Answer:
Step-by-step explanation:
(a) The average number of times a customer carries out banking transactions per week is 13 give or take 5 or so.
(b) The 90% confidence interval for the average number of banking operations per week is given by:
(Assuming a normal distribution of the number of banking operations)
CI=\overline{X}\pm 1.645\times \sigma/\sqrt{n}
CI=13\pm 1.645\times 5/\sqrt{350}
CI=13\pm 0.439645
CI=(12.560355, 13.439645)
The 99% confidence interval for the average number of banking operations per week is given by:
(Assuming a normal distribution of the number of banking operations)
CI=\overline{X}\pm 2.576\times \sigma/\sqrt{n}
CI=13\pm 2.576\times 5/\sqrt{350}
CI=13\pm 0.688465
CI=(12.311535, 13.688465)
The difference in the margin of error is 0.688465-0.439645=0.24882
The difference in the margin of error will make the confidence interval wider in the second case(99% confidence interval) as compared to the first case(90% confidence interval).
(c) Since, both our confident interval contains the value 10 hence we have sufficient evidence to conclude that the apparent difference in banking habits between the nation (It is given that the national survey suggest that on an average 10 transactions are performed per week by a single person) and the customer of the bank is not significant or is not real and is just due to the random errors/variations in sampling.
(d) A 95% confidence interval gives a range of values for the (A) Population Average which are plausible according to the observed data.
(The confidence interval always gives the range of possible values for the population values)
(e) The sample standard deviation measures how far (A) number of bank operations is from sample average.
The standard deviation for the sample average measures how far (B) Average number of bank operations is from the population average for typical (D) samples.
A survey of 865 voters in one state reveals that 408 favor approval of an issue before the legislature.
Construct the 95 % confidence interval for the true proportion of all voters in the state who favor approval.
A) 0.435 < p < 0.508
B) 0.444 < p < 0.500
C) 0.471 < p < 0.472
D) 0.438 < p < 0.505
Answer: D) 0.438 < p < 0.505
Step-by-step explanation:
We know that the confidence interval for population proportion is given by :-
[tex]\hat{p}\pm z^*\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex]
, where n= sample size
[tex]\hat{p}[/tex] = Sample proportion.
z* = critical value.
Given : A survey of 865 voters in one state reveals that 408 favor approval of an issue before the legislature.
i.e. n= 865
[tex]\hat{p}=\dfrac{408}{865}\approx0.4717[/tex]
Two-tailed critical avlue for 95% confidence interval : z* = 1.96
Then, the 95 % confidence interval for the true proportion of all voters in the state who favor approval will be :-
[tex]0.4717\pm (1.96)\sqrt{\dfrac{0.4717(1-0.4717)}{865}}\\\\\approx0.4717\pm 0.03327\\\\=(0.4717-0.03327,\ 0.4717+0.03327)=(0.43843,\ 0.50497)\approx(0.438,\ 0.505)[/tex]
Thus, the required 95% confidence interval : (0.438, 0.505)
Hence, the correct answer is D) 0.438 < p < 0.505
A researcher selects a sample from a population with μ = 30 and uses the sample to evaluate the effect of a treatment. After treatment, the sample has a mean of M = 32 and a variance of s2 = 6. Which of the following would definitely increase the likelihood of rejecting the null hypothesis?
Question options:
a.
Decrease the sample variance
b.
Increase the sample mean
c.
Increase the sample size
d.
All of the other options will increase the likelihood of rejecting the null hypothesis
Answer:
Option b) Increase the sample mean
Step-by-step explanation:
Given that a researcher selects a sample from a population with μ = 30 and uses the sample to evaluate the effect of a treatment. After treatment, the sample has a mean of M = 32 and a variance of s2 = 6.
This is a paired test with test statistic
=mean diff/std error
Mean difference would increase if sample mean increases.
This would increase the test statistic
Or otherwise decrease in variance will increase the test statistic
Or Increase in sample size would also increase test statistic
Of all these the II option is definite in increasing the likelihood of rejecting the null hypothesis because this would definitely increase the chances of rejecting H0.
Others may also have effect but not as much direct as sample mean difference.
Because variance and sample size have influence only upto square root of the difference.
How would you describe the difference between the graphs of f(x) = x ^ 2 + 4 and g(y) = y ^ 2 + 4 OA. g(y) is a reflection of f(x) ) over the line y = 1 . O O B. g(y) is a reflection of f(x) ) over the line y=x 1 g(y) is a reflection of f(x) over the x-axis OD. g(y) is a reflection of f(x) over the .
x is replaced with Y, so it is a reflection across the line y = x
Answer:
y=x
Step-by-step explanation:
Evaluate the integral Integral ∫ from (1,2,3 ) to (5, 7,-2 ) y dx + x dy + 4 dz by finding parametric equations for the line segment from (1,2,3) to (5,7,- 2) and evaluating the line integral of of F = yi + x j+ 3k along the segment. Since F is conservative, the integral is independent of the path.
[tex]\vec F(x,y,z)=y\,\vec\imath+x\,\vec\jmath+3\,\vec k[/tex]
is conservative if there is a scalar function [tex]f(x,y,z)[/tex] such that [tex]\nabla f=\vec F[/tex]. This would require
[tex]\dfrac{\partial f}{\partial x}=y[/tex]
[tex]\dfrac{\partial f}{\partial y}=x[/tex]
[tex]\dfrac{\partial f}{\partial z}=3[/tex]
(or perhaps the last partial derivative should be 4 to match up with the integral?)
From these equations we find
[tex]f(x,y,z)=xy+g(y,z)[/tex]
[tex]\dfrac{\partial f}{\partial y}=x=x+\dfrac{\partial g}{\partial y}\implies\dfrac{\partial g}{\partial y}=0\implies g(y,z)=h(z)[/tex]
[tex]f(x,y,z)=xy+h(z)[/tex]
[tex]\dfrac{\partial f}{\partial z}=3=\dfrac{\mathrm dh}{\mathrm dz}\implies h(z)=3z+C[/tex]
[tex]f(x,y,z)=xy+3z+C[/tex]
so [tex]\vec F[/tex] is indeed conservative, and the gradient theorem (a.k.a. fundamental theorem of calculus for line integrals) applies. The value of the line integral depends only the endpoints:
[tex]\displaystyle\int_{(1,2,3)}^{(5,7,-2)}y\,\mathrm dx+x\,\mathrm dy+3\,\mathrm dz=\int_{(1,2,3)}^{(5,7,-2)}\nabla f(x,y,z)\cdot\mathrm d\vec r[/tex]
[tex]=f(5,7,-2)-f(1,2,3)=\boxed{18}[/tex]
Which function has (2,8) on its graph
Answer:
y = 2x^2
Step-by-step explanation:
The the coordinates (2,8) are expressed in terms of (x,y). Substitute the x with 2 and y with 8 and plug them into the formulas. The one that has the two sides equal to each other is the function that has (2,8) on its line.
Suppose that in a bowling league, the scores among all bowlers are normally distributed with mean µ = 182 points and standard deviation σ = 14 points. A trophy is given to each player whose score is at or above the 97th percentile. What is the minimum score needed for a bowler to receive a trophy?
Answer:
209 points
Step-by-step explanation:
Mean points scored (μ) = 182 points
Standard deviation (σ) = 14 points
The z-score for any given game score 'X' is defined as:
[tex]z=\frac{X-\mu}{\sigma}[/tex]
At, the 97th percentile of a normal distribution, the z-score, according to a z-score table, is 1.881.
Therefore, the minimum score, X, needed for a bowler to receive a trophy is:
[tex]1.881=\frac{X-182}{14}\\X=208.334[/tex]
Since only whole point scores are possible, X=209 points.
To find the minimum score at the 97th percentile for bowlers in a league, one must calculate the z-score for the 97th percentile and then apply the formula Score = μ + (z * σ) using the league's mean and standard deviation.
Explanation:To find the minimum score needed for a bowler to receive a trophy (which is at or above the 97th percentile), we need to use the normal distribution properties. With a mean (μ) of 182 points and a standard deviation (σ) of 14 points, we can find the z-score corresponding to the 97th percentile using a z-table or a calculator with normal distribution functions. Once we have the z-score, we can use the formula:
Score = μ + (z * σ)
to calculate the score that corresponds to the 97th percentile.
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A poll showed that 48 out of 120 randomly chosen graduates of California medical schools last year intended to specialize in family practice. What is the width of a 90 percent confidence interval for the proportion that plan to specialize in family practice? Select one:a. ± .0736b. ± .0447c. ± .0876d. ± .0894
Answer:
± 0.0736
Step-by-step explanation:
Data provided in the question:
randomly chosen graduates of California medical schools last year intended to specialize in family practice, p = [tex]\frac{48}{120}[/tex] = 0.4
Confidence level = 90%
sample size, n = 120
Now,
For 90% confidence level , z-value = 1.645
Width of the confidence interval = ± Margin of error
= ± [tex]z\times\sqrt\frac{p\times(1-p)}{n}[/tex]
= ± [tex]1.645\times\sqrt\frac{0.4\times(1-0.4)}{120}[/tex]
= ± 0.07356 ≈ ± 0.0736
Hence,
The correct answer is option ± 0.0736
The correct option is a. [tex]\±0.0736.[/tex] The width of a [tex]90\%[/tex] confidence interval for the proportion that plan to specialize in family practice
To find the width of a confidence interval for a proportion, we can use the formula:
[tex]\[ \text{Width} = Z \times \sqrt{\frac{p(1-p)}{n}} \][/tex]
where:
[tex]\( Z \)[/tex] is the Z-score corresponding to the desired confidence level,
[tex]\( p \)[/tex] is the sample proportion,
[tex]\( n \)[/tex] is the sample size.
Given:
[tex]Sample\ proportion \( p = \frac{48}{120} = \frac{2}{5} = 0.4 \)[/tex],
[tex]Sample\ size \( n = 120 \)[/tex]
[tex]Confidence \ level = 90\%[/tex], which corresponds to a Z-score of approximately [tex]1.645.[/tex]
Substitute the values into the formula:
[tex]\[ \text{Width} = 1.645 \times \sqrt{\frac{0.4 \times 0.6}{120}} \][/tex]
[tex]\[ \text{Width} = 1.645 \times \sqrt{\frac{0.24}{120}} \][/tex]
[tex]\[ \text{Width} = 1.645 \times \sqrt{0.002} \][/tex]
[tex]\[ \text{Width} = 1.645 \times 0.0447 \][/tex]
[tex]\[ \text{Width} = 0.0736 \][/tex]
The area of a rectangular plot 24 feet long and 16 feet wide will be doubled by adding an equal width to each side of the plot. Which equation can be used to find this added width?
(2x + 24)(2x + 16) = 768
(x + 24)(x + 16) = 384
(x + 24)(x + 16) = 768
(2x + 24)(2x + 16) = 384
Answer:
B
Step-by-step explanation:
Stainless steels are frequently used in chemical plants to handle corrosive fluids, however, these steels are especially susceptible to stress corrosion cracking in certain environments. In a sample of 295 steel alloy failures that occurred in oil refineries and petrochemical plants in Japan over the last 10 tears, 118 were caused by stress corrosion cracking and corrosion fatigue (Materials Performance, 1981). Construct a 95% confidence interval for the true proportion of alloy failures caused by stress corrosion cracking.
Answer: 95% confidence interval would be (0.344,0.456).
Step-by-step explanation:
Since we have given that
n = 295
x = 118
so, [tex]\hat{p}=\dfrac{x}{n}=\dfrac{118}{295}=0.4[/tex]
At 95% confidence, z = 1.96
So, margin of error would be
[tex]z\times \sqrt{\dfrac{p(1-p)}{n}}\\\\=1.96\times \sqrt{\dfrac{0.4\times 0.6}{295}}\\\\=0.056[/tex]
so, 95% confidence interval would be
[tex]\hat{p}\pm \text{margin of error}\\\\=0.4\pm 0.056\\\\=(0.4-0.056,0.4+0.056)\\\\ =(0.344,0.456)[/tex]
Hence, 95% confidence interval would be (0.344,0.456).
Answer:
95% confidence interval would be (0.344,0.456).Step-by-step explanation:
Find the absolute maximum and absolute minimum values of f on the given interval.
a) f(t)= t sqrt(36-t^2) [-1,6]
absolute max=
absolute min=
b) f(t)= 2 cos t + sin 2t [0,pi/2]
absolute max=
absolute min=
Answer:
absolute max= (4.243,18)
absolute min =(-1,-5.916)
absolute max=(pi/6, 2.598)
absolute min = (pi/2,0)
Step-by-step explanation:
a) [tex]f(t) = t\sqrt{36-t^2} \\[/tex]
To find max and minima in the given interval let us take log and differentiate
[tex]log f(t) = log t + 0.5 log (36-t^2)\\Y(t) = log t + 0.5 log (36-t^2)[/tex]
It is sufficient to find max or min of Y
[tex]y'(t) = \frac{1}{t} -\frac{t}{36-t^2} \\\\y'=0 gives\\36-t^2 -t^2 =0\\t^2 =18\\t = 4.243,-4.243[/tex]
In the given interval only 4.243 lies
And we find this is maximum hence maximum at (4.243,18)
Minimum value is only when x = -1 i.e. -5.916
b) [tex]f(t) = 2cost +sin 2t\\f'(t) = -2sint +2cos2t\\f"(t) = -2cost-4sin2t\\[/tex]
Equate I derivative to 0
-2sint +1-2sin^2 t=0
sint = 1/2 only satisfies I quadrant.
So when t = pi/6 we have maximum
Minimum is absolute mini in the interval i.e. (pi/2,0)
f(1)=-3
f(n)=-5*f(n-1)-7
f(2)=?
pls help lol.
Answer:
f(2)=8
Step-by-step explanation:
f(2)=(-5)*f(1)-7=
(-5)*(-3)-7=8
f(2)=8
Complete parts (a) through (c) below.
(a) Determine the critical value(s) for a right-tailed test of a population mean at the alphaequals0.01 level of significance with 10 degrees of freedom.
(b) Determine the critical value(s) for a left-tailed test of a population mean at the alphaequals0.10 level of significance based on a sample size of nequals15.
(c) Determine the critical value(s) for a two-tailed test of a population mean at the alphaequals0.01 level of significance based on a sample size of nequals12.
Answer:
a) t = 2.7638
b) t = - 2.6245
c) t = 3.1058 on the right side and
t = -3.1058 on the left
Step-by-step explanation:
a)Determine critical value for a right-tail test for α = 0.01 level of significance and 10 degrees of fredom
From t-student table we find:
gl = 10 and α = 0.01 ⇒ t = 2.7638
b)Determine critical value for a left-tail test for α = 0.01 level of significance and sample size n = 15
From t-student table we find:
gl = 14 and α = 0.01 gl = n - 1 gl = 15 - 1 gl = 14
t = - 2.6245
c) Determine critical value for a two tails-test for α = 0.01 level of significance the α/2 = 0.005 and sample size n = 12
Then
gl = 11 and α = 0.005
t = 3.1058 on the right side of the curve and by symmetry
t = - 3.1058
From t-student table we find:
Suppose a television news broadcast reports that the proportion of people in the United States who are living with a particular disease is 0.09. A team of biomedical students examined a random sample of 527 medical records and found that 34 of them had this disease. They constructed the following 95% z z‑confidence interval for the proportion, p p, of people in the United States who have this disease. 0.0435 < p < 0.0855 0.0435
Answer:
On this case the 0.09 value is not included on the interval so we can say that the statement on the television news broadcast reports, is a value away from the real proportion at least at 95% of confidence.
Step-by-step explanation:
1) Notation and definitions
[tex]X=34[/tex] number of people living at USA with a particular disease.
[tex]n=527[/tex] random sample taken
[tex]\hat p=\frac{34}{527}=0.0645[/tex] estimated proportion of people living at USA with a particular disease
[tex]p[/tex] true population proportion of people living at USA with a particular disease.
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 means 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.
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 population proportion have the following distribution
[tex]p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})[/tex]
In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2 =0.025[/tex]. And the critical value would be given by:
[tex]z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96[/tex]
The confidence interval for the mean is given by the following formula:
[tex]\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex]
If we replace the values obtained we got:
[tex]0.0645 - 1.96\sqrt{\frac{0.0645(1-0.0645)}{527}}=0.0435[/tex]
[tex]0.0645 + 1.96\sqrt{\frac{0.0645(1-0.0645)}{527}}=0.0855[/tex]
The 95% confidence interval would be given by (0.0435;0.0855)
On this case the 0.09 value is not included on the interval so we can say that the statement on the television news broadcast reports, is a value away from the real mean at least at 95% of confidence.
For each of the following scenarios state whether H0 should be rejected or not. State any assumptions that you make beyond the information that is given.
(a) H0 : µ = 4, H1 : µ 6= 4, n = 15, X = 3.4, S = 1.5, α = .05.
(b) H0 : µ = 21, H1 : µ < 21, n = 75, X = 20.12, S = 2.1, α = .10.
(c) H0 : µ = 10, H1 : µ 6= 10, n = 36, p-value = 0.061.
Answer:
a)[tex]H_0 :\mu = 4\\ H_1 : \mu \neq 4[/tex] , n = 15 , X=3.4 , S=1.5 , α = .05
Formula : [tex]t = \frac{x-\mu}{\frac{s}{\sqrt{n}}}[/tex]
[tex]t = \frac{3.4-4}{\frac{1.5}{\sqrt{15}}}[/tex]
[tex]t =-1.549[/tex]
p- value = 0.607(using calculator)
α = .05
p- value > α
So, we failed to reject null hypothesis
b)[tex]H_0 :\mu = 21\\ H_1 : \mu < 21[/tex] , n =75 , X=20.12 , S=2.1 , α = .10
Formula : [tex]t = \frac{x-\mu}{\frac{s}{\sqrt{n}}}[/tex]
[tex]t = \frac{20.12-21}{\frac{2.1}{\sqrt{75}}}[/tex]
[tex]t =-3.6290[/tex]
p- value = 0.000412(using calculator)
α = .1
p- value< α
So, we reject null hypothesis
(c) [tex]H_0 :\mu = 10\\ H_1 : \mu \neq 10[/tex], n = 36, p-value = 0.061.
Assume α = .05
p-value = 0.061.
p- value > α
So, we failed to reject null hypothesis
The decision to reject the null hypothesis in each scenario depends on comparing calculated test statistics (or given p-value) to critical values associated with the significance level. In (a) and (c), we may not reject H0, and in (b) we would reject H0 if our test statistic is larger than the critical value.
Explanation:In hypothesis testing, we compare the test statistic, often a t-value, to the critical value determined by our chosen significance level, α. If the test statistic is greater, we reject the null hypothesis (H0).
(a) In the first scenario, we must calculate the test statistic: t = (X - µ) / (S/√n) = (3.4 - 4) / (1.5/√15); if this absolute value is less than our critical value associated with α = .05 and degrees of freedom = 14, we do not reject H0.
(b) For the second, the test statistic would be (20.12 - 21) / (2.1/√75); compare its value to the critical value associated with α = .10 and degrees of freedom = 74. If it is larger, we reject H0 due to the lower tail test in this scenario.
(c) In the third scenario, the p-value is given. Since our p-value = 0.061 is larger than our α = .05, we would not reject H0.
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An electrical engineer wishes to compare the mean lifetimes of two types of transistors in an application involving high-temperature performance. A sample of 60 transistors of type A were tested and were found to have a mean lifetime of 1827 hours and a standard deviation of 176 hours. A sample of 180 transistors of type B were tested and were found to have a mean lifetime of 1658 hours and a standard deviation of 246 hours. Let μX represent the population mean for transistors of type A and μY represent the population mean for transistors of type B. Find a 95% confidence interval for the difference μX−μY . Round the answers to three decimal places.
Answer:
add the decimals thats all
Step-by-step explanation:
A survey of the mean number of cents off that coupons give was conducted by randomly surveying one coupon per page from the coupon sections of a recent San Jose Mercury News. The following data were collected: 20¢; 75¢; 50¢; 65¢; 30¢; 55¢; 40¢; 40¢; 30¢; 55¢; $1.50; 40¢; 65¢; 40¢. Assume the underlying distribution is approximately normal. (a) Determine the sample mean in cents (Round to 3 decimal places)
Answer:
53¢
Step-by-step explanation:
First, I'll put these in order.
20¢;30¢; 30¢;75¢;40¢;40¢;40¢;40¢;50¢;55¢55¢65¢;65¢; $1.50;
Then, I'll combine like terms.
30+30=60
40+40+40+40(or 40 x 4)=160
55+55=110
65+65=130
60+160+110+130+20+75+50+$1.50=$7.55/14=53¢
PLZ correct me if i'm wrong :-D
One method for measuring air pollution is to measure the concentration of carbon monoxide, or CO, in the air. Suppose Nina, an environmental scientist, wishes to estimate the CO concentration in Budapest, Hungary. She randomly selects 48 locations throughout the city measures the CO concentration at each location. Based on her 48 samples, she computes the margin of error for a 95% t-confidence interval for the mean concentration of CO in Budapest, in g/m3, to be 4.28 What would happen to the margin of error if Nina decreases the confidence level to 90%? Nina increases the confidence level to 99%? Nina decreases the sample size to 34 locations? Nina increases the sample size to 70 locations? Answer Bank Decrease Stay the sameIncrease
Answer:
a) Nina decreases the confidence level to 90%? (Decrease)
b) Nina decreases the sample size to 34 locations? (Increase)
c) Nina increases the sample size to 70 locations? (Decrease)
Step-by-step explanation:
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 means 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.
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".
[tex]\bar X[/tex] represent the sample mean for the sample
[tex]\mu[/tex] population mean (variable of interest)
s represent the sample standard deviation
n=48 represent the original sample size
Confidence =95% or 0.95
ME=4.28 represent the margin of error.
The confidence interval for the mean is given by the following formula:
[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)
In order to calculate the mean and the sample deviation we can use the following formulas:
[tex]\bar X= \sum_{i=1}^n \frac{x_i}{n}[/tex] (2)
[tex]s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}[/tex] (3)
And the margin of error is given by the following expression:
[tex]ME= t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (4)
Based on the formula (4) we can answer all the questions involved:
a) Nina decreases the confidence level to 90%?
On this case the value for [tex]t_{\alpha/2}[/tex] will also decrease so the margin of error would decrease.
b) Nina decreases the sample size to 34 locations?
If we analyze the original sample size of 48 we see that if we reduce the value of n to 34, the margin of error would increase, because n is on the denominator of the margin of error.
c) Nina increases the sample size to 70 locations?
If we analyze the original sample size of 48 we see that if we increase the value of n to 70, the margin of error would decrease, because n is on the denominator of the margin of error.
Which of the following may be used to check the conditions needed to perform a two sample test for mean (independent samples)?
I. Both populations are approximately normally distributed
II. Both sample sizes greater than 30
III. Population of differences is approximately normally distributed
A) I or III
B) II or III
C) I or II
D) I, II, or III
Answer:
Option D is right
Step-by-step explanation:
given that a two sample test for mean of independent samples to be done.
We create hypotheses as:
[tex]H_0: \ bar x = \bar y[/tex] vs alternate suitably right or left or two tailed according to the needs.
The conditions needed for conducting this test would be
I. Both populations are approximately normally distributed
II. Both sample sizes greater than 30
III. Population of differences is approximately normally distributed
i.e. either i, ii or III
Option D is right.
The Martian Colonies elect their government through a lottery. There are100,000 people living on Mars, and every year, a council of 99 co-equalleaders is randomly selected from the population. In how many ways canthe leadership be elected? Give your answer in terms of permutations orcombinations and explain your choice. You do not have to evaluate.
Answer:
The answer is a 100,000-choose-99 (a combination.)
[tex]P(100000, 99) = \displaystyle \left( \begin{array}{c}100000\cr 99\end{array}\right)[/tex].
Step-by-step explanation:
A combination [tex]C(n, r)[/tex] or equivalently [tex]\displaystyle \left(\begin{array}{c}n \cr r\end{array}\right)[/tex] gives the number of ways to choose [tex]r[/tex] out of [tex]n[/tex] elements.
A permutation [tex]P(n, r)[/tex] also gives the number of ways to choose [tex]r[/tex] out of [tex]n[/tex] elements. On top of that, it accounts for the order of the elements. Two elements in different order counts twice in a permutation, but only once in a combination.
The question emphasize that the council members are "co-equal." That implies that the order of the members don't really matter. Hence a combination with [tex]n = 100000[/tex] and [tex]r = 99[/tex] would be a more suitable choice.
For the given function, determine consecutive values of x between which each real zero is located. f(x) = –11x^4 – 5x^3 – 9x^2 + 12x + 10
Answer:
[-1, 0][0, 1]Step-by-step explanation:
Descartes' rule of signs tells you this function, with its signs {- - - + +}, having one sign change, will have one positive real root.
When odd-degree terms have their signs changed, the signs {- + - - +} have three changes, so there will be 1 or 3 negative real roots.
The constant term (10) tells us the y-intercept is positive. The sum of coefficients is -11 -5 -9 +12 +10 = -3, so f(1) < 0 and there is a root between 0 and 1.
When odd-degree coefficients change sign, the sum becomes -11 +5 -9 -12+10 = -17, so there is a root between -1 and 0.
__
Synthetic division by (x+1) gives a quotient of -11x^3 +6x^2 -15x +27 -17/(x+1), which has alternating signs, indicating -1 is a lower bound on real roots.
Real roots are located in the intervals [-1, 0] and [0, 1].
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The remaining roots are complex. All roots are irrational.
The attached graph confirms that roots are in the intervals listed here. Newton's method iteration is used to refine these to calculator precision. Dividing them from f(x) gives a quadratic with irrational coefficients and complex roots.
The position vector for particle A is cos(t)i, and the position vector for particle B is sin(t)j. What is the difference in acceleration (i.e. the relative acceleration) between particle A and B at any time t? The acceleration vector of a particle moving in space is the second derivative of the position vector
Answer:
sin(t)j - cos(t)i
Step-by-step explanation:
Let's start with A:
Position vector = cos(t)i
Velocity vector = -sin(t)i (differentiating the position vector)
acceleration vector = -cos(t)i (differentiating the velocity vector)
Then we go to B:
Position vector = sin(t)j
Velocity vector = cos(t)j
acceleration vector = -sin(t)j
Relative acceleration = -cos(t)i - (-sin(t)j) = sin(t)j - cos(t)i
The average price of homes sold in the U.S. in the past year was $220,000. A random sample of 81 homes sold this year showed a sample mean price of $210,000. It is known that the standard deviation of the population is $36,000. Using a 1% level of significance, test to determine if there has been a significant decrease in the average price homes. Use the p value approach. Make sure to show all parts of the test, including hypotheses, test statistic, decision rule, decision and conclusion.
Final answer:
To test if there has been a significant decrease in the average price of homes sold in the U.S., we will conduct a hypothesis test using the p-value approach. We will define the null and alternative hypotheses, calculate the Z-test statistic, compare the p-value to the significance level, and make a conclusion based on the results.
Explanation:
To test if there has been a significant decrease in the average price of homes sold in the U.S., we will conduct a hypothesis test using the p-value approach. Let's define our hypotheses:
Null hypothesis (H0): The average price of homes sold in the U.S. is not significantly different from $220,000.
Alternative hypothesis (Ha): The average price of homes sold in the U.S. is significantly less than $220,000.
Next, we need to calculate the test statistic. Since the population standard deviation is known, we can use a Z-test. The formula for the Z-test statistic is:
Z = (sample mean - population mean) / (population standard deviation / sqrt(sample size))
In this case, the sample mean is $210,000, the population mean is $220,000, the population standard deviation is $36,000, and the sample size is 81. Plugging these values into the formula:
Z = (210,000 - 220,000) / (36,000 / sqrt(81))
Calculating this gives us a Z-statistic of -1.6875.
The final step is to compare the p-value of the test statistic to the significance level. Since the significance level is 1%, the critical value is 0.01. We can use a Z-table or calculator to find the p-value corresponding to a Z-statistic of -1.6875. The p-value is approximately 0.0466.
Since the p-value is less than the significance level of 0.01, we reject the null hypothesis. This means that there is sufficient evidence to conclude that there has been a significant decrease in the average price of homes sold in the U.S.
Salaries of 49 college graduates who took a statistics course in college have a mean, x overbar, of $ 65 comma 300. Assuming a standard deviation, sigma, of $17 comma 805, construct a 95% confidence interval for estimating the population mean mu.
Answer: [tex]60,540< \mu<70,060[/tex]
Step-by-step explanation:
The confidence interval for population mean is given by :-
[tex]\overline{x}-z^*\dfrac{\sigma}{\sqrt{n}}< \mu<\overline{x}+z^*\dfrac{\sigma}{\sqrt{n}}[/tex]
, where [tex]\sigma[/tex] = Population standard deviation.
n= sample size
[tex]\overline{x}[/tex] = Sample mean
z* = Critical z-value .
Given : [tex]\sigma=\$17,000[/tex]
n= 49
[tex]\overline{x}= \$65,300[/tex]
Two-tailed critical value for 95% confidence interval = [tex]z^*=1.960[/tex]
Then, the 95% confidence interval would be :-
[tex]65,300-(1.96)\dfrac{17000}{\sqrt{49}}< \mu<65,300+(1.96)\dfrac{17000}{\sqrt{49}}[/tex]
[tex]=65,300-(1.96)\dfrac{17000}{7}< \mu<65,300+(1.96)\dfrac{17000}{7}[/tex]
[tex]=65,300-4760< \mu<65,300+4760[/tex]
[tex]=60,540< \mu<70,060[/tex]
Hence, the 95% confidence interval for estimating the population mean [tex](\mu)[/tex] :
[tex]60,540< \mu<70,060[/tex]
Let X be the number of material anomalies occurring in a particular region of an aircraft gas-turbine disk. The article "Methodology for Probabilistic Life Prediction of Multiple-Anomaly Materials"� proposes a Poisson distribution for X. Suppose that ? = 4. (Round your answers to three decimal places.) (a) Compute both P(X ? 4) and P(X < 4). (b) Compute P(4 ? X ? 5). (c) Compute P(5 ? X). (d) What is the probability that the number of anomalies does not exceed the mean value by more than one standard deviation?
Answer:
0.6284,0.4335,0.1953.0.9786
Step-by-step explanation:
Given that X the number of material anomalies occurring in a particular region of an aircraft gas-turbine disk is following a poisson distribution with parameter 4
a) [tex]P(X\leq 4)=0.6284\\P(X<4)=0.4335[/tex]
b) [tex]P(4\leq x<5)\\=P(4)=0.1953\\[/tex]
c) P([tex]5\leq x)[/tex]=0.8046
d) the probability that the number of anomalies does not exceed the mean value by more than one standard deviation
=[tex]P(0\leq X\leq 8)\\=F(8)-F(0)\\=0.9786[/tex]
Let C be the positively oriented square with vertices (0,0), (1,0), (1,1), (0,1). Use Green's Theorem to evaluate the line integral ∫C7y2xdx+8x2ydy.
Answer:
1/2
Step-by-step explanation:
The interior of the square is the region D = { (x,y) : 0 ≤ x,y ≤1 }. We call L(x,y) = 7y²x, M(x,y) = 8x²y. Since C is positively oriented, Green Theorem states that
[tex]\int\limits_C {L(x,y)} \, dx + {M(x,y)} \, dy = \int\limits^1_0\int\limits^1_0 {(Mx - Ly)} \, dxdy[/tex]
Lets calculate the partial derivates of M and L, Mx and Ly. They can be computed by taking the derivate of the respective value, treating the other variable as a constant.
Mx(x,y) = d/dx 8x²y = 16xyLy(x,y) = d/dy 7y²x = 14xyThus, Mx(x,y) - Ly(x,y) = 2xy, and therefore, the line ntegral is equal to the double integral
[tex] \int\limits^1_0\int\limits^1_0 {2xy} \, dxdy[/tex]
We can compute the double integral by applying the Barrow's Rule, a primitive of 2xy under the variable x is x²y, thus the double integral can be computed as follows
[tex]\int\limits^1_0\int\limits^1_0 {2xy} \, dxdy = \int\limits^1_0 {x^2y} |^1_0 \,dy = \int\limits^1_0 {y} \, dy = \frac{y^2}{2} \, |^1_0 = 1/2[/tex]
We conclude that the line integral is 1/2
To evaluate the line integral using Green's Theorem, we need to find the curl of the vector field and the area enclosed by the square. The line integral of the vector field along the square is equal to the double integral of the curl over the region enclosed by the square. Using this method, we can find the value of the line integral to be 1/3.
Explanation:To evaluate the line integral using Green's Theorem, we first need to find the curl of the vector field. In this case, the vector field is F(x, y) = 7y^2x i + 8x^2y j. Taking the partial derivatives of its components with respect to x and y, we get curl(F) = (8x^2 - 14xy^2) k.
Next, we need to find the area enclosed by the square C, which is 1 unit^2. Using Green's Theorem, the line integral of F along C is equal to the double integral of curl(F) over the region D enclosed by C. Integrating curl(F) with respect to y, we get -7xy^2 + 6x^2y. Integrating this with respect to x over the given limits, we find the value of the line integral to be 1/3.
Learn more about Green's Theorem here:https://brainly.com/question/30763441
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A news vendor sells newspapers and tries to maximize profits. The number of papers sold each day is a random variable. However, analysis of the past month's data shows the distribution of daily demand in Table 16. A paper costs the vendor 20C. The vendor sells the paper for 30C. Any unsold papers are returned to the publisher for a credit of IOC. Any unsatisfied demand is estimated to cost IOC in goodwill and lost profit. If the policy is to order a quantity equal to the preceding day's demand, determine the average daily profit of the news vendor by simulating this system. Assume that the demand for day 0 is equal to 32. Demand per day Probability30 .0531 .1532 .2233 .3834 .1435 .06
A deck of cards contains 52 cards. They are divided into four suits: spades, diamonds, clubs and hearts. Each suit has 13 cards: ace through 10, and three picture cards: Jack, Queen, and King. Two suits are red in color: hearts and diamonds. Two suits are black in color: clubs and spades.
Use this information to compute the probabilities asked for below and leave them in fraction form. All events are in the context that three cards are dealt from a well-shuffled deck without replacement.
a. The first and second cards are both hearts.
b. The third card is an eight.
c. None of the three cards is an ace.
Answer:
a) 13/52 and the second 12/51
b) Two solutions:
b.1 if we did not picked up an eight in the first two cards 4/50
b.2 there is an eight i the two previous 3/59
c ) (48/52)*(47/51)*(46/50)
Step-by-step explanation:
Condition: Cards are taken out without replacement
a) Probability of first card is heart
There are 52 cards and 4 suits with the same probability , so you can compute this probability in two ways
we have 13 heart cards and 52 cards then probability of one heart card is 13/52 = 0.25
or you have 4 suits, to pick up one specific suit the probability is 1/4 = 0,25
Now we have a deck of 51 card with 12 hearts, the probability of take one heart is : 12/51
b) There are 4 eight (one for each suit ) P = 4/50 if neither the first nor the second card was an eight of heart, if in a) previous we had an eight, then this probability change to 3/50
c) The probability of the first card different from an ace is 48/52 , the probability of the second one different of an ace is 47/51 and for the thirsd card is 46/50. The probability of none of the three cards is an ace is
(48/52)*(47/51)*(46/50)
Cadmium, a heavy metal, is toxic to animals. Mushrooms, however, are able to absorb and accumulate cadmium at high concentrations. The Czech and Slovak governments have set a safety limit for cadmium in dry vegetables at 0.5 ppm. Below are cadmium levels for a random sample of the edible mushroom Boletus pinicoloa.0.24 0.92 0.59 0.19 0.62 0.330.16 0.25 0.77 0.59 1.33 0.32a) Perform a hypothesis test at the 5% significance level to determine if the meancadmium level in the population of Boletus pinicoloa mushrooms is greater than thegovernment’s recommended limit of 0.5 ppm. Suppose that the standard deviation ofthis population’s cadmium levels is o( = 0.37 ppm. Note that the sum of the data is 6.31 ppm. For this problem, be sure to: State your hypotheses, compute your test statistic, give the critical value.(b) Find the p-value for the test.
Answer:
hi
Step-by-step explanation:
Need help
6n - 3(2n - 5)
SHOW ALL WORK!
6n - 3(2n-5)
mutiply the bracket by -3
(-3)(2n)= -6n
(-3)(-5)= 15
6n-6n+15
0+15
answer:
15
A farmer wants to fence a rectangular garden next to his house which forms the northern boundary. The fencingfor the southern boundary costs $6 per foot, and the fencing for the east and west sides costs $3 per foot. If hehas a budget of $120 for the project, what are the dimensions of the largest area the fence can enclose?
Answer:
10 ft x 10 ft
Area = 100 ft^2
Step-by-step explanation:
Let 'S' be the length of the southern boundary fence and 'W' the length of the eastern and western sides of the fence.
The total area is given by:
[tex]A=S*W[/tex]
The cost function is given by:
[tex]\$ 120 = \$3*2W+\$6*S\\20 = W+S\\W = 20-S[/tex]
Replacing that relationship into the Area function and finding its derivate, we can find the value of 'S' for which the area is maximized when the derivate equals zero:
[tex]A=S*(20-S)\\A=20S-S^2\\\frac{dA}{dS} = \frac{d(20S-S^2)}{dS}\\0= 20-2S\\S=10[/tex]
If S=10 then W =20 -10 = 10
Therefore, the largest area enclosed by the fence is:
[tex]A=S*W\\A=10*10 = 100\ ft^2[/tex]