the conditional PDF fx|A(x|A) is (4xy / 0.96) for (0 < x < 0.2, 0 ≤ y ≤ 1). The correlation between X and Y is E[XY] = 0.0427. The covariance between X and Y is COV[XY] = -0.0986.
Conditional PDF fx|A(X|A):
To find the conditional PDF, we need to determine the range of x and y values that satisfy event A: [0 < x < 0.2].
Since the joint PDF fx(x, y) is given as 4xy for 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1, we can calculate the conditional PDF by normalizing the joint PDF over the range of x and y values satisfying event A.
First, let's find the normalization constant:
∫∫fx(x, y) dy dx = 1
∫∫4xy dy dx = 1
∫[0.2,1] ∫[0,1] 4xy dy dx = 1
4∫[0.2,1] [x/2 * y^2] [0,1] dx = 1
4∫[0.2,1] (x/2) dx = 1
2[1/2 * x^2] [0.2,1] = 1
x^2 |[0.2,1] = 1
(1^2 - 0.2^2) = 1
0.96 = 1
The normalization constant is 1/0.96.
Now, let's calculate the conditional PDF:
fx|A(x|A) = (fx(x, y) / ∫∫fx(x, y) dy dx) for (0 < x < 0.2)
fx|A(x|A) = (4xy / 0.96) for (0 < x < 0.2, 0 ≤ y ≤ 1)
Correlation E[XY]:
The correlation between X and Y can be calculated using the joint PDF:
E[XY] = ∫∫xy * fx(x, y) dy dx
E[XY] = ∫[0,0.2] ∫[0,1] xy * 4xy dy dx
E[XY] = 4 * ∫[0,0.2] ∫[0,1] x^2y^2 dy dx
E[XY] = 4 * ∫[0,0.2] (1/3)x^2 dx
E[XY] = 4 * (1/3) * [x^3/3] [0,0.2]
E[XY] = 4 * (1/3) * [(0.2)^3/3 - 0^3/3]
E[XY] = 4 * (1/3) * (0.008/3)
E[XY] = 0.0427
Covariance COV[XY]:
The covariance between X and Y can be calculated using the joint PDF:
COV[XY] = E[XY] - E[X]E[Y]
To find E[X] and E[Y], we need to calculate the marginal PDFs of X and Y:
fx(x) = ∫fx(x, y) dy
fx(x) = ∫4xy dy
fx(x) = 2x * y^2 |[0,1]
fx(x) = 2x * (1^2 - 0^2)
fx(x) = 2x
fy(y) = ∫fx(x, y) dx
fy(y) = ∫4xy dx
fy(y) = 2y * x^2 |[0,1]
fy(y) = 2y * (1^2 - 0^2)
fy(y) = 2y
Now, we can calculate E[X] and E[Y]:
E[X] = ∫x * fx(x) dx
E[X] = ∫x * 2x dx
E[X] = 2 * ∫x^2 dx
E[X] = 2 * [x^3/3] [0,1]
E[X] = 2 * (1/3 - 0/3)
E[X] = 2/3
E[Y] = ∫y * fy(y) dy
E[Y] = ∫y * 2y dy
E[Y] = 2 * ∫y^2 dy
E[Y] = 2 * [y^3/3] [0,1]
E[Y] = 2 * (1/3 - 0/3)
E[Y] = 2/3
Now, we can calculate the covariance:
COV[XY] = E[XY] - E[X]E[Y]
COV[XY] = 0.0427 - (2/3)(2/3)
COV[XY] = 0.0427 - 4/9
COV[XY] = -0.0986
Conclusion:
Based on the calculations, the conditional PDF fx|A(x|A) is (4xy / 0.96) for (0 < x < 0.2, 0 ≤ y ≤ 1). The correlation between X and Y is E[XY] = 0.0427. The covariance between X and Y is COV[XY] = -0.0986.
To determine whether X and Y are independent, we can compare the covariance with zero. Since COV[XY] is not equal to zero (-0.0986 ≠ 0), we can conclude that X and Y are dependent variables.
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problem 1.3 let pxnqn0,1,... be a markov chain with state space s t1, 2, 3u and transition probability matrix p 0.5 0.4 0.1 0.3 0.4 0.3 0.2 0.3 0.5 . compute the stationary distribution π.
To compute the stationary distribution π of the given Markov chain, we need to solve the equation πP = π, where P is the transition probability matrix.
The stationary distribution represents the long-term probabilities of being in each state of the Markov chain.
Let's denote the stationary distribution as π = (π1, π2, π3), where πi represents the probability of being in state i. We can set up the equation πP = π as follows:
π1 * 0.5 + π2 * 0.4 + π3 * 0.1 = π1
π1 * 0.3 + π2 * 0.4 + π3 * 0.3 = π2
π1 * 0.2 + π2 * 0.3 + π3 * 0.5 = π3
Simplifying the equations, we have:
0.5π1 + 0.4π2 + 0.1π3 = π1
0.3π1 + 0.4π2 + 0.3π3 = π2
0.2π1 + 0.3π2 + 0.5π3 = π3
Rearranging the terms, we get:
0.5π1 - π1 + 0.4π2 + 0.1π3 = 0
0.3π1 + 0.4π2 - π2 + 0.3π3 = 0
0.2π1 + 0.3π2 + 0.5π3 - π3 = 0
Simplifying further, we have the system of equations:
-0.5π1 + 0.4π2 + 0.1π3 = 0
0.3π1 - 0.6π2 + 0.3π3 = 0
0.2π1 + 0.3π2 - 0.5π3 = 0
Solving this system of equations, we can find the values of π1, π2, and π3, which represent the stationary distribution π of the Markov chain.
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suppose f(x,y,z)=x2 y2 z2 and w is the solid cylinder with height 5 and base radius 5 that is centered about the z-axis with its base at z=−1. enter θ as theta.
Suppose [tex]f(x,y,z)=x²y²z²[/tex] and w is the solid cylinder with height 5 and base radius 5 that is centered about the z-axis with its base at z = −1.
Let us evaluate the triple integral[tex]∭w f(x, y, z) dV[/tex]by expressing it in cylindrical coordinates.
The cylindrical coordinates of a point in three-dimensional space are represented by (r, θ, z).Here, the base of the cylinder is at z = -1, and the cylinder is symmetric about the z-axis. As a result, the range for z is -1 ≤ z ≤ 4. Because the cylinder is centered about the z-axis, the range of θ is 0 ≤ θ ≤ 2π.
The radius of the cylinder is 5 units, and it is centered about the z-axis. As a result, r ranges from 0 to 5.
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Describe a data set that you could collect with ordinal level of
measurement. Include where and how you could get this data.
The data set can be collected through surveys, interviews, or observations of patient behavior, and the data could be used to evaluate the quality of medical care provided by hospitals.
Ordinal degree of estimation is a scale used to quantify factors with various classes, every one of which is given an inconsistent positioning in light of its relative position. This degree of estimation is especially valuable in getting information for consumer loyalty overviews, for example, eatery or lodging audits, as well as in estimating mental develops like misery and tension.
The patient's level of satisfaction with hospital medical care is one example of a data set that could be gathered using an ordinal level of measurement. This informational collection will quantify patient fulfillment utilizing scales that action angles like correspondence, tidiness, and idealness of care. The reactions from the patients will be positioned by their degree of understanding, which will go from firmly consent to differ emphatically. The information could be gathered from a clinical office or clinic.
A survey that could be given to the patients directly while they are in the hospital or distributed to them online can be used to collect the data. The data can also be gathered by interviewing patients after they have received treatment or by observing how they act while they are in the hospital. To summarize, patient satisfaction with hospital medical care is a data set that can be gathered using the ordinal level of measurement. The data set can be gathered through surveys, interviews, or observations of patient behavior, and it could be used to assess the quality of hospital medical care.
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suppose the random variables and have joint distribution as follows: find the marginal distributions.
To find the marginal distributions of two random variables with a joint distribution, we need to sum up the probabilities across all possible values of one variable while keeping the other variable fixed. In this case, we can calculate the marginal distributions by summing the joint probabilities along the rows and columns of the given joint distribution table.
The marginal distribution of a random variable refers to the probability distribution of that variable alone, without considering the other variables. In this case, let's denote the random variables as X and Y. To find the marginal distribution of X, we sum up the probabilities of X across all possible values while keeping Y fixed. This can be done by summing the values in each row of the joint distribution table. The resulting values will give us the marginal distribution of X.
Similarly, to find the marginal distribution of Y, we sum up the probabilities of Y across all possible values while keeping X fixed. This can be done by summing the values in each column of the table. The resulting values will give us the marginal distribution of Y.
By calculatijoint distributionng the marginal distributions, we obtain the individual probability distributions of X and Y, which provide information about the likelihood of each variable taking
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According to a recent survey, 69 percent of the residents of a certain state who are age 25 years or older have a bachelor’s degree. A random sample of 50 residents of the state, age 25 years or older, will be selected. Let the random variable B represent the number in the sample who have a bachelor’s degree. What is the probability that B will equal 40 ?
According to the given information, the random variable B represents the number of residents who are at least 25 years old and have a bachelor’s degree out of the sample of 50 residents selected randomly.
To find the probability that B will equal 40, we will use the formula for the binomial distribution which is given as:P(B = k) = (nCk) * p^k * q^(n-k)Where,Binomial probability is denoted by P.B is the random variable whose value we have to find.n is the number of independent trials.k is the number of successful trials.p is the probability of success.q is the probability of failure.nCk is the number of combinations of n things taken k at a time.
Now, let's substitute the given values in the formula:P(B = 40) = (50C40) * 0.69^40 * (1-0.69)^(50-40)Now,50C40 = (50!)/(40! * (50-40)!)50C40 = 1144130400/8472886094430.69^40 = 2.4483 × 10^(-7)(1-0.69)^(50-40) = 0.0904Substituting all the given values we get:P(B = 40) = (1144130400/847288609443) * 2.4483 × 10^(-7) * 0.0904P(B = 40) = 0.0343Therefore, the probability that B will equal 40 is 0.0343.
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Are births equally likely across the days of the week? A random sample of 150 births give the following sample distribution: (Day of the week) (Sunday) (Monday) (Tuesday) (Wednesday) (Thursday) (Friday) (Saturday) Count 11 27 23 26 21 29 13 a. State the appropriate hypotheses. b. Calculate the expected count for each of the possible outcomes. c. Calculate the value of the chi-square test statistic. d. Which degrees of freedom should you use? e. Use Table C to find the p-value. What conclusion would you make?
Based on the p-value, we can make a conclusion about the null hypothesis. If the p-value is below a certain significance level (e.g., 0.05), we would reject the null hypothesis and conclude that births are not equally likely across the days of the week.
a. State the appropriate hypotheses:
The appropriate hypotheses for this problem are:
Null hypothesis (H₀): Births are equally likely across the days of the week.
Alternative hypothesis (H₁): Births are not equally likely across the days of the week.
b. Calculate the expected count for each of the possible outcomes:
To calculate the expected count for each day of the week, we need to determine the expected probability for each day and multiply it by the sample size.
Total count: 11 + 27 + 23 + 26 + 21 + 29 + 13 = 150
Expected probability for each day: 1/7 (since there are 7 days in a week)
Expected count for each day: (1/7) * 150 = 21.43
c. Calculate the value of the chi-square test statistic:
The chi-square test statistic can be calculated using the formula:
χ² = Σ((Observed count - Expected count)² / Expected count)
Using the observed counts from the given sample distribution and the expected count calculated in step (b), we can calculate the chi-square test statistic:
χ² = [(11-21.43)²/21.43] + [(27-21.43)²/21.43] + [(23-21.43)²/21.43] + [(26-21.43)²/21.43] + [(21-21.43)²/21.43] + [(29-21.43)²/21.43] + [(13-21.43)²/21.43]
Calculating this expression will give the value of the chi-square test statistic.
d. Degrees of freedom:
The degrees of freedom for a chi-square test in this case would be (number of categories - 1). Since we have 7 days of the week, the degrees of freedom would be 7 - 1 = 6.
e. Use Table C to find the p-value:
Using the calculated chi-square test statistic and the degrees of freedom, we can find the corresponding p-value from Table C of the chi-square distribution.
Consulting Table C with 6 degrees of freedom, we can find the critical chi-square value that corresponds to the calculated test statistic. By comparing the test statistic to the critical value, we can determine the p-value.
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Solve the following LP problem graphically using level curves. (Round your answers to two decimal places.) MAX: 5X₁ + 7X₂ Subject to: 3X₁ + 8X₂ ≤ 48 12X₁ + 11X₂ ≤ 132 2X₁ + 3X₂ ≤
The calculated value of the maximum value of the objective function is 61.92
Finding the maximum possible value of the objective functionFrom the question, we have the following parameters that can be used in our computation:
Objective function, 5X₁ + 7X₂
Subject to
3X₁ + 8X₂ ≤ 48
12X₁ + 11X₂ ≤ 132
2X₁ + 3X₂ ≤ 24
Next, we plot the graph (see attachment)
The coordinates of the feasible region are
(6.86, 3.43), (8.38, 2.86) and (9.43, 1.71)
Substitute these coordinates in the above equation, so, we have the following representation
5(6.86) + 7(3.43) = 58.31
5(8.38) + 7(2.86) = 61.92
5(9.43) + 7(1.71) = 59.12
The maximum value above is 61.92 at (8.38, 2.86)
Hence, the maximum value of the objective function is 61.92
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5. Given PA() = 0.4, P(B) = 0.55 and P(A n B) = 0.1 Find: (a) P(A' B') (b) P(A' | B) (c) P(B' A') (d) P(B' |A)
For the given probabilities,
(a) P(A' B') = 0.15
(b) P(A' | B) ≈ 0.818
(c) P(B' A') = 0.15
(d) P(B' | A) = 0.75
(a) P(A' B') can be calculated using the complement rule:
P(A' B') = 1 - P(A ∪ B)
= 1 - [P(A) + P(B) - P(A ∩ B)]
= 1 - [0.4 + 0.55 - 0.1]
= 1 - 0.85
= 0.15
(b) P(A' | B) can be calculated using the conditional probability formula:
P(A' | B) = P(A' ∩ B) / P(B)
= [P(B) - P(A ∩ B)] / P(B)
= (0.55 - 0.1) / 0.55
= 0.45 / 0.55
≈ 0.818
(c) P(B' A') can be calculated using the complement rule:
P(B' A') = 1 - P(B ∪ A)
= 1 - [P(B) + P(A) - P(B ∩ A)]
= 1 - [0.55 + 0.4 - 0.1]
= 1 - 0.85
= 0.15
(d) P(B' | A) can be calculated using the conditional probability formula:
P(B' | A) = P(B' ∩ A) / P(A)
= [P(A) - P(B ∩ A)] / P(A)
= (0.4 - 0.1) / 0.4
= 0.3 / 0.4
= 0.75
Therefore,
(a) P(A' B') = 0.15
(b) P(A' | B) ≈ 0.818
(c) P(B' A') = 0.15
(d) P(B' | A) = 0.75
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A smart phone
manufacturer wants to find out what proportion of its customers are
dissatisfied with the service received from their local
distributor. The manufacturer surveys a random sample of 65
cu
Smartphone manufacturer conducts survey to determine customers' satisfaction with service A smartphone manufacturer can use a random sampling technique to determine the percentage of customers who are dissatisfied with the services received from the local distributor.
The survey should aim to represent all smartphone users who have purchased their devices from the local distributor. A survey is a method of collecting data from a population, and in this case, the target population is smartphone users who have bought their phones from the local distributor.
The smartphone manufacturer can use a sample size calculator to determine the sample size required to achieve a margin of error that meets the survey's purpose. The sample size calculator considers the population size, level of confidence, margin of error, and population proportion to determine the required sample size.
With a margin of error of 5% and a 95% level of confidence, a sample size of 65 would be sufficient to represent the entire population.With the survey results, the smartphone manufacturer can determine the percentage of customers who are dissatisfied with the services provided by the local distributor.
If a significant percentage of customers are not satisfied with the service, the smartphone manufacturer can take corrective measures such as finding a new local distributor or working with the existing distributor to improve the service quality.
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Suppose that A and B are events on the same sample space with PlA) = 0.5, P(B) = 0.2 and P(AB) = 0.1. Let X =?+1B be the random variable that counts how many of the events A and B occur. Find Var(X)
The variance of X is 0.09.
Formula used: Variance is the square of the standard deviation. T
he formula to calculate variance of a discrete random variable X is given by:
Var(X) = E[X²] - [E(X)]²Calculation:
P(B) = 0.2P(A)
= 0.5P(AB) =
0.1
By definition,
P(A U B) = P(A) + P(B) - P(AB)
⇒ P(A U B) = 0.5 + 0.2 - 0.1
⇒ P(A U B) = 0.6
Now,E[X] = E[1B + ?]
⇒ E[X] = E[1B] + E[?]
Since 1B can have two values 0 and 1.
So,E[1B] = 1*P(B) + 0*(1 - P(B))
= P(B)
= 0.2P(A/B)
= P(AB)/P(B)
⇒ P(A/B)
= 0.1/0.2
= 0.5
So, the conditional probability distribution of ? given B is:
P(?/B) = {0.5, 0.5}
⇒ E[?] = 0.5(0) + 0.5(1)
= 0.5⇒ E[X]
= 0.2 + 0.5
=0.7
Now,E[X²] = E[(1B + ?)²]
⇒ E[X²] = E[(1B)²] + 2E[1B?] + E[?]²
Now,(1B)² can take only 2 values 0 and 1.
So,E[(1B)²] = 0²P(B) + 1²(1 - P(B))= 0.8
Also,E[1B?] = E[1B]*E[?/B]⇒ E[1B?] = P(B)*E[?/B]= 0.2 * 0.5 = 0.1
Putting the values in the equation:E[X²] = 0.8 + 2(0.1) + (0.5)²= 1.21Finally,Var(X) = E[X²] - [E(X)]²= 1.21 - (0.7)²= 0.09
Therefore, the variance of X is 0.09.
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find an equation for the parabola that has its vertex at the origin and satisfies the given condition.
directrix y = 1/2
To find the equation of a parabola with its vertex at the origin and a directrix at [tex]\(y = \frac{1}{2}\),[/tex] we can use the standard form equation for a parabola with a vertical axis.
The standard form equation for a parabola with a vertex at [tex]\((h, k)\)[/tex] is:
[tex]\[y = a(x - h)^2 + k\][/tex]
In this case, the vertex is at the origin [tex]\((0, 0)\),[/tex] so we have:
[tex]\[y = a(x - 0)^2 + 0\]\\\\\y = ax^2\][/tex]
Now, let's consider the directrix. The directrix is a horizontal line at [tex]\(y = \frac{1}{2}\),[/tex] which means the distance from any point on the parabola to the directrix should be equal to the distance from that point to the vertex.
The distance from a point [tex]\((x, y)\)[/tex] on the parabola to the directrix[tex]\(y = \frac{1}{2}\) is \(|y - \frac{1}{2}|\).[/tex] The distance from [tex]\((x, y)\) to the vertex \((0, 0)\) is \(\sqrt{x^2 + y^2}.[/tex]
According to the definition of a parabola, these two distances should be equal. Therefore, we have the equation:
[tex]\[|y - \frac{1}{2}| = \sqrt{x^2 + y^2}\][/tex]
To simplify this equation, we can square both sides to remove the square root:
[tex]\[(y - \frac{1}{2})^2 = x^2 + y^2\][/tex]
Expanding and rearranging, we get:
[tex]\[y^2 - y + \frac{1}{4} = x^2 + y^2\][/tex]
Combining the terms, we have:
[tex]\[x^2 = -y + \frac{1}{4}\][/tex]
Finally, rearranging the equation to isolate \(y\), we obtain the equation of the parabola:
[tex]\[y = -x^2 + \frac{1}{4}\][/tex]
So, the equation of the parabola with its vertex at the origin and the directrix [tex]\(y = \frac{1}{2}\) is \(y = -x^2 + \frac{1}{4}\).[/tex]
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the driving time for an individual from his home to his work is uniformly distributed between 200 to 470 seconds.
The probability that his driving time is between 350 and 400 seconds is approximately 0.185.
GThe driving time for an individual from his home to his work is uniformly distributed between 200 to 470 seconds.
To find the probability that his driving time is between 350 and 400 seconds
Let X be the driving time in seconds from his home to work, then X follows a uniform distribution between a=200 and b=470.
The probability density function of a uniform distribution is given by;`f(x) = 1/(b-a)` for `a ≤ x ≤ b`
Otherwise, `f(x) = 0`The probability that his driving time is between 350 and 400 seconds is given by;`P(350 ≤ X ≤ 400)`
We know that the uniform distribution is equally likely over the entire range of values from a to b, thus the probability of X being between any two values will be given by the ratio of the length of the interval containing those values to the length of the whole interval.
So,`P(350 ≤ X ≤ 400) = (length of the interval 350 to 400)/(length of the whole interval 200 to 470)
`Now,`Length of the interval 350 to 400 = 400 - 350 = 50 seconds``
Length of the whole interval 200 to 470 = 470 - 200 = 270 seconds`
Hence,`P(350 ≤ X ≤ 400) = (50)/(270)``P(350 ≤ X ≤ 400) ≈ 0.185`
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Here are summary statistics for randomly selected weights of newborn girls: n=174, x= 30.7 hg, s = 7.7 hg. Construct a confidence interval estimate of the mean. Use a 98% confidence level. Are these r
Therefore, we are 98% confident that the true mean weight of the newborn girl lies between 29.3306 and 32.0694 hg.
The solution to the given problem is as follows; Given summary statistics are; N = 174x = 30.7 hgs = 7.7 hg
To construct the confidence interval estimate of the mean, we will use the following formula;` CI = x ± t_(α/2) * (s/√n)` Where,α = 1 - confidence level = 1 - 0.98 = 0.02 Degrees of freedom = n - 1 = 174 - 1 = 173 (as t-value depends on the degrees of freedom)distribution is normal (because sample size > 30)
Now, to get the t-value we use the t-table which gives the critical values of t for a given confidence level and degrees of freedom. The t-value for a 98% confidence interval with 173 degrees of freedom is found in the row of the table corresponding to 98% and the column corresponding to 173 degrees of freedom.
This gives us a t-value of 2.3449. Calculating the interval estimate of the mean weight of the newborn girl;` CI = 30.7 ± 2.3449 * (7.7 / √174)`CI = 30.7 ± 2.3449 * 0.5852CI = 30.7 ± 1.3694CI = (29.3306, 32.0694)
The 98% confidence interval is (29.3306, 32.0694).
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Write the equation of the line passing through the origin and point (−3,5).
Simplifying:y = (-5/3)x The final answer is:y = (-5/3)x
To find the equation of the line passing through the origin and point (-3,5), we can use the point-slope form of the equation of a line, which is:y - y1 = m(x - x1)
where (x1, y1) is the given point and m is the slope of the line.
We know that the line passes through the origin, which is the point (0, 0).
Therefore, we have:x1 = 0 and y1 = 0We also need to find the slope of the line.
The slope of a line passing through two points (x1, y1) and (x2, y2) is given by:m = (y2 - y1)/(x2 - x1)Using the given points, we have:
x1 = 0, y1 = 0, x2 = -3, y2 = 5Substituting these values into the formula, we get:m = (5 - 0)/(-3 - 0) = -5/3
Therefore, the equation of the line passing through the origin and point (-3, 5) is:y - 0 = (-5/3)(x - 0)Simplifying:y = (-5/3)x
The final answer is:y = (-5/3)x
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The probability distribution for the random variable x follows. x 21 25 32 36 a. Is this probability distribution valid? Explain. - Select your answer - b. What is the probability that x = 32 (to 2 de
a. The probability distribution is valid since the sum of the probabilities is equal to 1, which means that the probabilities of all the possible events must add up to 1.
To check the distribution’s validity, it is necessary to add up the probability values of all the possible events. This is because a probability value that is less than 0 or more than 1 makes no sense and hence is not valid. The probabilities must also be non-negative.
Thus, we add the given probabilities together.
P(21) + P(25) + P(32) + P(36) = 0.15 + 0.25 + 0.3 + 0.15 = 0.85.
Hence, the probability distribution is valid.
b. To find the probability that x = 32 .
The probability of the random variable being equal to 32 is given as
P(x = 32) = 0.30
Therefore, the probability that x = 32 is 0.30.
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A spinner is divided into 4 sections. The spinner is spun 100 times.
The probability distribution shows the results.
What is P(2 ≤ x ≤ 4)?
Is my answer correct?
A spinner is divided into 4 sections. The spinner is spun 100 times and the probability distribution is given as follows:
Outcome 1234 Probability 0.450.200.250.10
Using the cumulative probability,
P(2 ≤ x ≤ 4) is:
P(2 ≤ x ≤ 4) = P(x = 2) + P(x = 3) + P(x = 4)P(2 ≤ x ≤ 4) = 0.2 + 0.25 + 0.1P(2 ≤ x ≤ 4) = 0.55
Therefore, the probability that the spinner lands on 2, 3 or 4 is 0.55. The answer is correct.P.S.: The question does not provide any information on how many sections the spinner has, but it gives the probability distribution of the spinner landing on each of the sections.
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hi
im not sure how to solve this one , the answers in purple are right
, i just dont know how to calculate n understand them
A survey randomly sampled 25 college students in California and asked about their opinions about online social networking 15 of them prefer the digital way of communicating with friends and family. 10
The standard error, in this case, is approximately 0.0979.
Based on the given information, we have:
Sample size (n): 25
Number of students who prefer online social networking (successes): 15
To calculate the sample proportion (p-hat), which represents the proportion of students who prefer online social networking, we divide the number of successes by the sample size:
p-hat = successes / n = 15 / 25 = 0.6
The sample proportion, in this case, is 0.6 or 60%.
To calculate the standard error (SE) of the sample proportion, we use the formula:
SE = √(p-hat * (1 - p-hat) / n)
SE = √(0.6 * (1 - 0.6) / 25) = √(0.6 * 0.4 / 25) = √(0.024 / 25) = 0.0979
The standard error, in this case, is approximately 0.0979.
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For each of the following journal articles, briefly describe the research methodology used based on the following headings: research philosophy, research approach to theory development, methodological choice, research strategy, time horizon, data analysis and presentation methods, and reliability and validity/trustworthiness [100 marks]
1. Fowler et al. (2014)
2. Chikerema & Makanyeza (2021)
3. Makanyeza & Chikazhe (2017)
4. Makanyeza & Du Toit (2017)
5. Makanyeza & Mutambayashata (2018)
6. Makanyeza (2017)
7. Musenze & Mayende (2019)
8. McEachern (2015)
9. Manyati & Mutsau (2021)
10. Makanyeza, Chitambara & Kakava (2018)
The primary data collected from reliable sources and checked for accuracy of the data.
1.Fowler et al. (2014):
Research Philosophy: Constructivist
Research Approach to Theory Development: Qualitative investigation
Methodological Choice: Grounded Theory
Research Strategy: Semi-structured interviews
Time Horizon: Cross-sectional
Data Analysis and Presentation Methods: Open and axial coding with narrative analysis for reporting results
Reliability and Validity/Trustworthiness: Participant and researcher triangulation used to increase credibility
2.Chikerema & Makanyeza (2021):
Research Philosophy: Constructivist
Research Approach to Theory Development: Qualitative exploration
Methodological Choice: Phenomenological inquiry
Research Strategy: Interviews and focus group discussions combined with document review and observation
Time Horizon: Cross-sectional
Data Analysis and Presentation Methods: Thematic analysis with painting of synthesized interpretations
Reliability and Validity/Trustworthiness: Using participant and researcher triangulation to test initial and emergent findings
3.Makanyeza & Chikazhe (2017):
Research Philosophy: Constructivist
Research Approach to Theory Development: Qualitative exploration
Methodological Choice: Narrative inquiry
Research Strategy: Interviews
Time Horizon: Cross-sectional
Data Analysis and Presentation Methods: Thematic analysis with reporting of the narratives presented
Reliability and Validity/Trustworthiness: Self-check and investigator triangulation to evaluate the accuracy of the data
4.Makanyeza & Du Toit (2017):
Research Philosophy: Constructivist
Research Approach to Theory Development: Qualitative investigation
Methodological Choice: Grounded Theory
Research Strategy: Interviews and document review
Time Horizon: Cross-sectional
Data Analysis and Presentation Methods: Open, axial, and selective coding to identify themes and patterns
Reliability and Validity/Trustworthiness: Member checking and researcher triangulation to promote trustworthiness of the results
5.Makanyeza & Mutambayashata (2018):
Research Philosophy: Constructivist
Research Approach to Theory Development: Qualitative exploration
Methodological Choice: Participatory action research
Research Strategy: Semi-structured interviews, focus group discussions, and classroom observation
Time Horizon: Cross-sectional
Data Analysis and Presentation Methods: Thematic analysis involving open coding and reduction of data into core themes
Reliability and Validity/Trustworthiness: Peer review and researcher triangulation to increase credibility of the results.
6.Makanyeza (2017):
Research Philosophy: Constructivist
Research Approach to Theory Development: Qualitative exploration
Methodological Choice: Ethnography
Research Strategy: Participant observation, semi-structured interviews, and focus group discussions
Time Horizon: Cross-sectional
Data Analysis and Presentation Methods: Open coding for generating categories and themes for analysis before developing a thematic framework
Reliability and Validity/Trustworthiness: Multiple data sources and triangulation of findings for enhancing validity and reliability.
7.Musenze & Mayende (2019):
Research Philosophy: Constructivist
Research Approach to Theory Development: Qualitative investigation
Methodological Choice: Grounded Theory
Research Strategy: Interviews and document review
Time Horizon: Cross-sectional
Data Analysis and Presentation Methods: Open coding, axial coding, and analytical memoing for identifying and testing themes
Reliability and Validity/Trustworthiness: Combining participant and researcher triangulation to increase reliability and credibility of results.
8.McEachern (2015):
Research Philosophy: Postpositivist
Research Approach to Theory Development: Quantitative exploration
Methodological Choice: Panel regression analysis
Research Strategy: Secondary data analysis
Time Horizon: Longitudinal
Data Analysis and Presentation Methods: Panel regression analysis to analyse relationships between key variables over time
Reliability and Validity/Trustworthiness: Primary data collected from reliable sources and checked for accuracy of the data.
9.Manyati & Mutsau (2021):
Research Philosophy: Postpositivist
Research Approach to Theory Development: Quantitative investigation
Methodological Choice: Structural equation modelling
Research Strategy: Questionnaire survey
Time Horizon: Cross-sectional
Data Analysis and Presentation Methods: Structural equation modelling for prediction of behavioural intentions
Reliability and Validity/Trustworthiness: Reliability and validity of the underlying scales/instruments used were assessed.
Hence, the primary data collected from reliable sources and checked for accuracy of the data.
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Compute the exact value of the expression: sin( 7 ) cot ( 7 ) – 2 cos( 7 ) =
We need to find the value of this expression. In order to compute the value of the given expression, we need to first find the values of sin(7), cot(7), and cos(7).Let's find the value of sin(7) using the unit circle. Sin is defined as the ratio of the side opposite to the angle and the hypotenuse in a right-angled triangle with respect to an angle.
Given expression is sin(7) cot(7) – 2 cos(7)
We need to find the value of this expression. In order to compute the value of the given expression, we need to first find the values of sin(7), cot(7), and cos(7).Let's find the value of sin(7) using the unit circle. Sin is defined as the ratio of the side opposite to the angle and the hypotenuse in a right-angled triangle with respect to an angle. When an angle of 7 degrees is formed with the x-axis, the x and y-coordinates of the point on the unit circle are (cos 7°, sin 7°). Hence, sin(7) = 0.12 (approx.) Let's find the value of cot(7) using the definition of cotangent.
cot(7) = cos(7) / sin(7)cos(7) can be found using the unit circle.
cos(7) = 0.99 (approx.)
cot(7) = cos(7) / sin(7) = 0.99 / 0.12 = 8.25 (approx.)
Let's find the value of cos(7) using the unit circle. cos(7) = 0.99 (approx.)
Now, substituting these values in the given expression, we get:
sin(7) cot(7) – 2 cos(7)= 0.12 × 8.25 - 2 × 0.99= 0.99 (approx.)
Therefore, the value of the given expression is approximately equal to 0.99. The value of sin(7), cot(7) and cos(7) were found using the definition of sin, cot and cos and unit circle. The expression sin(7) cot(7) – 2 cos(7) was evaluated using the above values of sin(7), cot(7), and cos(7).
sin is defined as the ratio of the side opposite to the angle and the hypotenuse in a right-angled triangle with respect to an angle. When an angle of 7 degrees is formed with the x-axis, the x and y-coordinates of the point on the unit circle are (cos 7°, sin 7°). Hence, sin(7) = 0.12 (approx.)
cot(7) can be defined as the ratio of the adjacent side and opposite side of an angle in a right-angled triangle. Hence, cot(7) = cos(7) / sin(7). Cosine of an angle is defined as the ratio of the adjacent side and hypotenuse of an angle in a right-angled triangle. When an angle of 7 degrees is formed with the x-axis, the x and y-coordinates of the point on the unit circle are (cos 7°, sin 7°). Hence, cos(7) = 0.99 (approx.). Finally, substituting these values in the given expression sin(7) cot(7) – 2 cos(7), we get,0.12 × 8.25 - 2 × 0.99= 0.99 (approx.) Therefore, the value of the given expression is approximately equal to 0.99.
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where would a value separating the top 15% from the other values on the gaph of a normal distribution be found? O A. the right side of the horizontal scale of the graph O B. the center of the horizontal scale of the graph O C. the left side of the horizontal scale of the graph OD, onthe top of the curve
The correct option is A) the right side of the horizontal scale of the graph. The values separating the top 15% from the other values on the graph of a normal distribution would be found on the right side of the horizontal scale of the graph.
The normal distribution is a symmetric distribution that describes the possible values of a random variable that cluster around the mean. It is characterized by its mean and standard deviation.A standard normal distribution has a mean of zero and a standard deviation of 1. The top 15% of the values of the normal distribution would be found to the right of the mean on the horizontal scale of the graph, since the normal distribution is a bell curve symmetric about its mean.
The values on the horizontal axis are standardized scores, also known as z-scores, which represent the number of standard deviations a value is from the mean.
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limh→0f(8 h)−f(8)h, where f(x)=3x 2. if the limit does not exist enter dne.
The limit of the given function f(x)=3x² exists & its value at limh→0f(8+h)−f(8) / h is 48.
We are given the function f(x) = 3x².
We are required to calculate the following limit:
limh→0f(8+h)−f(8) / h
To solve the above limit problem, we have to substitute the values of f(x) in the limit expression.
Here, f(x) = 3x²
So, f(8+h) = 3(8+h)²
= 3(64 + 16h + h²)
= 192 + 48h + 3h²
f(8) = 3(8)²
= 3(64)
= 192
Now, we substitute these values in the limit expression:
limh→0{[3(64 + 16h + h²)] - [3(64)]} / h
limh→0{192 + 48h + 3h² - 192} / h
limh→0(48h + 3h²) / h
limh→0(3h(16 + h)) / h
limh→0(3(16 + h))
= 48
Thus, the value of the limit is 48.
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← HW 7.2 Question 4 of 12 < View Policies Current Attempt in Progress Solve the given triangle. a = 7, c = 10, ß = 21° Round your answers to one decimal place. a≈ i Y≈ b≈ 4240 i ME
We can solve the triangle using the law of sines and law of cosines. To use the law of sines, we know that sin β/ b = sin γ/ c and
sin α/ a = sin γ/ c where α, β, and γ are the angles in the triangle.
We have two equations, so we can solve for b and γ: sin β/ b = sin γ/ c
=> sin 21°/ b = sin γ/10
=> sin γ = 10sin 21° / b.
sin α/ a = sin γ/ c
=> sin α/7 = sin γ/10
=> sin γ = 10sin α/7.
Therefore, 10sin 21° / b = 10sin α/7, and we can solve for α:
sin α = 7sin 21°/b
=> α = sin-1(7sin 21°/b).
We can then use the fact that the sum of angles in a triangle is 180° to solve for γ: γ = 180° - α - β.
To use the law of cosines, we know that a² = b² + c² - 2bc cos α.
We have a, c, and α, so we can solve for b: b² = a² + c² - 2ac cos β.
We have a, b, and c, so we can solve for the perimeter, P = a + b + c, and the semiperimeter, s = P/2.
Once we have the perimeter, we can use Heron's formula to find the area: A = sqrt(s(s - a)(s - b)(s - c)).
The approximate values of a, b, and γ, rounded to one decimal place, are:a ≈ 7.6b ≈ 4.2γ ≈ 138.0°
The area is approximately A ≈ 24.2.
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help please
The company from Example IV takes three hours to interview an unqualified applicant and five hours to interview a qualified applicant. Calculate Will Murray's Probability, XIV. Negative Binomial Distr
Note that the mean is 4 hours
The standard deviation is 2.236 hours.
How is this so?The mean time to conduct all the interviews =
(3 hours/unqualified applicant) * (0.5) + (5 hours/qualified applicant) * (0.5)
= 4 hours
The standard deviation of the time to conduct all the interviews is
√((3 hours)² * (0.5)² + (5 hours)² * (0.5)²)
= 2.236 hours
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Full Question:
Although part of your question is missing, you might be referring to this full question:
The company from Example IV takes three hours to interview an unqualified applicant and five hours to interview a qualified applicant. Calculate the mean and standard deviation of the time to conduct all the interviews.
If △STU ~ △XYZ, which statements must be true? Check all that apply. ∠S ≅ ∠X ∠T ≅ ∠Y ST = XY SU = XZ
If ΔSTU ~ ΔXYZ, the statements that must be true are ∠S ≅ ∠X, ∠T ≅ ∠Y and ST/XY = SU/XZ.Two triangles are said to be similar when their corresponding angles are equal and their corresponding sides are proportional.
The symbol ≅ denotes congruence, and ~ denotes similarity. Since it is given that ΔSTU ~ ΔXYZ, it can be deduced that corresponding angles are equal, and corresponding sides are proportional. That means, ∠S ≅ ∠X and ∠T ≅ ∠Y. The similarity ratio can be expressed as the ratio of corresponding sides. In this case, the ratio of sides can be expressed as ST/XY = SU/XZ. These ratios will be equal as the corresponding sides are proportional. Hence, the statement ST = XY is false but ST/XY = SU/XZ is true. Thus, the statements that must be true are ∠S ≅ ∠X, ∠T ≅ ∠Y and ST/XY = SU/XZ.
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find the surface area of the portion of the bowl z = 6 − x 2 − y 2 that lies above the plane z = 3.
Here's the formula written in LaTeX code:
To find the surface area of the portion of the bowl [tex]\(z = 6 - x^2 - y^2\)[/tex] that lies above the plane [tex]\(z = 3\)[/tex] , we need to determine the bounds of integration and set up the surface area integral.
The given surfaces intersect when [tex]\(z = 6 - x^2 - y^2 = 3\)[/tex] , which implies [tex]\(x^2 + y^2 = 3\).[/tex]
Since the bowl lies above the plane \(z = 3\), we need to find the surface area of the portion where \(z > 3\), which corresponds to the region inside the circle \(x^2 + y^2 = 3\) in the xy-plane.
To calculate the surface area, we can use the surface area integral:
[tex]\[ \text{{Surface Area}} = \iint_S dS, \][/tex]
where [tex]\(dS\)[/tex] is the surface area element.
In this case, since the surface is given by [tex]\(z = 6 - x^2 - y^2\)[/tex] , the normal vector to the surface is [tex]\(\nabla f = (-2x, -2y, 1)\).[/tex]
The magnitude of the surface area element [tex]\(dS\)[/tex] is given by [tex]\(\|\|\nabla f\|\| dA\)[/tex] , where [tex]\(dA\)[/tex] is the area element in the xy-plane.
Therefore, the surface area integral can be written as:
[tex]\[ \text{{Surface Area}} = \iint_S \|\|\nabla f\|\| dA. \][/tex]
Substituting the values into the equation, we have:
[tex]\[ \text{{Surface Area}} = \iint_S \|\|(-2x, -2y, 1)\|\| dA. \][/tex]
Simplifying, we get:
[tex]\[ \text{{Surface Area}} = 2 \iint_S \sqrt{1 + 4x^2 + 4y^2} dA. \][/tex]
Now, we need to set up the bounds of integration for the region inside the circle [tex]\(x^2 + y^2 = 3\)[/tex] in the xy-plane.
Since the region is circular, we can use polar coordinates to simplify the integral. Let's express [tex]\(x\)[/tex] and [tex]\(y\)[/tex] in terms of polar coordinates:
[tex]\[ x = r\cos\theta, \][/tex]
[tex]\[ y = r\sin\theta. \][/tex]
The bounds of integration for [tex]\(r\)[/tex] are from 0 to [tex]\(\sqrt{3}\)[/tex] , and for [tex]\(\theta\)[/tex] are from 0 to [tex]\(2\pi\)[/tex] (a full revolution).
Now, we can rewrite the surface area integral in polar coordinates:
[tex]\[ \text{{Surface Area}} = 2 \iint_S \sqrt{1 + 4x^2 + 4y^2} dA= 2 \iint_S \sqrt{1 + 4r^2\cos^2\theta + 4r^2\sin^2\theta} r dr d\theta. \][/tex]
Simplifying further, we get:
[tex]\[ \text{{Surface Area}} = 2 \iint_S \sqrt{1 + 4r^2} r dr d\theta. \][/tex]
Integrating with respect to \(r\) first, we have:
[tex]\[ \text{{Surface Area}} = 2 \int_{\theta=0}^{2\pi} \int_{r=0}^{\sqrt{3}} \sqrt{1 + 4r^2} r dr d\theta. \][/tex]
Evaluating this double integral will give us the surface area of the portion of
the bowl above the plane [tex]\(z = 3\)[/tex].
Performing the integration, the final result will be the surface area of the portion of the bowl [tex]\(z = 6 - x^2 - y^2\)[/tex] that lies above the plane [tex]\(z = 3\)[/tex].
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When studying radioactive material, a nuclear engineer found that over 365 days, 1,000,000 radioactive atoms decayed to 977,647 radioactive atoms, so 22,353 atoms decayed during 365 days. a. Find the
The half-life of the radioactive material is approximately 242.37 days.
When studying radioactive material, a nuclear engineer found that over 365 days, 1,000,000 radioactive atoms decayed to 977,647 radioactive atoms, so 22,353 atoms decayed during 365 days. A. Find the half-life of the radioactive material. When studying radioactive material, the half-life of the material refers to the amount of time it takes for half of the radioactive material to decay.
Thus, we can determine the half-life of the radioactive material from the given data as follows:
First, we can determine the number of radioactive atoms left after half-life as:
Atoms left after one half-life = 1,000,000/2 = 500,000 atoms.
Let T represent the half-life of the material. We can use the given data to determine the amount of time it takes for half of the radioactive material to decay as follows:
977,647 = 1,000,000 (1/2)^(365/T)
Rearranging the equation above: (1/2)^(365/T) = 0.977647
Taking the natural log of both sides:
ln (1/2)^(365/T) = ln 0.977647
Using the rule that ln (a^b) = b ln (a), we can simplify the left side of the equation as:
(365/T) ln (1/2) = ln 0.977647
Solving for T, we get:
T = -365/ln (1/2) x ln (0.977647)T ≈ 242.37 days
The half-life of the radioactive material is approximately 242.37 days.
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Find f(1) for the
piece-wise function.
f(x) = -
x-2 if x <3
x-1
if x>3
· f(1) = [ ? ]
The value of f(1) for the given piece-wise Function is 1.
The piece-wise function f(x), we need to evaluate the function at x = 1. Let's consider the two cases based on the given conditions.
1. If x < 3:
In this case, f(x) = -(x - 2).
Substituting x = 1 into this expression, we have:
f(1) = -(1 - 2) = -(-1) = 1.
2. If x > 3:
In this case, f(x) = x - 1.
Since x = 1 is not greater than 3, this case does not apply to f(1).
Since x = 1 satisfies the condition x < 3, we can conclude that f(1) = 1.
Therefore, the value of f(1) for the given piece-wise function is 1.
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6) Convert a WAIS-IV IQ (Mean = 100, s = 15) of 95 to a z-score: a) -0.05 b) -0.33 c) -0.95 d) 6.33 7) A z-score of 0.5 is at what percentile? a) 25th b) 50th c) 69th d) 84th 8) Abdul obtains a score
The correct answer is c) 69th. A z-score of 0.5 corresponds to a percentile of approximately 69.15%. This means that approximately 69.15% of the data falls below the given z-score.
To convert an IQ score of 95 to a z-score, we need to use the formula:
z = (x - μ) / σ
where:
x = IQ score
μ = mean
σ = standard deviation
Given:
x = 95
μ = 100
σ = 15
Plugging in the values into the formula, we get:
z = (95 - 100) / 15
z = -0.33
Therefore, the correct answer is b) -0.33.
To determine the percentile corresponding to a z-score of 0.5, we can refer to the standard normal distribution table or use a statistical calculator.
A z-score of 0.5 corresponds to a percentile of approximately 69.15%. This means that approximately 69.15% of the data falls below the given z-score.
Therefore, the correct answer is c) 69th.
The question regarding Abdul's score seems to be incomplete. Please provide the missing information or details related to Abdul's score so that I can assist you further.
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Given below is the stem-and-leaf display representing the amount of syrup used in fountain soda machines in a day by 25 McDonald's restaurants in Northern Virginia. 911, 4, 7 100, 2, 2, 3, 8 11/1, 3, 5, 5, 6, 6, 7,7,7 12/2, 2, 3, 4, 8, 9 13|0, 2 If a percentage histogram for the amount of syrup is constructed using "9.0 but less than 10.0" as the first class, what percentage of restaurants use at least 10 gallons of syrup in a day? 24 68 80 88 O None of the above are correct.
The correct answer is: None of the above are correct.
The percentage of restaurants that use at least 10 gallons of syrup in a day is 8%.
To determine the percentage of restaurants that use at least 10 gallons of syrup in a day based on the given stem-and-leaf display, we need to analyze the data and interpret the stem-and-leaf plot.
The stem-and-leaf display represents the amount of syrup used in fountain soda machines in a day by 25 McDonald's restaurants in Northern Virginia.
Each stem represents a tens digit, and each leaf represents a ones digit. The "|" separates the stems from the leaves.
Looking at the stem-and-leaf plot, we can see that the stem "9" has one leaf, which represents the value 1.
This means that there is one restaurant that uses syrup in the range of 9.0 to 9.9 gallons.
The stem "10" has two leaves, representing the values 0 and 2.
This indicates that two restaurants use syrup in the range of 10.0 to 10.9 gallons.
To find the percentage of restaurants that use at least 10 gallons of syrup, we need to calculate the proportion of restaurants that have a stem-and-leaf value of 10 or greater.
In this case, there are two restaurants out of a total of 25 that fall into this category.
The percentage can be calculated as (number of restaurants with 10 or greater / total number of restaurants) [tex]\times[/tex] 100:
Percentage = (2 / 25) [tex]\times[/tex] 100 = 8%
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A Bayesian search was conducted by the US Navy in 1968 to locate the lost submarine, USS Scorpion. Suppose you are in charge of searching for the lost submarine. Based on its last known location, the search area has been partitioned into the following three zones: 12 3 0.5 0.35 0.15 Before the search is conducted, the probabilities that the submarine is in Zone 1, 2, or 3 are, respectively, 0.5, 0.35, and 0.15. It is possible that we do not find the submarine when we search the zone where it is located. If the submarine is in Zone 1 and we search Zone 1, there is a 0.35 probability that we do not find it. Similarly, the probabilities for Zone 2 and Zone 3 are, respectively, 0.05 and 0.15. Assume that the search team is only able to search one zone per day and that the submarine stays in the same zone for the duration of the search. The search team cannot find the submarine if they search the zone where it is not located. (a) Which zone should we search on Day 1 to maximize the probability of finding the submarine on Day 1? (b) Update the probabilities that the submarine is in Zone 1, 2, or 3 given that we searched Zone 1 on Day 1 and did not find the submarine. (c) Suppose we know that the submarine is located in Zone 1 and so Zone 1 is searched each day until the submarine is found. On what day of the search can we expect to find the submarine?
For the chance of finding the submarine on Day 1, we should search Zone 1, as it has the highest initial probability of containing the submarine (0.5).
a. To maximize the probability of finding the submarine on Day 1, we should search Zone 1, as it has the highest initial probability of containing the submarine (0.5).
b. To update the probabilities, we can use Bayes' theorem. Let A be the event of not finding the submarine in Zone 1. Given that A occurred, we update the probabilities using P(A|Zone 1) = 0.35. Using Bayes' theorem, we can calculate the updated probabilities for Zone 1, 2, and 3.
c. If the submarine is known to be located in Zone 1 and we search Zone 1 every day until it is found, the expected day of finding the submarine depends on the probability of finding it each day. However, the provided information does not specify the probability of finding the submarine in Zone 1. Without that information, we cannot determine the specific day on which we can expect to find the submarine.
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