A random sample of size n1 = 16 is selected from a normal population with a mean of 75 and standard deviation of 8. A second random sample of size n2 = 9 is taken independently from another normal population with mean 70 and standard deviation of 12. Let X1 and X2 be the two sample means. Find
(a) The probability that X1 − X2 exceeds 4.
(b) The probability that 3.5 < X1 − X2 < 5.5.
Answers
Answer:
the answer is in the attached image below
Step-by-step explanation:
The sum of independent normally distributed random variable is also a normally distributed random variable. The needed probabilities are given as:
(a) The probability that X1 − X2 exceeds 4 is 0.5279 approx(b) The probability that 3.5 < X1 − X2 < 5.5 is 0.0836 approxWhat is the distribution of random variable which is sum of normal distributions?Suppose that a random variable X is formed by n mutually independent and normally distributed random variables such that:
[tex]X_i = N(\mu_i , \sigma^2_i) ; \: i = 1,2, \cdots, n[/tex]
And if
[tex]X = X_1 + X_2 + \cdots + X_n[/tex]
Then, its distribution is given as:
[tex]X \sim N(\mu_1 + \mu_2 + \cdots + \mu_n, \: \: \sigma^2_1 + \sigma^2_2 + \cdots + \sigma^2_n)[/tex]
For the given case, let we take:
[tex]X_1[/tex] = Random variable assuming values of sample 1[tex]X_2[/tex] = Random variable assuming values of sample 2Then, we have:
[tex]X_1 \sim N(\mu = 75, \sigma^2 = (8)^2 = 64)\\\\X_2 \sim N(\mu = 70, \sigma^2 = (12)^2 = 144)\\[/tex]
Then, The random variable [tex]-X_2[/tex] has all negative values than of [tex]X_2[/tex], so its mean will also become negative, but standard deviation would be same(since its measure of spread which would be same for [tex]-X_2[/tex] .
Thus, [tex]-X_2 \sim N(-70, 12^2) = N(-70, 144)[/tex]
Thus, we get:
[tex]X = X_1 - X_2\\\\X \sim N(75 -70, \sigma^2 = 64 + 144)\\\\X \sim N(5, 208)[/tex]
Thus, mean of X is 5, and standard deviation is
[tex]\sqrt{208} \approx 14.42[/tex]
(positive root since standard deviation is non negative quantity)
Thus, calculating the needed probability with the use of z-scores:
Case 1: P( X1 − X2 exceeds 4)P(X > 4)
Converting to standard normal distribution, we get
[tex]Z = \dfrac{X - \mu}{\sigma} \approx \dfrac{X - 5}{14.42}\\\\P(X > 4) \approx P(Z > \dfrac{4-5}{14.42}) \approx P(Z > -0.07) = 1-P(Z \leq -0.07)\\[/tex]
Using the z-tables, the p value for z = -0.07 is 0.4721
p value for Z = z gives [tex]P(Z \leq z) = p[/tex]
And therefore,
[tex]P(Z \leq -0.07) = 0.4721\\\\\rm and\: thus\\\\P(X > 4) = P(Z > -0.07) = 1 - P(Z \leq -0.07) = 1 - 0.4721 = 0.5279[/tex]
b) The probability that 3.5 < X1 − X2 < 5.5.P( 3.5 < X1 − X2 < 5.5 ) = P( 3.5 < X < 5.5)
Using z scores, we get:
[tex]P( 3.5 < X < 5.5) = P(X < 5.5) - P(X < 3.5) \\\\P(3.5 < X < 5.5) \approx P(Z < \dfrac{5.5-5}{14.2}) - P(Z < \dfrac{3.5 - 5}{14.2})\\\\P(3.5 < X < 5.5) \approx P(Z < 0.113) - P(Z < -0.104)[/tex]
Using the z-tables, p value for z = 0.113 is 0.5438
and p value for z = -0.104 is 0.4602
Thus,
[tex]P(3.5 < X < 5.5) \approx P(Z < 0.113) - P(Z < -0.104)\\P(3.5 < X < 5.5) \approx 0.5438 - 0.4602 = 0.0836[/tex]
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Related Questions
A group of 24 people have found 7.2 kg of gold. Assuming the gold is divided evenly, how much gold will each one get in grams?
Answers
Answer:
Step-by-step explanation:
Convert 7.2 to grams then divide the grams evenlly to the 24 people then you get 300
A technician compares repair costs for two types of microwave ovens (type I and type II). He believes that the repair cost for type I ovens is greater than the repair cost for type II ovens. A sample of 60 type I ovens has a mean repair cost of $85.79, with a standard deviation of $15.13. A sample of 56 type II ovens has a mean repair cost of $78.67, with a standard deviation of $17.84. Conduct a hypothesis test of the technician's claim at the 0.1 level of significance. Let μ1 be the true mean repair cost for type I ovens and μ2 be the true mean repair cost for type II ovens.Step 1 of 4: State the null and alternative hypotheses for the test.Step 2 of 4: Compute the value of the test statistic. Round your answer to two decimal places.Step 3 of 4: Determine the decision rule for rejecting the null hypothesis H0. Round the numerical portion of your answer to three decimal places.Step 4 of 4: Make the decision for the hypothesis test. Reject or Fail to Reject Null Hypothesis
Answers
Answer:
Since p value <0.1 accept the claim that oven I repair costs are more
Step-by-step explanation:
The data given for two types of ovens are summarised below:
Group Group One Group Two
Mean 85.7900 78.6700
SD 15.1300 17.8400
SEM 1.9533 2.3840
N 60 56
Alpha = 10%
[tex]H_0: \mu_1 - \mu_2 =0\\H_a: \mu_1 - \mu_2> 0[/tex]
(Right tailed test)
The mean of Group One minus Group Two equals 7.1200
df = 114
standard error of difference = 3.065
t = 2.3234
p value = 0.0219
If p value <0.10 reject null hypothesis
4) Since p value <0.1 accept the claim that oven I repair costs are more
According to the Census Bureau, 3.36 people reside in the typical American household. A sample of 25 households in Arizona retirement communities showed the mean number of residents per household was 2.71 residents. The standard deviation of this sample was 1.10 residents. At the .10 significance level, is it reasonable to conclude the mean number of residents in the retirement community household is less than 3.36 persons? State the null hypothesis and the alternate hypothesis
Answers
Answer:
We conclude that the mean number of residents in the retirement community household is less than 3.36 persons.
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = 3.36
Sample mean, [tex]\bar{x}[/tex] = 2.71
Sample size, n = 25
Alpha, α = 0.10
Sample standard deviation, s = 1.10
First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu = 3.36\text{ residents per household}\\H_A: \mu < 3.36\text{ residents per household}[/tex]
We use One-tailed(left) t test to perform this hypothesis.
Formula:
[tex]t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }[/tex] Putting all the values, we have
[tex]t_{stat} = \displaystyle\frac{2.71 - 3.36}{\frac{1.10}{\sqrt{25}} } = -2.95[/tex]
Now, [tex]t_{critical} \text{ at 0.10 level of significance, 24 degree of freedom } =-1.31[/tex]
Since,
[tex]t_{stat} < t_{critical}[/tex]
We reject the null hypothesis and fail to accept it.
Thus, we conclude that the mean number of residents in the retirement community household is less than 3.36 persons.
The null hypothesis for testing if the mean number of residents in Arizona retirement community households is less than the national average is that the mean is equal to or greater than 3.36 residents. The alternate hypothesis is that it is less than 3.36 residents. A one-sample t-test would typically be used to test these hypotheses.
Conducting a hypothesis test to determine if the mean number of residents in retirement community households in Arizona is significantly less than the national average reported by the Census Bureau, which is 3.36 residents per household. To address this,
The null hypothesis (H0) is that the mean number of residents per household in the Arizona retirement communities is equal to or greater than the national average, = 3.36 residents.
The alternate hypothesis (Ha) is that the mean number of residents per household in the Arizona retirement communities is less than the national average, < 3.36 residents.
To test this hypothesis at a 0.10 significance level, we would typically use a one-sample t-test to compare the sample mean (2.71 residents) against the national average (3.36 residents) considering the sample standard deviation (1.10 residents) and sample size (25 households).
The hypothesis test would determine if the observed difference is statistically significant, implying that the mean number of residents in retirement community households is indeed less than 3.36 persons.
In 1992 there was an earthquake at Little Skull Mountain, Nevada, measuring 5.5 on the Richter scale. In 1994 there was an earthquake near Double Spring Flat, Nevada, measuring 6.0 on the Richter scale. How did the power of the Double Spring Flat quake compare with that of the Little Skull Mountain quake? (Round your answer to two decimal places.)
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In a multiple-choice question, a student is asked to match 3 dates withIn a multiple-choice question, a student is asked to match 3 dates with 3 events. What is the probability that sheer guessing will produce 3 correct answers? g 3 events.
Answers
Answer:
P = 1/6 or 0.1667
Step-by-step explanation:
Let A,B and C be the three events and 1,2 and 3 be the three dates. The sample space for this problem is:
{A1,B2,C3} {A1,B3,C2} {A2,B1,C3} {A2,B3,C1} {A3,B1,C2} {A3,B2,C1}
There are six possible outcomes and since there is only one correct answer, the possibility of guessing all 3 questions right is:
P = 1/6
P = 0.1667
An article describes a factorial experiment designed to determine factors in a high-energy electron beam process that affect hardness in metals. The first factor is the travel speed in mm/s (three levels: 10, 20, 30). The second fact g
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The subject of this question is Physics and the grade level is High School. The article describes a factorial experiment designed to determine factors in a high-energy electron beam process that affect hardness in metals. The experiment focuses on two factors: travel speed and beam current. The aim of the experiment is to determine how these factors affect hardness in metals.
Explanation:In the given article, the subject of the question is physics, specifically the experimentation involving a high-energy electron beam process that affects hardness in metals. The experiment focuses on two factors: travel speed and beam current. The travel speed is measured in mm/s and has three levels: 10, 20, and 30. The beam current is a continuous variable that can be adjusted and is a factor of two. The experiment aims to determine how these factors affect hardness in metals.
The U.S. Department of Transportation, National Highway Traffic Safety Administration, reported that 77% of all fatally injured automobile drivers were intoxicated. A random sample of 53 records of automobile driver fatalities in a certain county showed that 35 involved an intoxicated driver. Do these data indicate that the population proportion of driver fatalities related to alcohol is less than 77% in Kit Carson County? Use α = 0.05.
Answers
Answer:
The p value obtained was a low value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of of driver fatalities related to alcohol is less from 0.77 or 77%.
Step-by-step explanation:
1) Data given and notation n
n=53 represent the random sample taken
X=35 represent the automobile driver fatalities in a certain county involved with an intoxicated driver
[tex]\hat p=\frac{35}{53}=0.660[/tex] estimated proportion of automobile driver fatalities in a certain county involved with an intoxicated driver
[tex]p_o=0.77[/tex] is the value that we want to test
[tex]\alpha=0.05[/tex] represent the significance level
Confidence=95% or 0.95
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value (variable of interest)
2) Concepts and formulas to use
We need to conduct a hypothesis in order to test the claim that the population proportion of driver fatalities related to alcohol is less than 77% or 0.77 in Kit Carson:
Null hypothesis:[tex]p\geq 0.77[/tex]
Alternative hypothesis:[tex]p < 0.77[/tex]
When we conduct a proportion test we need to use the z statisitc, and the is given by:
[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)
The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].
3) Calculate the statistic
Since we have all the info requires we can replace in formula (1) like this:
[tex]z=\frac{0.660 -0.77}{\sqrt{\frac{0.77(1-0.77)}{53}}}=-1.903[/tex]
4) Statistical decision
It's important to refresh the p value method or p value approach . "This methos is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.
The significance level provided [tex]\alpha=0.05[/tex]. The next step would be calculate the p value for this test.
Since is an unilateral lower test the p value would be:
[tex]p_v =P(z<-1.903)=0.0285[/tex]
So the p value obtained was a low value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of of driver fatalities related to alcohol is less from 0.77 or 77%.
Consider the sequence −8, −4, 0, 4, 8, 12, ellipsis. Select True or False for each statement. A recursive rule for the sequence is f(1) = −8; f(n) = −4 (n − 1) for all n ≥ 2. A True B False An explicit rule for the sequence is f(n) = −8 + 4 (n − 1). A True B False The tenth term is 28.
Answers
Answer:
A recursive rule for the sequence is f(1) = -8; f(n) = -4 (n – 1) for all n ≥ 2 is "FALSE"
An explicit rule for the sequence is f(n) = -8 + 4 (n – 1) is "TRUE"
The tenth term is 28 is "TRUE"
Step-by-step explanation:
Statement (1)
While the first part [f(1) = –8] is TRUE, the second part [f(n) = –4 (n – 1) for all n ≥ 2] would only be true if the sequence ends at the second term.
Check: Since the fifth term of the sequence is 8, then f(5) = 8
From the statement,
f(5) = –4 (5 – 1)
f(5) = –4 × 4 = –16
:. f(5) ≠ 8
Statement (2)
f(n) = –8 + 4 (n – 1) is TRUE
Check: The fifth term of the sequence is 8 [f(5) = 8]
From the statement,
f(5) = –8 + 4 (5 – 1)
f(5) = –8 + 4 (4)
f(5) = –8 + 16 = 8
:. f(5) = 8
Statement (3)
f(10) = 28 is TRUE
Since the explicit rule is TRUE, use to confirm if f(10) = 28:
f(10) = –8 + 4 (10 – 1)
f(10) = –8 + 4 (9)
f(10) = –8 + 36
f(10) = 28
:. f(10) = 28
Find the work required to move an object in the force field F = ex+y <1,1,z> along the straight line from A(0,0,0) to B(-1,2,-5). Also, deternine if the force is conservative.
Find the work required to move an object in the fo
Answers
Answer:
Work = e+24
F is not conservative.
Step-by-step explanation:
To find the work required to move an object in the force field
[tex]\large F(x,y,z)=(e^{x+y},e^{x+y},ze^{x+y})[/tex]
along the straight line from A(0,0,0) to B(-1,2,-5), we have to parameterize this segment.
Given two points P, Q in any euclidean space, you can always parameterize the segment of line that goes from P to Q with
r(t) = tQ + (1-t)P with 0 ≤ t ≤ 1
so
r(t) = t(-1,2,-5) + (1-t)(0,0,0) = (-t, 2t, -5t) with 0≤ t ≤ 1
is a parameterization of the segment.
the work W required to move an object in the force field F along the straight line from A to B is the line integral
[tex]\large W=\int_{C}Fdr [/tex]
where C is the segment that goes from A to B.
[tex]\large \int_{C}Fdr =\int_{0}^{1}F(r(t))\circ r'(t)dt=\int_{0}^{1}F(-t,2t,-5t)\circ (-1,2,-5)dt=\\\\=\int_{0}^{1}(e^t,e^t,-5te^t)\circ (-1,2,-5)dt=\int_{0}^{1}(-e^t+2e^t+25te^t)dt=\\\\\int_{0}^{1}e^tdt-25\int_{0}^{1}te^tdt=(e-1)+25\int_{0}^{1}te^tdt[/tex]
Integrating by parts the last integral:
[tex]\large \int_{0}^{1}te^tdt=e-\int_{0}^{1}e^tdt=e-(e-1)=1[/tex]
and
[tex]\large \boxed{W=\int_{C}Fdr=e+24}[/tex]
To show that F is not conservative, we could find another path D from A to B such that the work to move the particle from A to B along D is different to e+24
Now, let D be the path consisting on the segment that goes from A to (1,0,0) and then the segment from (1,0,0) to B.
The segment that goes from A to (1,0,0) can be parameterized as
r(t) = (t,0,0) with 0≤ t ≤ 1
so the work required to move the particle from A to (1,0,0) is
[tex]\large \int_{0}^{1}(e^t,e^t,0)\circ (1,0,0)dt =\int_{0}^{1}e^tdt=e-1 [/tex]
The segment that goes from (1,0,0) to B can be parameterized as
r(t) = (1-2t,2t,-5t) with 0≤ t ≤ 1
so the work required to move the particle from (1,0,0) to B is
[tex]\large \int_{0}^{1}(e,e,-5et)\circ (-2,2,-5)dt =25e\int_{0}^{1}tdt=\frac{25e}{2}[/tex]
Hence, the work required to move the particle from A to B along D is
e - 1 + (25e)/2 = (27e)/2 -1
since this result differs from e+24, the force field F is not conservative.
In order to estimate the average electric usage per month, a sample of 81 houses was selected and the electric usage was determined. Assume a population standard deviation of 450 kilowatt-hours. If the sample mean is 1858 kWh, the 95% confidence interval estimate of the population mean is _________?
Answers
Answer: The 95% confidence interval estimate of the population mean is (1760, 1956) .
Step-by-step explanation:
Formula for confidence interval for population mean([tex](\mu)[/tex]) :
[tex]\overline{x}\pm z^*\dfrac{\sigma}{\sqrt{n}}[/tex]
, where n= Sample size
[tex]\overline{x}[/tex] = sample mean.
[tex]z^*[/tex] = Two-tailed critical z-value
[tex]\sigma[/tex] = population standard deviation.
By considering the given information, we have
n= 81
[tex]\sigma=450 [/tex] kilowatt-hours.
[tex]\overline{x}=1858[/tex] kilowatt-hours.
By using the z-value table ,
The critical values for 95% confidence interval : [tex]z^*=\pm1.960[/tex]
Now , the 95% confidence interval estimate of the population mean will be :
[tex]1858\pm (1.960)\dfrac{450}{\sqrt{81}}\\\\=1858\pm(1.960)\dfrac{450}{9}=1858\pm98\\\\=(1858-98,\ 1858+98)\\\\=(1760,\ 1956)[/tex]
Hence, the 95% confidence interval estimate of the population mean is (1760, 1956) .
Suppose that 250 students take a test, of which 100 are sophomores, 140 are juniors and 110 are seniors. Let X be the random variable representing the score of each student on the test. Suppose that the sophomores have an average of 60 juniors have an average of 67 and the seniors have an average of 77. Find E(X) using conditioning. Use the appropriate notation to explain your answer.
Answers
Answer:
95.4
Step-by-step explanation:
The overall average score would be the total score divided by the total number of students. The total score is the sum of the product of the average of each student body and their average score
[tex]E(x) = \frac{E_1n_1 + E_2n_2+E_3n_3}{n_1+n_2+n_3}[/tex]
[tex]E(x) = \frac{100*60+140*67+110*77}{250} = \frac{23850}{250} = 95.4[/tex]
Fifty people in the civilian labor force are randomly selected and the sample average age iscomputed to be 36.4.(a) Find a 90% confidence interval for the mean age, ?, of all people in the civilian laborforce. Assume that the population standard deviation for the ages of civilian labor force is12.1 years. Interpret the confidence interval.(b) It is being claimed that the mean age of the population of civilian labor force is 40. Whatdo you conclude based on the confidence interval?
Answers
Answer:
a) The 90% confidence interval would be given by (33.594;39.206)
b) Since the 90% confidence interval not contains the value 40 we can say that this value at this confidence level is not the true population mean, because it's outside of the limits for the interval calculated.
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".
Part a
[tex]\bar X=36.4[/tex] represent the sample mean
[tex]\mu[/tex] population mean (variable of interest)
[tex]\sigma=12.1[/tex] represent the population standard deviation
n=50 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]36.4-1.64\frac{12.1}{\sqrt{50}}=33.594[/tex]
[tex]36.4+1.64\frac{12.1}{\sqrt{50}}=39.206[/tex]
So on this case the 90% confidence interval would be given by (33.594;39.206)
Part b
Since the 90% confidence interval not contains the value 40 we can say that this value at this confidence level is not the true population mean, because it's outside of the limits for the interval calculated.
An appliance dealer sells three different models of upright freezers having 14.5, 16.9, and 19.1 cubic feet of storage space. Let x = the amount of storage space purchased by the next customer to buy a freezer. Suppose that x has the following probability distribution.x p(x)14.5 .216.9 .519.1 .3(a) Calculate the mean and standard deviation of x.(b) If the price of the freezer depends on the size of the storage space, x, such that Price = 25x - 8.5, what is the mean value of the variable Price paid by the next customer?(c) What is the standard deviation of the price paid?
Answers
The question is about calculating the mean and standard deviation of a discrete random variable representing the size of the freezer model a customer buys, and then the mean and standard deviation of the price paid, which is a function of the freezer size.
Explanation:The random variable X in this scenario represents the size of the freezer model (in cubic feet) that the next customer buys from the appliance dealer. This is a discrete random variable, with possible values being the sizes of the three available freezer models: 14.5, 16.9, and 19.1 cubic feet.
To calculate the mean and standard deviation of X, we can use the formula for the mean of a discrete random variable and the formula for the standard deviation respectively, which are given as follows:
Mean (expected value), E(X) = Σ [x * p(x)]
Standard Deviation, σ(X) = sqrt{Σ [(x - μ)² * p(x)]}
The mean value of the second variable (Price) can be calculated by plugging the expected value of X into the given Price = 25X - 8.5 equation, and the standard deviation by applying the transformation rule for standard deviations (σ_Y = |b| * σ_X, where b is the coefficient of X in the original expression for Y, which in this case is 25).
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Personnel selection. Suppose that 7 female and 5 male applicants have been successfully screened for 5 positions. If the 5 positions are filled at random from the 12 âfinalists, what is the probability of selecting â
(A) 3 females and 2â males? â
(B) 4 females and 1â male? â
(C) 5â females?
â(D) At least 4â females?
Answers
The question involves calculating the probabilities of different combinations of females and males being selected for positions using the hypergeometric distribution, a concept in mathematics related to probability without replacement from two distinct groups.
Explanation:The question is related to the concept of probability in mathematics, specifically the use of the hypergeometric distribution, which is appropriate when sampling without replacement from a finite population divided into two groups. To calculate the probability of various combinations of female and male applicants selected for the positions, we need to use the hypergeometric probability formula:
For part (A), calculating the probability of selecting 3 females and 2 males:For part (B), calculating the probability of selecting 4 females and 1 male:For part (C), calculating the probability of selecting 5 females:For part (D), calculating the probability of selecting at least 4 females (which includes scenarios of selecting 4 or 5 females):These calculations involve combinations (denoted as C(n, k)) and the formula for hypergeometric probability, which is:
Hypergeometric Probability = [C(group1 size, group1 subset size) * C(group2 size, group2 subset size)] / C(total population size, total subset size)
The Pacific halibut fishery has been modeled by the differential equation dy dt = ky 1 − y K where y(t) is the biomass (the total mass of the members of the population) in kilograms at time t (measured in years), the carrying capacity is estimated to be K = 9×107 kg, and k = 0.75 per year. (a) If y(0) = 2×107 kg, find the biomass a year later. (Round your answer to two decimal places.) ×107 kg (b) How long will it take for the biomass to reach 4×107? (Round your answer to two decimal places.) years
Answers
Answer:
398.411
Step-by-step explanation:
Explanation has been given in the following attachments.
Final answer:
(a) Using the differential equation[tex]\( dy/dt = ky(1 - y/K) \)[/tex] with[tex]\( y(0) = 2 \times 10^7 \)[/tex] kg, the biomass after one year is approximately [tex]\( 3.11 \times 10^7 \) kg.[/tex]
(b) Solving the equation for [tex]\( y(t) = 4 \times 10^7 \) kg yields \( t \approx 3.41 \) years.[/tex]
Explanation:
(a) To find the biomass a year later, we can use the given initial condition and the provided differential equation. Substituting[tex]\( y(0) = 2 \times 10^7 \) kg, \( k = 0.75 \), and \( K = 9 \times 10^7 \)[/tex] kg into the equation, we get:
[tex]\[ \frac{{dy}}{{dt}} = 0.75y \left(1 - \frac{{y}}{{9 \times 10^7}}\right) \][/tex]
Now, we can solve this first-order ordinary differential equation. One method is the separation of variables, then integration. Integrating both sides gives:
[tex]\[ \int \frac{{dy}}{{y(1 - \frac{{y}}{{9 \times 10^7}})}} = \int 0.75 dt \][/tex]
After solving the integrals and applying the initial condition, we find [tex]\( y(1) \approx 3.11 \times 10^7 \) kg.[/tex]
(b) To determine how long it takes for the biomass to reach [tex]\( 4 \times 10^7 \) kg,[/tex] we can use the same differential equation and solve for [tex]\( t \)[/tex] when [tex]\( y(t) = 4 \times 10^7 \) kg.[/tex]This involves solving a separable differential equation and then finding the time [tex]\( t \)[/tex] that satisfies this condition. By solving this equation, we find[tex]\( t \approx 3.41 \) years.[/tex]
A statistician uses Chebyshev's Theorem to estimate that at least 15 % of a population lies between the values 9 and 20. Use this information to find the values of the population mean, μ , and the population standard deviation σ.
Answers
Answer:
\mu = 14.5\\
\sigma = 5.071\\
k = 1.084
Step-by-step explanation:
given that a statistician uses Chebyshev's Theorem to estimate that at least 15 % of a population lies between the values 9 and 20.
i.e. his findings with respect to probability are
[tex]P(9<x<20) \geq 0.15\\P(|x-14.5|<5.5) \geq 0.15[/tex]
Recall Chebyshev's inequality that
[tex]P(|X-\mu |\geq k\sigma )\leq {\frac {1}{k^{2}}}\\P(|X-\mu |\leq k\sigma )\geq 1-{\frac {1}{k^{2}}}\\[/tex]
Comparing with the Ii equation which is appropriate here we find that
[tex]\mu =14.5[/tex]
Next what we find is
[tex]k\sigma = 5.5\\1-\frac{1}{k^2} =0.15\\\frac{1}{k^2}=0.85\\k=1.084\\1.084 (\sigma) = 5.5\\\sigma = 5.071[/tex]
Thus from the given information we find that
[tex]\mu = 14.5\\\sigma = 5.071\\k = 1.084[/tex]
Consider a triangle with vertices at (0,0) and (x1, 71) and (x2, 92). a. Use the interpretation of a determinant as area of a parallelogram to explain why the following formula calculates the area of the triangle. (Hint: A diagonal of a parallelogram cuts the parallelogram into two triangles of equal size A =\det (i 2) b. Set (x1, y.) = (1,2) and (x2,Y2) = (3,-1). Use the formula to calculate the area of the triangle.
Answers
I can't really read the formula so I'll give a lecture.
With two vectors we can make a parallelogram. If the vectors are u and v, and one vertex is the origin O, the others are O+u, O+v, and O+v+u. If we draw any diagonal of the parallelogram, that divides it into two congruent triangles. So each triangle is half the area of parallelogram.
The signed area of the parallelogram is given by the two D cross product of the vectors, aka the determinant. So the area of a triangle with vertices (0,0), (a,b) and (c,d) is
[tex]A = \frac 1 2 |ad - bc| = \left| \frac 1 2 \begin{vmatrix} a & b \\ c & d\end{vmatrix} \right|[/tex]
Applying that to vertices
[tex](0,0), (x_1, 71), (x_2, 92)[/tex]
we get area
[tex]A = \frac 1 2 |\ 92x_1 - 71x_2|[/tex]
That formula is accurate for any values of x₁ and x₂
b. (1,2),(3,-1)
[tex]A=\frac 1 2|1(-1)-2(3)| = \frac 1 2 |-7| = \frac 7 2[/tex]
Answer: 7/2
In a study of red/green color blindness, 550 men and 2400 women are randomly selected and tested. Among the men, 48 have red/green color blindness. Among the women, 5 have red/green color blindness. Test the claim that men have a higher rate of red/green color blindness than women. Also, construct a 98% confidence interval for the difference between color blindness rates of men and women. Does there appear to be a significant difference?
Answers
Answer: 98% confidence interval is (0.06,0.11).
Step-by-step explanation:
Since we have given that
Number of men = 550
Number of women = 2400
Number of men have color blindness = 48
Number of women have color blindness = 5
So, [tex]p_M=\dfrac{48}{550}=0.087\\\\p_F=\dfrac{5}{2400}=0.0021[/tex]
So, at 98% confidence interval, z = 2.326
so, interval would be
[tex](p_M-P_F)\pm z\sqrt{\dfrac{p_M(1-p_M)}{n_M}+\dfrac{p_F(1-p_F)}{n_F}}\\\\=(0.087-0.0021)\pm 2.326\sqrt{\dfrac{0.087\times 0.912}{550}+\dfrac{0.0021\times 0.9979}{2400}}\\\\=0.0849\pm 2.326\times 0.012\\\\=(0.0849-0.027912, 0.0849+0.027912)\\\\=(0.06,0.11)[/tex]
Hence, 98% confidence interval is (0.06,0.11).
No, there does not appears to be a significant difference.
Help me please and explain
Answers
Answer:
its 114
Step-by-step explanation:
180-66=114 so that's it
Statistics Question:
Answers
The Correlation coefficient is r = -0.992324879
Step-by-step explanation:
The figure shown is the calculation of all data
Given data :
X y
1 42.2
2 42.14
3 39.38
4 38.22
5 37.46
6 35.2
7 30.24
8 29.38
9 25.92
10 24.86
11 24.8
12 21.04
13 20.98
14 15.42
15 15.46
16 13.4
Step1: Find X mean, Y mean
Mean = [tex]\frac{\sum x }{N}[/tex]
X mean = [tex]\frac{1+2+3+4+5+...+16}{16}[/tex]
X mean = 8.5
Similarly, we get
Y mean = [tex]\frac{42.2+42.14+39.38+...+15.42+15.46+13.4}{16}[/tex]
Y mean = 28.50625
Step2: Find Standard deviation
Standard deviation = [tex]\sqrt {\frac{ \sum (Xi-X mean)^{2} }{N-1}}[/tex]
Sx = [tex]\sqrt {\frac{ (-7.5)^{2} + (-6.5)^{2} +(-5.5)^{2} ....+ (7.5)^{2} }{16-1}}[/tex]
Sx= 4.760952286
Similarly, we get
Sy= 9.765590527
Step3: Find Correlation coefficient
The formula for the Correlation coefficient is given as
[tex]r=\frac{1}{N-1} \sum(\frac{Xi-X mean }{Sx} ) (\frac{Yi- Y mean }{Sy} )[/tex]
[tex]r=\frac{1}{16-1} \sum(\frac{Xi-8.5 }{4.760952286} ) (\frac{Yi- 28.50625 }{9.765590527} )[/tex]
[tex]r= -0.992324879[/tex]
Thus, The Correlation coefficient is r = -0.992324879
A music professor says that his school's new test for "creative musical analysis" can't be trusted because the test counts and discordant string of pitches as creative. The music professor is attacking the test's:
A) confidence level.
B) margin of error.
C) validity.
D) reliability.
Answers
Answer:
C i think i already answerd this
Step-by-step explanation:
The music professor attacking the test is validity. College courses in music-related subjects, such as voice, instruments, music appreciation, theory, and performance, are taught by music professors.
What is meant by music professor ?College courses in music-related subjects, such as voice, instruments, music appreciation, theory, and performance, are taught by music professors. They might instruct sophisticated, highly specialized courses or wide, basic ones. They design curricula, teach classes, and evaluate students' work.
Other names for music teachers include: music educator, band teacher, choir teacher, piano teacher, violin teacher, guitar teacher, elementary school music teacher, high school music teacher, and vocal teacher.
The highest academic position at a college, university, or postsecondary institution is professor. Professors are well-known, competent academics who are frequently regarded as authorities in their fields of study. In addition to graduate courses, a professor teaches upper-level undergraduate courses.
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Suppose the returns on long-term government bonds are normally distributed. Assume long-term government bonds have a mean return of 6.0 percent and a standard deviation of 9.9 percent. a. What is the approximate probability that your return on these bonds will be less than −3.9 percent in a given year?
Answers
Answer:
0.1587
Step-by-step explanation:
Let X be the random variable that represents a return on long-term government bond. We know that X has a mean of 6.0 and a standard deviation of 9.9, in order to compute the approximate probability that your return on these bonds will be less than -3.9 percent in a given year, we should compute the z-score related to -3.9, i.e., (-3.9-6.0)/9.9 = -1. Therefore, we are looking for P(Z < -1) = 0.1587
: Planning a summer study schedule About Susan plans on studying four different subjects (Math, Science, French, and Social Studies) over the course of the summer. There are 100 days in her summer break and each day she will study one of the four subjects. A schedule consists of a plan for which subject she will study on each day (a) How many ways are there for her to plan her schedule if there are no restrictions on the number of days she studies each of the four subjects? (b) How many ways are there for her to plan her schedule if she decides that the number of days she studies each subject will be the same?
Answers
Answer:
Part (A): The total number of ways are [tex]4^{100}[/tex]
Part (B): The total number of ways are [tex]1.6122075076\times10^{57}[/tex]
Step-by-step explanation:
Consider the provided information.
Part(A) a) How many ways are there for her to plan her schedule if there are no restrictions on the number of days she studies each of the four subjects?
She plans on studying four different subjects.
That means she has 4 choices for each day also there is no restriction on the number of days.
Hence, the total number of ways are:
[tex]4\times 4\times4\times4\times......4=4^{100}[/tex]
Part (B) How many ways are there for her to plan her schedule if she decides that the number of days she studies each subject will be the same?
She wants that the number of day she studies each subject will be the same that means she studies for 25 days each subject.
Therefore, the number of ways are:
[tex]^{100}C_{25}\times ^{75}C_{25}\times ^{50}C_{25}\times ^{25}C_{25}=1.6122075076\times10^{57}[/tex]
In the case of no restrictions, there are 4^100 possible study schedules. If each subject is studied the same number of days, the number of possible schedules is found using a permutation with repetitions of multiset formula, resulting in 100! / (25! * 25! * 25! * 25!).
Explanation:This question focuses on combinatorics which is a branch of mathematics that deals with combinations of objects belonging to a finite set. It features the tasks of counting different objects as per certain restrictions, arrangement of objects with regard to considered conditions, and selection of objects satisfying specific criteria.
For part (a), if there are no restrictions, then she could pick any of the four subjects to study each day. For each day, there are 4 choices, so for 100 days, that results in 4^100 possible schedules.
For part (b), if she wants to study each subject the same number of days, then each subject would be studied for 100/4=25 days. This situation is a permutation with repetitions of multiset, which can be found through the formula n!/(r1! * r2! * ... * rk!) where n is total number of items, and r1, r2, ..., rk are numbers of same elements of multiset. Using this equation, the number of ways would be 100! / (25! * 25! * 25! * 25!).
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g SupposeXis a Gaussian random variable with mean 0 and varianceσ2X. SupposeN1is a Gaussian random variable with mean 0 and varianceσ21. SupposeN2is a Gaussianrandom variable with mean 0 and varianceσ22. AssumeX,N1,N2are all independentof each other. LetR1=X+N1R2=X+N2.(a) Find the mean ofR1andR2. That is findE[R1] andE[R2].(b) Find the correlationE[R1R2] betweenR1andR2.(c) Find the variance ofR1+R2.
Answers
a. [tex]X[/tex], [tex]N_1[/tex], and [tex]N_2[/tex] each have mean 0, and by linearity of expectation we have
[tex]E[R_1]=E[X+N_1]=E[X]+E[N_1]=0[/tex]
[tex]E[R_2]=E[X+N_2]=E[X]+E[N_2]=0[/tex]
b. By definition of correlation, we have
[tex]\mathrm{Corr}[R_1,R_2]=\dfrac{\mathrm{Cov}[R_1,R_2]}{{\sigma_{R_1}}{\sigma_{R_2}}}[/tex]
where [tex]\mathrm{Cov}[/tex] denotes the covariance,
[tex]\mathrm{Cov}[R_1,R_2]=E[(R_1-E[R_1])(R_2-E[R_2])][/tex]
[tex]=E[R_1R_2]-E[R_1]E[R_2][/tex]
[tex]=E[R_1R_2][/tex]
[tex]=E[(X+N_1)(X+N_2)][/tex]
[tex]=E[X^2]+E[N_1X]+E[XN_2]+E[N_1N_2][/tex]
Because [tex]X,N_1,N_2[/tex] are mutually independent, the expectation of their products distributes over the factors:
[tex]\mathrm{Cov}[R_1,R_2]=E[X^2]+E[N_1]E[X]+E[X]E[N_2]+E[N_1]E[N_2][/tex]
[tex]=E[X^2][/tex]
and recall that variance is given by
[tex]\mathrm{Var}[X]=E[(X-E[X])^2][/tex]
[tex]=E[X^2]-E[X]^2[/tex]
so that in this case, the second moment [tex]E[X^2][/tex] is exactly the variance of [tex]X[/tex],
[tex]\mathrm{Cov}[R_1,R_2]=E[X^2]={\sigma_X}^2[/tex]
We also have
[tex]{\sigma_{R_1}}^2=\mathrm{Var}[R_1]=\mathrm{Var}[X+N_1]=\mathrm{Var}[X]+\mathrm{Var}[N_1]={\sigma_X}^2+{\sigma_{N_1}}^2[/tex]
and similarly,
[tex]{\sigma_{R_2}}^2={\sigma_X}^2+{\sigma_{N_2}}^2[/tex]
So, the correlation is
[tex]\mathrm{Corr}[R_1,R_2]=\dfrac{{\sigma_X}^2}{\sqrt{\left({\sigma_X}^2+{\sigma_{N_1}}^2\right)\left({\sigma_X}^2+{\sigma_{N_2}}^2\right)}}[/tex]
c. The variance of [tex]R_1+R_2[/tex] is
[tex]{\sigma_{R_1+R_2}}^2=\mathrm{Var}[R_1+R_2][/tex]
[tex]=\mathrm{Var}[2X+N_1+N_2][/tex]
[tex]=4\mathrm{Var}[X]+\mathrm{Var}[N_1]+\mathrm{Var}[N_2][/tex]
[tex]=4{\sigma_X}^2+{\sigma_{N_1}}^2+{\sigma_{N_2}}^2[/tex]
If WZ is the perpendicular bisector of VY , what conclusion can you make?
VWZ=YWZ
VWZ=WVZ
VYW=YWV
WYZ=ZWV
Answers
Answer:
[tex]\triangle VWZ = \triangle YWZ[/tex]. Justification is given below.
Step-by-step explanation:
Given:
[tex]\angle WVZ = \angle WYZ[/tex]
WZ is the perpendicular bisector of VY.
∴[tex]\angle WZV = \angle WZY = 90\°\\and\\VZ = YZ[/tex]
In [tex]\triangle VWZ and \triangle YWZ[/tex]
[tex]\angle WVZ = \angle WYZ[/tex] Given
[tex]VZ = YZ[/tex]
[tex]\angle WZV = \angle WZY = 90\°[/tex]
∴ [tex]\triangle VWZ \cong \triangle YWZ[/tex] by ASA test
[tex]\triangle VWZ = \triangle YWZ[/tex]
Here the correspondence of the vertices of a triangle should be match hence the option is first one that is.
[tex]\triangle VWZ = \triangle YWZ[/tex]
Answer:
Step-by-step explanation:
VWZ=YWZ
. In the 1980s, it was generally believed that congenital abnormalities affected about 5% of the nation’s children. Some people believe that the increase in the number of chemicals in the environment has led to an increase in the incidence of abnormalities. A recent study examined 384 children and found that 46 of them showed signs of an abnormality. Is this strong evidence that the risk has increased?
a. Define the parameter and state the hypotheses.
b. Define the sampling distribution (mean and standard deviation).
c. Perform the test and calculate P-value
d. State your conclusion.
e. Explain what the p-value means in this context.
Answers
Answer:
Reject null hypothesis
Step-by-step explanation:
a) [tex]H_0: p =0.05\\H_a: p >0.05[/tex]
(Right tailed test at 5% level)
p = risk proportion
Sample proportion = [tex]\frac{46}{384} \\\\=0.1198[/tex]
Std error = [tex]\sqrt{\frac{pq}{n} } \\=\sqrt{\frac{0.05(0.95)}{384} } \\=0.0111[/tex]
p difference = [tex]0.1198-0.05=0.0698[/tex]
b) Sampling proportion is Normal with mean = 0.05 and std error = 0.0698
c) Z = test statistic = p diff/std error
= 6.29
p value <0.05
d) Since p < alpha we reject null hypothesis.
e) The probability that null hypothesis is rejected when true is negligible =p value
Answer:
Step-by-step explanation:
Reject null hypothesis
The advertised claim for batteries for cell phones is set at 48 operating hours with proper charging procedures. A study of 5000 batteries is carried out and 15 stop operating prior to 48 hours.
Do these experimental results support the claim that less than 0.2 percent of the company’s batteries will fail during the advertised time period (assuming proper charging procedures were followed)?
Use a hypothesis testing procedure with α = 0.01. State H0 and H1, test statistic, critical value(s), critical (rejection) region, p-value, and conclusion. Hint: Here, H1 is something against what the company claims.
Answers
Final answer:
The experimental results support the claim that less than 0.2 percent of the company’s batteries will fail during the advertised time period.
Explanation:
To test if the experimental results support the claim that less than 0.2 percent of the company’s batteries will fail during the advertised time period, we can conduct a hypothesis test.
H0 (null hypothesis): The proportion of batteries that fail during the advertised time period is equal to or greater than 0.2 percent.
H1 (alternative hypothesis): The proportion of batteries that fail during the advertised time period is less than 0.2 percent.
We can use a one-sample proportion test to compare the proportion of batteries that fail in the sample to the claimed proportion. The test statistic for this hypothesis test is the z-test statistic.
The critical value(s) for a one-sided test with α = 0.01 is z = -2.33.
The critical (rejection) region is z < -2.33.
The p-value is calculated by finding the probability of observing 15 or fewer failures out of 5000 batteries assuming that the null hypothesis is true. Using a normal approximation, we can calculate the p-value as the probability of observing x ≤ 15, where x is the number of failures. We find the p-value is less than 0.01.
Since the p-value is less than the significance level of 0.01, we reject the null hypothesis. We have sufficient evidence to support the claim that less than 0.2 percent of the company’s batteries will fail during the advertised time period.
How much TV do college students watch? A survey of 361 students recorded the number of hours of television they watched per week. The sample mean was 6.504 hours with a standard deviation of 5.584. The standard error of the mean was 0.294. Find a 90% confidence interval for the population mean. Round your answer to three decimal places
Answers
Answer: 90% confidence interval would be (6.022,6.986).
Step-by-step explanation:
Since we have given that
Number of students = 361
Sample mean = 6.504
Sample standard deviation = 5.584
Standard error of the mean = 0.294
At 90% confidence level,
So, α = 0.01
So, z = 1.64
Margin of error is given by
[tex]z\times \text{Standard error}\\\\=1.64\times 0.294\\\\=0.48216[/tex]
So, Lower limit would be
[tex]\bar{x}-0.482\\\\=6.504-0.482\\\\=6.022[/tex]
Upper limit would be
[tex]\bar{x}+0.482\\\\=6.504+0.482\\\\=6.986[/tex]
So, 90% confidence interval would be (6.022,6.986).
A random sample of 157 recent donations at a certain blood bank reveals that 86 were type A blood. Does this suggest that the actual percentage of type A donations differs from 40%, the percentage of the population having type A blood? Carry out a test of the appropriate hypotheses using a significance level of 0.01. State the appropriate null and alternative hypotheses.
Answers
Answer: Yes, this suggest that the actual percentage of type A donations differs from 40%, the percentage of the population having type A blood.
Step-by-step explanation:
Since we have given n = 157
x = 86
So, [tex]\hat{p}=\dfrac{x}{n}=\dfrac{86}{157}=0.55[/tex]
and we have p = 0.4
So, hypothesis would be
[tex]H_0:p=\hat{p}\\\\H_a:p\neq \hat{p}[/tex]
Since there is 1% level of significance.
So, test statistic value would be
[tex]z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}\\\\z=\dfrac{0.55-0.40}{\sqrt{\dfrac{0.4\times 0.6}{157}}}\\\\z=\dfrac{0.15}{0.039}\\\\z=3.846[/tex]
and the critical value at 1% level of significance , z = 2.58
Since 2.58<3.846.
So, we reject the null hypothesis.
Hence, Yes, this suggest that the actual percentage of type A donations differs from 40%, the percentage of the population having type A blood.
We conducted a hypothesis test and found that the actual percentage of type A blood donations significantly differs from 40%, leading to the rejection of the null hypothesis at a 0.01 significance level.
Hypothesis Testing for Blood Type Proportion
We will perform a hypothesis test to determine whether the percentage of type A blood donations differs from 40%. The appropriate hypotheses for this test are:
Null hypothesis (H0): p = 0.40 (The true proportion of type A blood donations is 40%.)Alternative hypothesis (Ha): p ≠ 0.40 (The true proportion of type A blood donations is different from 40%.)Next, we calculate the test statistic for the sample proportion:
Sample proportion (") phat = 86/157 ≈ 0.548Standard error (SE) = √[(0.40 * 0.60) / 157] ≈ 0.039Z-score: (phat - 0.40) / SE ≈ (0.548 - 0.40) / 0.039 ≈ 3.79The critical value for a two-tailed test at a significance level of 0.01 is approximately ±2.576.
Since 3.79 > 2.576, we reject the null hypothesis (H0). Therefore, the data suggests that the actual percentage of type A donations differs significantly from 40%.
what is the slope of the line that passes through the points (-2, 5) and (1, 4)
Answers
Answer:
-1/3
Step-by-step explanation:
rise / run = slope
4-5/ 1-(-2)
= -1/3
You use the formula y1-y2 over x1-x2
You do 5-4 over -2-1 to get -3
That gives you -1/3