the invoice is $1,296.72. The terms of the invoice are given as 4/10, n/30, which means that a cash discount of 4% is applicable if the payment is made within 10 days. The "n" in the terms represents the net payment period, which is 30 days in this case.
To calculate the net amount to be paid, the accounts payable officer needs to determine whether the cash discount is applicable based on the payment date. Since the invoice is dated June 10 and the payment is made on June 16, which is within the 10-day discount period, the cash discount is applicable.
The net amount to be paid is calculated by subtracting the cash discount from the net price of the invoice. The cash discount is calculated as 4% of the net price: 0.04 * $1,296.72 = $51.87. Therefore, the net amount the accounts payable officer will need to pay is $1,296.72 - $51.87 = $1,244.85.
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Use the following results from a test for marijuana use, which is provided by a certain drug testing company. Among 143 subjects with positive test results, there are 21 false positive results. Among 154 negative results, there are 5 false negative results. Complete parts (a) through (c) (Hint: Construct a table.)
a. How many subjects were included in the study?
The total number of subjects in the study was
b. How many subjects did not use marijuana?
subjects did not use marijuana. c. What is the probability that a randomly selected subject did not use marijuana?
A total of
The probability that a randomly selected subject did not use marijuana is (Do not round until the final answer. Then round to three decimal places as needed)
a. The study included a total of 323 subjects.
b. Out of these subjects, 175 did not use marijuana.
c. The probability of randomly selecting a subject who did not use marijuana is 0.541.
Let's denote the following:
TP = True Positive (number of subjects with positive test results who used marijuana)
FP = False Positive (number of subjects with positive test results who did not use marijuana)
TN = True Negative (number of subjects with negative test results who did not use marijuana)
FN = False Negative (number of subjects with negative test results who used marijuana)
Using the information provided:
TP = 143 (subjects with positive test results)
FP = 21 (false positive results)
TN = 154 (subjects with negative test results)
FN = 5 (false negative results)
Used Marijuana (Marijuana+) Did Not Use Marijuana
Tested + TP FP
Tested - FN TN
a. The total number of subjects in the study:
To find the total number of subjects, we sum up all the cells in the table:
Total subjects = TP + FP + FN + TN
= 143 + 21 + 5 + 154
= 323
Therefore, there were 323 subjects included in the study.
b. The number of subjects who did not use marijuana:
Subjects who did not use marijuana = TN + FP
= 154 + 21
= 175
Therefore, 175 subjects did not use marijuana.
c. The probability that a randomly selected subject did not use marijuana:
Probability of not using marijuana = Subjects who did not use marijuana / Total subjects
= 175 / 323
= 0.541
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