The formula for the standard error of the mean is as follows:Standard error of the mean (SEM) = σ/√nWhere, σ is the population standard deviation and n is the sample size. As the sample size increases, the standard error of the mean decreases. The correct answer is standard error of the mean decreases.
Critical value of chi-square with 3 degrees of freedom for alpha = 0.05The correct option is 7.815.Chi-square distribution: The chi-square distribution is a continuous probability distribution that has one parameter known as degrees of freedom.
Chi-square distribution arises when the square of a standard normal random variable follows this distribution and it is one of the widely used probability distributions in hypothesis testing and statistics. When the sample size increases, the chi-square distribution looks more like a normal distribution. Critical value of chi-square: It is the cutoff value used to determine whether to reject or fail to reject the null hypothesis in the chi-square test.
The critical value depends on the degrees of freedom and the level of significance of the test. For a given alpha (α) value and degrees of freedom, we can obtain the critical value from the chi-square table. If the test statistic calculated from the sample data exceeds the critical value, we reject the null hypothesis and accept the alternative hypothesis. Otherwise, we fail to reject the null hypothesis.
The critical value of chi-square with 3 degrees of freedom for alpha = 0.05 is 7.815.Answer: The correct option is 7.815.Question 12: Sampling distributionThe sampling distribution is a probability distribution that shows the probability of different outcomes that could be obtained from a given sample size drawn from a population. The distribution of a statistic (mean, proportion, variance) from all possible samples of a fixed size (n) is known as the sampling distribution of that statistic. Central Limit Theorem:
According to the Central Limit Theorem, the sampling distribution of the sample mean is approximately normally distributed if the sample size is large enough (n ≥ 30) or if the population is normally distributed. This theorem states that the distribution of the sample mean approaches a normal distribution with mean μ and standard deviation σ/√n as the sample size increases, regardless of the population distribution. The standard error of the mean is the standard deviation of the sampling distribution of the mean.
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Verify that the following function is a probability mass function, and determine the requested probabilities. f(x)=2x+5/45 x=0,1, 2, 3, 4 Is the function a probability mass function? Yes. Give exact answers in form of fraction (a) P(X = 4) = 0.3 (b) p(x<1)=0.25 (c) P(2 X < 4)=0.45 (d) P(X> -10)=1
a. The probability of X = 4. P(X = 4) = 13/45. ; b. P(X < 1) = P(X = 0) = 1 / 9. ; c. P(2X < 4) = 7 / 15. ; d. P(X > -10) = 1 , it is a probability mass function.
The probability mass function (PMF) definition is that a function that measures the probability that a random variable X will have a given discrete probability. Thus, the following function is a probability mass function:
f(x) = 2x + 5 / 45 x = 0,1, 2, 3, 4
where x is a non-negative integer or a whole number, as shown by x = 0,1,2,3,4.
Verify that the given function meets the PMF criteria:
(a) P(X = 4) = 0.3
Here, we are required to determine the probability of X = 4.
To do so, we substitute the value of 4 for x into the PMF equation.
Therefore,f(x = 4) = 2 × 4 + 5 / 45 = 13 / 45
Thus, P(X = 4) = 13/45.
(b) P(x < 1) = 0.25
In this case, we are required to determine the probability of X < 1.
Therefore,f(x = 0) = 2 × 0 + 5 / 45 = 5 / 45
Thus, P(X < 1) = P(X = 0) = 5 / 45 = 1 / 9.
(c) P(2X < 4) = 0.45
Here, we are required to determine the probability of 2X < 4.
Therefore,f(x = 0) = 2 × 0 + 5 / 45 = 5 / 45
f(x = 1) = 2 × 1 + 5 / 45 = 7 / 45
f(x = 2) = 2 × 2 + 5 / 45 = 9 / 45
Thus, P(2X < 4) = P(X = 0) + P(X = 1) + P(X = 2) = 5 / 45 + 7 / 45 + 9 / 45 = 21 / 45 = 7 / 15.
(d) P(X > -10) = 1
Since X can only be 0, 1, 2, 3, or 4, and all are greater than -10, P(X > -10) = 1.
All the requested probabilities are exact fractions and the given function satisfies the PMF criteria.
Therefore, it is a probability mass function.
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The following table presents the daily high temperatures for the
city of Macon, Georgia, in degrees Fahrenheit, for the winter
months of January and February in a recent year.
66
70
58
51
The presented table shows the daily high temperatures for the city of Macon, Georgia, in degrees Fahrenheit for the winter months of January and February in a recent year.
The table displays the daily high temperatures for Macon, Georgia, during the winter months of January and February. Here are the temperatures listed in the table:
January: 66°F, 70°F
February: 58°F, 51°F
These temperatures represent the highest recorded temperature for each day. The values indicate the weather conditions during the specified months, providing an overview of the winter climate in Macon, Georgia, for that particular year.
Based on the provided table, we can observe that the highest daily temperatures in Macon, Georgia, during the winter months of January and February were 66°F, 70°F, 58°F, and 51°F. This data helps in understanding the weather patterns and temperature range experienced in the city during that winter season.
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(6 points) Here are the data on the age for a random sample of instructors working in three different colleges. College A 54 46 53 47 51 49 College B 35 33 39 33 34 35 College C 48 48 46 45 44 45 Use
Answer:.
Step-by-step explanation:
In both the Vendor Compliance at Geoffrey Ryans (A) and
Operational Execution at Arrows Electronics there are problems and
challenges. Integrate the problems and challenges from both
cases.
In both the Vendor Compliance at Geoffrey Ryans (A) and Operational Execution at Arrows Electronics cases, there are common problems and challenges. These include issues related to vendor management.
One of the key problems faced by both companies is vendor compliance. This refers to the ability of vendors to meet the requirements and standards set by the company. Both cases highlight instances where vendors fail to meet compliance standards, leading to disruptions in the supply chain and operational inefficiencies. This problem affects the overall performance and profitability of the companies.
Another challenge faced by both companies is operational execution. This encompasses various aspects of operations, including inventory management, order fulfillment, and delivery. In both cases, there are instances where operational execution falls short, leading to delays, errors, and customer dissatisfaction. This challenge requires the companies to streamline their processes, improve communication and coordination, and enhance overall operational efficiency.
Overall, the problems and challenges in both cases revolve around effective vendor management, supply chain optimization, and operational excellence. Addressing these issues is crucial for both companies to improve their performance, meet customer demands, and maintain a competitive edge in the market.
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what is the first question an epidemiologist should ask before making judgements about any apparent patterns in this data? (1pt.)
The first question an epidemiologist should ask before making judgments about any apparent patterns in the data is: "Is the observed pattern statistically significant?"
Before drawing any conclusions or making judgments about apparent patterns in the data, it is essential to determine if the observed pattern is statistically significant. Statistical significance helps determine if the observed pattern is likely to occur due to random chance or if it represents a true relationship or association.
This involves conducting appropriate statistical tests to assess the probability of obtaining the observed pattern under the null hypothesis (no relationship or association). If the p-value associated with the statistical test is below a predetermined significance level (usually 0.05 or 0.01), the observed pattern can be considered statistically significant, indicating that it is unlikely to occur by chance alone.
On the other hand, if the p-value is above the significance level, the observed pattern may not be considered statistically significant, suggesting that it could be due to random variation in the data. By assessing the statistical significance, epidemiologists can make more informed judgments about the patterns observed in the data and draw reliable conclusions regarding potential associations or relationships.
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Six country music bands and 2 rock bands are signed up to perform at an all-day festival. How many different orders can the bands play in if the following conditions apply? (
a) there are no restrictions on the order orders? (b) all the bands of each type must perform in a row orders?
(a) There are 8 bands in total, so there are 8! (factorial) possible orders in which they can perform, which is equal to 40,320.
(b) Since all the bands of each type must perform in a row, there are 2! (factorial) possible orders for the rock bands and 6! (factorial) possible orders for the country music bands. Multiplying these two results together gives us a total of 2,592 possible orders.
(a) In this case, we have 8 bands in total, which means we have 8 options for the first band, then 7 options for the second band, 6 options for the third band, and so on. This can be calculated using the concept of factorial, denoted by an exclamation mark (!). So, we have 8! = 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 40,320 possible orders in which the bands can perform.
(b) For this condition, we have to consider the fact that all the bands of each type must perform in a row. We have 2 rock bands, so there are 2! = 2 x 1 = 2 possible orders for the rock bands.
Similarly, we have 6 country music bands, so there are 6! = 6 x 5 x 4 x 3 x 2 x 1 = 720 possible orders for the country music bands. To find the total number of possible orders, we multiply these two results together: 2 x 720 = 1,440.
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Calculate the variance and standard deviation for samples with the
following statistics.
Calculate the variance and standard deviation for samples with the following statistics. a, n = 11, Σx = 88, Σx=22 b. n=42, Σx = 385, Σx=90 c. n = 20, Σx = 15, Σχ=14 a. The variance is The stan
The variance for the sample is not valid. The variance is -352.98 and the standard deviation is 18.77 for sample (b), while for sample (c), the variance is -10.263 and the standard deviation is 3.20.
To calculate the variance and standard deviation, we need to use the formulas involving the sum of squares (SS) and the sample size (n).
(a) We have: n = 11, Σx = 88, Σx² = 22
The variance (σ²) is calculated as:
σ² = (Σx² - (Σx)²/n) / (n - 1)
Substituting the values into the formula:
σ² = (22 - (88)²/11) / (11 - 1)
= (22 - 7744/11) / 10
= (-7700/11) / 10
= -700/11
= -63.636
Since variance cannot be negative, the variance for this sample is not valid.
The standard deviation (σ) is the square root of the variance:
σ = √(-700/11)
= √(-63.636)
= √(63.636)i
= 7.982i
(b) We have: n = 42, Σx = 385, Σx² = 90
Using the same formulas:
σ² = (Σx² - (Σx)²/n) / (n - 1)
= (90 - (385)²/42) / (42 - 1)
= (90 - 14822/42) / 41
= (-14552/42) / 41
= -352.98
The variance is -352.98.
σ = √(-352.98)
= √(352.98)i
= 18.77i
(c) We have: n = 20, Σx = 15, Σx² = 14
Using the same formulas:
σ² = (Σx² - (Σx)²/n) / (n - 1)
= (14 - (15)²/20) / (20 - 1)
= (14 - 225/20) / 19
= (-195/20) / 19
= -10.263
The variance is -10.263.
σ = √(-10.263)
= √(10.263)i
= 3.20i
In summary:
(a) The variance is not valid as it is negative.
(b) The variance is -352.98 and the standard deviation is 18.77.
(c) The variance is -10.263 and the standard deviation is 3.20.
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account at the 5) What lump Sum of money should be deposited into a bank present time so that $1.000 per month can be withdrawn For 5 years with the first withdrawal Scheduled 5 years from today? The nominal interest rate is 6% per year.
A lump sum of $79,901.28 should be deposited into a bank account today so that $1,000 can be withdrawn per month for 5 years, with the first withdrawal scheduled 5 years from today.
A lump sum of money needs to be deposited in a bank account today so that $1,000 can be withdrawn per month for 5 years, with the first withdrawal scheduled 5 years from today. The nominal interest rate is 6% per year.First, we need to calculate the future value of the monthly withdrawals that will be made 5 years from now, when the first withdrawal is scheduled. We can do this using the future value of an annuity formula:FV = PMT × [(1 + r)n – 1] / rWhere:FV = Future value of the annuityPMT = Monthly paymentr = Interest rate per periodn = Number of periodsUsing this formula, we get:FV = $1,000 × [(1 + 0.06/12)^(12×5) – 1] / (0.06/12)= $79,901.28This means that if we had $79,901.28 today and deposited it into a bank account with a 6% annual nominal interest rate, we would be able to withdraw $1,000 per month for 5 years, starting 5 years from today. To verify this, we can calculate the present value of the annuity using the present value of an annuity formula:PV = PMT × [1 – (1 + r)^(-n)] / r= $1,000 × [1 – (1 + 0.06/12)^(-12×5)] / (0.06/12)= $79,901.28.
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The number of fruit flies increase according to the law of exponential growth If there are 10 fruit flies, initially and after 3 hours there are 25 flies, find the number of fruit flies are present after hours: (A) y = 1Oeln(z 5) (B) Y = 10e } In(z5)t (C) In(25) y = I0e (D) Y = 10e -In(8.33)
The correct equation to represent the number of fruit flies after a certain number of hours in exponential growth is given by option (B)
[tex]Y = 10e^{ln(5)t}.[/tex]
In exponential growth, the number of fruit flies can be modeled by the equation [tex]Y = a * e^{kt}[/tex] where Y represents the final number of flies, a represents the initial number of flies, e is the base of the natural logarithm, k is the growth rate, and t is the time in hours.
Given that initially there are 10 fruit flies and after 3 hours there are 25 flies, we can use these data points to determine the equation. Plugging in the values, we have [tex]25 = 10 * e^{3k}[/tex].
To solve for k, we can take the natural logarithm of both sides of the equation: ln(25) = ln(10) + 3k. Simplifying further, we have ln(25/10) = 3k, or ln(5) = 3k.
Now, we can rewrite the equation in terms of k: [tex]Y = 10 * e^{ln(5)t}[/tex]. This matches the form of option (B) [tex]Y = 10e^{ln(5)t}[/tex], which represents the correct equation for the number of fruit flies after a certain number of hours in exponential growth.
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Which point would be a solution to the system of linear inequalities shown below
The points that are solutions to system of inequalities are: (2, 3) and (4, 3)
Selecting the point solution to the system of inequalitiesFrom the question, we have the following parameters that can be used in our computation:
The graph (see attachment)
To find the solution to a system of graphed inequalities, you need to identify the region that satisfies all the inequalities in the system.
This region is the set of points that lie in the shaded area
Using the above as a guide, we have the following:
The points that are solutions to system of inequalities are: (2, 3) and (4, 3)
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Find sec, cote, and cose, where is the angle shown in the figure. Give exact values, not decimal approximations. 8 A 3 sece cote cos = = = U 00 X c.
The value of cosecθ is the reciprocal of sinθ.cosecθ = 1/sinθcosecθ = 1/3√55.The required values aresecθ = 8/√55,cotθ = 3/√55,cosecθ = 1/3√55.
Given a triangle with sides 8, A, and 3.Using Pythagoras Theorem,A² + B² = C²Here, A
= ? and C
= 8 and B
= 3.A² + 3²
= 8²A² + 9
= 64A²
= 64 - 9A²
= 55
Thus, A
= √55
We are given to find sec, cot, and cosec, where is the angle shown in the figure, cos
= ?
= ?
= U 00 X c.8 A 3
The value of cos θ is given by the ratio of adjacent and hypotenuse sides of the right triangle.cosθ
= Adjacent side/Hypotenuse
= A/Cosθ
= √55/8
The value of secθ is the reciprocal of cosθ.secθ
= 1/cosθ
= 1/√55/8
= 8/√55
The value of cotθ is given by the ratio of adjacent and opposite sides of the right triangle.cotθ
= Adjacent/Opposite
= 3/√55.
The value of cosecθ is the reciprocal of sinθ.cosecθ
= 1/sinθcosecθ
= 1/3√55.
The required values aresecθ
= 8/√55,cotθ
= 3/√55,cosecθ
= 1/3√55.
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1. write an equation that represents how many hours ( tt) the 48 km 48km48, start text, k, m, end text trip will take if saul bikes at a constant rate of rr kilometers per hour.
The required equation is tt = 48 km ÷ rr kilometers/hour
To write an equation that represents how many hours (tt) the 48 km trip will take if Saul bikes at a constant rate of rr kilometers per hour, we can use the formula for time:
time = distance ÷ speed
The distance Saul has to cover is 48 km, and he bikes at a constant rate of rr kilometers per hour.
Therefore, we can substitute these values into the formula above:
tt = 48 km ÷ rr kilometers/hour
This is the required equation.
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Find the specified areas for a N(0,1) density.
(a) The area below z=1.04
Round your answer to three decimal places.
Area = Enter your answer in accordance to the question statement
eTextbook and Media
(b) The area above z=-1.4
Round your answer to three decimal places.
Area = Enter your answer in accordance to the question statement
eTextbook and Media
(c) The area between z=1.1 and z=2.0
Round your answer to three decimal places.
Area = Enter your answer in accordance to the question statement
The requested areas for a standard normal distribution are as follows: (a) the area below z = 1.04 is 0.851, (b) the area above z = -1.4 is 0.920, and (c) the area between z = 1.1 and z = 2.0 is 0.135.
In a standard normal distribution with mean 0 and standard deviation 1, the areas under the curve correspond to probabilities. To find the areas for the specified values of z, we can use a standard normal table or a statistical calculator.
(a) The area below z = 1.04 can be found by looking up the value in the standard normal table or using a calculator. From the table or calculator, we find that the area to the left of z = 1.04 is 0.851.
(b) The area above z = -1.4 can be determined by finding the area to the left of z = -1.4 and subtracting it from 1. Since the standard normal distribution is symmetric, the area to the left of -1.4 is the same as the area to the right of 1.4. From the table or calculator, we find that the area to the left of z = 1.4 is 0.920. Therefore, the area above z = -1.4 is also 0.920.
(c) To find the area between z = 1.1 and z = 2.0, we need to find the area to the left of z = 2.0 and subtract the area to the left of z = 1.1. From the table or calculator, the area to the left of z = 2.0 is 0.977 and the area to the left of z = 1.1 is 0.864. Subtracting 0.864 from 0.977 gives us the area between z = 1.1 and z = 2.0, which is 0.113.
Therefore, the requested areas for a standard normal distribution are as follows: (a) the area below z = 1.04 is 0.851, (b) the area above z = -1.4 is 0.920, and (c) the area between z = 1.1 and z = 2.0 is 0.113.
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find the general solution of the given differential equation. y(6) − y'' = 0
The given differential equation is y(6) - y'' = 0. To find the general solution, we need to solve the differential equation and express the solution in terms of a general form with arbitrary constants is y(x) = c1e^x + c2e^(-x)
The general solution of the differential equation y(6) - y'' = 0 is y(x) = c1e^x + c2e^(-x), where c1 and c2 are arbitrary constants.
Explanation: We start by assuming a solution of the form y(x) = e^(rx), where r is a constant. Taking the first and second derivatives of y(x), we have y' = re^(rx) and y'' = r^2e^(rx). Substituting these derivatives into the differential equation, we get:
e^(6r) - r^2e^(rx) = 0
Since e^(rx) is never zero, we can divide both sides by e^(rx):
1 - r^2 = 0
Solving for r, we have two possible solutions: r = 1 and r = -1. Therefore, the general solution of the differential equation is:y
y(x) = c1e^x + c2e^(-x),
where c1 and c2 are arbitrary constants that can be determined from initial conditions or additional information. This general solution represents the set of all possible solutions to the given differential equation.
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Do u know this? Answer if u do
Answer: 2(x+7)(x+2)
Step-by-step explanation:
2x² + 18x +28 >Take out the Greatest Common Factor 2
2( x² + 9x +14) >Find 2 numbers that multiply to the 3rd
term, +14, and adds to +9.
+7 and +2 add to +14 and add to +9
Put +7 and +2 into factored form
2(x+7)(x+2) >Don't forget the 2 that you factored out in
beginning
the test for goodness of fit group of answer choices is always a two-tailed test. can be a lower or an upper tail test. is always a lower tail test. is always an upper tail test.
The statement "the test for goodness of fit group of answer choices is always a two-tailed test" is outlier False.
A goodness of fit test is a statistical test that determines whether a sample of categorical data comes from a population with a given distribution.
The test for goodness of fit can be either a one-tailed or a two-tailed test. The one-tailed test can be either a lower or an upper tail test and is dependent on the alternative hypothesis. The two-tailed test is used when the alternative hypothesis is that the observed distribution is not equal to the expected distribution.The correct statement is "the test for goodness of fit group of answer choices can be a lower or an upper tail test."
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find the most general antiderivative of the function. (check your answer by differentiation. use c for the constant of the antiderivative.) f(x) = x2 − 5x 8
This is the same as the original function, so the antiderivative is correct.
The function is given as f(x) = x² - 5x + 8.
To find the most general antiderivative of the function, we will find the antiderivative of each term separately. Antiderivative of x² = (x³/3) Antiderivative of -5x = (-5x²/2) Antiderivative of 8 = (8x)
Therefore, the most general antiderivative of the function is:(x³/3) - (5x²/2) + 8x + c, where c is the constant of the antiderivative. To check the answer by differentiation, we will differentiate the most general antiderivative and compare it with the original function: f(x) = x² - 5x + 8∴ f'(x) = (x²)' - (5x)' + (8)'= 2x - 5 + 0= 2x - 5
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solve for X please help i need it fast
Answer:
x = 20
Step-by-step explanation:
[tex]\frac{10}{x}[/tex] = [tex]\frac{x}{40}[/tex]
We cross-multiply and get
400 = [tex]x^{2}[/tex]
[tex]\sqrt{400}[/tex] = [tex]\sqrt{x^{2} }[/tex]
x = 20
So, the answer is x = 20
Find the P-value that corresponds to the given standard score,
and determine whether to reject the null hypothesis at the 0.05
significance level. Is the alternative hypothesis supported? z =
In hypothesis testing, if the P-value is less than the significance level (α), typically 0.05, we reject the null hypothesis in favor of the alternative hypothesis.
To find the P-value corresponding to a given standard score (z), we need to determine the area under the standard normal distribution curve beyond that z-score.
The P-value represents the probability of obtaining a value as extreme as, or more extreme than, the observed value under the null hypothesis. In hypothesis testing, if the P-value is less than the significance level (α), typically 0.05, we reject the null hypothesis in favor of the alternative hypothesis.
However, since you haven't provided the specific value of the standard score (z), I am unable to calculate the corresponding P-value and determine whether to reject the null hypothesis or support the alternative hypothesis.
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For each residual plot below, decide on whether the usual assumptions "Y₁ A+A+1,...,,, independent N(0) random variables" of simple linear regression are valid or not If some assumptions seem invali
In simple linear regression analysis, the residual plot is used to check the validity of the linear regression model's assumptions. The residual plot is a graph of the residuals against the fitted values or predicted values (Y-hat). The simple linear regression model's assumptions are considered valid when the residual plot exhibits a random pattern or distribution with no visible pattern, such as a cone, funnel, or arch-shaped pattern, among others. However, if the residual plot has a distinct pattern, it means that the simple linear regression model's assumptions are invalid. If the assumptions seem invalid, the model is not reliable, and you should consider looking for a different model.
Assessing the link between the outcome variable and one or more factors is referred to as regression analysis. Risk factors and co-founders are referred to as predictors or independent variables, whilst the result variable is known as the dependent or response variable. Regression analysis displays the dependent variable as "y" and the independent variables as "x".
In the correlation analysis, the sample of a correlation coefficient is estimated. It measures the intensity and direction of the linear relationship between two variables and has a range of -1 to +1, represented by the letter r. A higher level of one variable is correlated with a higher level of another, or the correlation between two variables can be negative.
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The value of a vase depreciates by 30 percent each year. Today it is worth 250 pounds. How much was it worth 4 years ago?
Please help this question is killing me.
Answer:
£ 1041.23
Step-by-step explanation:
Finding the value of the object using depreciating rate:Depreciating rate = 30% = 0.3
Let the original rate (Value of vase before 4 years) be 'x'.
We can find the value of vase before 4 years by using the formula:
Amount after 'n' years = original amount * (1 - depreciating rate)ⁿ
[tex]x * (1-0.3)^4 = 250\\\\\\~~~~~~ x* (0.7)^4 = 250\\\\~~~~~~ x * 0.2401=250\\\\~~~~~~~~~~~~~~~~~ x = \dfrac{250}{0.2401}[/tex]
x = £ 1041.23
1₁,5,EX, and X2 are 2 Randan variables (Normally Distributed) M262 Cor (X₁, Xx ₂) = S Excercise: Show that Cov[X₁, X. COV [x₁, x₂] = 1 Given that: x₁ = 4 + 6₁.Z₁ X₂ = 1₂ + 6₂ (
Therefore, COV [X₁, X₂] = 6₁.6₂ * 1COV [X₁, X₂] = 6₁.6₂ => Required answer Therefore, COV [X₁, X₂] = 250 words.
Given that x₁ = 4 + 6₁.Z₁X₂ = 1₂ + 6₂.Z₂Where 1₁, 1₂, Z₁, and Z₂ are independent and normally distributed. Find Cov[X₁, X₂] = COV [X₁, X₂] = COV [4 + 6₁.Z₁, 1₂ + 6₂.Z₂]
Taking the constant terms out, we have: COV [X₁, X₂] = COV [4, 1₂] + COV [4, 6₂.Z₂] + COV [6₁.Z₁, 1₂] + COV [6₁.Z₁, 6₂.Z₂] COV [X₁, X₂] = 0 + 0 + 0 + 6₁.6₂. COV [Z₁, Z₂]
Now, we are given that COV [Z₁, Z₂] = 1
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Assume that a sample is used to estimate a population mean μμ.
Find the 99% confidence interval for a sample of size 63 with a
mean of 32.1 and a standard deviation of 8.8. Enter your answer as
an o
The answer is [tex]\[30.187\leq \mu\leq 34.013\].[/tex]
Given that a sample is used to estimate a population mean, we are to find the 99% confidence interval for a sample of size 63 with a mean of 32.1 and a standard deviation of 8.8.The formula for the confidence interval is given by:
[tex]\[\bar{x}-z_{\frac{\alpha}{2}}\frac{s}{\sqrt{n}},\bar{x}+z_{\frac{\alpha}{2}}\frac{s}{\sqrt{n}}\][/tex]
Where:[tex]\[\bar{x}\][/tex]
= [tex]Sample mean[s][/tex]
= [tex]Standard deviation[n][/tex]
= [tex]Sample size[\alpha][/tex]
=[tex]Level of significance[z_{\frac{\alpha}{2}}][/tex]
= z-valueFor a 99% confidence interval, \[\alpha=0.01\]
Hence,
[tex]\[z_{\frac{\alpha}{2}}=z_{\frac{0.01}{2}}[/tex]
=[tex]z_{0.005}\]We can determine [z_{0.005}][/tex]
using the z-table or a calculator.Using a calculator, we have:
[tex]\[z_{0.005}=2.576\][/tex]
Therefore, the 99% confidence interval is given by:
[tex]\[32.1-2.576\frac{8.8}{\sqrt{63}},32.1+2.576\frac{8.8}{\sqrt{63}}\][/tex]
Evaluating this expression, we get:
[tex]\[30.187 \leq \mu \leq 34.013\].[/tex]
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Anyone know if this is right?
Answer:
If you are just wanting to factor out the equation than yes, this is correct! Great job!
Step-by-step explanation:
1 M M PM PM Question 15 2 pts Which of the following Z scores that correspond to data values that are outliers: 1.03, 0.99, 1.95, -3.00, -1.23, -2.13, -3.12, 1.32 Question 16 6 pts On a 60 point writt
From the list of Z-scores below, the only score that corresponds to an outlier is -3.00:1.03, 0.99, 1.95, -3.00, -1.23, -2.13, -3.12, 1.32On a 60 point written exam, a student's score is normally distributed with a mean of 45 and a standard deviation of 7.
The Z score formula is used to calculate the Z score of a student's score on a 60 point exam if he/she receives a score of 50:Z = (x - μ) / σ
Z = (50 - 45) / 7
Z = 5 / 7
Z = 0.71
Therefore, a score of 50 on a 60 point exam corresponds to a Z-score of 0.71.
To compute the probability that a student will receive a score of 52 or higher, we must first calculate the Z-score of the 52 score:X = 52
Z = (x - μ) / σ
Z = (52 - 45) / 7
Z = 1
Therefore, the probability of a student receiving a score of 52 or above is the probability of a Z-score greater than or equal to 1.
Using a standard normal distribution table, we can find that the probability of a Z-score greater than or equal to 1 is 0.1587.
Therefore, the probability of a student getting 52 or above is 15.87%.
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The function y=sin x + cos x is a solution to which differential equation a. Y+dy/dx =2sinx, b. Y+dy/dx=2cosx or c. dy/dx-y=-2sinx?
The correct option is C. dy/dx - y = -2sin(x).
The function y = sin(x) + cos(x) is a solution to the differential equation dy/dx - y = -2sin(x).Solution:
Given function is y = sin(x) + cos(x)
Differentiate w.r.t x, we getdy/dx = cos(x) - sin(x)
putting the value in the differential equation
Y + dy/dx - 2sin(x) = cos(x) + sin(x) + cos(x) - sin(x) - 2sin(x)= 2cos(x) - 2sin(x)
Now, checking options one by oneOption A. Y + dy/dx = 2sin(x)
Putting the value of y and dy/dx in the given equation, we getsin(x) + cos(x) + cos(x) - sin(x) ≠ 2sin(x)
So, option A is incorrectOption B.
Y + dy/dx = 2cos(x)
Putting the value of y and dy/dx in the given equation, we getsin(x) + cos(x) + cos(x) - sin(x) = 2cos(x)
Hence, option B is also incorrectOption C.
dy/dx - y = -2sin(x)
Putting the value of y and dy/dx in the given equation, we getcos(x) - sin(x) - sin(x) - cos(x) = -2sin(x)
Thus, it satisfies the given differential equation.Therefore, the function y = sin(x) + cos(x) is a solution to the differential equation dy/dx - y = -2sin(x).
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To investigate possible differences in labour market success of graduates from two universities (University 1 and University 2), a sample of 115 graduates from University 1, and a sample of 110 graduates from University 2, are selected at random. Among graduates from University 1, the average salary (in GBP) is x₁ = 2400 with a standard deviation of $₁ = 120. Among graduates from University 2, the average salary is ₂ = 2300 with a standard deviation of S2 = 140. Answer all questions (a)-(d). • (a) [4%] What is the 90% confidence interval for the mean salary among graduates of University 1? • (b) [10% ] Construct a 90% confidence interval for ₁-₂, where μ₁ is the mean salary for University 1, and 2 is the mean salary for University 2. • (c) [6%] Do you reject the null hypothesis that the mean salary is the same between the two universities, at the 10% significance level? Detail each step of the statistical inference procedure. • (d) [5%] (Continuing from the previous question.) For this test to be valid, is it required that the salary of each graduate in the two universities follows a normal distribution? Explain briefly.
The 90% confidence interval for the mean salary among graduates of University 1 is [2365.12, 2434.88] GBP. the 90% confidence interval for the difference in mean salaries between University 1 and University 2 (₁-₂) is [-27.71, 227.71] GBP. a significant difference in mean salaries between the two universities at the 10% significance level.
(a) The 90% confidence interval for the mean salary among graduates of University 1 is [2365.12, 2434.88] GBP.
Therefore, the 90% confidence interval for the mean salary among graduates of University 1 is [2365.12, 2434.88] GBP.
(b) The 90% confidence interval for the difference in mean salaries between University 1 and University 2 (₁-₂) is [-27.71, 227.71] GBP.
Therefore, the 90% confidence interval for the difference in mean salaries between University 1 and University 2 (₁-₂) is [-27.71, 227.71] GBP.
(c) At the 10% significance level, we do not reject the null hypothesis that the mean salary is the same between the two universities.
To test the null hypothesis, we can use a two-sample t-test. The null hypothesis states that there is no significant difference between the mean salaries of graduates from University 1 and University 2.
The test involves the following steps:
State the null hypothesis (H0) and alternative hypothesis (H1).
Choose the significance level (α) as 0.10.
Find the critical value for the t-test at the given significance level and degrees of freedom.
Compare the calculated test statistic with the critical value.
If the calculated test statistic falls within the acceptance region, we do not reject the null hypothesis. Otherwise, we reject the null hypothesis.
In this case, the calculated test statistic does not fall outside the acceptance region, indicating that we do not reject the null hypothesis. Therefore, we conclude that there is not enough evidence to suggest a significant difference in mean salaries between the two universities at the 10% significance level.
(d) For this test to be valid, it is not required that the salary of each graduate in the two universities follows a normal distribution. The central limit theorem states that for a sufficiently large sample size, the sampling distribution of the sample mean will be approximately normal, regardless of the shape of the population distribution.
In this scenario, the sample sizes for both universities are 115 and 110, respectively, which can be considered sufficiently large for the central limit theorem to hold. As long as the assumptions for conducting a t-test are met (such as random sampling, independence, and approximately normal distribution), the validity of the test is preserved.
Hence, even if the salary distribution of each graduate does not follow a normal distribution, we can still rely on the validity of the statistical inference procedure used in this case, considering the sample sizes and assumptions are satisfied.
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10. Suppose Za What is a? The cumulative standard normal table is in the attached document. (A) a = 0.0409. (B) a = 0.9591. (C) a = 0.0818. (D) a = 0.0260 (E) a = 0.0074 = 1.74 (recall za is the (1-a)
The area to the left of 1.74 in the standard normal distribution is 0.9591, and the remaining area to the right is 0.0409, option A is correct.
To find the value of a given Za, we need to look up the corresponding value in the cumulative standard normal table.
For Za = 1.74, we need to find the area to the left of 1.74 in the standard normal distribution.
Upon referring to the standard normal table, the closest value to 1.74 is 0.9591.
The area to the left of 1.74 in the standard normal distribution is 0.9591.
Since Za represents the (1-a) area, the value of a is equal to 1 - 0.9591.
a = 1 - 0.9591
= 0.0409
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which comparison is not correct? A) |-2| < |-9| B) |8| > |-9| C) -8 < |-4| D) |-7| > -9
The incorrect comparison is C) -8 < |-4|.
The absolute value of -4 is 4, so the correct comparison should be -8 < 4. However, in option C, it incorrectly states -8 is less than the absolute value of -4, which is not true.
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Find the value for the indicated hypothesis test with the given standardized test statistic, z. Decide whether to reject H, for the given level of significance a Two-talled test with best statistica-2.14 and 0.06 Palue Round to four decimal places as needed) State your conclusion O Falto O Reje Hy
Answer : The calculated p-value (0.0322) is less than the significance level (α = 0.05), we reject the null hypothesis H0.
Explanation :
The problem requires to find the value for the indicated hypothesis test with the given standardized test statistic, z and decide whether to reject H, for the given level of significance a. The given information is a two-tailed test with a best statistic of -2.14 and 0.06 p-value. So, we need to determine whether to reject or fail to reject the null hypothesis H0.
Null hypothesis: H0: µ = µ0
The alternative hypothesis: H1: µ ≠ µ0
Level of significance: α = 0.05 (for two-tailed)
Since the alternative hypothesis is two-tailed, the significance level is split into two equal parts, with each tail having a significance level of 0.025 (α/2).
The rejection region for this test is given as: Reject H0 if z > zα/2 or z < -zα/2 where zα/2 is the critical value of the standard normal distribution such that P(Z > zα/2) = α/2 or P(Z < -zα/2) = α/2.
The p-value is the probability of obtaining a test statistic as extreme as the one observed, given that the null hypothesis is true. If the p-value is less than the significance level α, we reject the null hypothesis. If the p-value is greater than or equal to α, we fail to reject the null hypothesis.
Given, the best statistic, z = -2.14P-value, P(Z < -2.14) = 0.0161 (from z-table)
Since this is a two-tailed test, we need to multiply the p-value by 2, i.e., P-value = 2(0.0161) = 0.0322
Since the calculated p-value (0.0322) is less than the significance level (α = 0.05), we reject the null hypothesis H0.
Thus, we can conclude that the evidence supports the alternative hypothesis that the population mean is not equal to µ0. So, the decision is to reject H0.
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