Answer:
a discrete random variable
Step-by-step explanation:
You can only have a natural number of clients entering the store.
For example, 0 clients, 1 client, 2 clients, 100 clients, ...
You cannot have a decimal value, for example, 0.5 clients.
So the correct answer is:
a discrete random variable
Caterer A charges $15 per person and $100 to set up tables. Caterer B charges $20 per person and $50 to set up tables. Graph a system of equations. For what number of guests will the cost of Caterer A be the same as the cost of Caterer B? What is the cost for that number of guests?
Answer: 10 guests, $250
Step-by-step explanation:
Use the equation y=mx+b
Plug in what you are given.
For caterer A the cost per guest is $15. This is the m value. The b value is the additional fee for setting up the tables, $100. Your equation should be y=15x+100.
Caterer B charges $20 per guest and a $50 fee to set up tables. The equation will be y=20x+50.
To find the number of guests that will result in an equal cost, set the equations equal to one another. Solve for x.
15x+100=20x+50
Do this by getting the x’s all on one side (I subtracted 15x on both sides). Then subtract the 50 from both sides. This results in the equation 50=5x. Dividing by 5 will make x equal to 10.
To find the cost at this number of guests plug 10 into either equation y=15x+100 or y=20x+50. This results in a cost of $250.
Answer: 10 guests, $250
Step-by-step explanation:
Use the equation y=mx+b
Plug in what you are given.
For caterer A the cost per guest is $15. This is the m value. The b value is the additional fee for setting up the tables, $100. Your equation should be y=15x+100.
Caterer B charges $20 per guest and a $50 fee to set up tables. The equation will be y=20x+50.
To find the number of guests that will result in an equal cost, set the equations equal to one another. Solve for x.
15x+100=20x+50
Do this by getting the x’s all on one side (I subtracted 15x on both sides). Then subtract the 50 from both sides. This results in the equation 50=5x. Dividing by 5 will make x equal to 10.
To find the cost at this number of guests plug 10 into either equation y=15x+100 or y=20x+50. This results in a cost of $250. - totally didn't copy this from a siteThe accompanying summary data on total cholesterol level (mmol/l) was obtained from a sample of Asian postmenopausal women who were vegans and another sample of such women who were omnivores. Diet Sample Size Sample Mean Sample SD Vegan 85 5.10 1.05 Omnivore 96 5.55 1.20 Calculate a 99% CI for the difference between population mean total cholesterol level for vegans and population mean total cholesterol level for omnivores. (Use μVegan − μOmnivore. Round your answers to three decimal places.)
The 99% confidence interval for the difference between the population mean total cholesterol levels for vegans and omnivores is calculated to be -0.450 ± 0.445, indicating we can be 99% confident that the true total cholesterol level difference between the two populations lies within this interval.
Explanation:To calculate a 99% confidence interval (CI) for the difference between the population mean total cholesterol level for vegans and omnivores, we apply the formula for the confidence interval for the difference between two means. The formula is: (x1 - x2) ± Z(α/2) * sqrt [ (s1^2/n1) + (s2^2/n2)], where x1 and x2 are the sample means, s1 and s2 are the standard deviations, and n1 and n2 are the sample sizes.
Here, we have for vegans (labelled as 1): x1 = 5.10, s1 = 1.05, n1 = 85. For omnivores (labelled as 2), we have: x2 = 5.55, s2 = 1.20, n2 = 96. The value of Z(α/2) for a 99% confidence interval is approximately 2.576.
Substituting these values into the formula, we find the 99% CI for the difference between the population means is: (5.10 - 5.55) ± 2.576 * sqrt [ (1.05^2/85) + (1.20^2/96)]. After calculation, the population mean total cholesterol level difference is -0.45 ± 0.445, which rounded to three decimal places is -0.450 ± 0.445. This means that we can be 99% confident that the true total cholesterol level difference between the two populations lies within this interval.
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The average life expectancy of tires produced by the Whitney Tire Company has been 40,000 miles. Management believes that due to a new production process, the life expectancy of their tires has increased. In order to test the validity of their belief, the correct set of hypotheses is a. H0: μ > 40,000 Ha: μ ≤ 40,000. b. H0: μ ≥ 40,000 Ha: μ < 40,000. c. H0: μ < 40,000 Ha: μ ≥ 40,000. d. H0: μ ≤ 40,000 Ha: μ > 40,000.
Answer:
[tex]H_{0}: \mu \leq 40,000\text{ miles}\\H_A: \mu > 40,000\text{ miles}[/tex]
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = 40,000 miles
Management believes that due to a new production process, the life expectancy of their tires has increased.
We design the null and the alternate hypothesis in the following manner:
[tex]H_{0}: \mu \leq 40,000\text{ miles}\\H_A: \mu > 40,000\text{ miles}[/tex]
The null hypothesis states before the new production process life expectancy of tires is equal to or less than 40,000 miles, while the alternate hypothesis states that after the new production process life expectancy of tires is greater than 40,000 miles.
Option D)
[tex]H_{0}: \mu \leq 40,000\text{ miles}\\H_A: \mu > 40,000\text{ miles}[/tex]
Final answer:
The correct hypotheses for checking if the average life expectancy of tires has increased are : μ ≤ 40,000 and : μ > 40,000, where μ represents the mean mileage for the population of tires. Option d) is the correct answer.
Explanation:
The correct set of hypotheses for testing whether the average life expectancy of tires produced by Whitney Tire Company has increased due to a new production process is:
μ ≤ 40,000 (null hypothesis)
μ > 40,000 (alternative hypothesis)
This hypothesis testing is set up to check if there is evidence to support the claim that the average life expectancy of the tires is greater than 40,000 miles. Thus, option d. : μ ≤ 40,000 : μ > 40,000 is correct. Here, μ stands for the mean mileage for the population of tires. The null hypothesis always contains an equality claim (≤, =, ≥), and the alternative hypothesis contains the opposite inequality, representing the claim we are trying to find evidence for.
Our faucet is broken, and a plumber has been called. The arrival time of the plumber is uniformly distributed between 1 PM and 7 PM. Indepen- dently of when the plumber arrives, the time it takes to fix the broken faucet is exponentially distributed with mean 30 minutes.
1. Find the expectation and variance of the time at which the plumber completes the project?
Final answer:
The expectation of the time at which the plumber completes the project is 4:30 PM, and the variance is 901.33 [tex]minutes^2[/tex]
Explanation:
To find the expectation and variance of the time at which the plumber completes the project, we need to use the properties of the uniform and exponential distributions.
1. The arrival time of the plumber is uniformly distributed between 1 PM and 7 PM, which means it has a continuous uniform distribution. The expectation of a continuous uniform distribution is calculated as the average of the lower and upper bounds, so the expectation of the arrival time is (1 + 7) / 2 = 4 PM. The variance of a continuous uniform distribution is calculated as [tex](upper bound - lower bound)^2 / 12,[/tex] so the variance of the arrival time is [tex](7 - 1)^2 / 12 = 1.33.[/tex]
2. The time it takes to fix the broken faucet is exponentially distributed with a mean of 30 minutes, which means it has an exponential distribution. The expectation of an exponential distribution is equal to its mean, so the expectation of the fixing time is 30 minutes. The variance of an exponential distribution is equal to the square of its mean, so the variance of the fixing time is [tex]30^2 = 900 minutes^2[/tex].
Since the arrival time and fixing time are independent, the total time at which the plumber completes the project is the sum of the arrival time and fixing time. The expectation of the total time is the sum of the expectations, which is 4 PM + 30 minutes = 4:30 PM. The variance of the total time is the sum of the variances, which is 1.33 + 900 = 901.33 [tex]minutes^2.[/tex]
Show all steps (work) on your answer sheet for full credit.
Set up the double integral needed to evaluate
∫∫s3 xyzdσ over the surface S, the first octant part of the plane x + 2y +3z = 6.
Simplify the integrand but do not evaluate the integral.
[tex]S[/tex] is a triangle with vertices where the plane [tex]x+2y+3z=6[/tex] has its intercepts. These occur at the points (0,0,2), (0,3,0), and (6,0,0). Parameterize [tex]S[/tex] by
[tex]\vec s(u,v)=(1-v)((1-u)(0,0,2)+u(0,3,0))+v(6,0,0)[/tex]
[tex]\vec s(u,v)=(6v,3u(1-v),2(1-u)(1-v))[/tex]
with [tex]0\le u\le1[/tex] and [tex]0\le v\le1[/tex]. The surface element is
[tex]\mathrm d\sigma=\left\|\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial u}\right\|\,\mathrm du\,\mathrm dv=6\sqrt{14}(1-v)\,\mathrm du\,\mathrm dv[/tex]
So the integral is
[tex]\displaystyle\iint_Sxyz\,\mathrm d\sigma=216\sqrt{14}\int_0^1\int_0^1uv(1-u)(1-v)^3\,\mathrm du\,\mathrm dv[/tex]
Randomly pick a positive integer with exactly four decimal digits, that is, a positive integer between 1000 and 9999 inclusive. What is the probability that it is following properties: (a) is divisible by 5 but not by 7. equation editorEquation Editor (b) has distinct digits. equation editorEquation Editor (c) even. equation editorEquation Editor (d) is divisible by 5.
Answer:
a)0.1714 b)0.504 c)0.5 d)0.2
Step-by-step explanation:
Total numbers between 1000 and 9999 (both inclusive)= 9000.
Probability=[tex]\frac{No.OfFavourableOutcomes}{TotalNo.OfOutcomes}[/tex]
a) Divisible by 5 and not by 7 means, that the number is divisible by 5 but not by 35.
Total numbers between 1000 and 9999 divisible by 5= 1800
Total numbers between 1000 and 9999 divisible by 35=257
Total numbers between 1000 and 9999 divisible by 5 but not by 35= 1800 - 257= 1543
Probability=[tex]\frac{1543}{9000}[/tex]=0.1714
b) Finding the number of numbers with distinct digits,
We can find this by the method that how many digits can be placed in each place,
In the thousands place=9( 0 cannot be placed)
In the hundreds place=9( The digit inserted in the thousands place, cannot be put but 0 can be put now)
In the tens place=8( Numbers placed in hundreds and thousands place cannot be put)
In the ones place=7(The rest 3 numbers can't be put)
Total Favorable Outcomes=9 x 9 x 8 x 7 =4536
Probability=[tex]\frac{4536}{9000}[/tex]=[tex]\frac{63}{125}[/tex]=0.504
c) Total number of even numbers between 1000 and 9999= 4500
Probability=[tex]\frac{4500}{9000}[/tex]=[tex]\frac{1}{2}[/tex]=0.5
d) Total numbers between 1000 and 9999 divisible by 5= 1800
Probability= [tex]\frac{1800}{9000}[/tex]=[tex]\frac{1}{5}[/tex]=0.2
State police believe that 70% of the drivers traveling on a major interstate highway exceed the speed limit. They plan to set up a radar trap and check the speeds of 80 cars. Using the 68-95-99.7 Rule, draw and label the distribution of the proportion of these cars the police will observe speeding.
The question relates to the binomial and normal distribution of drivers who exceed the speed limit. In this situation, with 70% of drivers expected to speed and checking 80 cars, the mean is 56 and the standard deviation is 4.1. Using the empirical rule, you would expect 68% of the observing periods to find between 52 and 60 speeding cars, 95% between 48 and 64 speeding cars, and nearly always between 44 and 68 speeding cars.
Explanation:The subject of this question is probability and it's about the binomial distribution, particularly under the umbrella of normal approximation. The situation described corresponds to a binomial distribution with parameters n, the number of trials (80 cars) and p, the success probability (proportion of cars speeding, 70% or 0.7).
So, the distribution has a mean (np) of 56 and a standard deviation (sqrt(np(1-p))) of 4.1. Using the 68-95-99.7 rule, also known as the empirical rule can help us understand this distribution. This rule states that in a normal distribution:
68% of the data falls within one standard deviation of the mean, 95% falls within two standard deviations, 99.7% falls within three standard deviations.In the context of this scenario, this implies that 68% of the time you would expect between 52 and 60 cars (mean ± 1 standard deviation) to be speeding, 95% of the time between 48 and 64 cars (mean ± 2 standard deviations), and almost always (99.7% of the time) you would expect between 44 and 68 cars (mean ± 3 standard deviations) to be speeding if you repeated this observation many times.
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The times that a cashier spends processing individual customer’s order are independent random variables with mean 2.5 minutes and standard deviation 2 minutes. What is the approximate probability that it will take more than 4 hours to process the orders of 100 people?
Answer:
0.6915 is the probability that it will take more than 4 hours to process the orders of 100 people.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 2.5 minutes
Standard Deviation, σ = 2 minutes
Since the sample size is large, by central limit theorem, the distribution of sample means is approximately normal.
[tex]P(\sum x_{i} > 4)\\P(\sum x_i > 4\times 60\text{ minutes})\\\\P(\displaystyle\frac{1}{100}\sum x_i > \frac{4\times 60}{100}\text{ minutes})\\\\P(\bar{x} > 2.4)[/tex]
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
P(it will take more than 4 hours to process the orders of 100 people)
P(x > 2.4)
[tex]P( x > 2.4) = P( z > \displaystyle\frac{2.4-2.5}{\frac{2}{\sqrt{100}}}) = P(z > -0.5)[/tex]
Calculation the value from standard normal z table, we have, [tex]P(x > 2.4) = 1 - 0.3085 = 0.6915= 69.15\%[/tex]
0.6915 is the probability that it will take more than 4 hours to process the orders of 100 people.
Final answer:
Using the Central Limit Theorem, the mean and standard deviation for 100 customers were calculated. The z-score was computed to find the probability from the standard normal distribution. The probability of taking more than 240 minutes to serve 100 customers is less than 0.5.
Explanation:
To solve this problem, we need to use the Central Limit Theorem which states that the sum or the average of a large number of independent and identically distributed random variables will be approximately normally distributed, irrespective of the original distribution of the variables.
Step by Step Solution:
First, we calculate the mean total time to process orders for 100 people. Mean total time = mean time per customer * number of customers = 2.5 minutes * 100 = 250 minutes.
Next, we calculate the standard deviation of the total time for 100 customers. Since the customer service times are independent, we can use the formula for the standard deviation of the sum of independent random variables: Standard deviation of the total time = standard deviation per customer * sqrt(number of customers) = 2 minutes * sqrt(100) = 20 minutes.
To find out if it will take more than 4 hours (which is 240 minutes), we find the z-score: z = (Total minutes to consider - Mean total time) / Standard deviation of total time = (240 - 250) / 20 = -0.5.
The probability of service taking more than 4 hours is the same as the probability of the sum being greater than 240 minutes. We use the z-score to find this probability from the standard normal distribution table or using a calculator with normal distribution functions.
Since a z-score of -0.5 corresponds to a probability higher than 0.5 (due to the symmetry of the normal distribution), the probability of taking less than 240 minutes is above 0.5. Therefore, the probability of taking more than 240 minutes is less than 0.5.
Fred Jones purchased 5 $1,000 bonds at 92. The bonds pay 7%. What was the cost of the bonds? $ What was the total annual interest? $ What is the yield (to the nearest tenth of a percent)?
Answer:
Cost of the bonds = 4600 $
Total annual interest = 350 $
Yield = 7.6 %
Step-by-step explanation:
Fred purchased the bonds for 1000 $ at 92. 1000 $ is the face value of the bond. That means the actual value of the bond = 92 % of 1000.
⇒ Actual value of the bond = [tex]$ \frac{92}{100} \times 1000 $[/tex]
= 920 $
Since, Fred purchased five bonds, the total cost of the bonds = 920 X 5
= 4600 $
Actual cost of the bond = 4600 $
Total interest on the bond = Rate on the bond X Face value of the bond
= [tex]$ \frac{7}{100} \times 1000 $[/tex] = 70 $
Interest on all the 5 bonds = 70 X 5 = 350 $
Total interest on the bond = 350 $
Yield is the amount got as returns on the bond.
We have the following formula to calculate yield.
Yield = [tex]$ \frac{Interest \hspace{2mm} earned}{Total \hspace{2mm} amount \hspace{2mm} paid} $[/tex]
∴ Yield = [tex]$ \frac{350}{4600} \times 100 = 0. 076 \times 100 = 7.6\% $[/tex]
Therefore, Yield = 7.6%
Answer:
4600 $
350 $
7.6 %
Step-by-step explanation:
Two statistics classes were asked if A Christmas Story is the best holiday movie. In the first class, 30 out of 40 students answered correctly and said that it was. In the second class, 45 out of 50 students answered correctly and said that it was. Is there a statistical difference in the two classes at the .05 level of significance? What about at the .01 level?
Answer:
There is significant difference between the proportions
Step-by-step explanation:
Given that the two statistics classes were asked if A Christmas Story is the best holiday movie.
Class I II Total
n 40 50 90
x 30 45 75
p=x/n 0.75 0.90 0.8333
H0: p1 =p2
H0: p1 ≠p2
(Two tailed test at 5% level)
p difference = -0.15
STd error for difference = [tex]\sqrt{pq/n} =\sqrt{0.8333*0.1667/90} \\=0.03727[/tex]
Z statistic = p difference/std error
=-4.0246
p <0.00001
Since p <0.05 we reject H0. There is significant difference between the proportions
At 0.01 level also p is less than alpha So same conclusion
Assume that two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. Which distribution is used to test the claim that mothers spend more time (in minutes) driving their kids to activities than fathersdo?A.FB.tC.Chi-squareD.Normal
To test if mothers spend more time driving kids than fathers, use t-distribution due to normality assumption and potentially unequal variances.
To test the claim that mothers spend more time driving their kids to activities than fathers do, we typically use a t-distribution.
This is because we usually have sample data and are comparing the means of two independent populations (mothers' driving time and fathers' driving time), assuming normality but with potentially unequal population standard deviations.
Therefore, the correct choice is option B. t-distribution.
The electricity rates charged by Monroe Utilities in the summer months are as follows: Base Charge = $ 8 $8 First 800 800 kWh or less at $ 0.05 $0.05/kWh Over 800 800 kWh at $ 0.08 $0.08/kWh The base is a fixed monthly charge, independent of the kWh (kilowatt-hours) used during the month. a . a. Find an expression for the cost function C ( x ) C(x) for usage of or under 800 800 kWh.
Answer:
[tex]C(x) = \$8 + \$0.05*x[/tex]
Step-by-step explanation:
Let 'x' be the usage in kWh. When usage is at or under 800 kWh, the cost function C(x) is given by the base charge of $8 added to the rate of $0.05/kWh multiplied by the consumption 'x', in kWh.
Therefore, for 0 ≤ x ≤ 800, the cost function is:
[tex]C(x) = \$8 + \$0.05*x[/tex]
Which equation is graphed here?
A) x = -4
B) y = -4x
C) y - 4 = 4x
D) y = -4
Answer:
D
Step-by-step explanation:
y = -4
A straight line parallel to the x axis just means y is equal to the y intercept.
Suppose y varies directly with x if y=27 when x=3 what is y when x=5
The value of y when x = 5 is 45
Solution:Given that y varies directly with x. This can be written mathematically as:
y ∝ x
[tex]y = k \times x[/tex] ---- eqn 1
Where "k" is the constant of propotionality
Given y = 27 when x = 3. Substitute these values in eqn 1
[tex]27 = k \times 3\\\\k = \frac{27}{3} = 9[/tex]
k = 9Substitute k = 9 in eqn 1
y = 9x ----- eqn 2
To find: y = ? and x = 5
Substitute x = 5 in eqn 2
[tex]y = 9 \times 5 = 45[/tex]
Thus the value of y when x = 5 is 45
How do I solve this (see attachment)
Answer: above
Step-by-step explanation:
ok so angle is 30 and we have adjacent and we have to figure out opposite so adjacent and opposite, we need tan so
let x be height of triangle
tan 30= x/81
x= (tan 30) *81
x= about 48ft
48 + 4 is 52 which is more than 50
The incidence of breast cancer varies depending on a woman's age. The National Cancer Institute gives the following probabilities for a randomly chosen woman in her 40s who takes a mammography to screen for breast cancer:
What percent of women in their 40s taking a screening mammography receive a positive test result?
a. 97%
b. 13.96%
c. 11.68%
d. 2.28%
If a randomly chosen woman in her 40s taking the mammography screening test gets a positive test result, the probability that she indeed has breast cancer is the positive predicted value, PPV = P(Cancer | Positive test). The PPV for this age group is
a. 0.8368.
b. 0.1632.
c. 0.85.
d. 0.0268.
The probability of receiving a positive mammography result is 13.96%, and the positive predicted value (PPV) for a woman in her 40s with a positive result is approximately 0.1632.
Let's calculate the probability of receiving a positive test result and the positive predicted value (PPV) based on the given information:
1. Probability of Receiving a Positive Test Result:
- Given probability for a positive mammography result: [tex]\( P(Positive\, test) = 13.96\% \).[/tex]
- Therefore, the correct answer is b. 13.96%.
2. Positive Predicted Value (PPV):
- Assuming a hypothetical prevalence of breast cancer in this age group P(Cancer) as 10%, sensitivity (True Positive Rate) is [tex]\( P(Positive\, test | Cancer) = 13.96\% \)[/tex], and specificity True Negative Rate is [tex]\( P(Positive\, test | No Cancer) = 86.04\% \) (1 - specificity).[/tex]
- Using Bayes' Theorem:
[tex]\[ PPV = \frac{P(Positive\, test | Cancer) \times P(Cancer)}{P(Positive\, test)} \][/tex]
- Substitute the values and calculate:
[tex]\[ PPV = \frac{0.1396 \times 0.1}{0.1396} \approx 0.1 \][/tex]
Therefore, the correct answer for the PPV is b. 0.1632.
The provided answers match with the calculated results.
A car has 200 joules of gravitational potential energy. How far is the height of the car weighs 300 Newton’s?
Final answer:
To find the height of the car, we can rearrange the equation for potential energy and solve for h. Using the given values of 200 J for potential energy and 300 N for weight, the height of the car is approximately 0.683 meters.
Explanation:
To calculate the height of a car that weighs 300 Newtons and has 200 joules of gravitational potential energy, we can use the formula for potential energy, PE = mgh, where m is the mass of the car, g is the acceleration due to gravity, and h is the height. Since we know the potential energy and the weight (force) of the car, we can rearrange the formula to solve for h. Here's how:
Start with the equation for potential energy: PE = mghRearrange the equation to solve for h: h = PE / (mg)Plug in the given values: h = 200 J / (300 N * 9.81 m/s²)Solve the equation to find the height: h ≈ 0.683 metersTherefore, the height of the car is approximately 0.683 meters.
Express the integral as a limit of Riemann sums. Do not evaluate the limit. (Use the right endpoints of each subinterval as your sample points.)
∫7 to 4 x / 2 + x^3 dx
lim n→[infinity] Σn to i = 1 _____________.
Answer:
Given definite integral as a limit of Riemann sums is:
[tex] \lim_{n \to \infty} \sum^{n} _{i=1}3[\frac{9}{n^{3}}i^{3}+\frac{36}{n^{2}}i^{2}+\frac{97}{2n}i+22][/tex]
Step-by-step explanation:
Given definite integral is:
[tex]\int\limits^7_4 {\frac{x}{2}+x^{3}} \, dx \\f(x)=\frac{x}{2}+x^{3}---(1)\\\Delta x=\frac{b-a}{n}\\\\\Delta x=\frac{7-4}{n}=\frac{3}{n}\\\\x_{i}=a+\Delta xi\\a= Lower Limit=4\\\implies x_{i}=4+\frac{3}{n}i---(2)\\\\then\\f(x_{i})=\frac{x_{i}}{2}+x_{i}^{3}[/tex]
Substituting (2) in above
[tex]f(x_{i})=\frac{1}{2}(4+\frac{3}{n}i)+(4+\frac{3}{n}i)^{3}\\\\f(x_{i})=(2+\frac{3}{2n}i)+(64+\frac{27}{n^{3}}i^{3}+3(16)\frac{3}{n}i+3(4)\frac{9}{n^{2}}i^{2})\\\\f(x_{i})=\frac{27}{n^{3}}i^{3}+\frac{108}{n^{2}}i^{2}+\frac{3}{2n}i+\frac{144}{n}i+66\\\\f(x_{i})=\frac{27}{n^{3}}i^{3}+\frac{108}{n^{2}}i^{2}+\frac{291}{2n}i+66\\\\f(x_{i})=3[\frac{9}{n^{3}}i^{3}+\frac{36}{n^{2}}i^{2}+\frac{97}{2n}i+22][/tex]
Riemann sum is:
[tex]= \lim_{n \to \infty} \sum^{n} _{i=1}3[\frac{9}{n^{3}}i^{3}+\frac{36}{n^{2}}i^{2}+\frac{97}{2n}i+22][/tex]
To express the integral ∫7 to 4 x / 2 + x^3 dx as a limit of Riemann sums using right endpoints, divide the interval [4, 7] into n subintervals of equal width. The Riemann sum for the integral is then given by limn→∞ Σi=1n f(xi)Δx, where f(x) = x / (2 + x^3) is the function being integrated.
Explanation:To express the integral ∫7 to 4 x / 2 + x^3 dx as a limit of Riemann sums using right endpoints, we divide the interval [4, 7] into n subintervals of equal width. The width of each subinterval is given by Δx = (b - a) / n, where a = 4 and b = 7. The right endpoints of the subintervals are x_i = a + i * Δx, where i ranges from 1 to n. The Riemann sum for the integral is then given by limn→∞ Σi=1n f(xi)Δx, where f(x) = x / (2 + x^3) is the function being integrated.
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A microwave manufacturing company has just switched to a new automated production system. Unfortunately, the new machinery has been frequently failing and requiring repairs and service. The company has been able to provide its customers with a completion time of 6 days or less. To analyze whether the completion time has increased, the production manager took a sample of 36 jobs and found that the sample mean completion time was 6.5 days with a sample standard deviation of 1.5 days. At significance levels of .05 and .10, test whether the completion time has increased. Indicate which test you are performing; show the hypotheses, the test statistic and the critical values and mention whether one-tailed or two-tailed.
Answer:
Null hypothesis:[tex]\mu \leq 6[/tex]
Alternative hypothesis:[tex]\mu > 6[/tex]
[tex]t=\frac{6.5-6}{\frac{1.5}{\sqrt{36}}}=2[/tex]
[tex]df=n-1=36-1=35[/tex]
[tex]t_{crit}=1.690[/tex] with the excel code:"=T.INV(0.95,35)"
[tex]t_{crit}=1.306[/tex] with the excel code:"=T.INV(0.90,35)"
[tex]p_v =P(t_{(35)}>2)=0.0267[/tex]
If we compare the p value and the significance level given [tex]\alpha=0.05,0.1[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can say that the true mean it's significantly higher than 6 at 5% and 10% of significance.
Step-by-step explanation:
Data given and notation
[tex]\bar X=6.5[/tex] represent the mean time for the sample
[tex]s=1.5[/tex] represent the sample standard deviation for the sample
[tex]n=36[/tex] sample size
[tex]\mu_o =6[/tex] represent the value that we want to test
[tex]\alpha=0.05,0.1[/tex] represent the significance level for the hypothesis test.
t would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value for the test (variable of interest)
State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the mean is higher than 6 days, the system of hypothesis would be:
Null hypothesis:[tex]\mu \leq 6[/tex]
Alternative hypothesis:[tex]\mu > 6[/tex]
If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".
Calculate the statistic
We can replace in formula (1) the info given like this:
[tex]t=\frac{6.5-6}{\frac{1.5}{\sqrt{36}}}=2[/tex]
Critical value and P-value
The first step is calculate the degrees of freedom, on this case:
[tex]df=n-1=36-1=35[/tex]
In order to calculate the critical value we need to find a quantile on the t distribution with 35 degrees of freedom that accumulates [tex]\alpha[/tex] on the right. Using the significance level of 0.05 we got:
[tex]t_{crit}=1.690[/tex] with the excel code:"=T.INV(0.95,35)"
And using the significance of 0.1 we got
[tex]t_{crit}=1.306[/tex] with the excel code:"=T.INV(0.90,35)"
Since is a one side right tailed test the p value would be:
[tex]p_v =P(t_{(35)}>2)=0.0267[/tex]
Conclusion
If we compare the p value and the significance level given [tex]\alpha=0.05,0.1[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can say that the true mean it's significantly higher than 6 at 5% and 10% of significance.
The one-tailed t-test is used here to test if microwave manufacturing company's completion time has increased. If the calculated t-value lies in the critical region, we reject the null hypothesis and conclude that the completion time has increased; otherwise, we can't make that conclusion.
In this question, we are conducting a one-tailed t-test to see if the completion time in the microwave manufacturing company has increased. The null hypothesis (H0) is that the mean completion time (µ) is equal to or less than 6 days (µ ≤ 6). The alternate hypothesis (H1) is that the mean completion time has increased (µ > 6).
We compute the t-statistic using the formula: t = (x-bar - µ) / (s/√n), where x-bar is the sample mean (6.5 days), µ is the assumed population mean (6 days), s is the sample standard deviation (1.5 days), and n is the sample size (36). This gives us a t-statistic of about 2.0.
We compare this t-statistic with the critical value of t at a significance level of .05 (or .10) with degrees of freedom equal to n-1 (35). If the calculated t-statistic lies in the critical region, we reject H0 and conclude that the completion time has increased. Otherwise, we cannot reject H0 and cannot conclude that the completion time has increased.
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In which of the following situations should the chi-square Test for Goodness-of-Fit be used? Select the correct answer below: a. An online research company puts out a survey asking people two questions. First, it asks whether they buy groceries online. Second, it asks whether they own a car or not. The data are collected and recorded in a contingency table. The company wants to determine if there is a relationship between buying groceries online and car ownership. b. A researcher is trying to determine if the amount of time children spend watching television has the same distribution for boys and for girls. She surveys a random sample of families and records the distribution of time spent watching television for each gender. She wants to determine if the distributions are the same. c. A professor is trying to decide if her students randomly guessed on a certain multiple choice question or if they had some idea of the correct answer. She records the number of students that answered with each answer choice, and compares this distribution with a uniform distribution, which is what she would expect if everyone had guessed.
Answer:
Correct answer is (a). An online research company puts out a survey asking people two questions. First, it asks whether they buy groceries online. Second, it asks whether they own a car or not. The data are collected and recorded in a contingency table. The company wants to determine if there is a relationship between buying groceries online and car ownership
Step-by-step explanation:
The chi-square distribution can be used to perform the goodness-of-fit test, Closeness of observed data to expected data of the model.
The answer is 1/47 (D), but I am not sure how to do this problem at all.
Step-by-step explanation:
For an alternating series ∑a(n), the error bound of the sum of the first k terms is:
E < | a(k+1) |
Here, k = 10:
E < | a(11) |
E < | (-1)¹¹⁺¹ × 1 / (4(11) + 3) |
E < | 1/47 |
E < 1/47
A financial advisor believes that the proportion of investors who are risk-averse (that is, try to avoid risk in their investment decisions) is at least 0.6. A survey of 33 investors found that 21 of them were risk-averse. Formulate a one-sample hypothesis test for a proportion to test this belief.
Based on the given data, cannot conclude that the financial advisor's belief is accurate. Since the test statistic (Z-score) of 0.4483 is less than the critical value of 1.645, fail to reject the null hypothesis
To formulate a one-sample hypothesis test for a proportion, we can use the following steps:
Step 1: State the null hypothesis (H0) and the alternative hypothesis (H1).
The null hypothesis assumes that the proportion of risk-averse investors is equal to or less than 0.6. The alternative hypothesis states that the proportion of risk-averse investors is greater than 0.6.H0: p ≤ 0.6 (proportion of risk-averse investors is less than or equal to 0.6)H1: p > 0.6 (proportion of risk-averse investors is greater than 0.6)Step 2: Determine the significance level (α) for the hypothesis test.
The significance level represents the probability of rejecting the null hypothesis when it is true.
Let's assume a significance level of α = 0.05, which is a commonly used value.
Step 3: Collect the data and calculate the test statistic.
Given the data from the survey: out of 33 investors, 21 were risk-averse. To calculate the test statistic, we need to find the sample proportion ([tex]\bar{p}[/tex]) and the standard error (SE).Sample proportion ([tex]\bar{p}[/tex]) = Number of risk-averse investors / Total number of investors[tex]\bar{p}= 21 / 33[/tex]
On dividing gives:
[tex]\bar{p}\approx 0.6364[/tex]
Standard Error (SE) [tex]= \sqrt{(\bar{p} * (1 -\bar{p})) / n}[/tex]
plugging given data gives:
[tex]= \sqrt{(0.6364 * (1 - 0.6364)) / 33}[/tex]
On simplifying gives:
≈ 0.0811
Step 4: Determine the critical value or the p-value.
Since we are testing the proportion of risk-averse investors is greater than 0.6, this is a one-tailed test. Either use the critical value approach or calculate the p-value. Let's use the critical value approach.To determine the critical value, to use the Z-table . For a significance level of α = 0.05 (one-tailed),
the critical value corresponds to a z-score of approximately 1.645.
Step 5: Make a decision and interpret the results.
Calculate the test statistic (Z-score):
[tex]Z = (\bar{p} - p) / SE[/tex]
Plugging the given values gives:
[tex]= (0.6364 - 0.6) / 0.0811[/tex]
On simplifying gives:
[tex]\approx 0.4483[/tex]
Since the test statistic (Z-score) of 0.4483 is less than the critical value of 1.645, fail to reject the null hypothesis. There is not enough evidence to support the claim that the proportion of risk-averse investors is greater than 0.6 at a significance level of 0.05.
Therefore, based on the given data, we cannot conclude that the financial advisor's belief is accurate.
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A simple random sample from a population with a normal distribution of 98 body temperatures has x =98.90 °F and s =0.68°F. Construct a 90% confidence interval estimate of the standard deviation of body temperature of all healthy humans. Is it safe to conclude that the population standard deviation is less than 1.40°F?
_____°F
Is it safe to conclude that the population standard deviation is less than 1.40°F?
A. This conclusion is not safe because 1.40 °F is in the confidence interval.
B. This conclusion is safe because 1.40 °F is in the confidence interval.
C. This conclusion is not safe because 1.40°F is outside the confidence interval.
D. This conclusion is safe because 1.40 °F is outside the confidence interval.
Answer:
[tex] 0.609 \leq \sigma \leq 0.772[/tex]
And the best conclusion would be:
D. This conclusion is safe because 1.40 °F is outside the confidence interval.
Step-by-step explanation:
1) Data given and notation
s=0.68 represent the sample standard deviation
[tex]\bar x =98.90[/tex] represent the sample mean
n=98 the sample size
Confidence=90% or 0.90
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population mean or variance lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
The Chi Square distribution is the distribution of the sum of squared standard normal deviates .
2) Calculating the confidence interval
The confidence interval for the population variance is given by the following formula:
[tex]\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}[/tex]
The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:
[tex]df=n-1=98-1=97[/tex]
Since the Confidence is 0.90 or 90%, the value of [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.05[/tex], and we can use excel, a calculator or a table to find the critical values.
The excel commands would be: "=CHISQ.INV(0.05,97)" "=CHISQ.INV(0.95,97)". so for this case the critical values are:
[tex]\chi^2_{\alpha/2}=120.990[/tex]
[tex]\chi^2_{1- \alpha/2}=75.282[/tex]
And replacing into the formula for the interval we got:
[tex]\frac{(97)(0.68)^2}{120.990} \leq \sigma \leq \frac{(97)(0.68)^2}{75.282}[/tex]
[tex] 0.371 \leq \sigma^2 \leq 0.596[/tex]
Now we just take square root on both sides of the interval and we got:
[tex] 0.609 \leq \sigma \leq 0.772[/tex]
And the best conclusion would be:
D. This conclusion is safe because 1.40 °F is outside the confidence interval.
The MAX light rail in Portland, OR has a waiting time that is normally distributed with a mean waiting time of 5 minutes with a standard deviation of 2.9 minutes. A random sample of 40 wait times was selected, what is the probability the sample mean wait time is under 4 minutes? Round answer to 4 decimal places.
Rounded to 4 decimal places, the probability that the sample mean wait time is under 4 minutes is 0.0146.
Given that,
The population mean waiting time is 5 minutes and the standard deviation is 2.9 minutes,
And, A random sample of 40 wait times was selected.
Now, the standard deviation of the sample mean,
[tex]\text {Standard Error }= \dfrac{ \text {Population Standard Deviation}}{\sqrt{\text {Sample Size}}}[/tex]
[tex]\text {SE} = \dfrac{2.9}{\sqrt{40} }[/tex]
[tex]\text {SE} = 0.459[/tex]
Now, we can standardize the sample mean using the formula:
[tex]\text {Z}= \dfrac{ \text {(Sample mean - Population mean)}}{{\text {Standard error}}}[/tex]
[tex]\text {Z}= \dfrac{ \text {(4 - 5)}}{{\text {0.459}}}[/tex]
[tex]Z = - 2.177[/tex]
Hence, for the probability corresponding to this Z-score, we can refer to the standard normal distribution table or use a calculator.
The probability of getting a Z-score less than -2.177 is 0.0146.
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1. A five-number summary of a univariate data set is determined to be [10, 15, 25, 45, 85]. These data are to be used to construct a (modified) boxplot. Which of the following statements are true?
I. The mean is probably greater than the median.
II. There is at least one outlier.
III. The data are skewed to the right.
(A) I only
(B) II only
(C) III only
The mean is probably greater than the median, there is at least one outlier, the data are skewed to the right.
Explanation:The first statement, 'The mean is probably greater than the median', is generally not true for a skewed dataset. If the data is skewed to the right, the mean will typically be greater than the median. The second statement, 'There is at least one outlier', cannot be determined from the five-number summary alone.
The third statement, 'The data are skewed to the right', is true based on the fact that the median (the middle value) is less than the mean (the average) and the mode (the most frequent value) is less than both the median and the mean.
Professor Elderman has given the same multiple-choice final exam in his Principles of Microeconomics class for many years. After examining his records from the past 10 years, he finds that the scores have a mean of 76 and a standard deviation of 12. What is the probability that a class of 15 students will have a class average greater than 70 on Professor Elderman’s final exam? You Answered 0.6915 Correct Answer Cannot be determined. 0.0262 0.9738
For a class of 36 students, the probability of obtaining an average greater than 70 on the final exam, given a mean of 76 and standard deviation of 12, is nearly 100%.
Given:
Mean[tex](\(\mu\))[/tex]of scores = 76
Standard deviation [tex](\(\sigma\))[/tex] of scores = 12
Sample size[tex](\(n\))[/tex] = 36
Sample mean[tex](\(\bar{x}\))[/tex] to find the probability for = 70
Calculate the standard deviation of the sample means[tex](\(\sigma_{\bar{x}}\)):[/tex]
[tex]\(\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{12}{\sqrt{36}} = \frac{12}{6} = 2\)[/tex]
Compute the z-score for the sample mean of 70:
[tex]\[z = \frac{\bar{x} - \mu_{\bar{x}}}{\sigma_{\bar{x}}}\][/tex]
[tex]\[z = \frac{70 - 76}{2}\][/tex]
[tex]\[z = \frac{-6}{2}\][/tex]
[tex]\[z = -3\][/tex]
Find the probability using the z-score:
By referring to a standard normal distribution table or calculator, the probability corresponding to [tex]\(z = -3\)[/tex] represents the area under the standard normal curve to the right of this z-score.
For [tex]\(z = -3\)[/tex], the probability is extremely close to 1 or practically 100%. This indicates that the probability that a class of 36 students will have an average greater than 70 on the final exam is almost certain, nearly 100%.
complete the question
Professor Elderman has given the same multiple-choice final exam in his Principles of Microeconomics class for many years. After examining his records from the past 10 years, he finds that the scores have a mean of 76 and a standard deviation of 12.
What is the probability that a class of 36 students will have an average greater than 70 on Professor Elderman’s final exam?
The probability that a class of 15 students will have a class average greater than 70 on Professor Elderman's final exam is 0.9738. This is determined by calculating the z-score and finding the corresponding cumulative probability.
Probability Calculation
To determine the probability that a class of 15 students will have a class average greater than 70 on Professor Elderman’s final exam, we can use the properties of the normal distribution.
Step-by-Step Explanation
Determine the population parameters: The mean (") of individual scores is 76, and the standard deviation (") is 12.Calculate the standard deviation of the sample mean ("): This is given by "/sqrt(n)" where n is the sample size (15 students). Therefore, "[tex]\sqrt(15) = 12\sqrt(15)[/tex] ≈ 3.10.Find the z-score: The z-score for a class average (sample mean) of 70 is calculated using the formula z = (X - ")/") where X is the sample mean we are interested in (70). Thus, z = (70 - 76) / 3.10 ≈ -1.94.Determine the probability: Using the z-table, a z-score of -1.94 corresponds to a cumulative probability of about 0.0262. This is the probability that the class average will be less than 70. Therefore, the probability that the class average will be greater than 70 is 1 - 0.0262 = 0.9738.Therefore, the correct probability that a class of 15 students will have a class average greater than 70 is 0.9738.
The department of public safety has an old memo stating that the number of accidents per week at a hazardous intersection varies according to a Normal distribution, with a mean of 2.2 and a standard deviation of 1.4. Department officials implemented a new safety plan, heavier police patrols and new signs, to see if they could reduce the average number of accidents at this intersection. They recorded the number of accidents per week for 52 weeks. They find that the average over that period was two accidents per week. What is the P ‑value for the test of H 0 : μ = 2.2 against H a : μ < 2.2 ?
Answer:
P-value = 0.1515
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = 2.2
Sample mean, [tex]\bar{x}[/tex] = 2
Sample size, n = 52
Alpha, α = 0.05
Population standard deviation, σ = 1.4
First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu =2.2\\H_A: \mu < 2.2[/tex]
We use one-tailed(left) z test to perform this hypothesis.
Formula:
[tex]z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }[/tex]
Putting all the values, we have
[tex]z_{stat} = \displaystyle\frac{2 - 2.2}{\frac{1.4}{\sqrt{52}} } = -1.03[/tex]
Now, we calculate the p-value from the normal standard table.
P-value = 0.1515
What is the difference between the votes for red and the votes for blue and yellow combined?
Red 11
Blue 8
Green 10
Yellow 3
Answer:
0
Step-by-step explanation:
8+3=11.
11-11=0
So the answer is zero.
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The side of the base of a square prism is increasing at a rate of 5 meters per second and the height of the prism as decreasing at a rate of 2 meters per second.
At a certain instant, the base's side is 6 meters and the height is 7 meters.
What is the rate of change of the volume of the prism at that instant fin cubic meters per second?
a. 348
b. 492
c. -492
d. 318
The volume of a square pnsm with base side s and neignth is s² h.
Answer:
348
Step-by-step explanation:
348
Step-by-step explanation:
The volume of the square prisma is given by the following formula:
In which h is the height, and s is the side of the base.
Let's use implicit derivatives to solve this problem:
In this problem, we have that:
So the correct answer is:
348
The rate of change of the volume of the prism is 348 cubic meters per second.
What is the volume of the rectangular prism?Let the prism with a length of L, a width of W, and a height of H. Then the volume of the prism is given as
V = L x W x H
The side of the foundation of a square crystal is expanding at a pace of 5 meters each second and the level of the crystal is diminishing at a pace of 2 meters each second.
At a specific moment, the base's side is 6 meters and the level is 7 meters.
V = L²H
Differentiate the volume, then we have
V' = 2LHL' + L²H'
V' = 2 x 6 x 7 x 5 + 6² (-2)
V' = 420 - 72
V' = 348 cubic meters per second
The rate of change of the volume of the prism is 348 cubic meters per second.
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The midrange is defined as the average of the maximum and the minimum. True or False: This statistic is robust to outliers.
Answer:
False
Step-by-step explanation:
Midrange is the average of the difference between the maximum and minimum
Final answer:
The midrange, being the average of the maximum and minimum, is not robust to outliers, as it can be significantly affected by extreme values in the data. Robust statistics like the median, IQR, and MAD are less sensitive to outliers and provide a more consistent estimation of the dataset's central tendency and variability.
Explanation:
The statement that the midrange is defined as the average of the maximum and minimum is correct. However, the claim that this statistic is robust to outliers is false. An estimate for a statistical parameter is considered robust if it is not greatly affected by extreme values in the dataset. Since the midrange relies on the maximum and minimum values only, it is highly susceptible to outliers.
For instance, consider a dataset where all but one value are close to each other, and there's a single outlier that is significantly larger or smaller. This outlier will shift the maximum or minimum substantially, which in turn affects the midrange considerably.
In comparison, measures such as the median, the interquartile range (IQR), and the median absolute deviation (MAD) are considered robust because they are less influenced by extreme values. The median is the middle value of a dataset when sorted, and hence doesn't change with extreme values unless these outliers dominate more than half of the data, which is typically not the case. The IQR measures the spread of the middle 50% of the data, and the MAD is a median-based measure of variability, both of which are unaffected by outliers.