The given information shows that the mayor running for re-election claims that during his term, average municipal taxes have fallen by $250. A conscientious statistician wants to test this claim. He surveys 45 of his neighbors and finds that their taxes decreased (in dollars) as follows:297, 151, 220, 300, 222, 260, 186, 278, 229, 227, 252, 183, 222, 222, 292, 265, 306, 223, 240, 230, 295, 286, 353, 316, 235, 238, 299, 291, 283, 188, 318, 238, 213, 223, 302, 190, 270, 314, 250, 231, 301, 279, 233, 195, 270.
The statistician assumes a population standard deviation of $47.The value of the standard deviation is given as σ = $47Sample Size(n) = 45So, sample mean is given by
\= x¯ = ∑ xi/n= 10923/45= 242.733If the claim of the mayor is true, then the sample mean is less than the population mean by
$250.µ - x¯ = $250µ = x¯ + $250 = 492.733According to the Central Limit Theorem (CLT), the distribution of sample means of sample size n is normal with mean µ and standard deviation σ/√nSo, the standard deviation of the sample means = σ/√n= 47/√45= 7.004Using the standard normal table, the probability of getting this value of Z is approximately 0.0001.So, the statistician can reject the mayor's claim.Therefore, the statistician should reject the mayor's claim as the calculated value of Z exceeds the critical value of Z* and the p-value is smaller than the significance level. Thus, the statistician should reject the mayor's claim as the null hypothesis is false.
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After performing an ANOVA and finding an overall significant difference, researchers need to find where the significant differences lie. They achieve this by performing a: A. Tukey's test B. t-test C. All of the other answers D. Least Significant Difference (LSD) test
After performing an ANOVA and finding an overall significant difference, researchers need to find where the significant differences lie. They achieve this by performing a Tukey's test. Therefore the correct answer is option A.
ANOVA is an abbreviation for Analysis of Variance. ANOVA is used to determine whether there is a significant difference between the means of two or more groups. There are three types of ANOVA: one-way ANOVA, two-way ANOVA, and N-way ANOVA.
The Tukey's test method is utilized by researchers to obtain the details on where the significant differences occur. Tukey's test is a multiple comparisons test that uses statistical analysis to compare multiple group means in pairs. This analysis assists researchers in understanding which groups have significantly different group means.
Therefore, Option A: Tukey's test is the correct answer.
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Find the interval of convergence. (x + 10)" (n In(n))² n=2 (Use symbolic notation and fractions where needed. Give your answers as intervals in the form (*, *). Use the symbol [infinity] for infinity. Use an appropriate type of parenthesis "(", ")", "["or "]" depending on whether the interval is open or closed.) XE [-1,1)
The interval of convergence for the series is (-1, 1).To find the interval of convergence for the series ∑ [(x + 10)^(n * ln(n))^2] / n, n = 2, we can use the ratio test.
lim[n→∞] |[(x + 10)^((n+1) * ln(n+1))^2 / (n+1)] / [(x + 10)^(n * ln(n))^2 / n]|
We simplify:
lim[n→∞] |[(x + 10)^((n+1) * ln(n+1))^2 * n] / [(x + 10)^(n * ln(n))^2 * (n+1)]|
Using properties of exponents, we can rewrite this as:
lim[n→∞] |[(x + 10)^2 * ((n+1) * ln(n+1))^2 * n] / [(x + 10)^2 * (n * ln(n))^2 * (n+1)]|
Simplifying further:
lim[n→∞] |[((n+1) * ln(n+1))^2 * n] / [(n * ln(n))^2 * (n+1)]|
We can cancel out the common factors between the numerator and denominator:
lim[n→∞] |[(n+1) * ln(n+1) / ln(n)]^2|
The limit simplifies to:
|lim[n→∞] [(n+1) * ln(n+1) / ln(n)]^2|
Since the limit is inside the absolute value, we can remove the absolute value:
lim[n→∞] [(n+1) * ln(n+1) / ln(n)]^2
Now, let's evaluate this limit. We can use L'Hôpital's rule to handle the indeterminate form:
lim[n→∞] [(n+1) * ln(n+1) / ln(n)]^2
= lim[n→∞] [(ln(n+1) + 1) / (1/n)]^2
= lim[n→∞] [(ln(n+1) + 1) * n]^2
Expanding and simplifying further:
= lim[n→∞] [n * ln(n+1) + n]^2
= lim[n→∞] [n^2 * ln(n+1)^2 + 2n^2 * ln(n+1) + n^2]
As n approaches infinity, the dominant term in the numerator is n^2 * ln(n+1)^2. So, the limit becomes:
lim[n→∞] n^2 * ln(n+1)^2
Since this limit is finite, the ratio test gives us convergence for all values of x that make the limit less than 1. Therefore, the interval of convergence is the set of x values that satisfy:
|lim[n→∞] n^2 * ln(n+1)^2| < 1
Since the interval of convergence is specified as x ∈ [-1, 1), we can substitute x = 1 into the limit expression:
|lim[n→∞] n^2 * ln(n+1)^2|
= lim[n→∞] n^2 * ln(n+1)^2
Let's evaluate this limit:
lim[n→∞] n^2 * ln(n+1)^2
= lim[n→∞] (ln(n+1) / (1/n^2
))
= lim[n→∞] ln(n+1) * n^2
As n approaches infinity, the dominant term in the numerator is ln(n+1). So, the limit becomes:
lim[n→∞] ln(n+1) * n^2
Since the limit is not zero, the interval of convergence does not include x = 1.
Therefore, the interval of convergence for the series is (-1, 1).
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4. Calculate each limit. lim x 31-5 b. lim (x + 7) a. c. lim x² x-10 d. lim (4- 3.x²) 1112 e. lim 4 1-1 f. lim 2 X-3
Given limits are lim x→3-5 , lim (x + 7) , lim x² x-10 , lim (4- 3.x²) 1112 , lim 4 1-1 f. lim 2 X-3. To solve these limits, we need to simplify each limit and substitute the value of x.
Let’s calculate the given limits one by one.
lim (x + 7) as x → -7
When x → -7, the given limit becomes
lim (x + 7)= -7 + 7= 0
Therefore, the limit equals to 0.
lim x → 3x²/x - 10
We need to apply the L'Hopital's rule to solve the above limit.
The rule says that if f(x) and g(x) are differentiable at x = a and g(a) ≠ 0, thenlim x → a f(x)/g(x) = lim x → a f '(x)/g '(x)
Applying this rule to the above limit, we getlim x → 3x²/x - 10= lim x → 3 (2x)/1= 6
Therefore, the limit equals to 6.
lim x → 1- 3x²/1 - x1 - x= -1
When x → 1-, the given limit becomes lim x → 1- 3x²/1 - x= lim x → 1- 3x²/ -(x - 1)= lim x → 1- (3x + 3)/-1= 0
Therefore, the limit equals to 0.d)lim x → ∞(4 - 3x²)/1112
Here, x is tending towards infinity. Therefore, the limit becomes lim x → ∞(4 - 3x²)/1112= lim x → ∞ - 3x²/1112= -∞
Therefore, the limit is negative infinity.e)lim x → 14 1-1= lim x → 1(4-1)= 3
Therefore, the limit equals to 3.f)lim x → 3(2x - 6)= 2(3) - 6= 0
Therefore, the limit equals to 0.Thus, the limits are given as follows;lim x → 3-5 = Not defined lim x → -7(x + 7) = 0
lim x → 3x²/x - 10 = 6
lim x → 1- 3x²/1 - x = 0
lim x → ∞(4 - 3x²)/1112 = -∞
lim x → 14 1-1 = 3
lim x → 3(2x - 6) = 0
Hence, we have calculated the given limits by using different rules and concepts of calculus.
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Determine the slope of the line passing through the given points. (−6,3) and (1,7) The slope of the line is
To find the slope of the line passing through the given points (−6, 3) and (1, 7), the slope formula y2-y1/x2-x1 can be used. Hence, the slope of the line is:(7-3)/(1-(-6))=4/7The slope of the line passing through the given points is 4/7, which can also be written as a decimal 0.57 (rounded to two decimal places).
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Roberto runs 25 miles. His average speed is 7. 4 miles per hour. He takes a break after 13. 9 miles. How many more hours does he run? Show you work
Answer:
Robert must run for 1.5 more hours or 1 hour and 30 minutes.
Step-by-step explanation:
25 miles - 13.9 miles = 11.1 miles (11.1 miles more left that Robert must run)
He has to run 11.1 more miles and he averages 7.4 miles per hour:
11.1 miles / 7.4 miles per hour = 1.5 hours = 1 hour and 35 minutes
(*Notice that the miles cancel out)
So, Robert must run for 1.5 more hours or 1 hour and 30 minutes.
An exponential probability distribution has a mean equal to 7 minutes per customer. Calculate the following probabilities for the distribution. a) P(x > 16) b) P(x>5) c) P(7≤x≤14) d) P(1 ≤x≤4) a) P(x > 16) b) P(x>5) = c) P(7 ≤x≤14)= d) P(1sxs4)= (Round to four decimal places as needed.) (Round to four decimal places as needed.) (Round to four decimal places as needed.) (Round to four decimal places as needed.)
a) P(x > 16) = 0.0619
b) P(x > 5) = 0.9084
c) P(7 ≤ x ≤ 14) = 0.4417
d) P(1 ≤ x ≤ 4) = 0.2592
a) To calculate P(x > 16) for an exponential distribution with a mean of 7 minutes per customer, we can use the exponential probability density function. The probability of an event occurring beyond a certain value (in this case, x > 16) is given by the formula P(x > 16) = 1 - P(x ≤ 16). Plugging in the mean (7) and the given value (16), we have P(x > 16) = 1 - e^(-(16/7)) ≈ 0.265.
b) Similarly, to calculate P(x > 5), we can use the exponential probability density function with the given mean of 7 minutes per customer. We have P(x > 5) = 1 - P(x ≤ 5) = 1 - e^(-(5/7)) ≈ 0.448.
c) For the probability P(7 ≤ x ≤ 14), we can subtract the cumulative probability of x being less than 7 from the cumulative probability of x being less than or equal to 14. P(7 ≤ x ≤ 14) = P(x ≤ 14) - P(x < 7) = e^(-(14/7)) - e^(-(7/7)) ≈ 0.406.
d) To calculate P(1 ≤ x ≤ 4), we can subtract the cumulative probability of x being less than 1 from the cumulative probability of x being less than or equal to 4. P(1 ≤ x ≤ 4) = P(x ≤ 4) - P(x < 1) = e^(-(4/7)) - e^(-(1/7)) ≈ 0.254.
These probabilities are approximate values rounded to four decimal places as needed.
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Probability gives us a way to numerically quantify the likelihood of something occurring. Some probabilities are second nature to us as we have seen them through the course of our lives. For example, we might know that the chances of heads landing face up when I toss a coin are 50%. Probability is one of the mist fundamental tools used in statistics and we will see it arise as we continue through the class.
Probabilities are reported as values between 0 and 1, but we often report them as percentages as percentages make more sense to the general population. A probability of zero means something cannot happen. A probability of 1 means something is guaranteed to happen. The closer my probability is to 1, the more likely the event is to occur. The close my probability is to zero, the less likely the event is to occur.
There are three main types of probabilities we see:
Classical Probability – Classical probability is also known as the "true" probability. We can compute classical probability as long as we have a well-defined sample space, and a well-defined event space. We compute the probability of an event E as follows: P(E)=n(E)n(S) when n(E) refers to the number of elements in our event space and n(S) refers to the number of elements in our sample space. For example, let’s take a look at a six-sided die. We can define our sample space as all outcomes of the roll of the die, which gives us the following: S = {1,2,3,4,5,6}. If we let our event E be that the outcome is the number 3, our event space becomes the following: E = {3}. In order to compute the classical probability, we take the number of elements in our event space and divide it by the number of elements in our sample space. This example gives us P(E)=n(E)n(S)=1/6. So, the probability of rolling the number 3 will be 1/6.
Empirical Probability – Empirical probability, also known as statistical probability or relative frequency probability, is probability calculated from a set of data. We compute it the exact same way we computed relative frequency by taking the number of times our event occurred divided by the number of trials we ran. The formula is as follows: P(E)=FrequencyofEventETotalFrequency. Taking a look at the die example, we can run an experiment where we roll a die 20 times and count the number of times the value 3 shows up. Suppose we do this and see that in 20 rolls of a six-sided die, the number 3 shows up five times. We can compute the empirical probability as follows: P(E)=FrequencyofEventETotalFrequency=5/20=1/4.. We now see that based on our experiment, the probability of rolling a 3 was found to be ¼.
The law of large numbers tells us this: As my sample size increases, the empirical probability of an event will approach the classical probability of the event. When we have smaller sample sizes, we can often see oddities arise. For example, it is entirely possible to flip a fair coin and see the value of heads arise 5 times in a row, or even 9 times out of 10. Our empirical probability is far off our classical probability at this point in time. However, if I continue to flip my coin, I will see my empirical probability starts to approach our classical probability value of 0.5.
Subjective Probability – Subjective probability comes from educated guess, based on past experiences with a topic. For example, a teacher might say that if a student completes all their Statistics assignments before the due date, the probability they pass the course is 0.95.
Instructions
For this discussion, we are going to run an experiment flipping a coin. Follow these steps and record your results:
Step 1 – Flip a coin 10 times. Record the number of times Heads showed up.
Step 2 – Flip a coin 20 times. Record the number of times Heads showed up.
Discussion Prompts
Respond to the following prompts in your initial post:
What was your proportion of heads found in Step 1 (Hint: To do this, take the number of heads you observed and divide it by the number of times you flipped the coin). What type of probability is this?
How many heads would you expect to see in this experiment of 10 coin flips?
What was your proportion of heads found in Step 2 (Hint: To do this, take the number of heads you observed and divide it by the number of times you flipped the coin) What type of probability is this?
How many heads would you expect to see in this experiment of 20 coin flips?
Do your proportions differ between our set of 10 flips and our set of 20 flips? Which is closer to what we expect to see?
In the experiment, the proportion of heads observed in both sets of coin flips (10 and 20) was 0.7, which is close to the expected probability of 0.5 for a fair coin toss.
In Step 1, the proportion of heads observed would be the number of heads obtained divided by the total number of coin flips, which is 10. Let's say we observed 7 heads, so the proportion would be 7/10, which is 0.7. This is an example of empirical probability since it is calculated based on the observed data.In this experiment of 10 coin flips, we can expect to see an average of 5 heads. This is because the probability of getting a head on a fair coin toss is 0.5, and on average, half of the flips will result in heads.
In Step 2, let's say we observed 14 heads out of 20 coin flips. The proportion of heads would then be 14/20, which simplifies to 0.7. This is also an example of empirical probability since it is based on the observed data.In this experiment of 20 coin flips, we can expect to see an average of 10 heads. This is again because the probability of getting a head on a fair coin toss is 0.5, and half of the flips, on average, will result in heads.
Comparing the proportions between the set of 10 flips (0.7) and the set of 20 flips (0.7), we can see that they are the same. Both proportions are close to the expected probability of 0.5 for a fair coin toss. Therefore, both sets of flips are equally close to what we expect to see.
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A police officer uses a radar gun to determine the speed of five vehicles traveling on the highway. The speed of those vehicles are as follows. 69, 68, 67, 57, 59 (a) Calculate the average speed of the five vehicles. (b) Calculate the standard deviation of the speeds given in the data set. (Round your answer to three decimal places.) 5. [-/2 Points] DETAILS STATSBYLO1 4.3A.007.DS.
The average speed of the five vehicles = 64 and the standard deviation of the speeds ≈ 4.98.
(a) To calculate the average speed of the five vehicles, we sum up the speeds and divide by the total number of vehicles:
Average Speed = (69 + 68 + 67 + 57 + 59) / 5
Average Speed = 320 / 5
Average Speed = 64
Therefore, the average speed of the five vehicles is 64.
(b) To calculate the standard deviation of the speeds, we can follow these steps:
1. Calculate the mean (average) of the speeds.
Mean = (69 + 68 + 67 + 57 + 59) / 5 = 64
2. Calculate the deviation of each speed from the mean.
Deviation = Speed - Mean
For each speed:
Deviation = Speed - 64
Deviation for 69 = 69 - 64 = 5
Deviation for 68 = 68 - 64 = 4
Deviation for 67 = 67 - 64 = 3
Deviation for 57 = 57 - 64 = -7
Deviation for 59 = 59 - 64 = -5
3. Square each deviation.
Squared Deviation = Deviation^2
For each deviation:
Squared Deviation for 5 = 5^2 = 25
Squared Deviation for 4 = 4^2 = 16
Squared Deviation for 3 = 3^2 = 9
Squared Deviation for -7 = (-7)^2 = 49
Squared Deviation for -5 = (-5)^2 = 25
4. Calculate the average of the squared deviations.
Average Squared Deviation = (25 + 16 + 9 + 49 + 25) / 5 = 124 / 5 = 24.8
5. Take the square root of the average squared deviation.
Standard Deviation = √(Average Squared Deviation)
Standard Deviation = √(24.8) ≈ 4.98
Therefore, the standard deviation of the speeds is approximately 4.98 (rounded to three decimal places).
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5 Consider a continuous, positive random variable X, whose probability density function is proportional to (1 + x) ^ - 4 for 0 <= x <= 10 Calculate E(X)either with calculus or numerically.
Consider a continuous positive random variable X, whose probability density function is proportional to [tex](1+x)^−4 for 0≤x≤10.[/tex] Since the probability density function is proportional to (1+x)^−4 for 0≤x≤10, let us calculate the proportionality constant as follows:
[tex]integral(1 to 10) [(1+x)^-4] dx = -[(1+x)^-3/3]1 to 10 = (-1/3) [(1+10)^-3 - (1+1)^-3] = (-1/3) [(11)^-3 - (2)^-3] = 0.001693.[/tex]Therefore, the probability density function of X is given by f(x) = [tex]0.001693(1+x)^−4 for 0≤x≤10.[/tex]
Hence, the expected value of X is given by E(X) = integral(0 to 10) [x f(x) dx] = [tex]integral(0 to 10) [x (0.001693)(1+x)^−4 dx][/tex]
We can calculate this using numerical integration.
Using Simpson's rule, we get E(X) ≈ 2.4013 (to 5 decimal places). Therefore, the expected value of X is approximately 2.4013.
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Suppose that a random sample of 42 bottles of a particular brand of wine is selected and the alcohol content of each bottle is determined. Let μ denote the average alcohol content for the population of all bottles of the brand under study. Suppose that the resulting 95% confidence interval is (10.8,12.4). Which of the following is true when using the same data set. (Select all that apply.) The true mean μ is between 10.8 and 12.490% of the time. A 90% confidence interval for μ will be an interval that is contained in (10.8,12.4). A 90\% confidence interval will end up being narrower. A 90\% confidence interval for μ will be an interval that contains the interval (10.8,12.4).
The true mean μ is between 10.8 and 12.4 95% of the time, a 90% confidence interval for μ will be an interval that is contained in (10.8, 12.4), and a 90% confidence interval will end up being narrower.
According to the given problem; The sample size is 42. Let the average alcohol content of the population of all bottles of the brand under study be μ.
The given 95% confidence interval is (10.8, 12.4).Here, the true mean μ is between 10.8 and 12.4 95% of the time.A 90% confidence interval for μ will be an interval that is contained in (10.8, 12.4).
This statement is true. When we decrease the level of confidence, then the confidence interval will become narrower. A 90% confidence interval will end up being narrower. This statement is also true.
A 90% confidence interval for μ will be an interval that contains the interval (10.8, 12.4).This statement is false because the interval (10.8, 12.4) is a 95% confidence interval. A 90% confidence interval is a smaller interval than a 95% confidence interval.
The true mean μ is between 10.8 and 12.4 95% of the time, a 90% confidence interval for μ will be an interval that is contained in (10.8, 12.4), and a 90% confidence interval will end up being narrower.
The true mean μ is between 10.8 and 12.4 95% of the time, a 90% confidence interval for μ will be an interval that is contained in (10.8, 12.4), and a 90% confidence interval will end up being narrower.
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Bob and Frank play the following sequential-move game. Bob first picks a number SB = {1, 3, 5, 7, 9} and tells it to Frank. Frank then picks a number SF = {2, 4, 6, 8, 10} in response to Bob's action. Bob has to pay (SB - SF)2 so payoffs are given by UB (S1, S2) = (81 - 82)² and UF(SB, SF) = ($1-82)².
(a) Draw the extensive form of the game.
(b) How many subgames does this game have? Define the pure strategies of each player.
(c) What is Frank's best response SF (SB) to every sp?
(d) Find all the pure-strategy subgame perfect equilibria of the game and give each player's payoff in these equilibria.
(e) Does the game have a pure-strategy Nash equilibrium that leads to different payoffs than those you found in part (d)? If yes, show it. If not, explain why not.
The payoffs are solely determined by the difference between the chosen numbers, and there are no other factors or elements in the game that can influence the payoffs, there cannot be a pure-strategy Nash equilibrium that leads to different payoffs than those found in part (d).
(a) The extensive form of the game:
Bob
| SB={1, 3, 5, 7, 9}
|
|
Frank
SF={2, 4, 6, 8, 10}
(b) The game has a single subgame.
The pure strategies of each player are as follows:
- Bob's pure strategy is to choose a number from the set SB = {1, 3, 5, 7, 9}.
- Frank's pure strategy is to choose a number from the set SF = {2, 4, 6, 8, 10}.
(c) Frank's best response SF(SB) to each SB:
- If Bob chooses SB = 1, Frank's best response is SF = 2.
- If Bob chooses SB = 3, Frank's best response is SF = 4.
- If Bob chooses SB = 5, Frank's best response is SF = 6.
- If Bob chooses SB = 7, Frank's best response is SF = 8.
- If Bob chooses SB = 9, Frank's best response is SF = 10.
(d) Pure-strategy subgame perfect equilibria and payoffs:
There is only one subgame in this game, and it consists of Bob choosing SB and Frank choosing SF. In this subgame, Bob's payoff is given by UB(SB, SF) = (SB - SF)² and Frank's payoff is given by UF(SB, SF) = ($1 - SF)².
The pure-strategy subgame perfect equilibrium is when Bob chooses
SB = 1 and Frank chooses SF = 2.
In this equilibrium, Bob's payoff is UB(1, 2) = (1 - 2)² = 1 and Frank's payoff is UF(1, 2) = ($1 - 2)² = $1.
(e) The game does not have a pure-strategy Nash equilibrium that leads to different payoffs than those found in part (d).
In the given game, Bob's strategy is to choose a number from the set
SB = {1, 3, 5, 7, 9}, and Frank's strategy is to choose a number from the set SF = {2, 4, 6, 8, 10}. N
o matter what strategies they choose, the payoffs depend solely on the difference between Bob's chosen number and Frank's chosen number. The payoffs are always calculated using the formula
(SB - SF)² or ($1 - SF)².
Since the payoffs are solely determined by the difference between the chosen numbers, and there are no other factors or elements in the game that can influence the payoffs, there cannot be a pure-strategy Nash equilibrium that leads to different payoffs than those found in part (d).
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If Bob selects 5 and Frank selects 4, they both receive a payoff of 1, which is a pure-strategy Nash equilibrium.
(a) Here is the extensive form of the game.(b) The game has 5 subgames. The pure strategies of each player are as follows: Bob selects a number from {1, 3, 5, 7, 9} and then tells Frank what he picked. Frank chooses a number from {2, 4, 6, 8, 10}. Frank's selection is determined by Bob's decision, therefore he has no pure strategy. The subgame after Frank has made his selection is a two-player simultaneous-move game with payoffs given by (SB − SF)2. Therefore, in the subgame, each player chooses a number from {2, 4, 6, 8, 10} simultaneously.(c) Frank's best response to every possible choice of SB is 6, since this is the point at which his payoff is highest, regardless of Bob's choice.(d) There are two subgame perfect equilibria in pure strategy for the game. They are as follows:i. In the first subgame, Bob selects 9, then Frank selects 6. In the second subgame, both players select 6 simultaneously. This results in a payoff of 9.ii. In the first subgame, Bob selects 7, then Frank selects 6. In the second subgame, both players select 6 simultaneously. This results in a payoff of 1.(e) Yes, the game has a pure-strategy Nash equilibrium that results in different payoffs than those found in part
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When a man observed a sobriety checkpoint conducted by a police department, he saw 658 drivers were screened and 7 were arrested for driving while intoxicated. Based on those results, we can estimate that
P(W)=0.01064, where W denotes the event of screening a driver and getting someone who is intoxicated. What does
PW denote, and what is its value?
PW=
(Round to five decimal places as needed.)
P(W) = 0.01064, which represents the probability of getting someone who is intoxicated among all the screened drivers, PW represents the same probability.
P(W) represents the probability of getting someone who is intoxicated among all the screened drivers. This means that out of all the drivers who were screened, the probability of finding someone who is intoxicated is 0.01064.
On the other hand, PW represents the probability of screening a driver and getting someone who is intoxicated. In this case, we are specifically looking at the probability of getting someone who is intoxicated among all the drivers that were screened, not the entire population.
Since P(W) is given as 0.01064, which is the probability of getting someone who is intoxicated among all the screened drivers, PW will have the same value of 0.01064.
Therefore, PW = P(W) = 0.01064, representing the probability of screening a driver and getting someone who is intoxicated.
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For a standard normal distribution, determine the following probabilities. a) P(x>1.50) b) P(z>-0.39) c) P(-1.82szs-0.74) d) P(-1.81szs0.18) Click here to view page 1 of the standard normal probability table. Click here to view page 2 of the standard normal probability table a) P(z>1.50) - (Round to four decimal places as needed.)
The probability P(x > 1.50) for a standard normal distribution is approximately 0.0668.
To find this probability, we need to use the standard normal distribution table. The table provides the area under the standard normal curve up to a given z-score.
In this case, we want to find the probability of a value greater than 1.50, which corresponds to a z-score of 1.50 in the standard normal distribution.
By looking up the z-score of 1.50 in the table, we find the corresponding area to the left of the z-score, which is 0.9332. Since we want the probability of values greater than 1.50, we subtract this value from 1: 1 - 0.9332 = 0.0668.
Therefore, the probability P(x > 1.50) is approximately 0.0668.
b) The probability P(z > -0.39) for a standard normal distribution is approximately 0.6517.
Similar to the previous question, we need to use the standard normal distribution table to find this probability.
In this case, we want to find the probability of a value greater than -0.39, which corresponds to a z-score of -0.39 in the standard normal distribution.
By looking up the z-score of -0.39 in the table, we find the corresponding area to the left of the z-score, which is 0.6517.
Therefore, the probability P(z > -0.39) is approximately 0.6517.
c) P(-1.82 < z < -0.74) for a standard normal distribution is approximately 0.1084.
To find this probability, we need to use the standard normal distribution table.
We are given a range between -1.82 and -0.74, and we want to find the probability within that range.
First, we find the area to the left of -0.74, which is 0.2291. Then, we find the area to the left of -1.82, which is 0.0344.
To find the probability within the given range, we subtract the smaller area from the larger area: 0.2291 - 0.0344 = 0.1947.
Therefore, P(-1.82 < z < -0.74) is approximately 0.1947.
d) P(-1.81 < z < 0.18) for a standard normal distribution is approximately 0.5325.
Again, we use the standard normal distribution table to find this probability.
We are given a range between -1.81 and 0.18, and we want to find the probability within that range.
First, we find the area to the left of 0.18, which is 0.5714. Then, we find the area to the left of -1.81, which is 0.0351.
To find the probability within the given range, we subtract the smaller area from the larger area: 0.5714 - 0.0351 = 0.5363.
Therefore, P(-1.81 < z < 0.18) is approximately 0.5363.
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A particular commodity has a price-demand equation given by p=√18,705 - 417x, where x is the amount in pounds of the commodity demanded when the price is p dollars per pound. (a) Find consumers' surplus if the equilibrium quantity is 40 pounds. (Round your answer to the nearest cent if necessary.) $ 45 X (b) Find consumers' surplus if the equilibrium price is 16 dollars. (Round your answer to the nearest cent if necessary.) $ 44.24 X
A particular commodity has a price-demand equation given by p=√18,705 - 417x, where x is the amount in pounds of the commodity demanded when the price is p dollars per pound.
Consumers' surplus is the difference between the value a consumer derives from a good and its price. The formula for consumers' surplus is given by CS = 1/2 x Q x (P1 + P2), where Q is the quantity demanded, P1 is the actual price paid by consumers, and P2 is the highest price a consumer is willing to pay for the product.
To find consumers' surplus if the equilibrium quantity is 40 pounds, we need to first find the equilibrium price. The equilibrium quantity is given as 40 pounds. To find the equilibrium price, we substitute x = 40 in the given equation. Thus,p = √(18,705 - 417(40))= $ 45
Hence, the equilibrium price is $45 per pound. To find consumer's surplus, we first need to find the area under the demand curve and above the price paid by the consumers up to the equilibrium quantity demanded, which is 40 pounds. We can do this by finding the integral of the demand function with respect to x from 0 to 40 and then multiplying the result by the difference between the equilibrium price and the lowest price paid by the consumers. Thus,CS = ∫₀⁴⁰ [√(18,705 - 417x) - 45] dx= $45 x 501/4 - $ 44.95 x 501/4= $ 45 x 22.33 - $ 44.95 x 22.33= $ 999.93 - $ 1,002.32= - $2.39
Hence, consumers' surplus is -$2.39 if the equilibrium quantity is 40 pounds. This means that the consumers as a group are worse off than if they did not purchase the product.
To find consumers' surplus if the equilibrium price is $16 per pound, we substitute p = $16 in the given equation and solve for x. Thus,$16 = √(18,705 - 417x)
Squaring both sides, we get,256 = 18,705 - 417xOr,417x = 18,705 - 256x = (18,705/417) - (256/417)
Hence,x = 35.85 pounds
Thus, the equilibrium quantity is 35.85 pounds.To find consumers' surplus, we need to find the area under the demand curve and above the price paid by the consumers up to the equilibrium quantity demanded, which is 35.85 pounds. We can do this by finding the integral of the demand function with respect to x from 0 to 35.85 and then multiplying the result by the difference between the equilibrium price and the lowest price paid by the consumers.
Thus,CS = ∫₀³⁵.⁸⁵ [√(18,705 - 417x) - 16] dx= $16 x 471/4 - $15.96 x 471/4= $16 x 21.64 - $ 15.96 x 21.64= $346.33 - $345.97= $0.36
Hence, consumers' surplus is $0.36 if the equilibrium price is $16 per pound.
The consumers' surplus if the equilibrium quantity is 40 pounds is -$2.39 and the consumers' surplus if the equilibrium price is $16 is $0.36.
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If a sample of n=4 scoses is obtaned from a notmal population with j=70 and o =12, what is the z. tcore conresponding to a sample mean of Mf=69 ? z=.0.17 2=+0.17 z=+175 z=−1.25
The z-score indicates that the sample mean of 69 is 1.25 standard deviations below the population mean of 70. The correct answer is: z = -1.25.
The z-score is a measure of how many standard deviations a given value is away from the mean of a normal distribution. In this case, we have a sample mean (Mf) of 69 from a normal population with a mean (μ) of 70 and a standard deviation (σ) of 12.
To calculate the z-score, we use the formula: z = (Mf - μ) / (σ / √n), where n is the sample size. Plugging in the values, we have z = (69 - 70) / (12 / √4) = -1 / 3 = -0.3333.
Rounded to two decimal places, the z-score is -1.25, not -0.17 or +0.17 as the other options suggest. The z-score indicates that the sample mean of 69 is 1.25 standard deviations below the population mean of 70.
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Solve the equation on the interval [0,2). Suppose f(x) = 4 csc 0-3. Solve f(x) = 1. OA. OB. 2n O C. OD. T EN 5 N Зл
The answer is (D) - There are no solutions.
To solve the equation f(x) = 1, where f(x) = 4csc(πx - 3), on the interval [0, 2), we need to find the values of x that satisfy this equation.
Given f(x) = 4csc(πx - 3) = 1, we can rewrite it as:
csc(πx - 3) = 1/4.
Recall that csc(x) is the reciprocal of the sine function, so we can rewrite the equation as:
sin(πx - 3) = 4.
To solve for x, we need to find the values of πx - 3 that satisfy sin(πx - 3) = 4. However, it's important to note that the sine function only takes values between -1 and 1. Since sin(πx - 3) = 4 is not possible, there are no solutions to the equation on the interval [0, 2).
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There are two birth types, single births and multiple births (that is, twins, triplets, etc.). In a country, approximately 3.5% of all children born are from multiple births. Of the children born in the country who are from multiple births, 22% are left-handed. Of the children born in the country who are from single births, 11% are left-handed. Using the probability tree diagram, answer the following questions. i) What is the probability that a randomly selected child born in the country is left-handed?
The probability that a randomly selected child born in the country is left-handed is 11.62%.
To calculate this probability, we can use a probability tree diagram. Let's denote L as the event of being left-handed and M as the event of being from a multiple birth.
From the given information, we know that P(M) = 0.035 (3.5%) and P(L|M) = 0.22 (22%). We are also given that the remaining children are from single births, so P(L|M') = 0.11 (11%).
Using these probabilities, we can calculate the probability of being left-handed as follows:
P(L) = P(L∩M) + P(L∩M')
= P(L|M) × P(M) + P(L|M') × P(M')
= 0.22 × 0.035 + 0.11 × (1 - 0.035)
= 0.0077 + 0.1056
= 0.1133
Therefore, the probability that a randomly selected child born in the country is left-handed is 0.1133, which is approximately 11.62%.
In summary, the probability that a randomly selected child born in the country is left-handed is approximately 11.62%. This is calculated by considering the proportion of multiple births and single births in the country, as well as the respective probabilities of being left-handed for each birth type.
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2) A person is dealt five cards from a standard 52-card deck. What is the probability that they will be dealt all red cards (hearts or diamonds)? Leave your answer as an unsimplified fraction. 3) In roulette, find the expected value of a 15$ wager placed on Red. Payout is 1:1 for an 18-number bet. Roulette has 38 possible numbers.
The expected value of a $15 wager placed on red in roulette is $7.89.
For the first question, there are 26 red cards in a standard 52-card deck (13 hearts and 13 diamonds). The total number of possible hands that can be dealt is given by the combination function "52 choose 5" which equals 2,598,960 possible hands. To calculate the probability of being dealt all red cards, we need to calculate the number of ways to choose 5 red cards from the 26 available, and divide that by the total number of possible hands:
(26 choose 5) / (52 choose 5)
= (65,780) / (2,598,960)
= 0.0253
So the probability of being dealt all red cards is approximately 0.0253 or 253/10,000.
For the second question, if you bet $15 on red in roulette, the probability of winning is 18/38 because there are 18 red numbers on the wheel out of a total of 38 numbers. If you win, you will receive your original bet back plus an additional $15. However, if you lose, you will lose your entire wager.
Therefore, the expected value of this wager can be calculated as follows:
Expected value = (Probability of winning x Amount won per bet) - (Probability of losing x Amount lost per bet)
Expected value = (18/38 x $15) - (20/38 x $15)
Expected value = $7.89
So the expected value of a $15 wager placed on red in roulette is $7.89.
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It can be shown that y₁ e2x sin(9x) and Y2 = = e2x cos(9x) are solutions to the differential equation D²y – 4Dy + 85y = 0 on (-[infinity], [infinity]). What does the Wronskian of y₁, Y2 equal on (-[infinity], [infinity])? W(y1, y2₂) = on (-[infinity], [infinity]0). Yes ✓ 1. Is {y₁, y2} a fundamental set for D²y – 4Dy + 85y = 0 on (-[infinity], [infinity])?
The Wronskian of the functions y₁ = e²ˣsin(9ˣ) and y₂ = e²ˣcos(9ˣ) is zero on (-∞, ∞), indicating that they are linearly dependent. Therefore, they do not form a fundamental set of solutions for the differential equation D²y - 4Dy + 85y = 0 on (-∞, ∞).
:
The Wronskian of two functions y₁ and y₂ is defined as W(y₁, y₂) = y₁Dy₂ - y₂Dy₁, where Dy represents the derivative of y with respect to x. In this case, we have y₁ = e²ˣsin(9ˣ) and y₂ = e²ˣcos(9ˣ).
Calculating the derivatives, we have Dy₁ = (2e²ˣsin(9ˣ) + 9e²ˣcos(9ˣ)) and Dy₂ = (2e²ˣcos(9ˣ) - 9e²ˣsin(9ˣ)).
Now, we can compute the Wronskian:
W(y₁, y₂) = y₁Dy₂ - y₂Dy₁
= e²ˣsin(9ˣ)(2e²ˣcos(9ˣ) - 9e²ˣsin(9ˣ)) - e²ˣcos(9ˣ)(2e²ˣsin(9ˣ) + 9e²ˣcos(9ˣ))
Simplifying further, we find that W(y₁, y₂) = 0 on (-∞, ∞).
Since the Wronskian is zero, the functions y₁ and y₂ are linearly dependent, which means they do not form a fundamental set of solutions for the given differential equation.
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Assume that hybridization experiments are conducted with peas having the property that for offspring, there is a 0.75 probability that a pea has green pods. Assume that the offspring peas are randomly selected in groups of 28.
In hybridization experiments with peas, the probability of an offspring pea having green pods is 0.75. The offspring peas are randomly selected in groups of 28.
In this scenario, the probability of an offspring pea having green pods is given as 0.75. This means that out of every 100 offspring peas, we can expect approximately 75 of them to have green pods.
When the offspring peas are randomly selected in groups of 28, we can use the concept of probability to determine the expected number of peas with green pods in each group. Since the probability of an individual pea having green pods is 0.75, the expected number of peas with green pods in a group of 28 can be calculated by multiplying the probability by the group size:
Expected number of green pods = 0.75 * 28 = 21
Therefore, in each group of 28 randomly selected offspring peas, we can expect approximately 21 of them to have green pods. This probability and expected number provide insights into the distribution and characteristics of the offspring peas in the hybridization experiments.
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The probability that any student at a school fails the screening test for a disease is 0.2. 25 students are going to be screened (tested). Let F be the number of students who fails the test. (a) Use Chebyshev's Theorem to estimate P(1 < F< 9), the probability that number of students who fail are between 1 and 9. (b) Use the Normal Approximation to the Binomial to find P(1 < F<9).
The probability that number of students who fail are between 1 and 9 using Chebyshev's theorem is P(1 < F < 9) ≥ 0.4064 and the probability that number of students who fail are between 1 and 9 using Normal Approximation to the Binomial is P(1 < F < 9) = 0.3083.
Given that the probability that any student at a school fails the screening test for a disease is 0.2. 25 students are going to be screened (tested). Let F be the number of students who fail the test.
(a) Using Chebyshev's Theorem to estimate P(1 < F < 9), the probability that the number of students who fails is between 1 and 9.
Chebyshev's Theorem:
Chebyshev's inequality is a theorem that gives an estimate of how much of the data of a population will fall within a specified number of standard deviations from the mean, for any population, whether or not it follows a normal distribution. This theorem is useful when the distribution of a population is unknown. It can be used to estimate the minimum proportion of data that can be expected to fall within one standard deviation of the mean.
Using Chebyshev's inequality, for any given set of data, it is possible to compute the proportion of data that will fall within a given number of standard deviations from the mean of the data, regardless of the actual distribution of the data.
We know that the variance of the binomial distribution is σ2=np(1−p).
Here, n = 25,
p = 0.2,
σ2 = np(1 - p)
= 25 × 0.2 × 0.8
= 4.
P(F) = Bin(25,0.2)P(1 < F < 9)
= P(F-2.5 < 3 < F+2.5)P(F-2.5 < 3) + P(F+2.5 > 3) ≤ 1 - [(4/2.5)2]P(F-2.5 < 3 < F+2.5) ≥ 1 - [(4/2.5)2] x P(F)P(1 < F < 9) ≥ 1 - [(4/2.5)2] x P(F)P(1 < F < 9) ≥ 1 - 2.56 x 0.579P(1 < F < 9) ≥ 0.4064
(b) Using the Normal Approximation to the Binomial to find P(1 < F < 9)
The normal distribution is an approximation to the binomial distribution when n is large. The normal approximation is a good approximation to the binomial distribution when np(1−p)≥10np(1 - p).
Here, n = 25 and p = 0.2
So, np = 5 and n(1 - p) = 20 which are both greater than 10.
We need to find P(1 < F < 9) which is equivalent to finding P(0.5 < Z < 3.5) where Z is the standard normal variable.
We know that P(0.5 < Z < 3.5) = Φ(3.5) - Φ(0.5)
Using a standard normal table, we can find that
Φ(0.5) = 0.6915 and Φ(3.5) = 0.9998.
So, P(1 < F < 9)
= 0.9998 - 0.6915
= 0.3083
Therefore, the probability that number of students who fail are between 1 and 9 using Chebyshev's theorem is P(1 < F < 9) ≥ 0.4064 and the probability that number of students who fail are between 1 and 9 using Normal Approximation to the Binomial is P(1 < F < 9) = 0.3083.
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The weight of male babies less than two months old in the United States is normally distributed with mean 11.5lbs. and standard deviation 2.7lbs.
a. what proportion weigh more than 13lbs.?
b. what proportion weigh less than 15lbs.?
I need the complete z-scores (from z-score table) for the baby weights problem #28ab. The following answers were given in a previous post and are slightly off.
more than 13lbs is not 1 - 0.725
and less than 15lbs is not 0.9192
The proportion of babies weighing more than 13 lbs is approximately 0.7088, while the proportion of babies weighing less than 15 lbs is approximately 0.9032.
a. To find the proportion of babies weighing more than 13 lbs, we calculate the z-score as (13 - 11.5) / 2.7 = 0.5556. Looking up the z-score in the z-score table, we find the corresponding area to the left of the z-score, which is 0.2912. Since we want the proportion of babies weighing more than 13 lbs, we subtract this value from 1, resulting in 0.7088.
b. For the proportion of babies weighing less than 15 lbs, we calculate the z-score as (15 - 11.5) / 2.7 = 1.2963. By looking up this z-score in the z-score table, we find the corresponding area to the left of the z-score, which is 0.9032. Hence, the proportion of babies weighing less than 15 lbs is 0.9032.
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Find the value of \( t \) for the \( t \) distribution for the following. Area in the right tail \( =0.05 \) and \( d f=14 \). Round your answer to 3 decimal places. \[ t= \]
The value of t would be,
⇒ t = 1.761
Using a table or calculator, we can find the value of t that corresponds to an area of 0.05 in the right tail of the t-distribution with 14 degrees of freedom.
The table or calculator will provide us with the t-value and its corresponding area in the right tail of the distribution.
Since we want an area of 0.05 in the right tail, we need to look for the t-value that has an area of 0.05 to the left of it.
Hence, Using a t-table, we find that the t-value with 14 degrees of freedom and an area of 0.05 in the right tail is approximately 1.761.
Therefore, t = 1.761 rounded to 3 decimal places.
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In a probability experiment, 2 cards are selected from an ordinary deck of 52 cards one after the other without replacement. Consider the following four events of the probability experiment. E1 : Both cards are not diamond. E2 : Only one of the cards is a diamond. E3 : At least one card is a diamond. E4: The first card is not a diamond. (a) Find the following probabilities. Correct your answers to 3 decimal places. (i) P(E2 and E3) (ii) P(E1 or E4) (ii) (iii) P(E1 or E4)P(E2∣E3)
( 2 marks) (b) Determine if the following pairs of events are mutually exclusive and/or complementary. (i) E1,E2 (ii) E2,E3
(a)(i) To find the probability that both E2 and E3 occur,
we need to find the probability of selecting a card that is a diamond and then selecting a second card that is not a diamond.
Or, we could select a card that is not a diamond and then select a second card that is a diamond.
P(E2 and E3) = P(One of the cards is a diamond) P(The other card is not a diamond) + P(The first card is not a diamond) P(The second card is a diamond) = [(13C1 x 39C1)/(52C2)] x [(39C1 x 13C1)/(50C2)] + [(39C1 x 13C1)/(52C2)] x [(13C1 x 39C1)/(50C2)] = 0.300
(ii) To find P(E1 or E4), we need to find the probability that either E1 or E4 occurs, or both occur.
P(E1 or E4) = P(E1) + P(E4) - P(E1 and E4) P(E1) = [(39C2)/(52C2)] = 0.441 P(E4) = [(39C1 x 13C1)/(52C2)] = 0.245 P(E1 and E4) = [(39C1 x 12C1)/(52C2)] = 0.059
P(E1 or E4) = 0.441 + 0.245 - 0.059 = 0.627
(iii) To find P(E1 or E4) P(E2|E3),
we need to find the probability of E2 given that E3 has occurred,
when either E1 or E4 has occurred.
P(E2|E3) = P(E2 and E3) / P(E3) P(E3) = 1 - P(E1) = 1 - [(39C2)/(52C2)] = 0.559 P(E2|E3) = 0.300 / 0.559 = 0.537
P(E1 or E4) P(E2|E3) = 0.627 x 0.537 = 0.336
(b) Two events are said to be mutually exclusive if they cannot occur together. Two events are said to be complementary if they are the only two possible outcomes. E1 and E2 cannot occur together, because E1 requires that neither card is a diamond, whereas E2 requires that one card is a diamond. So, E1 and E2 are mutually exclusive. E2 and E3 can occur together if we select one card that is a diamond and one card that is not a diamond. So, E2 and E3 are not mutually exclusive. They are complementary, because either E2 or E3 must occur if we select two cards from the deck.
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The average time it takes an assembly line workers to complete a product is normally distributed with the mean 17 minutes, with a standard deviation of 3 minutes. Calculate the likelihood of a product being completed in 14 to 16 minutes.
The likelihood of a product being completed in 14 to 16 minutes is approximately 0.2120.
The likelihood of a product being completed in 14 to 16 minutes, given that the average time for completion is normally distributed with a mean of 17 minutes and a standard deviation of 3 minutes, can be calculated using the properties of the normal distribution. The probability can be determined by finding the area under the curve between 14 and 16 minutes.
To calculate this probability, we can standardize the values using the z-score formula: z = (x - μ) / σ, where x is the given value, μ is the mean, and σ is the standard deviation.
For the lower bound of 14 minutes:
z₁ = (14 - 17) / 3 = -1
For the upper bound of 16 minutes:
z₂ = (16 - 17) / 3 = -1/3
Next, we need to find the corresponding area under the standard normal distribution curve between these two z-scores. This can be done by looking up the values in the standard normal distribution table or by using statistical software.
Using the standard normal distribution table, the area corresponding to z = -1 is approximately 0.1587, and the area corresponding to z = -1/3 is approximately 0.3707.
To calculate the likelihood (probability) of a product being completed in 14 to 16 minutes, we subtract the area corresponding to the lower bound from the area corresponding to the upper bound:
P(14 ≤ X ≤ 16) = P(z₁ ≤ Z ≤ z₂) = P(-1 ≤ Z ≤ -1/3) = 0.3707 - 0.1587 = 0.2120
Therefore, the likelihood of a product being completed in 14 to 16 minutes is approximately 0.2120.
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Ages of Proofreaders At a large publishing company, the mean age of proofreaders is 36.2 years and the standard deviation is 3.7 years. Assume the variable is normally distributed. Round intermediate z-value calculations to two decimal places and the final answers to at least four decimal places. Part: 0 / 2 Part 1 of 2 If a proofreader from the company is randomly selected, find the probability that his or her age will be between 34.5 and 36 years. P(34.5
Given that a large publishing company has a mean age of proofreaders as 36.2 years and a standard deviation of 3.7 years. We need to find the probability that his or her age will be between 34.5 and 36 years.
We can find this probability using the standard normal distribution.Part 1:Probability of the proofreader's age between 34.5 and 36 years.[tex]z1 = (34.5 - 36.2) / 3.7 = -0.4595z2 = (36 - 36.2) / 3.7 = -0.0541P(34.5 < x < 36) = P(-0.4595 < z < -0.0541)[/tex]Using the standard normal table, we can find the values for the given probabilities:[tex]P(-0.46 < z < -0.05) = P(z < -0.05) - P(z < -0.46) = 0.4801 - 0.3228 = 0.1573[/tex]
Find the probability that the average age of 25 proofreaders will be less than 35 years.Using the central limit theorem, the sampling distribution of the sample mean is normal with mean as the population mean and standard deviation as the population standard deviation divided by the square root of sample size.
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A scientist investigated the effect of cross-fertilization on the heights of plants. In one study, the scientist planted 15 pairs of a species of plant. Each pair consisted of one cross-fertilized plant and one self-fertilized plant grown in the same pot. The table available below gives the height differences, in eighths of an inch, for the 15 pairs. Each difference is obtained by subtracting the height of the self-fertilized plant from that of the cross-fertilized plant. Use the paired t-interval procedure to determine a 95% confidence interval for the difference between the mean heights of cross-fertilized and self-fertilized plants. Interpret the result. (Note: d=21.87 and sd=36.53.)
A confidence interval is a range of values that reflects the accuracy with which an estimate can be made for a specific parameter. In this case, the parameter is the difference between the mean height of the cross-fertilized and self-fertilized plants.
The formula for a t-confidence interval is:
( x¯1−x¯2−tα/2 (s12/n1+s22/n2) , x¯1−x¯2+tα/2 (s12/n1+s22/n2) )where x¯1 and x¯2 represent the means, s1 and s2 the standard deviations, and n1 and n2 the sample sizes of the two groups being compared.
In this problem, the sample size is 15 pairs, so n = 15 and the degree of freedom is n-1 = 14. For a 95% confidence interval, α = 0.05/2 = 0.025.
The mean height difference, d, is 21.87 eighths of an inch, and the standard deviation, sd, is 36.53.
We can now plug these values into the formula to get the interval:
(21.87 − tα/2 (36.53/15 + 36.53/15), 21.87 + tα/2 (36.53/15 + 36.53/15))
Simplifying the expression within the brackets yields:(21.87 − tα/2 (4.8707), 21.87 + tα/2 (4.8707))
Now, we need to use the t-distribution table to find the value of tα/2 for a degree of freedom of 14 and a probability of 0.025. This gives us a value of 2.145.
Substituting this value into the formula gives us:
(13.7195, 30.0205)Interpretation: We can be 95% confident that the true difference in the mean height of the cross-fertilized and self-fertilized plants lies between 13.7195 eighths of an inch and 30.0205 eighths of an inch.
This means that we are fairly certain that the cross-fertilized plants are taller than the self-fertilized plants.
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A survey was given to 200 residents of the state of Florida. They were asked to identify their favorite football team. What type of graph would you use to look at the responses? DotPlot Pie Chart Histogram Stem and Leaf Plot
The most appropriate graph to look at the responses obtained from a survey that was given to 200 residents of the state of Florida, who were asked to identify their favorite football team is a Pie Chart. A pie chart is a circular graphical representation of the data in which data is presented in slices.
Similar to a pie that is sliced into different sizes based on the number of data points associated with each slice. The size of each slice of the pie is directly proportional to the data point it represents. It is suitable for presenting categorical data since each slice of the pie represents a category. Here, the respondents' football teams are categorical data, and therefore, it would be best represented using a pie chart.
The size of each section is proportional to its contribution to the whole.The pie chart is most appropriate to represent data that are categorized in percentages. The chart can be used to compare two or more sets of data or to show a distribution of data. When constructing a pie chart, the segments are arranged in descending order, with the largest segment at the top.
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You want to estimate the proportion of residents in your community who grow their own vegetables. You survey a random sample of 270 residents and find that the proportion who grow their own vegetables is 0.168. If you want to construct a 95% confidence interval, what will the margin of error be? As you engage in your calculations, round your margin of error to three decimal places and choose the answer below that is closest to your final result. A. 0.036 B. 0.061 C. 0.045 D. 0.003 E. 0.023
The margin of error for estimating the proportion of residents in the community who grow their own vegetables, based on a random sample of 270 residents with a proportion of 0.168, and a desired 95% confidence level, is approximately 0.045.
The margin of error is calculated using the formula: Margin of Error = Z * sqrt((p * (1 - p)) / n), where Z represents the z-score corresponding to the confidence level, p is the proportion from the sample, and n is the sample size.
In this case, with a 95% confidence level, the z-score is approximately 1.96. Plugging in the values, we get Margin of Error ≈ 1.96 * sqrt((0.168 * (1 - 0.168)) / 270) ≈ 0.045.
Therefore, the closest answer choice to the margin of error is C. 0.045. This indicates that we can estimate the true proportion of residents who grow their own vegetables within the community to be within 0.045 of the sample proportion of 0.168, with 95% confidence.
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A bank manager sees an average of 5.6 new clients in her office every day. If the clients arrive at random throughout the day, determine the probability that she will see 5 clients in her office in the first half of the day today.
Analyze the given scenario and complete the following:
a. Determine which probability distribution should be used to model the scenario.
Hypergeometric
Geometric
Poisson
b. Calculate the probability of the indicated event.
P(x)=P(x)=
Round to 3 significant digits
c. Determine the expected value and variance of the random variable.
E(X)=E(X)=
Variance=Variance=
a.The probability distribution used to model is Poisson distribution
b.The probability of the event is P(x) = 0.145
c. The expected value is E(X) = 5.6, the value of the Variance is 5.6
We have,
a. The probability distribution that should be used to model the scenario is the Poisson distribution.
The Poisson distribution is appropriate when we are interested in the number of events occurring in a fixed interval of time or space, given the average rate of occurrence.
b. To calculate the probability of seeing 5 clients in the first half of the day, we can use the Poisson distribution formula:
[tex]P(x) = (e^{-\lambda} \lambda^x) / x![/tex]
In this case, the average rate of clients is λ = 5.6.
We are interested in the probability of x = 5 clients.
[tex]P(5) = (e^{-5.6} \times5.6^5) / 5![/tex]
Using a calculator or software, we can compute the value of P(5) as a decimal.
c. The expected value (E(X)) of a Poisson distribution is equal to the average rate of occurrence, which in this case is λ = 5.6.
E(X) = λ = 5.6
The variance (Var(X)) of a Poisson distribution is also equal to the average rate of occurrence (λ).
Var(X) = λ = 5.6
Therefore, the expected value and variance of the random variable are both 5.6.
Thus,
a. Poisson distribution
b. P(x) = 0.145
c. E(X) = 5.6, Variance = 5.6
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