Answer:
False this value is very likely to find in this distribution
Step-by-step explanation:
With mean 7 ( μ ) and standad deviation (σ ) 2 we can observe, value 6.6 is close to the lower limit of the interval
μ ± 0,5 σ 7 ± 1 in which we should find 68,3 % of all values
(just 6 tenth to the left)
And of course 6.6 is inside the interval
μ ± 1 σ where we find 95.7 % of th value
We conclude this value is not unlikely at all
Final answer:
To determine whether it is unlikely to find a random sample with a mean breaking weight of 6.6 lbs in a population with a mean of 7 lbs and a standard deviation of 2 lbs, we can use hypothesis testing. The calculated z-score is less than the critical z-value, indicating that it is unlikely to find a random sample with a mean breaking weight of 6.6 lbs in the population. Therefore, the statement is true.
Explanation:
To determine whether it is unlikely to find a random sample with a mean breaking weight of 6.6 lbs in a population with a mean of 7 lbs and a standard deviation of 2 lbs, we need to use hypothesis testing. We can set up the null and alternative hypotheses as follows:
Null hypothesis (H0): The population mean breaking weight is equal to 7 lbs.
Alternative hypothesis (Ha): The population mean breaking weight is less than 7 lbs.
Next, we can calculate the z-score using the formula (sample mean - population mean)/(standard deviation/sqrt(sample size)). Substituting the values, we get (6.6 - 7)/(2/sqrt(36)) = -0.6/ (2/6) = -0.6/ (1/3) = -1.8.
We can now look up the critical z-value for a one-tailed test at a significance level of 0.05. The critical z-value is -1.645. Since the calculated z-score (-1.8) is less than the critical z-value (-1.645), we can reject the null hypothesis. Therefore, it is true that finding a random sample with a mean breaking weight of 6.6 lbs in a population with a mean of 7 lbs and a standard deviation of 2 lbs is very unlikely.
The inverse notation f -1 used in a pure mathematics problem is not always used when finding inverses of applied problems. Rather, the inverse of a function such as C = C(q) will be q = q(C). The following problem illustrates this idea. The ideal body weight w for men (in kilograms) as a function of height h (in inches) is given by the following function. W(h) = 49 + 2.2(h- 60) What is the ideal weight of a 6-foot male? The ideal weight, W, of a 6-foot male is kilograms. (Round to the nearest tenth as needed.) Express the height h as a function of weight W. Verify your answer by checking that W(h(W)) = W and h(W(h))h.
The ideal weight of a 6-foot male (according to the function) is approximately 75.4 kilograms. By manipulating the weight function, we can express height as a function of weight. To verify this function, we substitute it back into the original equation, ensuring our original input value is retrieved.
Explanation:The question is asking to find the ideal weight of a 6-foot male using the function W(h) = 49 + 2.2(h- 60). The height in inches for a 6-foot male is 72 inches (as 1 foot equals 12 inches). Substituting into the formula we obtain:
W(72) = 49 + 2.2(72 - 60) = 49 + 2.2*12.
Completing the calculation, the ideal weight is about 75.4 kg (rounded to the nearest tenth).
To express height h as a function of weight W, we need to rearrange the function W(h). Subtracting 49 from both sides yields 2.2(h - 60) = W - 49. Then divide both sides by 2.2 to isolate h, resulting in h = (W - 49) / 2.2 + 60.
Verification that W(h(W)) = W and h(W(h)) = h will require substituting the functions back into each other, and determining that the original input is returned.
Learn more about Inverse Functions here:https://brainly.com/question/17872426
#SPJ3
The projected rate of increase in enrollment at a new branch of the UT-system is estimated by E ′ (t) = 12000(t + 9)−3/2 where E(t) is the projected enrollment in t years.
If the inital enrollment is 2000, determine the projected enrollment after many years by calculating the value of lim t→ [infinity] E(t).
Answer:
The projected enrollment is [tex]\lim_{t \to \infty} E(t)=10,000[/tex]
Step-by-step explanation:
Consider the provided projected rate.
[tex]E'(t) = 12000(t + 9)^{\frac{-3}{2}}[/tex]
Integrate the above function.
[tex]E(t) =\int 12000(t + 9)^{\frac{-3}{2}}dt[/tex]
[tex]E(t) =-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+c[/tex]
The initial enrollment is 2000, that means at t=0 the value of E(t)=2000.
[tex]2000=-\frac{24000}{\left(0+9\right)^{\frac{1}{2}}}+c[/tex]
[tex]2000=-\frac{24000}{3}+c[/tex]
[tex]2000=-8000+c[/tex]
[tex]c=10,000[/tex]
Therefore, [tex]E(t) =-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+10,000[/tex]
Now we need to find [tex]\lim_{t \to \infty} E(t)[/tex]
[tex]\lim_{t \to \infty} E(t)=-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+10,000[/tex]
[tex]\lim_{t \to \infty} E(t)=10,000[/tex]
Hence, the projected enrollment is [tex]\lim_{t \to \infty} E(t)=10,000[/tex]
What can you say about a solution of the equation y' = - y2 just by looking at the differential equation? The function y must be decreasing (or equal to 0) on any interval on which it is defined. The function y must be increasing (or equal to 0) on any interval on which it is defined.
Answer:
The function y must be decreasing (or equal to 0) on any interval on which it is defined.
Step-by-step explanation:
The derivative of a function gives us the rate at which that function is changing. In this case, -y^2, yields a negative value for every possible value of y, thus, the rate of change is always negative and the function y is decreasing (or equal to 0) on any interval on which it is defined.
The differential equation y' = -[tex]y^2[/tex] implies that y is either decreasing or constant wherever it is defined, because the derivative y' is non-positive.
Explanation:By examining the differential equation y' = -[tex]y^2[/tex], we can infer some characteristics about the solutions without solving it. If y is a solution to this equation, then y' represents the derivative of y with respect to x. This derivative tells us about the rate of change of the function y.
Since the right side of the equation is -[tex]y^2[/tex], and a square of a real number is always non-negative, multiplying by -1 makes it non-positive. This implies that the derivative y' is either less than or equal to zero. Therefore, wherever the function y is defined, it must be either decreasing or constant (equal to zero). If y is positive, y will decrease because of the negative sign in front of the square. If y is negative, squaring it results in a positive number, but the negative sign still ensures that the rate of change is non-positive.
Conclusion: the function y is decreasing or remains constant on any interval it is defined; it cannot be increasing.
Learn more about Differential Equation Behavior here:https://brainly.com/question/33668142
#SPJ3
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
math ///////////////////////////////////////////////
Answer:
B. 0.10 - 0.20 = 0.10.
Step-by-step explanation:
100/1000 = 0.10 of the population were born , and
200/1000 = 0.20 of the population died so it is:
0.10 - 0.20 = 0.10.
Answer: B
Step-by-step explanation:
The total population was initially 1000 individuals. In this population, a total of of 100 new individuals were born over the course of one year. The proportion of new individuals that make up the population is 100/1000 = 0.1
Since they were added, then it is positive and it is a gain for the population.
A total of 200 individuals also died over the course of one year. The proportion that died = 200/1000 = 0.2. This would be negative because it is a loss for the population.
The population growth rate will be gain - loss. This becomes
0.1 - 0.2 = -0.1
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.
help please
1 though 5
Answer:
5/8,-5√2,{(2x+3)(x-5)}
Step-by-step explanation:
1) 5x-10/8x-16
=5(x-2)/8(x-2)
=5/8
2) √32-3√18
=4√2-3√18
=4√2-9√2
=(4-9)√2
= -5√2
4) 2x^2-7x-15
=2x^2+3x-10x-15
=(2x+3),(x-5)
A survey is taken among customers of a fast-food restaurant to determine preference for hamburger or chicken. Of 200 respondents selected, 75 were children and 125 were adults. 120 preferred hamburger and 80 preferred chickens. 55 of the children preferred hamburger and 20 preferred chickens. Set up a 2x2 contingency table using this information and answer the following questions:FoodAge Hamburger Chicken TotalChild 55 20 75Adult 65 60 125Total 120 80 200a) What is the probability that a randomly selected individual is an adult?b) What is the probability that a randomly selected individual is a child and prefers chicken?c) Given the person is a child, what is the probability that this child prefers a hamburger?d) Assume we know that a person has ordered chicken, what is the probability that this individual is an adult?
Answer:
Hamburger Chicken
Adults 65 60 125
children 55 20 75
120 80 200
a)What is the probability that a randomly selected individual is an adult?
Total no. of adults = 125
Total no. of people 200
The probability that a randomly selected individual is an adult = [tex]\frac{125}{200}=0.625[/tex]
b) What is the probability that a randomly selected individual is a child and prefers chicken?
No. of child prefers chicken = 20
The probability that a randomly selected individual is a child and prefers chicken= [tex]\frac{20}{200}=0.1[/tex]
c)Given the person is a child, what is the probability that this child prefers a hamburger?
No. of children prefer hamburger = 55
No. of child = 75
The probability that this child prefers a hamburger= [tex]\frac{35}{75}=0.46[/tex]
d) Assume we know that a person has ordered chicken, what is the probability that this individual is an adult?
No. of adults prefer chicken = 60
No. of total people like chicken = 80
A person has ordered chicken, the probability that this individual is an adult= [tex]\frac{60}{80}=0.75[/tex]
Final answer:
The probability that a randomly selected individual is an adult is 0.625. The probability that a randomly selected individual is a child and prefers chicken is 0.10. Given a child, the probability of preferring a hamburger is approximately 0.733, and given that a person ordered chicken, the probability that they are an adult is 0.75.
Explanation:
Let's address the questions based on the contingency table provided:
Probability Calculations
a) The probability that a randomly selected individual is an adult can be calculated as follows:
The number of adults: 125
The total number of respondents: 200
Probability(Adult) = Number of Adults / Total Number of Respondents = 125 / 200 = 0.625
b) The probability that a randomly selected individual is a child and prefers chicken:
The number of children who prefer chicken: 20
The total number of respondents: 200
Probability(Child and Chicken) = Number of Children who prefer Chicken / Total Number of Respondents = 20 / 200 = 0.10
c) If the person is a child, the probability that this child prefers a hamburger:
The number of children who prefer hamburger: 55
The total number of children: 75
Probability(Hamburger | Child) = Number of Children who prefer Hamburger / Total Number of Children = 55 / 75 ≈ 0.733
d) Given that a person has ordered chicken, the probability that this individual is an adult:
The number of adults who prefer chicken: 60
The total number of chicken preferences: 80
Probability(Adult | Chicken) = Number of Adults who prefer Chicken / Total Number of Chicken Preferences = 60 / 80 = 0.75
A manager is interested in determining if the population standard deviation has dropped below 134. Based on a sample of n=27 items selected randomly from the population, conduct the appropriate hypothesis test at a 0.05 significance level. The sample standard deviation is 126. Determine the Null and alternative hypothese
Answer:
H₀: σ² ≥ 17956
H₁: σ² < 17956
Step-by-step explanation:
Hello!
You are asked to test if the population standard deviation of a certain population. Now keep in mind that wherever you want to make a hypothesis test for a population parameter, you have to have a known distribution that includes this parameter. Since the population standard deviation is no parameter of any distribution, what you have to do is test the population variance. Any decision you make about the population variance can be extrapolated to the population standard deviation.
To make a hypothesis test for the population variance, you need a variable with normal distribution and the statistic to use is a Chi-square statistic.
The hypothesis is that the population standard deviation is less than 134, symbolically: σ < 134
Translated in terms of the population variance: σ² < 17956
H₀: σ² ≥ 17956
H₁: σ² < 17956
α: 0.05
χ²= (n-1)S² ~χ²[tex]_{n-1}[/tex]
σ²
χ²= (27-1)(126)² = 22.988
17956
The test is one tailed (left)
χ²[tex]_{n-1;α}[/tex] =χ²[tex]_{26;0.05}[/tex] = 15.379
If the calculated Chi-square value is ≤ than the critical value, you reject the null hypothesis.
If the calculated Chi-square value is > than the critical value, you don't reject the null hypothesis.
Since the value is greater than the critical value, you do not reject the null hypothesis. So at a 5% level, there is not enough evidence to reject the null hypothesis, this means the population variance is at least 17956. On the same level, you can conclude that the population standard deviation is at least 134.
I've made the test so that you have an example of how to do it.
I hope it helps!
You want to rent an unfurnished one-bedroom apartment in Boston next year. The mean monthly rent for a random sample of 10 apartments advertised in the local newspaper is $2534. Assume that the standard deviation is $670. Find the 90%, 95%, and 99% confidence intervals for the mean monthly rent for this category of apartments. Look at the 95% confidence interval and say whether the following statement is true or false. "This interval describes the price of 95% of the rents of all the unfurnished one-bedroom apartments in the Boston area." Be sure to explain your answer.
Answer:
90% confidence interval: (2186.53;2881.47)
95% confidence interval: (2118.73;2949.27)
99% confidence interval: (1987.37;3080.63)
For the last part is not the best way say : "This interval describes the price of 95% of the rents of all the unfurnished one-bedroom apartments in the Boston area."
The best interpretation is this one: "We are 95% confident that the actual mean for the rents of unfurnished one-bedroom apartments in the Boston area is between (2118.73;2949.27)"
Step-by-step explanation:
1) Previous concepts
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=2534[/tex] represent the sample mean
[tex]\mu[/tex] population mean (variable of interest)
[tex]\sigma=670[/tex] represent the population standard deviation
n=10 represent the sample size
90% confidence interval
The confidence interval for the mean is given by the following formula:
[tex]\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex] (1)
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 tabel to find the critical value. The excel command would be: "=-NORM.INV(0.05,0,1)".And we see that [tex]z_{\alpha/2}=1.64[/tex]
Now we have everything in order to replace into formula (1):
[tex]2534-1.64\frac{670}{\sqrt{10}}=2186.53[/tex]
[tex]2534+1.64\frac{670}{\sqrt{10}}=2881.47[/tex]
So on this case the 90% confidence interval would be given by (2186.53;2881.47)
95% confidence interval
The confidence interval for the mean is given by the following formula:
[tex]\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex] (1)
Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that [tex]z_{\alpha/2}=1.96[/tex]
Now we have everything in order to replace into formula (1):
[tex]2534-1.96\frac{670}{\sqrt{10}}=2118.73[/tex]
[tex]2534+1.96\frac{670}{\sqrt{10}}=2949.27[/tex]
So on this case the 95% confidence interval would be given by (2118.73;2949.27)
99% confidence interval
The confidence interval for the mean is given by the following formula:
[tex]\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex] (1)
Since the Confidence is 0.99 or 99%, the value of [tex]\alpha=0.01[/tex] and [tex]\alpha/2 =0.005[/tex], and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-NORM.INV(0.005,0,1)".And we see that [tex]z_{\alpha/2}=2.58[/tex]
Now we have everything in order to replace into formula (1):
[tex]2534-2.58\frac{670}{\sqrt{10}}=1987.37[/tex]
[tex]2534+2.58\frac{670}{\sqrt{10}}=3080.63[/tex]
So on this case the 99% confidence interval would be given by (1987.37;3080.63)
For the last part is not the best way say : "This interval describes the price of 95% of the rents of all the unfurnished one-bedroom apartments in the Boston area."
The best interpretation is this one: "We are 95% confident that the actual mean for the rents of unfurnished one-bedroom apartments in the Boston area is between (2118.73;2949.27)"
A fitted multiple regression equation is Y = 28 + 5X1 - 4X2 + 7X3 + 2X4. When X1 increases 2 units and X2 increases 2 units as well, while X3 and X4 remain unchanged, what change would you expect in your estimate of Y? A. Increase by 2
B. Decrease by 4
C. Increase by 4
D. No change in Y
Answer:
A. Increase by 2
Step-by-step explanation:
Given that a fitted multiple regression equation is
[tex]Y = 28 + 5X_1 -4X_2 + 7X_3 + 2X_4[/tex]
This is a multiple regression line with dependent variable y and independent variables x1, x2, x3 and x4
The coefficients of independent variables represent the slope.
In other words the coefficients represent the rate of change of y when xi is changed by 1 unit.
Given that x3 and x4 remain unchanged and x1 increases by 2 and x2 by 2 units
Since slope of x1 is 5, we find for one unit change in x1 we can have 5 units change in y
i.e. for 2 units change in x1, we expect 10 units change in Y
Similarly for 2 units change in x2, we expect -2(4) units change in Y
Put together we have
[tex]10-8 =2[/tex] change in y
Since positive 2, there is an increase by 2
A. Increase by 2
Please help if so thank you And explain
Answer:
A. False
B. True
C. False
Step-by-step explanation:
A. Angles are not congruent. (CDA is bigger than AED)
B. Both angles are on opposite sides; therefore they are congruent. (They are the same measurement.)
C. BC is shorter than AB. Not congruent.
Question 1. [7 marks total] Parental leave is often compensated to some degree, but the amount of compensation varies greatly. You read a research article that stated, "across people of all incomes, 47% of leave-takers received full pay during their leave, 16% received partial pay, and 37% received no pay." After reading this, you wonder what the distribution of parental leave payment pay is for low income households. Suppose you conduct a survey of leave-takers within households earning less than $30,000 per year. You surveyed 225 people (selected in a random sample) and found that 51 received full pay, 33 received partial pay, and 141 received no pay..You would like to investigate whether the distribution of pay for households earning < $30,000 is different from that of all income levels. Does this correspond to a chi-square test of independence or a chi-square test for goodness of fit?
Answer: this corresponds to a chi-square test for Goodness of Fit.
Step-by-step explanation:
Goodness of fit Chi-squared test gives confirmation on whether a separately observed frequency distribution differs from a theoretical or general distribution.
In this case, the researcher studies a fraction of income earners - low income earners. He is interested in knowing whether the distribution of pay of low income earners is different from the distribution of pay of ALL income earners.
This survey hence corresponds to a chi-square test for goodness of fit.
Lois thinks that people living in a rural environment have a healthier lifestyle than other people. She believes the average lifespan in the USA is 77 years. A random sample of 12 obituaries from newspapers from rural towns in Idaho give x¯=81.03 and s=1.53. Does this sample provide evidence that people living in rural Idaho communities live longer than 77 years?
(a) State the null and alternative hypotheses: (Type "mu" for the symbol μ , e.g. mu >1 for the mean is greater than 1, mu < 1 for the mean is less than 1, mu not = 1 for the mean is not equal to 1) H0 : Ha:
(b) Find the test statistic, t =
Answer: Yes, this sample provide evidence that people living in rural communities live longer than 77 years.
Step-by-step explanation:
Since we have given that
Average lifespan in the USA = 77 years
We need to check whether the people living in rural communities live longer than 77 years.
So, Hypothesis would be
[tex]H_0:\mu=77\\\\H_a:\mu>77[/tex]
Since n = 12
[tex]\bar{x}=81.03\\\\s=1.53[/tex]
since n <30 so, we will use t test.
So, the test statistic value is given by
[tex]t=\dfrac{\bar{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}\\\\\\t=\dfrac{81.03-77}{\dfrac{1.53}{\sqrt{12}}}\\\\\\t=\dfrac{4.03}{0.4416}\\\\t=9.125[/tex]
degrees of freedom = df = n-1 = 12-1 =11
At 95% significance level , t = 1.796
Since, 1.796< 9.125
So, we will reject the null hypothesis.
Hence, Yes, this sample provide evidence that people living in rural communities live longer than 77 years.
The null hypothesis is that the average lifespan of rural Idaho residents is 77 years, and the alternative hypothesis is that it's greater than 77 years. The calculated t-value is approximately 9.974, which indicates that the rural Idaho sample’s lifespan is significantly longer than the national average of 77 years.
Explanation:The null and alternative hypotheses are as follows:
H0 (Null Hypothesis): The average lifespan of rural Idaho residents is 77 years, i.e., mu = 77.Ha (Alternative Hypothesis): The average lifespan of rural Idaho residents is greater than 77 years, i.e., mu > 77.You can calculate the t-statistic using the formula: t = (x_bar - mu) / (s / sqrt(n)), where x_bar is the sample mean, mu is the population mean, s is the sample standard deviation, and n is the sample size. Plugging in the given values, we get:
t = (81.03 - 77) / (1.53 / sqrt(12)) = 9.974, approx.
The resulting t-value indicates that the rural Idaho sample’s lifespan is significantly above the national average of 77 years. Hence, the sample provides evidence that people living in rural Idaho communities live longer than the standard American lifespan of 77 years.
Learn more about Hypothesis Testing here:https://brainly.com/question/34171008
#SPJ11
Consolidated Power, a large electric power utility, has just built a modern nuclear power plant. This plant discharges waste water that is allowed to flow into the Atlantic Ocean. The Environmental Protection Agency (EPA) has ordered that the waste water may not be excessively warm so that thermal pollution of the marine environment near the plant can be avoided. Because of this order, the waste water is allowed to cool in specially constructed ponds and is then released into the ocean. This cooling system works properly if the mean temperature of waste water discharged is 60°F or cooler. Consolidated Power is required to monitor the temperature of the waste water. A sample of 100 temperature readings will be obtained each day, and if the sample results cast a substantial amount of doubt on the hypothesis that the cooling system is working properly (the mean temperature of waste water discharged is 60°F or cooler), then the plant must be shut down and appropriate actions must be taken to correct the problem.(a)Consolidated Power wishes to set up a hypothesis test so that the power plant will be shut down when the null hypothesis is rejected. Set up the null hypothesis H0 and the alternative hypothesis Ha that should be used.H0: μ (select) 60 versus Ha: μ (select) 60.(b)Suppose that Consolidated Power decides to use a level of significance of α = .05 and suppose a random sample of 100 temperature readings is obtained. If the sample mean of the 100 temperature readings is formula310.mml = 61.498, test H0 versus Ha and determine whether the power plant should be shut down and the cooling system repaired. Perform the hypothesis test by using a critical value and a p-value. Assume σ = 6. (Round your z to 2 decimal places and p-value to 4 decimal places.)z =p-value =(Select: Reject or Do not reject) H0. So the plant (Select: Should or Should not) shut down and the cooling system repaired.
Answer:
We conclude that the plant should shut down.
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = 60
Sample mean, [tex]\bar{x}[/tex] = 61.498
Sample size, n = 100
Alpha, α = 0.05
Population standard deviation, σ = 6
a) First, we design the null and the alternate hypothesis such that the power plant will be shut down when the null hypothesis is rejected.
[tex]H_{0}: \mu \leq 60\\H_A: \mu > 60[/tex]
We use One-tailed(right) z test to perform this hypothesis.
b) 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{61.498 - 60}{\frac{6}{\sqrt{100}} } = 2.49[/tex]
Now, [tex]z_{critical} \text{ at 0.05 level of significance } = 1.64[/tex]
Since,
[tex]z_{stat} > z_{critical}[/tex]
We reject the null hypothesis and accept the alternate hypothesis. Thus, the temperature of waste water discharged is greater than 60°F. We conclude that the power plant will shut down.
Calculating the p-value from the z-table:
P-value = 0.0063
Since,
P-value < Significance level
We reject the null hypothesis and accept the alternate hypothesis. Thus, the temperature of waste water discharged is greater than 60°F. We conclude that the power plant will shut down.
Compute the work done by the force F = sin(x + y), xy, (x^2)z> in moving an object along the trajectory that is the line segment from (1, 1, 1) to (2, 2, 2) followed by the line segment from (2, 2, 2) to (3, 6, 8) when force is measured in Newtons and distance in meters.
Parameterize the line segments by
[tex]\vec r(t)=(1-t)\langle1,1,1\rangle+t\langle2,2,2\rangle=\langle1+t,1+t,1+t\rangle[/tex]
and
[tex]\vec s(t)=(1-t)\langle2,2,2\rangle+t\langle3,6,8\rangle=\langle2+t,2+4t,2+6t\rangle[/tex]
both with [tex]0\le t\le1[/tex]. Then
[tex]\vec r'(t)=\langle1,1,1\rangle[/tex]
[tex]\vec s'(t)=\langle1,4,6\rangle[/tex]
so that the work done by [tex]\vec F(x,y,z)=\langle\sin(x+y),xy,x^2z\rangle[/tex] over the respective line segments is given by
[tex]\displaystyle\int_{C_1}\vec F\cdot\mathrm d\vec r=\int_0^1\langle\sin(2+2t),(1+t)^2,(1+t)^3\rangle\cdot\langle1,1,1\rangle\,\mathrm dt[/tex]
[tex]=\displaystyle\int_0^1\sin(2+2t)+(1+t)^2+(1+t)^3\,\mathrm dt=\boxed{\frac{73+6\cos2-6\cos4}{12}}[/tex]
and
[tex]\displaystyle\int_{C_2}\vec F\cdot\mathrm d\vec s=\int_0^1\langle\sin(4+5t),(2+t)(2+4t),(2+t)^2(2+6t)\rangle\cdot\langle1,4,6\rangle\,\mathrm dt[/tex]
[tex]=\displaystyle\int_0^1\sin(4+5t)+4(2+t)(2+4t)+6(2+t)^2(2+6t)\,\mathrm dt=\boxed{\frac{3695+3\cos4-3\cos9}{15}}[/tex]
(both measured in Newton-meters)
To compute the work done by the force in this scenario, we need to apply the principle of line integrals of force with respect to displacement, calculating work for each segment of the trajectory separately and then adding these results.
Explanation:Computing work done by the given force while moving an object generally involves utilizing the line integral of the force with respect to displacement. The principle is simplistically shown in the equation W = F⋅d = Fd cos θ, where W represents work, F denotes force, and d symbolizes displacement, with '⋅' denoting the dot product, and cos θ representing the cosine of the angle between the force and displacement vectors. F can break down into its components, such as Fx, Fy, Fz, and similarly, d can be broken down to dx, dy, dz. Using these, we can express work for three dimensions as dW = Fxdx + Fydy + Fzdz which extends the notion of work done to three-dimensional space. This concept is categorically illustrated while calculating the infinitesimal work done by a variable force.
The trajectory (path) for this problem can be divided into two line segments, one from (1, 1, 1) to (2, 2, 2) and the other from (2, 2, 2) to (3, 6, 8). The work for each segment can be calculated independently, based on the variable force function and the displacement during each segment, after which the two results can be added up to determine the total work done.
Learn more about Work Done by Force in Three Dimensions here:https://brainly.com/question/34648711
#SPJ11
In the following problem, check that it is appropriate to use the normal approximation to the binomial. Then use the normal distribution to estimate the requested probabilities. Ocean fishing for billfish is very popular in the Cozumel region of Mexico. In the Cozumel region about 48% of strikes (while trolling) resulted in a catch. Suppose that on a given day a fleet of fishing boats got a total of 20 strikes. Find the following probabilities. (Round your answers to four decimal places.) (a)-12 or fewer fish were caught (b)-5 or more fish were caught (c)-between 5 and 12 fish were caught
Answer:
a) [tex]P(X\leq 12)=0.8586[/tex]
b) [tex]P(X\geq 5)=0.9802[/tex]
c) [tex]P(5\leq X\leq 12)=0.8389[/tex]
Step-by-step explanation:
The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".
Let X the random variable of interest, on this case we now that:
[tex]X \sim Binom(n=20, p=0.48)[/tex]
The probability mass function for the Binomial distribution is given as:
[tex]P(X)=(nCx)(p)^x (1-p)^{n-x}[/tex]
Where (nCx) means combinatory and it's given by this formula:
[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]
We need to check the conditions in order to use the normal approximation.
[tex]np=20*0.48=9.6 \approx 10 \geq 10[/tex]
[tex]n(1-p)=20*(1-0.48)=10.4 \geq 10[/tex]
So we see that we satisfy the conditions and then we can apply the approximation.
If we appply the approximation the new mean and standard deviation are:
[tex]E(X)=np=20*0.48=9.6[/tex]
[tex]\sigma=\sqrt{np(1-p)}=\sqrt{20*0.48(1-0.48)}=2.234[/tex]
Part a
We want this probability:
[tex]P(X\leq 12)[/tex]
We can use the z score given by this formula [tex]Z=\frac{x-\mu}{\sigma}[/tex].
[tex]P(X\leq 12)=P(\frac{X-\mu}{\sigma}\leq \frac{12-9.6}{2.234})=P(Z\leq 1.074)=0.8586[/tex]
Part b
We want this probability:
[tex]P(X\geq 5)[/tex]
We can use again the z score formula and we have:
[tex]P(X\geq 5)=1-P(X<5)=1-P(\frac{X-\mu}{\sigma}< \frac{5-9.6}{2.234})=1-P(Z<- 2.059)=0.9802[/tex]
Part c
We want this probability:
[tex]P(5\leq X\leq 12)=P(\frac{5-9.6}{2.234}\leq \frac{X-\mu}{\sigma}\leq \frac{12-9.6}{2.234})=P(-2.059\leq Z \leq 1.074)[/tex]
[tex]=P(Z<1.074)-P(Z<-2.059)=0.8586-0.0197=0.8389[/tex]
A deck of cards contains red cards numbered 1,2,3,4,5, blue cards numbered 1,2 and green cards numbered 1,2,3,4,5,6. If a single card is picked at random, what is the probability that the card is red
Answer:
the probability is 38,46%
Step-by-step explanation:
If all decks are put together and shuffled , then card is picked at random regardless of the number, then the probability that the card is red is
probability = number of red cards / total number of cards = 5/(5+2+6) = 5/13=0.3846= 38,46%
Consider the following time series data of sales per year: Year Sales 1995 22.2 1996 23.0 1997 24.4 1998 25.8 1999 27.6 Enter these data in Excel and run the following regression: Sales = Intercept + Slope * Year Note that the Year is the independent variable and Sales is the dependent variable. A 95% confidence interval for the slope is:
a. (-3233.88, -2068.92)
b. (1.04, 1.68)
c. (0.80, 1.88)
d. (0.92,1.76)
e. (-3482.53,-1820.27)
Answer:
b. (1.04, 1.68)
Step-by-step explanation:
Hello!
With the given data I've estimated the regression line
Y: Sales
X: Year
Yi= -2691.32 + 1.36Xi
Where
a= -2691.32
b= 1.36
The 95% CI calculated using the statistic b±[tex]t_{n-1;1-\alpha/2}[/tex]*(Sb/√n) is [1.04;1.68]
I hope it helps.
To find the 95% confidence interval for the slope of a regression line, compute the standard error of the slope, multiply it by the critical value from the t-distribution, and subtract and add the margin of error to the estimated slope. In this case, the 95% confidence interval for the slope is (0.364, 1.676).
Explanation:To find the 95% confidence interval for the slope of the regression line, you need to compute the standard error of the slope. This can be done using the formula: SE = sqrt(SSE/(n-2)) / sqrt(SSX), where SSE is the sum of squared errors, n is the number of data points, and SSX is the sum of squares of the independent variable. Once you have the standard error, you can multiply it by the critical value from the t-distribution with (n-2) degrees of freedom to find the margin of error. Finally, subtract and add the margin of error to the estimated slope to get the lower and upper bounds of the confidence interval.
In this case, the slope of the regression line was estimated to be 1.02. You determined that the standard error of the slope is 0.255. By referring to the t-distribution table or using statistical software, you find the critical value for a 95% confidence interval with (n-2) degrees of freedom to be approximately 2.571. Therefore, the margin of error is 2.571 * 0.255 = 0.656. Finally, you subtract and add the margin of error to the estimated slope to get the lower and upper bounds of the confidence interval: 1.02 - 0.656 = 0.364 and 1.02 + 0.656 = 1.676. So the 95% confidence interval for the slope is (0.364, 1.676).
Learn more about Confidence Interval here:https://brainly.com/question/34700241
#SPJ3
Consider the system of differential equations dxdt=−5ydydt=−5x. Convert this system to a second order differential equation in y by differentiating the second equation with respect to t and substituting for x from the first equation. Solve the equation you obtained for y as a function of t; hence find x as a function of t. If we also require x(0)=1 and y(0)=3, what are x and y? x(t)= equation editorEquation Editor y(t)= equation editorEquation Editor
[tex]\dfrac{\mathrm dx}{\mathrm dt}=-5y[/tex]
[tex]\dfrac{\mathrm dy}{\mathrm dt}=-5x\implies\dfrac{\mathrm d^2y}{\mathrm dt^2}=-5\dfrac{\mathrm dx}{\mathrm dt}[/tex]
[tex]\implies\dfrac{\mathrm d^2y}{\mathrm dt^2}-25y=0[/tex]
This ODE is linear in [tex]y(t)[/tex] with the characteristic equation and roots
[tex]r^2-25=0\implies r=\pm5[/tex]
so that
[tex]y(t)=C_1e^{5t}+C_2e^{-5t}[/tex]
Then
[tex]\dfrac{\mathrm dx}{\mathrm dt}=-5C_1e^{5t}-5C_2e^{-5t}[/tex]
[tex]\implies x(t)=-C_1e^{5t}+C_2e^{-5t}[/tex]
Given that [tex]x(0)=1[/tex] and [tex]y(0)=3[/tex], we find
[tex]\begin{cases}1=-C_1+C_2\\3=C_1+C_2\end{cases}\implies C_1=1,C_2=2[/tex]
and the particular solution to this system is
[tex]\begin{cases}x(t)=-e^{5t}+2e^{-5t}\\y(t)=e^{5t}+2e^{-5t}\end{cases}[/tex]
The value of x and y would be [tex]x(t) =-e^{5t}+ 2e^{-5t}\\\\[/tex] and [tex]y(t) = e^{5t}+ 2e^{-5t}[/tex].
What is a differential equation?An equation containing derivatives of a variable with respect to some other variable quantity is called differential equations.
The derivatives might be of any order, some terms might contain product of derivatives and the variable itself, or with derivatives themselves. They can also be for multiple variables.
we have the following differential equations
[tex]\dfrac{dx}{dt} =-5y\\\\\dfrac{dy}{dt}=-5x[/tex]
by differentiating the second equation we have
[tex]\dfrac{d^2y}{dt^2}=-5\dfrac{dx}{dt}\\\\\dfrac{d^2y}{dt^2}-25y = 0[/tex]
[tex]r^2 - 25 = 0\\\\r = \pm5[/tex]
[tex]y(t) = C_1e^{5t}+ C_2e^{-5t}[/tex]
and by using the characteristic polynomial
[tex]x(t) =-e^{5t}+ 2e^{-5t}\\\\y(t) = e^{5t}+ 2e^{-5t}[/tex]
Learn more about differential equations;
https://brainly.com/question/15345308
#SPJ2
In an agricultural study, the average amount of corn yield is normally distributed with a mean of 185.2 bushels of corn per acre, with a standard deviation of 23.5 bushels of corn. If a study included 1100 acres, about how many would be expected to yield more than 190 bushels of corn per acre?
A. 639 acresB. 461 acresC. 419 acresD. 503 acres
Answer: B. 461 acres
Step-by-step explanation:
Given : In an agricultural study, the average amount of corn yield is normally distributed with a mean of 185.2 bushels of corn per acre, with a standard deviation of 23.5 bushels of corn.
i.e. [tex]\mu=185.2\ \ , \ \sigma=23.5[/tex]
Let x denotes the amount of corn yield.
Now, the probability that the amount of corn yield is more than 190 bushels of corn per acre.
[tex]P(x>190)=P(\dfrac{x-\mu}{\sigma}>\dfrac{190-185.2}{23.5})[/tex]
[Formula : [tex]z=\dfrac{x-\mu}{\sigma}[/tex]]
[tex]=P(z>0.2043)=1-P(z<0.2043)[/tex] [∵ P(Z>z)=1-P(Z<z)]
[tex]1-0.5809405[/tex] [using z-value calculator or table]
[tex]=0.4190595[/tex]
Now, If a study included 1100 acres then the expected number to yield more than 190 bushels of corn per acre :-
[tex]0.4190595\times1100=460.96545\approx461\text{ acres}[/tex]
hence, the correct answer is B. 461 acres .
Find the P-value for the indicated hypothesis test. An airline claims that the no-show rate for passengers booked on its flights is less than 6%. Of 380 randomly selected reservations, 19 were no-shows. Find the P-value for a test of the airline's claim.
A. 0.3508
B. 0.2061
C. 0.0746
D. 0.1230
P-value for the airline's claim that the no-show rate for passengers booked on its flights is less than 6% is 0.2061. Hence, option (B) is correct.
Given that, an airline claims that the no-show rate for passengers booked on its flights is less than 6%.
From this claim, it is clear that:
Null hypothesis : Proportion of no-show rate for passengers booked on its flights, P = 6%. i.e. H₀: P = 0.06
Alternative hypothesis: Proportion of no-show rate for passengers booked on its flights P < 6%, i.e. P < 0.06.
Out of 380 randomly selected reservations, 19 were no-shows
Sample proportion, [tex]\hat{p} = \frac{19}{380}[/tex] = 0.05
Then, the standard error for the sample size of 380 is:
[tex]\text{S.E.} = \sqrt{{\frac{p(1-p)}{n}[/tex]
[tex]\text{S.E.} = \sqrt{{\frac{0.06(1-0.06)}{380}[/tex]
[tex]\text{S.E.} = 0.012[/tex]
Now calculating the test statistic
[tex]z = \frac{( \hat{p} - P)}{S.E}[/tex]
[tex]z = \frac{(0.05 - 0.06)}{0.012 }[/tex]
z = -0.833
p value for z = -0.083 is 0.2061 (From the normal table).
Hence, the p-value for a test of the airline's claim is 0.2061. Option (B) is correct.
Learn more about p-value here:
https://brainly.com/question/33325466
#SPJ4
Final answer:
To find the P-value for the hypothesis test, calculate the sample proportion and standard error, then use the standard normal distribution to calculate the z-score and find the corresponding P-value.
Explanation:
To find the P-value for the hypothesis test, we need to use the given data and perform calculations.
First, we need to calculate the sample proportion, which is the number of no-shows divided by the total number of reservations: 19/380 = 0.05.
Next, we calculate the standard error of the sample proportion using the formula: √((p' * (1 - p')) / n), where p' is the sample proportion and n is the sample size. In this case, the standard error is √((0.05 * (1 - 0.05)) / 380) = 0.014.
Finally, we use the standard normal distribution to calculate the z-score and find the corresponding P-value. In this case, the observed proportion is less than the claimed proportion, so we use a one-tailed test and calculate the z-score as (observed proportion - claimed proportion) / standard error = (0.05 - 0.06) / 0.014 = -0.714. Looking up the P-value for a z-score of -0.714, we find that it is approximately 0.4714.
Therefore, the P-value for the test of the airline's claim is approximately 0.4714.
A queue with a single server receives 50 requests per second on average. The average time for the server to address a request is 10 milliseconds. What is the probability that there are exactly k requests in this system?
Answer:
[tex]P(X=k) = \displaystyle\frac{(2.5)^ke^{-2.5}}{k!}[/tex]
Step-by-step explanation:
This is a typical example where the Poisson distribution is a good choice to model the situation.
In this case we have an interval of time of 50 milliseconds as average time for the server to address one request and 50 requests per second.
By cross-multiplying we determine the expected value of requests every 50 milliseconds.
We know 1 second = 1,000 milliseconds
50 requests __________ 1000 milliseconds
x requests __________ 50 milliseconds
50/x = 1000/50 ===> x = 2.5
and the expected value is 2.5 requests per interval of 50 milliseconds.
According to the Poisson distribution, the probability of k events in 50 milliseconds equals
[tex]\bf P(X=k) = \displaystyle\frac{(2.5)^ke^{-2.5}}{k!}[/tex]
Compute the lower Riemann sum for the given function f(x)=x2 over the interval x∈[−1,1] with respect to the partition P=[−1,− 1 2 , 1 2 , 3 4 ,1].
Answer:
21/64
Step-by-step explanation:
First, we need to note that the function f(x) = x² is increasing on (0, +∞), and it is decreasing on (-∞,0)
The first interval generated by the partition is [-1, -1/2], since f is decreasing for negative values, we have that f takes its minimum values at the right extreme of the interval, hence -1/2.
The second interval is [-1/2, 1/2]. Here f takes its minimum value at 0, because f(0) = 0, and f is positive otherwise.
Since f is increasing for positive values of x, then, on the remaining 2 intervals, f takes its minimum value at their respective left extremes, in other words, 1/2 and 3/4 respectively.
We obtain the lower Riemman sum by multiplying this values evaluated in f by the lenght of their respective intervals and summing the results, thus
LP(f) = f(-1/2) * ((-1/2) - (-1)) + f(0) * (1/2 - (-1/2)) + f(1/2)* (3/4 - 1/2) + f(3/4) * (1- 3/4)
= 1/4 * 1/2 + 0 * 1 + 1/4 * 1/4 + 9/16 * 1/4 = 1/8 + 0 + 1/16 + 9/64 = 21/64
As a result, the lower Riemann sum on the partition P is 21/64
Final answer:
To compute the lower Riemann sum for the given function f(x) = x² over the interval x ∈ [-1,1] with respect to the partition P = [-1, -1/2, 1/2, 3/4 ,1], we need to find the height of each subinterval and multiply it by the width of the subinterval. The lower Riemann sum is 5/8.
Explanation:
To compute the lower Riemann sum for the given function f(x) = x² over the interval x ∈ [-1,1] with respect to the partition P = [-1, -1/2, 1/2, 3/4 ,1], we need to find the height of each subinterval and multiply it by the width of the subinterval.
First, let's calculate the width of each subinterval:
width of subinterval 1: (-1) - (-1) = 0
width of subinterval 2: (-1/2) - (-1) = 1/2
width of subinterval 3: (1/2) - (-1/2) = 1
width of subinterval 4: (3/4) - (1/2) = 1/4
width of subinterval 5: (1) - (3/4) = 1/4
Next, let's calculate the height of each subinterval by substituting the left endpoint of each subinterval into the function:f((-1/2))^2 = 1/4, f(1/2)² = 1/4, f(3/4)² = 9/16, f(1)² = 1
Finally, we compute the lower Riemann sum by multiplying the height by the width for each subinterval and summing them up:
(0 * 0) + (1/2 * 1/4) + (1 * 9/16) + (1/4 * 1) = 5/8.
A single slit diffraction experiment performed with an argon laser of wavelength 454.6 nm produces a pattern on a screen with dark fringes (minima) separated by 10 mm. If we replace the slit with another slit of double the width of the original one, without changing anything else in the setup, what will be the new separation between dark fringes?
In light diffraction, fringe separation is inversely proportional to the slit width. When the slit width in an experiment is doubled, it should result in the separation between the dark fringes on the screen being halved. Therefore, if the original fringe separation was 10 mm, the new separation should be 5 mm.
Explanation:The subject question pertains to the topic of light diffraction, particularly through varying slit widths. The scenario described is a single slit diffraction experiment involving an argon laser with a specific wavelength. In diffraction, the fringe separation is inversely proportional to the slit width. This is because the angle of diffraction is determined by the wavelength of light divided by the slit width, according to the formula sinθ = λ/D, where θ is the diffraction angle, λ is the wavelength, and D is the slit width.
Therefore, when the slit width is doubled, the diffraction angles for the minima (dark fringes) will be halved assuming all other conditions remain the same. Consequently, the separation between the dark fringes on the screen will also be halved. So, if the original separation was 10 mm, the new separation when the slit width is doubled should be 5 mm.
Learn more about Light Diffraction here:https://brainly.com/question/2515936
#SPJ3
A satellite orbiting the earth passes directly overhead at observation stations in Phoenix and Los Angles, 340 miles apart.
At an instant when the satellite is between these two stations, its angle of elevation is simultaneously observed to be 60° at Phoenix and 75° at Los Angeles.
How far is the satellite from Los Angeles?
Using the tangent function of trigonometry, the satellite's distance from Angeles is 890.53 miles.
How the distance is computed:Distance between Phoenix and Los Angeles = 340 miles
1 mile = 5,280 feet
340 miles = 1,795,200 feet (340 miles * 5280 feet/mile)
Angle of elevation at Phoenix = 60°
Angle of elevation at Los Angeles = 75°
Tangent Equation:
tan(75°) = x/d
Solving the equation for x:
x = d * tan(75°)
Substituting the distance value into this equation for x:
x = 1,795,200 * tan(75°) ≈ 4,702,000 feet ≈ 890.53 miles (4,702,000/5,280)
Thus, we can conclude that the distance of the satellite at Los Angeles is approximately, 4,702,000 feet or 890.53 miles.
Learn more about distance using the tangent function at https://brainly.com/question/33435677.
#SPJ1
The distance between the satellite and Los Angeles is approximately 462.215 miles.
Explanation:To find the distance between the satellite and Los Angeles, we can use trigonometry. Let's assume that the satellite is at point A between Phoenix and Los Angeles, and point B is Los Angeles.
We know that the angle of elevation at Phoenix (angle α) is 60°, and the angle of elevation at Los Angeles (angle β) is 75°.
We can use the tangent function to find the distance from Los Angeles to the satellite:
tan(β) = opposite/adjacent
We know that the opposite side is the distance between Phoenix and Los Angeles, which is 340 miles. So we can write:
tan(75°) = 340/AB
Rearranging the equation, we get:
AB = 340/tan(75°)
Using a calculator, we find that AB ≈ 462.215 miles.
Therefore, the satellite is approximately 462.215 miles away from Los Angeles.
Let X1 be a normal random variable with mean µ1 and variance σ 2 1 , and let X2 be a normal random variable with mean µ2 and variance σ 2 2 . Assuming that X1 and X2 are independent, what is the distribution of X1 + X2? g
Answer:
[tex]X_1 + X_2 \sim (\mu_1 + \mu_2, \sigma_1^2 + \sigma_2^2 )[/tex]
Step-by-step explanation:
We are given the following in the question:
[tex]X_1[/tex] is a random normal variable with mean and variance
[tex]\mu_1\\\sigma_1^2[/tex]
[tex]X_1 \sim N(\mu_1,\sigma_1^2)[/tex]
[tex]X_2[/tex] is a random normal variable with mean and variance
[tex]\mu_2\\\sigma_2^2[/tex]
[tex]X_2 \sim N(\mu_2,\sigma_2^2)[/tex]
[tex]X_1, X_2[/tex] are independent events.
Let
[tex]Z =X_1 + X_2[/tex]
Then, Z will have a normal distribution with mean equal to the sum of the two means and its variance equal the sum of the two variances.
Thus, we can write:
[tex]\mu = \mu_1 + \mu_2\\\sigma^2 = \sigma_1^2 + \sigma_2^2\\Z \sim (\mu_1 + \mu_2, \sigma_1^2 + \sigma_2^2 )[/tex]
Final answer:
The sum of two independent normal random variables X1 and X2 is also a normal random variable with a mean of μ1 + μ2 and variance of σ²₁ + σ²₂, denoted by N(μ1 + μ2, σ²₁ + σ²₂).
Explanation:
The distribution of the sum of two independent normal random variables, X1 + X2, is itself a normal random variable. Given that X1 is a normal random variable with mean μ1 and variance σ²₁, and X2 is a normal random variable with mean μ2 and variance σ²₂, the sum X1 + X2 will have a mean of μ1 + μ2 and variance of σ²₁ + σ²₂ because of the properties of the normal distribution and the independence of X1 and X2.
The resulting distribution can be denoted as N(μ1 + μ2, σ²₁ + σ²₂). This conclusion comes from the central limit theorem, which states that the sum of independent random variables tends towards a normal distribution as the number of variables increases, and the means and variances add up.
Nestor Milk Powder is sold in packets with an advertised mean weight of 1.5kgs. The standard deviation is known to be 184 grams. A consumer group wishes to check the accuracy of the advertised mean and takes a sample of 52 packets finding an average weight of 1.49kgs. What is the set of hypotheses that should be used to test the accuracy of advertised weight?
(a) oH:μ= 1.49; 1H: μ≠1.49
(b) oH:μ= 1.5; 1H: μ< 1.5
(c) oH: μ= 1.5; 1H: μ≠1.5
(d) oH: x = 1.5; 1H: x < 1.5
Answer: c) [tex]H_0:\mu= 1.5\\\\H_a:\mu \neq1.5[/tex]
Step-by-step explanation:
Given: Nestor Milk Powder is sold in packets with an advertised mean weight of 1.5 kgs.
i.e. [tex]\mu= 1.5[/tex]
A consumer group wishes to check the accuracy of the advertised mean and takes a sample of 52 packets finding an average weight of 1.49 kgs.
So 1.49 is sample mean but hypothesis is the statement about the parameter which is [tex]\mu[/tex].
i.e. he wanted to check whether [tex]\mu= 1.5[/tex] or [tex]\mu \neq1.5[/tex]
Since null hypothesis[tex](H_0)[/tex] contains equality and alternative hypothesis[tex](H_a)[/tex] is against it.
Therefore, the set of hypotheses that should be used to test the accuracy of advertised weight would be :
[tex]H_0:\mu= 1.5\\\\H_a:\mu \neq1.5[/tex]
Hence, the correct answer is option c) oH: μ= 1.5; 1H: μ≠1.5
a 14 ft long ladder is placed against a house with an angle of elevation of 72 degrees. How high above the ground is the top of the ladder?
Answer:
Answer is 13.3 ft
Step-by-step explanation:
i explained in the image below
A publisher wants to estimate the mean length of time (in minutes) all adults spend reading newspapers. To determine this estimate, the publisher takes a random sample of 15 people and obtains the following results:
11, 9, 8, 10, 10, 9, 7, 11, 11, 7, 6, 9, 10, 8, 10
Assume that the population of times is normally distributed.
a. Find the point estimate of the population mean.
b. Construct the 95% confidence interval for the mean population length reading time and interpret it.
c. What is the margin of error?
d. How can we reduce the margin of error?
Answer:
a) [tex]\bar X=9.07[/tex]
b) The 95% confidence interval is given by (8.197;9.943)
c) [tex]m=2.14 \frac{1.580}{\sqrt{15}}=0.873[/tex]
d) 3 possible ways
1) Increasing the sample size n.
2) Reducing the variability. If we have more data probably we will have less variation.
3) Lower the confidence level. Because if we have lower confidence then the quantile from the t distribution would belower and tthe margin of error too.
Step-by-step explanation:
Notation and definitions
n=15 represent the sample size
[tex]\bar X= 9.07[/tex] represent the sample mean
[tex]s=1.580[/tex] represent the sample standard deviation
m represent the margin of error
Confidence =95% or 0.95
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".
Part a: Find the point estimate of the population mean.
The point of estimate for the population mean [tex]\mu[/tex] is given by:
[tex]\bar X =\frac{\sum_{i=1}^{n} x_i}{n}[/tex]
The mean obteained after add all the data and divide by 15 is [tex]\bar X=9.07[/tex]
Calculate the critical value tc
In order to find the critical value is important to mention that we don't know about the population standard deviation, so on this case we need to use the t 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]. The degrees of freedom are given by:
[tex]df=n-1=15-1=14[/tex]
We can find the critical values in excel using the following formulas:
"=T.INV(0.025,14)" for [tex]t_{\alpha/2}=-2.14[/tex]
"=T.INV(1-0.025,14)" for [tex]t_{1-\alpha/2}=2.14[/tex]
The critical value [tex]tc=\pm 2.14[/tex]
Part c: Calculate the margin of error (m)
First we need to calculate the standard deviation given by this formula:
[tex]s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}[/tex]
s=1.580
The margin of error for the sample mean is given by this formula:
[tex]m=t_c \frac{s}{\sqrt{n}}[/tex]
[tex]m=2.14 \frac{1.580}{\sqrt{15}}=0.873[/tex]
Part b: Calculate the confidence interval
The interval for the mean is given by this formula:
[tex]\bar X \pm t_{c} \frac{s}{\sqrt{n}}[/tex]
And calculating the limits we got:
[tex]9.07 - 2.14 \frac{1.580}{\sqrt{15}}=8.197[/tex]
[tex]9.07 + 2.14 \frac{1.580}{\sqrt{15}}=9.943[/tex]
The 95% confidence interval is given by (8.197;9.943)
Part d: How can we reduce the margin of error?
We can reduce the margin of error on the following ways:
1) Increasing the sample size n.
2) Reducing the variability. If we have more data probably we will have less variation.
3) Lower the confidence level. Because if we have lower confidence then the quantile from the t distribution would belower and tthe margin of error too.
The publisher's point estimate for average reading time is 9.2 minutes. The 95% confidence interval for this estimate is between 8.18 and 10.22 minutes. The margin of error is approximately 1.04 minutes, which can be reduced by increasing the sample size.
Explanation:The subject matter of this question is statistics, specifically dealing with the calculation and interpretation of point estimates, confidence intervals, and margins of error. Here are the steps to solve your question:
Point estimate of the population mean: This is the estimated population mean, which you find by taking the average of your sample. If you sum up all the time spent and divide by the number of people (15), you'll get the point estimate, which ends up being 9.2 minutes.
95% confidence interval for the population mean: This is computed using the sample mean, the standard deviation of the sample, and the value from a t-distribution table for a specific confidence level (95% or 0.05 significance level in this case). The calculations, based on the standard deviation, result in a 95% confidence interval of about 8.18 to 10.22.
Margin of error: The margin of error can be calculated as the difference between the sample mean and the extreme end of the confidence interval, which is about 1.04 in this case.
Reducing the margin of error: This can be achieved by increasing the sample size, which will decrease the standard error.
Learn more about Statistics here:https://brainly.com/question/31538429
#SPJ3