Answer:
A
Step-by-step explanation:
The Hyperbolic Functions and their Inverses: For our purposes, the hyperbolic functions, such as sinhx=ex−e−x2andcoshx=ex+e−x2 are simply extensions of the exponential, and any questions concerning them can be answered by using what we know about exponentials. They do have a host of properties that can become useful if you do extensive work in an area that involves hyperbolic functions, but their importance and significance is much more limited than that of exponential functions and logarithms. Let f(x)=sinhxcoshx.
d/dx f(x) =_________
Answer:
[tex]\dfrac{d}{dx}f(x)=\dfrac{e^{2x}+e^{-2x}}{2}[/tex]
Step-by-step explanation:
It is given that
[tex]\sinh x=\dfrac{e^x-e^{-x}}{2}[/tex]
[tex]\cosh x=\dfrac{e^x+e^{-x}}{2}[/tex]
[tex]f(x)=\sinh x\cosh x=[/tex]
Using the given hyperbolic functions, we get
[tex]f(x)=\dfrac{e^x-e^{-x}}{2}\times \dfrac{e^x+e^{-x}}{2}[/tex]
[tex]f(x)=\dfrac{(e^x)^2-(e^{-x})^2}{4}[/tex]
[tex]f(x)=\dfrac{e^{2x}-e^{-2x}}{4}[/tex]
Differentiate both sides with respect to x.
[tex]\dfrac{d}{dx}f(x)=\dfrac{d}{dx}\left(\dfrac{e^{2x}-e^{-2x}}{4}\right )[/tex]
[tex]\dfrac{d}{dx}f(x)=\left(\dfrac{2e^{2x}-(-2)e^{-2x}}{4}\right )[/tex]
[tex]\dfrac{d}{dx}f(x)=\left(\dfrac{2(e^{2x}+e^{-2x})}{4}\right )[/tex]
[tex]\dfrac{d}{dx}f(x)=\dfrac{e^{2x}+e^{-2x}}{2}[/tex]
Hence, [tex]\dfrac{d}{dx}f(x)=\dfrac{e^{2x}+e^{-2x}}{2}[/tex].
The probability that an archer hits her target when it is windy is 0.4; when it is not windy, her probability of hitting the target is 0.7. On any shot, the probability of a gust of wind is 0.3. Find the probability that a. on a given shot there is a gust of wind and she hits her target.
b. she hits the target with her first shot.
c. she hits the target exactly once in two shots.
d. there was no gust of wind on an occasion when she missed.
Answer:
a) the probability is 0.12 (12%)
b) the probability is 0.61 (61%)
c) the probability is 0.476 (47.6%)
a) the probability is 0.538 (53.8%)
Step-by-step explanation:
a) denoting the event H= hits her target and G= a gust of wind appears hen
P(H∩G) = probability that a gust of wind appears * probability of hitting the target given that is windy = 0.3* 0.4 = 0.12 (12%)
b) for any given shot
P(H)= probability that a gust of wind appears*probability of hitting the target given that is windy + probability that a gust of wind does not appear*probability of hitting the target given that is not windy = 0.3*0.4+0.7*0.7 = 0.12+0.49 = 0.61 (61%)
c) denoting P₂ as the probability of hitting once in 2 shots and since the archer can hit in the first shot or the second , then
P₂ = P(H)*(1-P(H))+ (1-P(H))*P(H) = 2*P(H) *(1-P(H)) = 2*0.61*0.39= 0.476 (47.6%)
d) for conditional probability we can use the theorem of Bayes , where
M= the archer misses the shot → P(M) = 1- P(H) = 0.39
S= it is not windy when the archer shots → P(S) = 1- P(G) = 0.7
then
P(S/M) = P(S∩M)/P(M) = 0.7*(1-0.7)/0.39 = 0.538 (53.8%)
where P(S/M) is the probability that there was no wind when the archer missed the shot
A researcher interested in weight control wondered whether normal and overweight individuals differ in their reaction to the availability of food. Thus, normal and overweight participants were told to eat as many peanuts as they desired while working on a questionnaire. One manipulation was the proximity of the peanut dish (close or far from the participant); the second manipulation was whether the peanuts were shelled or unshelled. After filling out the questionnaire, the peanut dish was weighed to determine the amount of peanuts consumed.1. Identify the design (e.g., 2 X 2 factorial).2. Identify the total number of conditions.3. Identify the manipulated variable(s).4. Is this an IV X PV design? If so, identify the participant variable(s).5. Is this a repeated measures design? If so, identify the repeated variable(s).6. Identify the dependent variable(s).
Answer:
Complete question for your reference is below.
A researcher interested in weight control wondered whether normal and overweight individuals differ in their reaction to the availability of food. Thus, normal and overweight participants were told to eat as many peanuts as they desired while working on a questionnaire. One manipulation was the proximity of the peanut dish (close or far from the participant); the second manipulation was whether the peanuts were shelled or unshelled. After filling out the questionnaire, the peanut dish was weighed to determine the amount of peanuts consumed.
1. Identify the design (e.g., 2 X 2 factorial).
2. Identify the total number of conditions.
3. Identify the manipulated variable(s).
4. Is this an IV X PV design? If so, identify the participant variable(s).
5. Is this a repeated measures design? If so, identify the repeated variable(s).
6. Identify the dependent variable(s).
Step-by-step explanation:
Please find attached file for complete answer solution and explanation.
The research design is a 2x2 factorial with the manipulated or independent variables being the proximity of the peanut dish and whether the peanuts are shelled, resulting in four conditions. It is not an IV x PV design, nor a repeated measures design. The dependent variable in this study is the amount of peanuts consumed.
Explanation:The research design posed in this question appears to be a 2x2 factorial design. This is because there are two independent variables (the proximity of the peanut dish and whether the peanuts are shelled or not), each with two levels (close or far, and shelled or unshelled respectively).
Given this, there would be a total of four conditions in this study (close-shelled, close-unshelled, far-shelled, far-unshelled). The manipulated variables, also referred to as independent variables, in this study are the 'proximity of the peanut dish' and 'whether the peanuts are shelled or unshelled'.
This does not appear to be an IV X PV (Independent Variable x Participant Variable) design because there is no information about a participant variable being used here. It also does not seem to be a repeated measures design as individuals are not exposed to all conditions; rather they experience one specific condition. The dependent variable is the amount of peanuts consumed, as determined by weighing the peanut dish after participants have finished eating.
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what is the area of the shaded sector ?
Step-by-step explanation:
Find the area using a proportion.
A / (πr²) = θ / 360°
A / (π (20 ft)²) = 160° / 360°
A ≈ 558.5 ft²
Suppose that the warranty cost of defective widgets is such that their proportion should not exceed 5% for the production to be profitable. Being very cautious, you set a goal of having 0.05 as the upper limit of a 90% confidence interval, when repeating the previous experiment. What should the maximum number of defective widgets be, out of 1024, for this goal to be reached.
Answer: 63 defective widgets
Step-by-step explanation:
Given that the proportion should not exceed 5%, that is:
p< or = 5%.
So we take p = 5% = 0.05
q = 1 - 0.05 = 0.095
Where q is the proportion of non-defective
We need to calculate the standard error (standard deviation)
S = √pq/n
Where n = 1024
S = √(0.05 × 0.095)/1024
S = 0.00681
Since production is to maximize profit(profitable), we need to minimize the number of defective items. So we find the limit of defective product to make this possible using the Upper Class Limit.
UCL = p + Za/2(n-1) × S
Where a is alpha of confidence interval = 100 -90 = 10%
a/2 = 5% = 0.05
UCL = p + Z (0.05) × 0.00681
Z(0.05) is read on the t-distribution table at (n-1) degree of freedom, which is at infinity since 1023 = n-1 is large.
Z a/2 = 1.64
UCL = 0.05 + 1.64 × 0.00681
UCL = 0.0612
Since the UCL in this case is a measure of proportion of defective widgets
Maximum defective widgets = 0.0612 ×1024 = 63
Alternatively
UCL = p + 3√pq/n
= 0.05 + 3(0.00681)
= 0.05 + 0.02043 = 0.07043
UCL =0.07043
Max. Number of widgets = 0.07043 × 1024
= 72
The net worthw(t) of a company is growing at a rate of′(t) = 2000−121t2dollarsper year, wheretis in years since 1990.(a) If the company is worth $40,000 in 1990, how much is it worth in 2000
Answer:
The worth of the company in 2000 is $56,000.
Step-by-step explanation:
The growth rate of the company is:
[tex]f'(t)=2000-12t^{2}[/tex]
To determine the worth of the company in 2000, first compute the change in the net worth during the period 1990 (t = 0) to 2000 (t = 10) as follows:
[tex]\int\limits^{10}_{0} {2000-12t^{2}} \, dt =\int\limits^{10}_{0} {2000} \, dt-12\int\limits^{10}_{0} {t^{2}} \, dt=2000 |t|^{10}_{0}-12|\frac{t^{3}}{3}|^{10}_{0}\\=(2000\times10)-(4\times10^{3})\\=20000-4000\\=16000[/tex]
The increase in the company's net worth from 1990 to 2000 is $16,000.
If the company's worth was $40,000 in 1990 then the worth of the company in 2000 is:
Worth in 2000 = Worth in 1990 + Net increase in company's worth
[tex]=40000+16000\\=56000[/tex]
Thus, the worth of the company in 2000 is $56,000.
The timekeeper for a particular mile race uses a stopwatch to determine the finishing times of the racers. He then calculates that the mean time for the first three finishers was 5.75 minutes. After checking his stopwatch, he notices to his horror that the stopwatch begins timing at 45 seconds rather than at 0, resulting in scores each of which is 45 seconds too long. What is the correct mean time for the first three finishers?
Final answer:
The correct mean time for the first three finishers is 5 minutes, after subtracting the 45-second error from the initially recorded mean time of 5.75 minutes.
Explanation:
The correct mean time for the first three finishers, after adjusting for the stopwatch error, can be calculated by subtracting the 45 seconds from the initially recorded meantime. Since the mean time was calculated to be 5.75 minutes (or 345 seconds), we must correct each time by subtracting 45 seconds and then recalculating the mean.
To find the correct mean time, we do the following steps:
First, convert the mean time from minutes to seconds, so 5.75 minutes = 5 minutes and 45 seconds or 345 seconds.Subtract 45 seconds from each racer's finish time to correct the error: 345 seconds - 45 seconds = 300 seconds.Then, convert the corrected total time back to minutes, so 300 seconds is equal to 5 minutes.The correct mean time for the first three finishers is therefore 5 minutes.
A 99% confidence interval for the mean \muμ of a population is computed from a random sample and found to be 6 ± 3. We may conclude that Group of answer choices there is a 99% probability that \mu μ is between 3 and 9. there is a 99% probability that the true mean is 6, and there is a 99% chance that the true margin of error is 3. if we took many additional random samples, the mean would be 6 and the margin of error would be 3. if we obtained several more samples and calculated a confidence interval for each the margin of error would be 3 but the sample mean would be different.
Answer:
There is a 99% probability that \mu μ is between 3 and 9.
Step-by-step explanation:
For a confidence level of x% for a mean being found between a and b, it means that we are x% sure, that is, there is a x% probability that the true mean for the population is between a and b.
A 99% confidence interval for the mean μ of a population is computed from a random sample and found to be 6 ± 3.
99% probability that the true mean of the population is between 3 and 9.
So the correct answer is:
There is a 99% probability that \mu μ is between 3 and 9.
A 99% confidence interval of 6 ± 3 implies that we are 99% confident that the population mean is between 3 and 9. If further samples were taken, both the mean and margin of error could vary, depending on the specific data points and variability of each sample.
Explanation:A 99% confidence interval refers to the range of values within which we are 99% confident that the population mean (μ) resides. In this case, the confidence interval is 6 ± 3, which implies that we are 99% confident that the mean is anywhere between 3 and 9. It does not mean that there is a 99% probability that the true mean is 6, or that the true margin of error is 3. The margin of error is a function of your data's variability and your sample size, not a fixed number.
If we take more random samples, the mean could be different because each sample may contain different individuals or data points with different distributions. However, if the sampling process is properly randomized and unbiased, on average, the mean of the samples should be close to the population mean.
While the margin of error would be expected to remain around 3 for similar sample sizes and variability, it is not guaranteed to always be 3, as it depends on the specific data in each sample.
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Every morning Jack flips a fair coin ten times. He does this for anentire year. LetXbe the number of days when all the flips come out the same way(all heads or all tails).(a) Give the exact expression for the probabilityP(X >1).(b) Is it appropriate to approximateXby a Poisson distribution
The probability question from part (a) requires calculating the chance of getting all heads or all tails on multiple days in a year, which involves complex probability distributions. For part (b), using a Poisson distribution could be appropriate due to the rarity of the event and the high number of trials involved.
Explanation:The question pertains to the field of probability theory and involves calculating the probability of specific outcomes when flipping a fair coin. For part (a), Jack flips a coin ten times each morning for a year, counting the days (X) when all flips are identical (all heads or all tails). The exact expression for P(X > 1), the probability of more than one such day, requires several steps. First, we find the probability of a single day having all heads or all tails, then use that to calculate the probability for multiple days within the year. For part (b), whether it is appropriate to approximate X by a Poisson distribution depends on the rarity of the event in question and the number of trials. A Poisson distribution is typically used for rare events over many trials, which may apply here.
For part (a), the probability on any given day is the sum of the probabilities of all heads or all tails: 2*(0.5^10). Over a year (365 days), we need to calculate the probability distribution for this outcome occurring on multiple days. To find P(X > 1), we would need to use the binomial distribution and subtract the probability of the event not occurring at all (P(X=0)) and occurring exactly once (P(X=1)) from 1. However, this calculation can become quite complex due to the large number of trials.
For part (b), given the low probability of the event (all heads or all tails) and the high number of trials (365), a Poisson distribution may be an appropriate approximation. The mean (λ) for the Poisson distribution would be the expected number of times the event occurs in a year. Since the probability of all heads or all tails is low, it can be considered a rare event, and the Poisson distribution is often used for modeling such scenarios.
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The operations manager of a large production plant would like to estimate the mean amount of time a worker takes to assemble a new electronic component. Assume that the standard deviation of this assembly time is 3.6 minutes and is normally distributed.
a) After observing 120 workers assembling similar devices, the manager noticed that their average time was 16.2 minutes. Construct a 92% confidence interval for the mean assembly time.
b) How many workers should be involved in this study in order to have the mean assembly time estimated up to
We can calculate the 92% confidence interval for the mean assembly time to be approximately 15.572 to 16.828 minutes. The information provided, however, is insufficient to answer part b) of the question.
Explanation:In order to construct a 92% confidence interval for the mean assembly time, we first have to find the z-score corresponding to this level of confidence. The z-score for a 92% confidence interval is approximately 1.75. This value is found using a standard normal distribution table (or z-table).
So, the 92% confidence interval for the mean assembly time of 16.2 minutes can be computed as follows: 16.2 ± 1.75*(3.6/sqrt(120)). This gives a confidence interval of approximately 15.572 to 16.828 minutes.
Regarding the second part of your question (b), it seems unfinished and lacks enough information for me to provide a complete answer. Typically, a desired margin of error or a certain level of confidence would need to be specified to determine the sample size.
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a) The 92% confidence interval for the mean assembly time is (15.75 minutes, 16.65 minutes). b) About 993 workers are needed for the study to estimate the mean assembly time with a 0.2-minute margin of error.
To construct the confidence interval, use the formula for the confidence interval of the mean for a normal distribution: [tex]\[ \text{CI} = \bar{x} \pm z \times \frac{\sigma}{\sqrt{n}} \][/tex]
1. Calculate the standard error of the mean: [tex]\[ \text{SE} = \frac{\sigma}{\sqrt{n}} = \frac{3.6}{\sqrt{120}} \approx 0.329 \][/tex]
2. Find the critical z-value for a 92% confidence interval (using a z-table or calculator): [tex]\( z = 1.75 \) (approximately).[/tex]
3. Substitute the values into the formula: [tex]\[ \text{CI} = 16.2 \pm 1.75 \times 0.329 \][/tex]
4. Calculate the confidence interval: [tex]\[ \text{CI} = (16.2 - 0.575, 16.2 + 0.575) \][/tex]
[tex]\[ \text{CI} = (15.625, 16.775) \][/tex]
So, the 92% confidence interval for the mean assembly time is approximately (15.625 minutes, 16.775 minutes).
b) To estimate the mean assembly time with a desired margin of error, use the formula for the required sample size: [tex]\[ n = \left( \frac{z \times \sigma}{\text{ME}} \right)^2 \][/tex]
Given the desired margin of error (ME), let's say 0.2 minutes, and using \( z = 1.75 \) from the 92% confidence level:
[tex]\[ n = \left( \frac{1.75 \times 3.6}{0.2} \right)^2 \]\[ n = (31.5)^2 \]\[ n = 992.25 \][/tex]
So, approximately 993 workers should be involved in the study to estimate the mean assembly time with a margin of error of 0.2 minutes.
Explain why the difference in match win probabilities is significant. (ii) Explain why the number of challenges remaining in the set is significant. (iii) On what basis did the authors conclude that players challenge too few calls
Answer:
Step-by-step explanation:
The number of observations in model 1 are more compared with model 2 and model 3. And, the coefficient values increased from model 1 to model 3. Moreover, the standard error values are increased. Hence, the test statistic value increases. Therefore, the variable “difference in match win probabilities” is significant.
State the null and alternative Hypotheses:
Null Hypothesis:
H0: The difference in match win probabilities is not significant.
Alternative Hypothesis:
H1: The difference in match win probabilities is significant.
From the given information (model 3), the coefficient of “The difference in match win probabilities” is -0.550 and the corresponding standard error is 0.240. The sample size is 1973.
The test statistic is -2.2917 and it is calculated below: on the picture attached.
Statistics homework question answer, step 1, image 1
The number of sample observations is n=1973 and consider the level of significance as 0.05.
The degree of freedom is, n – 1 = 1973–1 = 1972.
Discuss the significance of the discontinuities of T to someone who uses the road. Because of the steady increases and decreases in the toll, drivers may want to avoid the highest rates at t = 7 and t = 24 if feasible. Because of the sudden jumps in the toll, drivers may want to avoid the higher rates between t = 0 and t = 7, between t = 10 and t = 16, and between t = 19 and t = 24 if feasible. The function is continuous, so there is no significance. Because of the sudden jumps in the toll, drivers may want to avoid the higher rates between t = 7 and t = 10 and between t = 16 and t = 19 if feasible
the answer is in the attachment
The lifetime of a certain transistor in a certain application has mean 900 hours and standard deviation 30 hours. Find the mean and standard deviation of the length of time that four bulbs will last
Answer:
[tex]E(T)=E(X1 + X2 + X3 + X4 =E(x1) + E(x2) + E(x3) + E(x4) [/tex]
[tex]E(T) =900+900+900+900 =3600[/tex]
And the mean would be:
[tex] \bar X = \frac{T}{4} = \frac{3600}{4}= 900[/tex]
And the standard deviation of total time would be:
[tex]SD(T)=\sqrt(Var(T)) = sqrt(Var(X1) + Var(x2) + Var(x3) + Var(x4))[/tex]
[tex]\sigma=sqrt(30^2+30^2+30^2+30^2) =\sqrt(3600) =60[/tex]
Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Solution to the problem
Let total life time of t four transistors T = X1 + X2 + X3 + X4 (where X1,X2,X3 and X4 are life time of individual transistors
For this case the mean length of time that four transistors will last
[tex]E(T)=E(X1 + X2 + X3 + X4 =E(x1) + E(x2) + E(x3) + E(x4) [/tex]
[tex]E(T) =900+900+900+900 =3600[/tex]
And the mean would be:
[tex] \bar X = \frac{T}{4} = \frac{3600}{4}= 900[/tex]
And the standard deviation of total time would be:
[tex]SD(T)=\sqrt(Var(T)) = sqrt(Var(X1) + Var(x2) + Var(x3) + Var(x4))[/tex]
[tex]\sigma=sqrt(30^2+30^2+30^2+30^2) =\sqrt(3600) =60[/tex]
The mean lifetime of four transistors is 3600 hours and the standard deviation of the lifetime is 60 hours.
Explanation:The mean and standard deviation are statistical concepts used to describe a dataset. The mean, or average, is the sum of all data points divided by the number of data points. In this case, the mean lifetime of a single transistor is 900 hours. If you have four transistors, their total lifetime would be four times the mean of a single transistor. So, the mean lifetime for four transistors is 4 * 900 = 3600 hours.
The standard deviation measures the amount of variation in a set of values. When you are considering the lifetime of multiple transistors, you add the variances - not the standard deviations - and then take the square root. The variance is the square of the standard deviation. So, the standard deviation for the lifetime of four transistors is sqrt(4) * 30 = 60 hours.
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A 20% solution of fertilizer is to be mixed with a 50% solution of fertilizer in order to get 180 gallons of a 40% solution. How many gallons of the 20% solution and 50%
solution should be mixed?
Answer: 60 gallons of the 20% solution and 120 gallons of the 50% solution should be mixed.
Step-by-step explanation:
Let x represent the number of gallons of 20% solution that should be mixed.
Let y represent the number of gallons of 50% solution that should be mixed.
A 20% solution of fertilizer is to be mixed with a 50% solution of fertilizer in order to get 180 gallons of a 40% solution. This means that
0.2x + 0.5y = 0.4×180
0.2x + 0.5y = 72- - - - - - - - - - - -1
Since the total number of gallons is 180, it means that
x + y = 180
Substituting x = 180 - y into equation 1, it becomes
0.2(180 - y) + 0.5y = 72
36 - 0.2y + 0.5y = 72
- 0.2y + 0.5y = 72 - 36
0.3y = 36
y = 36/0.3
y = 120
x = 180 - y = 180 - 120
x = 60
To get a 40% solution by mixing a 20% and 50% solution, you need 60 gallons of the 20% solution and 120 gallons of the 50% solution.
Explanation:To solve this problem, we can use the concept of mixtures. Let's represent the amount of 20% solution as x gallons and the amount of 50% solution as y gallons. From the given information, we can set up the following equations:
x + y = 180 (equation 1)
0.2x + 0.5y = 0.4 * 180 (equation 2)
Simplifying equation 2 gives us 0.2x + 0.5y = 72. To eliminate decimals, we can multiply both sides by 10 to get 2x + 5y = 720.
Now we have a system of equations. We can solve it by substitution or elimination. Let's use elimination:
Multiplying equation 1 by 2 gives us 2x + 2y = 360. Subtracting this from equation 2 gives us 3y = 360. Dividing both sides by 3 gives us y = 120. Substituting this value into equation 1 gives us x + 120 = 180. Subtracting 120 from both sides gives us x = 60.
Therefore, we need 60 gallons of the 20% solution and 120 gallons of the 50% solution.
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A process manufactures ball bearings with diameters that are normally distributed with mean 25.14 mm and standard deviation 0.08 mm. Specifications call for the diameter to be in the interval 2.5 ± 0.01 cm. What proportion of the ball bearings will meet the specification? A. Describe the distribution of ball diameters using proper statistical notation. B. Represent the situation described in the problem graphically. C. Calculate the proportion of ball bearings meeting the specifications
Answer:
a) Let X the random variable that represent the diameters of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(2.514 cm,0.008 cm)[/tex]
Where [tex]\mu=2.514[/tex] and [tex]\sigma=0.008[/tex]
b) For this case we can see the interval required on the figure attached,
c) [tex]P(2.49<X<2.51)=P(\frac{2.49-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{2.51-\mu}{\sigma})=P(\frac{2.49-2.514}{0.008}<Z<\frac{2.51-2.514}{0.008})=P(-3<z<-0.5)[/tex]
And we can find this probability with this difference:
[tex]P(-3<z<-0.5)=P(z<-0.5)-P(z<-3)[/tex]
And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.
[tex]P(-3<z<-0.5)=P(z<-0.5)-P(z<-3)=0.309-0.00135=0.307[/tex]
Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Part a
Let X the random variable that represent the diameters of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(2.514 cm,0.008 cm)[/tex]
Where [tex]\mu=2.514[/tex] and [tex]\sigma=0.008[/tex]
Part b
For this case we can see the interval required on the figure attached,
Part c
We are interested on this probability
[tex]P(2.49<X<2.51)[/tex]
And the best way to solve this problem is using the normal standard distribution and the z score given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
If we apply this formula to our probability we got this:
[tex]P(2.49<X<2.51)=P(\frac{2.49-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{2.51-\mu}{\sigma})=P(\frac{2.49-2.514}{0.008}<Z<\frac{2.51-2.514}{0.008})=P(-3<z<-0.5)[/tex]
And we can find this probability with this difference:
[tex]P(-3<z<-0.5)=P(z<-0.5)-P(z<-3)[/tex]
And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.
[tex]P(-3<z<-0.5)=P(z<-0.5)-P(z<-3)=0.309-0.00135=0.307[/tex]
Alice and Bob each picks an integer number uniformly at random between 1 and n. Assume that all possible combinations of two numbers are equally likely to be picked. What is the probability that Alice’s number is bigger than Bob’s?
Answer:
The answer is 1/n
Step-by-step explanation:
First number = Alice's number
Second Number = Bob's Number
There are already 2 good answers if the first number is AT LEAST 1 greater than the second.
However, what if the first number is EXACTLY 1 greater than the second?
I will assume the first number is removed from the array after being selected.
If the first number is 1, then it is impossible, because the second number cannot be less than 1.
If the first number is 2, then the second number could be 1, which has a probability of 1/(n-1)
If the first number is 3, then the second number could be 2, which has a probability of 1/(n-1)
etc…
There is a 1/n probability of each first number selected, so the answer would be:
0*1/n+1/n*1/(n−1)+1/n*1/(n−1)+⋯+1/n*1/(n−1)
There are n terms in that series.
That gives us:
0+1/n(n−1)+1/n(n−1)+⋯+1/n(n−1)
There are n-1 of those identical terms, because 1 of the terms is 0.
(n−1)/n(n−1)=1/n
So the answer is 1/n.
Researchers conducted a study of obesity in children. They measured body mass index (BMI), which is a measure of weight relative to height. High BMI is an indication of obesity. Data from a study published in the Journal of the American Dietetic Association shows a fairly strong positive linear association between mother’s BMI and daughter’s BMI (r = 0.506). This means that obese mothers tend to have obese daughters.
1. Based on this study, what proportion of the variation in the daughter BMI measurements is explained by the mother BMI measurements?
2. What are some of the other variables that explain the variability in the daughter BMI?
Answer:
Part a
For this case after do the operations we got a value for the correlation coeffcient of:
[tex] r =0.506[/tex]
With this value we can find the determination coefficient:
[tex] r^2 = 0.506^2 = 0.256[/tex]
And with this value we can analyze the proportion of variance explained by one variable and the other. So we can conclude that 25.6% of the mother's BMI variation is explained by the daugther's BMI.
Part a
Since the BMI is a relation between height and weight, other possible variables that can explain the variability are (weight , height, age)
Step-by-step explanation:
Previous concepts
The correlation coefficient is a "statistical measure that calculates the strength of the relationship between the relative movements of two variables". It's denoted by r and its always between -1 and 1.
And in order to calculate the correlation coefficient we can use this formula:
[tex]r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^2 -(\sum x)^2][n\sum y^2 -(\sum y)^2]}}[/tex]
Solution to the problem
Part a
For this case after do the operations we got a value for the correlation coeffcient of:
[tex] r =0.506[/tex]
With this value we can find the determination coefficient:
[tex] r^2 = 0.506^2 = 0.256[/tex]
And with this value we can analyze the proportion of variance explained by one variable and the other. So we can conclude that 25.6% of the mother's BMI variation is explained by the daugther's BMI.
Part a
Since the BMI is a relation between height and weight, other possible variables that can explain the variability are (weight , height, age)
One hundred eight Americans were surveyed to determine the number of hours they spend watching television each month. It was revealed that they watched an average of 151 hours each month with a standard deviation of 32 hours. Assume that the underlying population distribution is normal.
Construct a 99% confidence interval for the population mean hours spent watching television per month.
Answer:
The 99% confidence interval for the population mean hours spent watching television per month is between 143.07 hours and 158.93 hours.
Step-by-step explanation:
We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:
[tex]\alpha = \frac{1-0.99}{2} = 0.005[/tex]
Now, we have to find z in the Ztable as such z has a pvalue of [tex]1-\alpha[/tex].
So it is z with a pvalue of [tex]1-0.005 = 0.995[/tex], so [tex]z = 2.575[/tex]
Now, find M as such
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.
[tex]M = 2.575*\frac{32}{\sqrt{108}} = 7.93[/tex]
The lower end of the interval is the mean subtracted by M. So it is 151 - 7.93 = 143.07 hours
The upper end of the interval is the mean added to M. So it is 151 + 7.93 = 158.93 hours
The 99% confidence interval for the population mean hours spent watching television per month is between 143.07 hours and 158.93 hours.
Long-run classical model from Chapter 3. You must provide properly labeled graphs to get full credit!!!!!!! 3) A) Suppose there is a permanent increase in the labor force (L). a) What will be the impact on the real wage (W/P) and the real rental price of capital (R/P)
Answer:
Below the classical model, economic growth is necessarily achieved because of stability in the wage level. For instance, one case of unemployment predominates at a real wage (W / P)1.
Currently, the excessive labor supply would lower the actual wage level before labor supply equals its demand. Eventually, real wage rates would decline to (W / P)F, whereby aggregate labor demand is perfectly matches by aggregate labor supply.
Only the supply side of the production market for products defines the quantity of output & jobs in the classical model.
As the classical method is supply-determined, it states that equiproportional increases (or declines) will not alter the supply of labor in both the rate of money wage and the price level.
A bottling company uses a filling machine to fill plastic bottles with a popular cola. The bottles are supposed to contain 300 ml. In fact, the contents vary according to a normal distribution with mean μ = 298 ml and standard deviation σ = 3ml.
What is the probability that a randomly selected bottle contains less than 295 ml?
Answer:
0.16
Step-by-step explanation:
For a normal distribution, the 68-95-99.7 rule says that 68% of the distribution lies within 1 standard deviation from the mean, 95% within two standard deviations and 99.7% within three standard deviations.
From the question,
Mean μ = 298 ml
Standard deviation σ = 3 ml
A value of 295 ml is within one standard deviation = 68%.
Since it's on the lower side, it's within 68% ÷ 2 = 34%.
The lower half below the mean is 50% of the distribution. Hence, for a selection less than 1 standard deviation, the probability is
50% - 34% = 16% = 0.16
Final answer:
The probability that a randomly selected bottle contains less than 295 ml of cola is calculated using the normal distribution. With a z-score of -1, ths probability is approximately 15.87%.
Explanation:
To determine the probability that a randomly selected bottle contains less than 295 ml of cola, we can use the properties of the normal distribution.
Given that the mean is 298 ml and the standard deviation is 3 ml, we must first find the z-score associated with 295 ml.
The z-score formula is z = (X - mu) / SD where X is the value for which we want to find the probability.
Substituting the given values into the formula, we get:
z = (295 - 298) / 3 = -1.
A z-score of -1 tells us that 295 ml is one standard deviation below the mean.
To find the probability corresponding to this z-score, we refer to a standard normal distribution table or use relevant statistical software. If we look up the probability for z = -1 in the z-table, we find that the area to the left of z is approximately 0.1587.
Therefore, the probability that a randomly selected bottle contains less than 295 ml of cola is about 15.87%.
Isaac, a manager at a supermarket, is inspecting cans of pasta to make sure that the cans are neither dented nor have other defects. From past experience, he knows that 1 can in every 90 is defective. What is the probability that Isaac will find his first defective can among the first 50 cans
Answer:
0.0064 is the probability that Isaac will find his first defective can among the first 50 cans.
Step-by-step explanation:
W are given the following in the question:
Probability of defective can =
[tex]P(A) = \dfrac{1}{90}[/tex]
We have to find the probability that Isaac will find his first defective can among the first 50 cans.
Then the number of adults follows a geometric distribution, where the probability of success on each trial is p, then the probability that the kth trial (out of k trials) is the first success is
[tex]P(X=k) = (1-p)^{k-1}p[/tex]
We have to evaluate:
[tex]P(x = 50)\\= (1-\frac{1}{90})^{50-1}(\frac{1}{90})\\= 0.0064[/tex]
0.0064 is the probability that Isaac will find his first defective can among the first 50 cans.
Suppose that the current measurements in a strip of wire are assumed to follow a normal distribution with a mean of 10 millimeters and a standard deviation of 2 millimeters. What is the probability that a measurement exceeds 13 milliamperes
Answer: the probability that a measurement exceeds 13 milliamperes is 0.067
Step-by-step explanation:
Suppose that the current measurements in a strip of wire are assumed to follow a normal distribution, we would apply the formula for normal distribution which is expressed as
z = (x - µ)/σ
Where
x = current measurements in a strip.
µ = mean current
σ = standard deviation
From the information given,
µ = 10
σ = 2
We want to find the probability that a measurement exceeds 13 milliamperes. It is expressed as
P(x > 13) = 1 - P(x ≤ 13)
For x = 13,
z = (13 - 10)/2 = 1.5
Looking at the normal distribution table, the probability corresponding to the z score is 0.933
P(x > 13) = 1 - 0.933 = 0.067
Final answer:
To find the probability that a current measurement exceeds 13 milliamperes in a normally distributed set with mean 10 mA and SD 2 mA, calculate the Z-score and use a normal distribution table or software.
Explanation:
The student's question seems to mistakenly mix units (millimeters instead of milliamperes), but assuming the intent was to refer to electrical current and not physical measurements of wire, we will proceed on the basis that the actual question is about the probability of a current measurement exceeding 13 milliamperes.
To calculate the probability that a measurement exceeds 13 milliamperes when the current measurements in a strip of wire are normally distributed with a mean of 10 milliamperes and standard deviation of 2 milliamperes, we need to use the Z-score formula:
Z = (X - μ) / σ
where X is the value in question (13 milliamperes), μ is the mean (10 milliamperes), and σ is the standard deviation (2 milliamperes).
Plugging in the values:
Z = (13 - 10) / 2 = 1.5
After finding the Z-score, we would look up this value in a standard normal distribution table or use a statistical software to find the probability that Z exceeds 1.5, which gives us the probability that a measurement exceeds 13 milliamperes. For this Z-score, the probability is approximately 6.68%.
Find out for what value of the variable:
do the trinomial a2+7a+6 and the binomial a+1 have the same value?
do the trinomial 3x2−x+1 and the trinomial 2x2+5x−4 have the same value?
The Main Answer for:
1. The value of the variable for which the trinomial [tex]a^2+7a+6[/tex] and the binomial a+1 have the same value is [tex]a=-1[/tex] or [tex]a=-5[/tex].
2. The value of the variable for which the trinomial [tex]3x^2-x+1[/tex] and the trinomial [tex]2x^2+5x-4[/tex] have the same value is [tex]x=1[/tex] or [tex]x=5[/tex].
To find the value of the variable for which two polynomials have the same value, we set them equal to each other and solve for the variable.
1. For the trinomial [tex]a^2+7a+6[/tex] and the binomial [tex]a+1[/tex]:
[tex]a^2+7a+6=a+1[/tex]
First, let's rewrite the equation in standard form:
[tex]a^2+7a+6-(a+1)=0\\a^2+7a+6-a-1=0\\a^2+6a+5=0[/tex]
Now, we can solve this quadratic equation by factoring:
[tex]a^2+6a+5=0\\(a+5)(a+1)=0[/tex]
Setting each factor equal to zero:
[tex]a+5=0[/tex]⇒[tex]a=-5[/tex]
[tex]a+1=0[/tex]⇒[tex]a=-1[/tex]
So, the value of the variable for which the trinomial [tex]a^2+7a+6[/tex] and the binomial a+1 have the same value is [tex]a=-1[/tex] or [tex]a=-5[/tex].
2. For the trinomial [tex]3x^2-x+1[/tex] and the trinomial [tex]2x^2+5x-4[/tex]:
[tex]3x^2-x+1=2x^2+5x-4[/tex]
Again, let's rewrite the equation in standard form:
[tex]3x^2-x+1-(2x^2+5x-4)=0\\3x^2-x+1-2x^2-5x+4=0\\X^2-6x+5=0[/tex]
Now, let's solve this quadratic equation by factoring:
[tex]X^2-6x+5=0\\(x-5)(x-1)=0[/tex]
Setting each factor equal to zero:
[tex]x-5=0[/tex]⇒[tex]x=5[/tex]
[tex]x-1=0[/tex]⇒[tex]x=1[/tex]
So, the value of the variable for which the trinomial [tex]3x^2-x+1[/tex] and the trinomial [tex]2x^2+5x-4[/tex] have the same value is [tex]x=1[/tex] or [tex]x=5[/tex].
COMPLETE QUESTION:
Find out for what value of the variable:
1. Do the trinomial [tex]a^2+7a+6[/tex] and the binomial [tex]a+1[/tex] have the same value?
2. Do the trinomial [tex]3x^2-x+1[/tex] and the trinomial [tex]2x^2+5x-4[/tex] have the same value?
A random variableX= {0, 1, 2, 3, ...} has cumulative distribution function.a) Calculate the probability that 3 ≤X≤ 5.b) Find the expected value of X, E(X), using the fact that. (Hint: You will have to evaluate an infinite sum, but that will be easy to do if you notice that
Answer:
a) P ( 3 ≤X≤ 5 ) = 0.02619
b) E(X) = 1
Step-by-step explanation:
Given:
- The CDF of a random variable X = { 0 , 1 , 2 , 3 , .... } is given as:
[tex]F(X) = P ( X =< x) = 1 - \frac{1}{(x+1)*(x+2)}[/tex]
Find:
a.Calculate the probability that 3 ≤X≤ 5
b) Find the expected value of X, E(X), using the fact that. (Hint: You will have to evaluate an infinite sum, but that will be easy to do if you notice that
Solution:
- The CDF gives the probability of (X < x) for any value of x. So to compute the P ( 3 ≤X≤ 5 ) we will set the limits.
[tex]F(X) = P ( 3=<X =< 5) = [1 - \frac{1}{(x+1)*(x+2)}]\limits^5_3\\\\F(X) = P ( 3=<X =< 5) = [-\frac{1}{(5+1)*(5+2)} + \frac{1}{(3+1)*(3+2)}}\\\\F(X) = P ( 3=<X =< 5) = [-\frac{1}{(42)} + \frac{1}{(20)}}]\\\\F(X) = P ( 3=<X =< 5) = 0.02619[/tex]
- The Expected Value can be determined by sum to infinity of CDF:
E(X) = Σ ( 1 - F(X) )
[tex]E(X) = \frac{1}{(x+1)*(x+2)} = \frac{1}{(x+1)} - \frac{1}{(x+2)} \\\\= \frac{1}{(1)} - \frac{1}{(2)}\\\\= \frac{1}{(2)} - \frac{1}{(3)} \\\\= \frac{1}{(3)} - \frac{1}{(4)}\\\\= ............................................\\\\= \frac{1}{(n)} - \frac{1}{(n+1)}\\\\= \frac{1}{(n+1)} - \frac{1}{(n+ 2)}[/tex]
E(X) = Limit n->∞ [1 - 1 / ( n + 2 ) ]
E(X) = 1
A rocket is launched upward from a launching pad and the function h determines the rocket's height above the launching pad (in miles) given a number of minutes t since the rocket was launched.
What does the equality h(4) = 516 convey about the rocket in this context? Select all that apply.
(A) The rocket travels 516 miles every 4 minutes
(B) 14 minutes after the rocket was launched, the rocket is 516 miles above the ground
(C) The rocket is currently 516 miles above the ground
(D) 516 minutes after the rocket was launched, the rocket is 4 miles above the ground
Answer:
(B) 4 minutes after the rocket was launched, the rocket is 516 miles above the ground
Step-by-step explanation:
The function h(t) represents the height of the rocket above the launchpad after a time t minutes.
h(4) = 516 means that when t = 4 minutes, the height h is 516 miles above the launch pad. Note that the time t is measured from when the rocket is launched.
The first option indicates a rate of change, which is a velocity. This is not indicated in the question because velocity is a time derivative of the the height function.
Option C implies that the rocket is 516 miles currently but we do not know what time currently is from the time of launch.
The fourth option has reversed the roles of the variable by implying the time is 516 minutes while the height is 4 miles which is not what the function means.
A sample of 30 is taken from a population of Inconel weldments. The average thickness is 1.87 inches at a key location on the part. The sample standard deviation is 0.125 inches and will be used as an estimate of population standard deviation.Calculate the 99% confidence interval. (Hint: Z(a/2) is 2.58 for a 99% CI)a. (1.811, 1.929)b. (1.611, 1.729)c. (1.711, 1.829)d. (1.511, 1.629)
Answer:
a. (1.811, 1.929)
Step-by-step explanation:
We have to find find M as such
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
In which, z is Z(a/2), [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample. So
[tex]M = 2.58*\frac{0.125}{\sqrt{30}} = 0.059[/tex]
The lower end of the interval is the mean subtracted by M. So it is 1.87 - 0.059 = 1.811.
The upper end of the interval is the mean added to M. So it is 1.87 + 0.059 = 1.929
So the correct answer is:
a. (1.811, 1.929)
In a poll, respondents were asked if they have traveled to Europe. 68 respondents indicated that they have traveled to Europe and 124 respondents said that they have not traveled to Europe. If one of these respondents is randomly selected, what is the probability of getting someone who has traveled to Europe?
Answer:
T= A person selected in the poll travel to Europe
NT= A person selected in the poll NOT travel to Europe
For this case we have the following respondents for each event
n(T)= 68
n(NT) = 124
So then the total of people for the poll are:
[tex] n = n(T) + n(NT)= 68 +124= 192[/tex]
And we are interested on the probability of getting someone who has traveled to Europe, and we can use the empirical definition of probability given by:
[tex] p =\frac{Possible}{Total}[/tex]
And if we replace we got:
[tex] p = \frac{n(T)}{n}= \frac{68}{192}= 0.354[/tex]
So then the probability of getting someone who has traveled to Europe is 0.354
Step-by-step explanation:
For this case we define the following events:
T= A person selected in the poll travel to Europe
NT= A person selected in the poll NOT travel to Europe
For this case we have the following respondents for each event
n(T)= 68
n(NT) = 124
So then the total of people for the poll are:
[tex] n = n(T) + n(NT)= 68 +124= 192[/tex]
And we are interested on the probability of getting someone who has traveled to Europe, and we can use the empirical definition of probability given by:
[tex] p =\frac{Possible}{Total}[/tex]
And if we replace we got:
[tex] p = \frac{n(T)}{n}= \frac{68}{192}= 0.354[/tex]
So then the probability of getting someone who has traveled to Europe is 0.354
The time, in number of days, until maturity of a certain variety of tomato plant is Normally distributed, with mean μ and standard deviation s = 2.4. You select a simple random sample of four plants of this variety and measure the time until maturity. The sample yields ¯x = 65. You read on the package of seeds that these tomatoes reach maturity, on average, in 61 days. You want to test to see if your seeds are reaching maturity later than expected, which might indicate that your package of seeds is too old.
The appropriate hypotheses are:
a) H0 : μ = 61 , Hα : μ > 61 .
b) H0 : μ = 65 , Hα : μ < 65 .
c) H0 : μ = 61 , Hα : μ < 61 .
d) H0 : μ = 65 , Hα : μ > 65 .
Answer:
Option a) H0 : μ = 61 , Hα : μ > 61 .
Step-by-step explanation:
We are given that the time, in number of days, until maturity of a certain variety of tomato plant is Normally distributed, with mean μ and standard deviation s = 2.4.
You select a simple random sample of four plants of this variety and measure the time until maturity. The sample yields x bar = 65.
You read on the package of seeds that these tomatoes reach maturity, on average, in 61 days.
Now, since we know about;
sample size, n = 4
sample standard deviation, s = 2.4
sample, x bar = 65
The only line which represent information about the population mean is reading on the package of seeds that these tomatoes reach maturity, on average, in 61 days.
This shows that population mean , [tex]\mu[/tex] = 61 and in null and alternate hypothesis only population mean is tested.
So, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu[/tex] = 61
Alternate Hypothesis, [tex]H_1[/tex] : [tex]\mu[/tex] > 61 { we are asked for later than expected}
Given subspaces H and K of a vector space V, the sum of H and K, written as H+K, is the set of all vectors in V that can be written as the sum of two vectors, one in H and the other in K; that is, H+K={w:w=u+v for some u in H and some v in K}
a. Show that H+K is a subspace of V.
b. Show that H is a subspace of H+K and K is a subspace of H+K.
Answer:
a. H + K = {u + v}
b. See explanation below
Step-by-step explanation:
Given
H+K={w:w=u+v for some u in H and some v in K}
a.
From the (given) above,
We have that
0 = v1 + 0.u
v1 represents the zero vector in K
We also have that
0 = v + u1
u1 represents the zero vector in H
Up this point, we've shown that both H and K have vector factors, V
We'll use the available data to solve for the zero vector of H + K
This is given by
u1 + v1 + u2 + v2 ---- Rearrange
u1 + u2 + v1 + v2 ---- Let u3 = u1 + u2 and v3 = v1 + v2.
So, we have
u3 + v3 as the zero vector of H + K
H + K = {u + v}
So, H + K is a subspace of V
b
Since H, K are subspace of V, then they are
1. Both close under scalar multiplication
2. Both closed under vector addition
1 and 2 gives
c(u + v) where c is a constant
= c.u + c.v
We can then conclude that H is a subspace of H+K and K is a subspace of H+K because H + K is closed under scalar multiplication and vector addition
The sum of two subspaces H and K is a subspace of V as it contains the zero vector, is closed under addition, and under scalar multiplication. Additionally, H and K themselves also qualify as subspaces of this resultant sum subspace, H+K.
Explanation:To show that H+K is a subspace of V, we need to prove that it satisfies three properties: it contains the zero vector, it is closed under addition, and it is closed under scalar multiplication.
1. Zero Vector: Since H and K are subspaces, they both contain the zero vector. So, 0 = 0 + 0 represents a vector in H+K (with the first 0 from H and the second from K).
2. Closure under Addition: If w1 and w2 are vectors in H+K, w1 can be written as h1 + k1 (h1 in H, k1 in K) and w2 as h2 + k2 (h2 in H, k2 in K). Adding w1 and w2 gives: (h1 + k1) + (h2 + k2) = (h1 + h2) + (k1 + k2), where h1+h2 is in H and k1+k2 is in K, hence, the result is in H+K.
3. Closure under Scalar Multiplication: Let w be in H+K and c be any scalar. Then w = h + k, so cw = c(h + k) = ch + ck, where ch is in H and ck is in K, hence, cw is in H+K.
To show that H is a subspace of H+K and K is a subspace of H+K, note that for any vector h in H, it can be written as h+0 (with 0 as the zero vector of K), and similarly for any vector k in K as 0+k. Thus, H and K are subspaces of H+K.
Learn more about Vector Subspaces here:https://brainly.com/question/30031077
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A department store, on average, has daily sales of 28,651.79. the standard deviation pf sales is $1000. On Tuesday, the store sold $36,211.08 worth of goods. Find Tuesday's Z-score. Was Tuesday an unusually good day?
Answer:
Tuesday's Z-score is 7.56
Step-by-step explanation:
We are given that a department store, on average, has daily sales of 28,651.79 and the standard deviation of sales is $1000.
Also, it is given that on Tuesday, the store sold $36,211.08 worth of goods.
Let X = Daily sales of goods
So, X ~ N([tex]\mu = 28,651.79, \sigma^{2} =1000^{2}[/tex])
The z-score probability distribution is given by;
Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ standard normal N(0,1)
Now, Tuesday,s Z-score is given by;
Z = [tex]\frac{36,211.08-28,651.79}{1000}[/tex] = 7.56
Yes, Tuesday was an unusually good day as on this day more worth of sales takes place as compared to the average daily sales of $28,651.79 .