The 99% confidence interval for the percentage of girls born using this method is 86.7% - 93.3%.
Step-by-step calculation:
1. Calculate the sample proportion of girls:
Divide the number of girls by the total number of babies:
Sample proportion ([tex]\hat p[/tex]) = 270 girls / 300 babies = 0.9
2. Determine the z-score for a 99% confidence level:
Using a z-score table or calculator, find the z-score that corresponds to a 99% confidence level. This value is approximately 2.576.
3. Calculate the margin of error:
Margin of error (E) = z-score × [tex]\sqrt {(\hat p(1 - \hat p) / n)[/tex]
E = 2.576×√(0.9 × 0.1 / 300) ≈ 0.033
4. Construct the confidence interval:
Lower limit = [tex]\hat p[/tex] - E = 0.9 - 0.033 = 0.867
Upper limit = [tex]\hat p[/tex] + E = 0.9 + 0.033 = 0.933
5. Interpretation:
We are 99% confident that the true proportion of girls born using this method lies between 86.7% and 93.3%.
6. While the upper limit suggests a potential increase in girls, the confidence interval remains wide, and further studies are needed for a conclusive effectiveness claim.
The manufacturer of a new compact car claims the miles per gallon (mpg) for the gasoline consumption is approximately normal with mean 26 mpg and standard deviation 12 mpg. If a random sample of 36 such cars are chosen and tested, what is the probability the average mpg is less than 28 mpg?
Answer:
The probability the average mpg is less than 28 mpg is 0.8413.
Step-by-step explanation:
Given : The manufacturer of a new compact car claims the miles per gallon (mpg) for the gasoline consumption is approximately normal with
Mean [tex]\mu=26[/tex] mpg and standard deviation [tex]\sigma=12[/tex] mpg.
Number of sample n=36
To find : What is the probability the average mpg is less than 28 mpg?
Solution :
Applying z-score formula,
[tex]z=\dfrac{x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
The portability is given by, [tex]P(X<28)[/tex]
[tex]=P(\dfrac{x-\mu}{\frac{\sigma}{\sqrt{n}}}<\dfrac{28-26}{\frac{12}{\sqrt{36}}})[/tex]
[tex]=P(\dfrac{x-\mu}{\frac{\sigma}{\sqrt{n}}}<\dfrac{2}{\frac{12}{6}})[/tex]
[tex]=P(z<\dfrac{2}{2})[/tex]
[tex]=P(z<1)[/tex]
Using z-table,
[tex]=0.8413[/tex]
Therefore, the probability the average mpg is less than 28 mpg is 0.8413.
Answer:
he probability the average mpg is less than 28 mpg is 0.8413.
Step-by-step explanation:
Given : The manufacturer of a new compact car claims the miles per gallon (mpg) for the gasoline consumption is approximately normal with
Mean mpg and standard deviation mpg.
Number of sample n=36
To find : What is the probability the average mpg is less than 28 mpg?
Solution :
Applying z-score formula,
The portability is given by,
Using z-table,
Therefore, the probability the average mpg is less than 28 mpg is 0.8413.
Step-by-step explanation:
????⃗ (x,y)=(3x−4y)????⃗ +2x????⃗ F→(x,y)=(3x−4y)i→+2xj→ and ????C is the counter-clockwise oriented sector of a circle centered at the origin with radius 44 and central angle ????/3π/3. Use Green's theorem to calculate the circulation of ????⃗ F→ around ????C.
The circulation of [tex]\( \vec{F} \)[/tex] around [tex]\( C \)[/tex] is[tex]\( 3\pi \)[/tex], computed using Green's theorem by transforming the line integral into a double integral over the enclosed region.
To calculate the circulation of [tex]\( \vec{F} \)[/tex] around C, we first need to compute the line integral of [tex]\( \vec{F} \)[/tex] along C. Using Green's theorem, we can rewrite this line integral as a double integral over the region enclosed by C .
Given [tex]\( \vec{F}(x, y) = (3x - 4y)\vec{i} + 2x\vec{j} \)[/tex], we can compute the partial derivatives [tex]\( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \)[/tex], where[tex]\( P = 3x - 4y \) and \( Q = 2x \).[/tex]
[tex]\[\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 2 - (-4) = 6\][/tex]
Since the region enclosed by C is the counter-clockwise oriented sector of a circle with radius 2 and central angle [tex]\( \frac{\pi}{4} \)[/tex], we can parameterize the curve [tex]\( C \)[/tex] as [tex]\( x(t) = 2\cos(t) \) and \( y(t) = 2\sin(t) \)[/tex] where [tex]\( 0 \leq t \leq \frac{\pi}{4} \).[/tex]
Now, applying Green's theorem, we have:
[tex]\[\text{Circulation} = \iint_D (6) \, dA\][/tex]
Using polar coordinates, the double integral becomes:
[tex]\[\begin{aligned}\text{Circulation} &= \int_{0}^{\frac{\pi}{4}} \int_{0}^{2} (6) \cdot r \, dr \, dt \\&= \int_{0}^{\frac{\pi}{4}} \left[3r^2\right]_{0}^{2} \, dt \\&= \int_{0}^{\frac{\pi}{4}} 3(2)^2 \, dt \\&= \int_{0}^{\frac{\pi}{4}} 12 \, dt \\&= [12t]_{0}^{\frac{\pi}{4}} \\&= 12\left(\frac{\pi}{4}\right) - 12(0) \\&= 3\pi\end{aligned}\][/tex]
Therefore, the circulation of [tex]\( \vec{F} \)[/tex] around [tex]\( C \)[/tex] is [tex]\( 3\pi \).[/tex]
The question probable maybe:
Given in the attachment
(a) A random sample of 10 houses in a particular area, each of which is heated with natural gas, is selected and the amount of gas (therms) used during the month of January is determined for each house. The resulting observations are 153, 103, 125, 149, 118, 109, 86, 122, 138, 99. Let μ denote the average gas usage during January by all houses in this area. Compute a point estimate of μ.therms(b) Suppose there are 25,000 houses in this area that use natural gas for heating. Let τ denote the total amount of gas used by all of these houses during January. Estimate τ using the data of part (a).therms(c) Use the data in part (a) to estimate p, the proportion of all houses that used at least 100 therms.(d) Give a point estimate of the population median usage (the middle value in the population of all houses) based on the sample of part (a).therms
Answer:
a) [tex]\hat \mu = \bar x =\sum_{i=1}^{10} \frac{x_i}{10}=120.2[/tex]
b) [tex]\hat \tau =n\hat \mu =25000x120.2=3005000[/tex]
c) [tex]\hat p= \hat \theta =\frac{8}{10}=0.8[/tex]
d) Median =120
Step-by-step explanation:
1) Some important concepts
The mean refers to the "average that is used to derive the central tendency of data analyzed. It is determined by adding all the data points in a population and then dividing the total by the number of points".
Method of moments "involves equating sample moments with theoretical moments". For example the first sample moment about the origin is defined as [tex]M_1=\frac{1}{n} \sum_{i=1}^{n}x_i =\bar X [/tex]
The median is "the middle number in a sorted, ascending or descending, list of numbers and can be more descriptive of that data set than the average".
When we are trying to estimate the population proportion, p.
All estimation is based on the fact that the normal can be used to approximate the binomial distribution when np and nq are both at least 5. Where p is the probability of success and q the probability of failure.
2) Part a
Using the method of the moments a point of estimate for the [tex]\mu[/tex] is:
[tex]\hat \mu = \bar x =\sum_{i=1}^{10} \frac{x_i}{10}=120.2[/tex]
3) Part b
If [tex]\hat \mu[/tex] is an individual estimate for the average gas usage during January and [tex]\tau[/tex] represent "the total amount of gas used by all of these houses during January" then the estimation for the total would be given by:
[tex]\hat \tau =n\hat \mu =25000x120.2=3005000[/tex]
3) Part c
For this part we want to estimate p ="the proportion of all houses that used at least 100 therms". If X is the random variable who represent the number of houses that exceed the usage of 100, we see that 8 out of 10 values are above 100, so the random variable X would be distributed binomial
[tex]X \sim Bin(10,0.8)[/tex] where n=10 and
[tex]\hat p= \hat \theta =\frac{8}{10}=0.8[/tex]
4) Part d
In order to find the median we need to put the data in order first, like this:
86,99,103,109,118,122,125,138,149,153
Since we have 10 observations and this number is even the procedure that we need to use in order to find the median is:
a) Find the value at position[n/2]=[10/2]=[5] on the data set ordered. For this case the value at position [5] is 118
b) Find the value at position[n/2 +1]=[10/2 +1]=[6] on the data set ordered. For this case the value at position [6] is 122
c) Find the average from the values obtained on steps a) and b). for this case (118+122)/2=120
So the Median = 120
The mean of the data set is 120.2
The total gas used = 3005000
The proportion of at least therms = 0.8
The median of the set = 120
a. To get the average gas usage, we are asked to calculate the mean of the observation.
Average
[tex]\frac{ 153+103+125+149+118+109+86+122+138+99}{10} \\\\[/tex]
= 120.2
b. The question says that 25000 houses make use of natural gas, then
total gases used by these houses =
120.2 * 25000
= 3005000
c. The proportion of the houses that used above 100 therms,
The house above 100 here are, 125, 149, 118, 109, 122, 138
They are 8 in number.
8/10 = 0.8
0.8 is the proportion that used at least 100 therms.
d. We have to find the median for the set here. We arrange the details in ascending order.
86,99,103,109,118,122,125,138,149,153
Median = 118+122/2
= 240/2
= 120
A survey reported in Time magazine included the question ‘‘Do you favor a federal law requiring a 15 day waiting period to purchase a gun?"" Results from a random sample of US citizens showed that 318 of the 520 men who were surveyed supported this proposed law while 379 of the 460 women sampled said ‘‘yes"". Use this information to find a 95% confidence interval for the difference in the two proportions
the 95% confidence interval for the difference in proportions is approximately (0.1562, 0.2686).
To find the 95% confidence interval for the difference in proportions, we can use the formula:
[tex]\[\text{CI} = (\hat{p}_1 - \hat{p}_2) \pm z \times \sqrt{\frac{{\hat{p}_1(1 - \hat{p}_1)}}{n_1} + \frac{{\hat{p}_2(1 - \hat{p}_2)}}{n_2}}\][/tex]
Where:
- [tex]\(\hat{p}_1\) and \(\hat{p}_2\)[/tex] are the sample proportions.
- [tex]\(n_1\) and \(n_2\)[/tex] are the sample sizes.
- z is the z-score corresponding to the desired level of confidence.
Given:
- [tex]\(n_1 = 520\), \(n_2 = 460\)[/tex]
- [tex]\(\hat{p}_1 = \frac{318}{520}\), \(\hat{p}_2 = \frac{379}{460}\)[/tex]
- z = 1.96 for a 95% confidence interval
Let's plug in the values and calculate:
[tex]\[\hat{p}_1 = \frac{318}{520} \approx 0.6115\]\[\hat{p}_2 = \frac{379}{460} \approx 0.8239\]\[\text{CI} = (0.8239 - 0.6115) \pm 1.96 \times \sqrt{\frac{{0.6115 \times (1 - 0.6115)}}{520} + \frac{{0.8239 \times (1 - 0.8239)}}{460}}\]\[\text{CI} = (0.2124) \pm 1.96 \times \sqrt{\frac{{0.6115 \times 0.3885}}{520} + \frac{{0.8239 \times 0.1761}}{460}}\][/tex]
[tex]\[\text{CI} = (0.2124) \pm 1.96 \times \sqrt{0.000457 + 0.000368}\]\[\text{CI} = (0.2124) \pm 1.96 \times \sqrt{0.000825}\]\[\text{CI} = (0.2124) \pm 1.96 \times 0.0287\]\[\text{CI} = (0.2124) \pm 0.0562\][/tex]
Thus, the 95% confidence interval for the difference in proportions is approximately (0.1562, 0.2686).
The weather in a certain locale consists of alternating wet and dry spells. Suppose that the number of days in each rainy spell is a Poisson distribution with mean 2, and that a dry spell follows a geometric distribution with mean 7. Assume that the successive durations of rainy and dry spells are independent. What is the long-run fraction of time that it rains?
Answer:
2/9
Step-by-step explanation:
The Poisson’s distribution is a discrete probability distribution. A discrete probability distribution means that the events occur with a constant mean rate and independently of each other. It is used to signify the chance (probability) of a given number of events occurring in a fixed interval of time or space.
In the long run, fraction of time that it rains = E(Number of days in rainy spell) / {E(Number of days in a rainy spell) + E(Number of days in a dry spell)}
E(Number of days in rainy spell) = 2
E(Number of days in a dry spell) = 7
In the long run, fraction of time that it rains = 2/(2 + 7) = 2/9
Given the parameters of the rainy spell and dry spell, the long-run fraction of time that it rains can be calculated by dividing the mean of the rainy days by the sum of the average rainy and dry days. Hence, it rains roughly 22.22% of the time in the long-term.
Explanation:The question is asking about the long-run fraction of time that it rains, based on a rainy spell following a Poisson distribution with a mean of 2 days, and a dry spell following a geometric distribution with an average of 7 days, with the sequences being independent.
We are being asked to calculate the proportion of time that it rains in the long-run, given these distribution parameters. The Poisson and geometric distributions are often used in this type of probability assessment.
To tackle this, we need to divide the mean of the rainy days by the sum of the average rainy and dry days. Thus, the long-run fraction of time it rains is given by [tex]2/(2+7) = 2/9.[/tex]
So, in the long run, it rains roughly 22.22% (or 2/9) of the time.
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An object moves in simple harmonic motion with period 7 seconds and amplitude 3cm. At time =t0 seconds, its displacement d from rest is 0cm, and initially it moves in a negative direction. Give the equation modeling the displacement d as a function of time t.
The equation modeling the displacement d as a function of time t for an object in simple harmonic motion with a period of 7 seconds and amplitude of 3cm is d = -3cos(2pi/7 * t + pi).
Explanation:The displacement d of an object moving in simple harmonic motion can be modeled by the equation d=Acos(wt).
Given that the period T is 7 seconds, we can use the formula T=2pi/w to solve for the angular frequency w. Rearranging the equation, we have w = 2pi/T. Plugging in the given period T=7, we get w = 2pi/7.
Since the object initially moves in a negative direction, we would have a phase shift of pi in the cosine function. Therefore, the equation modeling the displacement d as a function of time t is d = -3cos(2pi/7 * t + pi).
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Consider the following binomial experiment: A study in a certain community showed that 5% of the people suffer from insomnia. If there are 10,400 people in this community, what is the standard deviation of the number of people who suffer from insomnia?
Answer:
22.23
Step-by-step explanation:
Given that in a binomial experiment study, in a certain community showed that 5% of the people suffer from insomnia
i.e p = 0.05
q=0.95
n=10400
[tex]np=10400*0.05=202\\npq =494[/tex]
Var(x)=[tex]npq=494\\std dev=\sqrt{494} \\=22.23[/tex]
the standard deviation of the number of people who suffer from insomnia
=22.23
The question is about calculating the standard deviation for the number of people suffering from insomnia in a particular community. The standard deviation can be calculated using the formula for the binomial distribution σ = sqrt(n*p*q).
Explanation:The question asks for determination of the standard deviation of the number of people suffering from insomnia in a community of 10,400 where 5% suffer from insomnia. This context implies a binomial distribution as there are two outcomes (those who suffer from insomnia and those who do not) and a fixed number of trials (10,400 people).
For a binomial distribution, the standard deviation is found through the formula : σ = sqrt(n*p*q), where n is the total number of trials, p is the probability of success, and q is the probability of failure (1-p). Here, n=10,400, p=0.05 and q=0.95.
Here the standard deviation would be calculated as σ = sqrt{10,400 * 0.05 * 0.95}.
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a sample of 546 boys aged 6–11 was weighed, and it was determined that 89 of them were overweight. A sample of 508 girls aged 6–11 was also weighed, and 74 of them were overweight. Can you conclude that the proportion of boys who are overweight differs from the proportion of girls who are overweight? Find the P-value and state a conclusion. Round the answer to four decimal places.
Answer:
p-value: 0.6527
Step-by-step explanation:
Hello!
You have two samples to study, from each sample the weight of each child was measured and counted the total of overweight kids (x: "success") in each group:
Sample 1 (Boys aged 6-11)
n₁= 546
x₁= 89
^p₁= x₁/n₁ = 89/546 ≅0.16
Sample 2 (girls aged 6-11)
n=508
x= 74
^p= x/n = 74/508 ≅ 0.15
If the hypothesis statement is "The proportion of boys that are overweight differs from the proportion of girls that are overweight", the test hypothesis is:
H₀: ρ₁ = ρ₂
H₁: ρ₁ ≠ ρ₂
This type of hypothesis leads to a two-tailed rejection region, then the p-value will also be two-tailed. To calculate the p-value you have to first calculate the value of the statistic under the null hypothesis, in this case, is a test for the difference between two proportions:
Z= (^ρ₁ - ^ρ₂) - (ρ₁ - ρ₂) ≈ N(0;1)
√(ρ` * (1 - ρ`) * (1/n₁ + 1/n₂))
ρ`= x₁ + x₂ = 89+74 = 0.154 ≅ 0.15
n₁ + n₂ 546 + 508
Z⁰ᵇ = (0.16-0.15) - (0)
√(0.15 * (1 - 0.15) * (1/546 + 1/508))
Z⁰ᵇ = 0.45
I've mentioned before that in this test you have a two-tailed p-value. The value calculated (0.45) corresponds to the right or positive tail and the left tail is symmetrical to it concerning the distribution mean, in this case, is 0, so it is -0.45. To obtain the p-value you need to calculate the probability of both values and add them:
P(Z>0.45) + P(Z<-0.45) = (1- P(Z<0.45)) + P(Z<-0.45) = (1-0.67364) + 0.32636 = 0.65272 ≅ 0.6527
p-value: 0.6527
Since there is no signification level in the problem, I'll use the most common to reach a decision. α: 0.05
Since the p-value is greater than α, you do not reject the null Hypothesis, in other words, there is no significative difference between the proportion of overweight boys and the proportion of overweight girls.
I hope it helps!
A running shoe company wants to sponsor the fastest 5% of runners. You know that in this race, the running times are normally distributed with a mean of 7.2 minutes and a standard deviation of 0.56 minutes.
How fast would you need to run to be sponsored by the company?
a) 6.3 minutes
b) 6.1 minutes
c) 8.3 minutes
d) 8.1 minutes
Answer:
a) 6.3 minutes
Step-by-step explanation:
Population mean (μ) = 7.2 minutes
Standard deviation (σ) = 0.56 minutes
The z-score for any running time 'X' is given by:
[tex]z=\frac{X-\mu}{\sigma}[/tex]
In this scenario, the company is looking for the top 5% runners, that is, runners at and below the 5-th percentile of the normal distribution. The equivalent z-score for the 5-th percentile is 1.645.
Therefore, the minimum speed, X, a runner needs to achieve in order to be sponsored is:
[tex]-1.645=\frac{X-7.2}{0.56}\\X= 6.3\ minutes[/tex]
The marginal cost of drilling an oil well depends on the depth at which you are drilling; drilling becomes more expensive, per meter, as you dig deeper into the earth. The fixed costs are 1,000,000 riyals (the riyal is the unit of currency of Saudi Arabia), and, if x is the depth in meters, the marginal costs are C' (x) = 4000 + 10x (Riyals/meter).Find the total cost of drilling a 500-meter well.
To calculate the total cost of drilling a 500-meter well, add the fixed cost to the sum of the marginal costs for every meter drilled. The total cost comes out to 3,500,000 riyals.
Explanation:The total cost of drilling a 500-meter well comprises both fixed costs and the marginal cost per meter of depth. We are given that the fixed costs amount to 1,000,000 riyals. The marginal cost function is given as C'(x) = 4000 + 10x, which means the cost per additional meter drilled increases linearly with depth.
To find the total cost of drilling a 500-meter well, we need to compute the integral (i.e., the area under the curve) of the marginal cost function from 0 to 500 and add the fixed costs. This calculation represents the sum of the increasing cost per meter for every meter drilled.
By evaluating the integral ∫ (4000 + 10x) dx from 0 to 500, we get 2,500,000 riyals. This is the total variable cost of drilling a 500m well. Adding it to the fixed cost (1,000,000 riyals), the grand total comes out to be 3,500,000 riyals.
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9. A judge hears the following arguments in a murder hearing. The DNA test that places the accused at the murder scene has a true positive rate of 90% (i.e. the probability that the test returning positive given that the accused was actually present at the scene is 0.9). Similarly, the DNA test has a false negative rate of 80% (i.e. the probability that the test returns negative given that the accused was not present at the scene is 0.8). Everyone in the town has a equal probability of being at the murder scene, and the town has a population of 10,000. Given the fact that the DNA test returned a positive result for the accused, what is the probability that the accused was at the murder scene?
Answer:
0.0004498
Step-by-step explanation:
Let us define the events:
A = The test returns positive.
B = The accused was present.
Since everyone in the town has an equal probability of being at the murder scene, and the town has a population of 10,000
P(B) = 1/10000 = 0.0001
We have that the probability the test returning positive given that the accused was actually present at the scene is 0.9
P(A | B) = 0.9
and the probability that the test returns negative given that the accused was not present at the scene is 0.8
[tex]\large P(A^c|B^c)=0.8[/tex]
where
[tex]\large A^c,\;B^c[/tex] are the complements of A and B respectively.
We want to determine the probability that the DNA test returned a positive result given that the accused was at the murder scene, that is, P(B | A).
We know that P(A | B) = 0.9, so
[tex]\large \frac{P(A\cap B)}{P(B)}=0.9\Rightarrow P(A\cap B)=0.9P(B)=0.9*0.0001\Rightarrow\\\\P(A\cap B)=0.00009[/tex]
Now, we have
[tex]\large P(B|A)=\frac{P(A\cap B)}{P(A)}=\frac{0.00009}{P(A)}[/tex]
So if we can determine P(A), the result will follow.
By De Morgan's Law
[tex]\large A^c\cap B^c=(A\cup B)^c[/tex]
so
[tex]\large 0.8=P(A^c|B^c)= \frac{P(A^c\cap B^c)}{P(B^c)}=\frac{P((A\cup B)^c)}{P(B^c)}=\frac{1-P(A\cup B)}{1-P(B)}\Rightarrow\\\\\frac{1-P(A\cup B)}{1-0.0001}=0.8\Rightarrow P(A\cup B)=1-0.8(1-0.0001)\Rightarrow\\\\P(A\cup B)=0.20008[/tex]
Using the formula
[tex]\large P(A\cup B)=P(A)+P(B)-P(A\cap B)[/tex]
and replacing the values we have found
[tex]\large 0.20008=P(A)+0.0001-0.00009\Rightarrow\\\\P(A)=0.20007[/tex]
and finally, the desired result is
[tex]\large P(B|A)=\frac{0.00009}{P(A)}=\frac{0.00009}{0.20007}\Rightarrow\\\\\boxed{P(B|A)=0.0004498}[/tex]
A graphics designer is designing an advertising brochure for an art show. Each page of the brochure is rectangular with an area of 42 insquared and a perimeter of 26 in. Find the dimensions of the brochure.
The longer side is __ in.
The shorter side is __ in.
(Type exact answers, using radicals as needed. Simplify your answers.)
Answer: The length of the loner side is 7 in. and the length of the shorter side is 6 in.
Step-by-step explanation: Given that a graphics designer is designing an advertising brochure for an art show. Each page of the brochure is rectangular with an area of 42 in squared and a perimeter of 26 in.
We are to find the dimensions of the brochure.
Let l and b represents the lengths of the longer side and shorter side respectively of each page of the brochure.
Then, according to the given information, we have
[tex]l\times b=42~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)[/tex]
and
[tex]2(l+b)=26\\\\\Rightarrow l+b=13\\\\\Rightarrow l=13-b~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)[/tex]
Substituting the value of l from equation (ii) in equation (i), we get
[tex](13-b)b=42\\\\\Rightarrow b^2-13b+42=0\\\\\Rightarrow b^2-6b-7b+42=0\\\\\Rightarrow (b-6)(b-7)=0\\\\\Rightarrow b-6=0,~~~b-7=0\\\\\Rightarrow b=6,7.[/tex]
Since b is the length of the shorter side, so b = 6 in.
From equation (ii), we get
[tex]l=13-6=7.[/tex]
Thus, the length of the loner side is 7 in. and the length of the shorter side is 6 in.
To find the dimensions of the brochure, we use the formulas for the area and perimeter of a rectangle. Solving the system of equations produced by these formulas, we find that the length of the brochure is 7 inches and the width is 6 inches.
Explanation:In order to find the dimensions of the brochure, we can use the formulas for the area and perimeter of a rectangle. Given area, A = 42 inches squared and perimeter, P = 26 inches. The formulas for the area and perimeter of the rectangle are A = length x width and P = 2(length + width).
Let's denote the length of the rectangle as 'l' and the width as 'w'. Now we know that:
l x w = 42 inches (according to area formula)
2(l + w) = 26 inches (according to perimeter formula)
This is a system of two equations which can be solved simultaneously. After solving these equations, we find that the length (longer side) is 7 inches, and the width (shorter side) is 6 inches.
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The probability that a university graduate will be offered no jobs within a month of graduation is estimated to be 10%. The probability of receiving one, two, and three job offers has similarly been estimated to be 43%, 34%, and 13%, respectively. Determine the following probabilities: A. P (A graduate is offered fewer than two jobs) B. P (A graduate is offered more than one job)
Answer:
a) P(A graduate is offered fewer than two jobs) = 0.53.
b) P(A graduate is offered more than one job) = 0.47.
Step-by-step explanation:
Let X be a random variable denoting the number of jobs offers that a university graduate gets within a month of graduation.
The probability that a university graduate will be offered no jobs within a month of graduation is estimated to be 10% i.e. [tex]P(X=0)=0.10[/tex]
The probability of receiving one job offers has similarly been estimated to be 43% i.e. [tex]P(X=1)=0.43[/tex]
The probability of receiving two job offers has similarly been estimated to be 34% i.e. [tex]P(X=2)=0.34[/tex]
The probability of receiving three job offers has similarly been estimated to be 13% i.e. [tex]P(X=3)=0.13[/tex]
a) P (A graduate is offered fewer than two jobs) i.e. P(X<2)
So, [tex]P(X<2)=P(X=0)+P(X=1)[/tex]
[tex]P(X<2)=0.10+0.43[/tex]
[tex]P(X<2)=0.53[/tex]
P(A graduate is offered fewer than two jobs) = 0.53.
b) P (A graduate is offered more than one job) i.e. P(X>1)
So, [tex]P(X>1)=P(X=2)+P(X=3)[/tex]
[tex]P(X>1)=0.34+0.13[/tex]
[tex]P(X>1)=0.47[/tex]
P(A graduate is offered more than one job) = 0.47.
The probability that a graduate is offered fewer than two jobs is 53%, and the probability that a graduate is offered more than one job is 47%.
Explanation:In the scenario given, you want to find two probabilities: A) the probability that a graduate is offered fewer than two jobs, and B) the probability that a graduate is offered more than one job. The total probability should add up to 100%, or a probability of 1.
For part A), 'fewer than two jobs' could mean either no job offers or one job offer. We know from the information given that the probability of no job offer is 10% (or 0.10) and the possibility of one job offer is 43% (or 0.43). So you add these two probabilities together: 0.10 + 0.43 = 0.53. Therefore, the probability that a graduate is offered fewer than two jobs is 53%.
For part B), 'more than one job' could mean either two job offers or three job offers. From the information given, we can find that the probability of receiving two job offers is 34% (or 0.34) and the probability for three job offers is 13% (or 0.13). Adding these two probabilities gives 0.34 + 0.13 = 0.47. Hence, the probability that a graduate is offered more than one job is 47%.
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Which of the following random variables is not discrete?
A) The number of classes taken in one semester by a student
B) The annual rainfall in a city
C) The attendance at a football game
D) The number of patients treated at an emergency room in a day
Answer:
b) The annual rainfall in a city
Step-by-step explanation:
Remember, a discrete variable is one that can only take a finite number of values between any two values of a characteristic and a continuous variable is one that can take an infinite number of values between any two values of a characteristic.
a) Observe that the variable x='classes taken in one semester' can take the values 0,1,2,...,n.
Then the variable x is discrete
b) Observe that the variable x='annual rainfall in a city' can take the values 2in, 1.6in, 5.1 in, 0.1in
Then, the variable x can be take a infinite number of values between two number. So x isn't a discrete variable.
c) The variable x='attendance at a football game' can take the values 3000,5000... n. And never will be a decimal number because There cannot be a personal decimal number. Therefore, x is a discrete variable.
d) The variable x='patients treated at an emergency room in a day' can take the values 1,2,3,...,n. And never will be a decimal number because There cannot be a personal decimal number. Therefore, x is a discrete variable.
An elevator can safely hold 3,500 lbs. A sign in the elevator limits the passenger count to 15. If the adult population has a mean weight of 180 lbs with a 25 lbs standard deviation, how unusual would it be, if the central limit theorem applied, that an elevator holding 15 people would be carrying more than 3,500 pounds? (Hint: if X is a random variable indicating a person’s weight, then assume X Normal( = 180; 2 = 252); use related d, p, q, and r functions to get the numerical answer.)
Answer:
[tex]1.75*10^{-27}[/tex]
Step-by-step explanation:
If, collectively, 15 people weigh more than 3500 pounds, that means each person must weigh more than 3500/15 = 233.33 pounds.
If the distribution for population weights is normal at mean = 180 and standard deviation = 25 lbs, that means the probability for 1 person to weigh higher than 233 lbs is
[tex]1 - P(x > 233, \mu = 180, \sigma = 25) = 1 - 0.984 = 0.016[/tex]
For all 15 people to have higher weigh than that then the probability is
[tex]0.016^{15} = 1.75*10^{-27}[/tex]
This is indeed very unlikely to happen
Final answer:
It would be highly unusual for 15 people in an elevator to have a combined weight of more than 3,500 lbs; the z-score of 8.27 reflects an extremely small probability, pointing to a rare event.
Explanation:
To determine how unusual it would be for 15 people in an elevator to have a combined weight of more than 3,500 lbs, we use the Central Limit Theorem. Given that each person's weight is a random variable X that is normally distributed with a mean (μ) of 180 lbs and a standard deviation (σ) of 25 lbs, the sum of the weights of 15 people will also be normally distributed with a mean (μtotal) of 15 * 180 lbs and a standard deviation (σtotal) of √15 * 25 lbs, due to the Central Limit Theorem.
The next step is to calculate these values:
μtotal = 15 * 180 = 2700 lbsσtotal = √15 * 25 ≈ 96.82 lbsNow we calculate the z-score to determine how many standard deviations away 3,500 lbs is from the mean:
Z = (X - μtotal) / σtotal
Z = (3500 - 2700) / 96.82 ≈ 8.27
Using standard normal distribution tables or a calculator, we find that the probability of a z-score of 8.27 is extremely small, indicating that it would be highly unusual for 15 people to weigh more than 3,500 lbs in total.
In a random sample of 13 microwave ovens, the mean repair cost was $85.00 and the standard deviation was $15.30. Using the standard normal distribution with the appropriate calculations for a standard deviation that is known, assume the population is normally distributed, find the margin of error and construct a 95% confidence interval for the population mean. A 95% confidence interval using the t-distribution was (75.8, 94.2 ). Compare the results.
Answer: Margin of error = 8.32, and Confidence interval using normal distribution is narrower than confidence interval using t-distribution.
Step-by-step explanation:
Since we have given that
n = 13
Mean repair cost = $85.00
Standard deviation = $15.30
At 95% confidence interval,
z= 1.96
Since it is normally distributed.
Margin of error is given by
[tex]z\times \dfrac{\sigma}{\sqrt{n}}\\\\=1.96\times \dfrac{15.30}{\sqrt{13}}\\\\=8.32[/tex]
95% confidence interval would be
[tex]\bar{x}\pm z\dfrac{\sigma}{\sqrt{n}}\\\\=85\pm 1.96\times \dfrac{15.30}{\sqrt{13}}\\\\=85\pm 8.32\\\\=(85-8.32,85+8.32)\\\\=(76.68,93.32)[/tex]
A 95% confidence interval using the t-distribution was (75.8, 94.2 ).
Confidence interval using normal distribution is narrower than confidence interval using t-distribution.
Final answer:
A 95% confidence interval for the mean repair cost of microwave ovens with a known standard deviation is calculated using the z-score. The margin of error is found to be approximately $8.31, resulting in a confidence interval of ($76.69, $93.31).
Explanation:
To calculate the 95% confidence interval for the population mean when the population standard deviation is known, we can use the z-score associated with the 95% confidence level, which is 1.96. The formula for the margin of error (EBM) is EBM = z * (σ/√n), where σ is the population standard deviation, n is the sample size, and z is the z-score. Given that the sample standard deviation is $15.30, we assume it to be the population standard deviation because the question states that it is known.
With a sample mean (μ) of $85.00, a standard deviation of $15.30, and a sample size of 13, the margin of error is calculated as follows:
EBM = 1.96 * (15.30/√13) = 1.96 * 4.24 ≈ $8.31
The 95% confidence interval is therefore ($85.00 - $8.31, $85.00 + $8.31) = ($76.69, $93.31). The results using the z-distribution are similar to those obtained using the t-distribution, but usually, the t-distribution would be used when the sample size is small and the population standard deviation is unknown, which results in a wider interval due to the extra uncertainty.
Air containing 0.04% carbon dioxide is pumped into a room whose volume is 6000 ft3. The air is pumped in at a rate of 2000 ft3/min, and the circulated air is then pumped out at the same rate. If there is an initial concentration of 0.4% carbon dioxide, determine the subsequent amount in the room at any time.What is the concentration at 10 minutes?
Answer:
the concentration at 10 minutes= 0.4+0.0133= 0.4133%
Step-by-step explanation:
Air containing 0.04% carbon dioxide
V, volume of room is 6000 ft3.
Q, rate of air 2000 ft3/min,
initial concentration of 0.4% carbon dioxide,
determine the subsequent amount in the room at any time.
What is the concentration at 10 minutes?
firstly, we find the time taken for air to completely filled the room
Q = V/t
t = V/Q = 6000/2000 = 3min
so, its take 3mins for air to be completely filled in the room and for exhaust air to move out.
there is an initial concentration of 0.4% carbon dioxide, and the air pump in is 0.04%.
therefore,
3mins = 0.04% of CO2
3*60 =180sec = 0.04%
1sec = 0.04/180 = 0.00022%/sec
so at any time the concentration of CO2 is 0.4 + 0.00022 =0.40022%/sec
What is the concentration at 10 minute
the concentration at 10minutes = the concentration for 1minute because at every minutes, the concentration moves in is moves out. = concentration for 2000ft3.
for 0.04% = 6000ft3
? = 2000ft3
= 2000* 0.04)/6000 =0.0133%
the concentration at 10 minutes= 0.4+0.0133= 0.4133%
A trucking firm suspects that the mean life of a certain tire. it uses is less than 33,000 miles. To check the claim, the firm randomly selects and tests 18 of these tires in gets a mean lifetime of 32, 450 miles with a standard deviation of 1200 miles. At α = 0.05, test the trucking firms claim.
a. State Hypothesis and Identify Claim.
b. Identify level of significance.
c. Choose correct probability distribution, locate critical values.identify rejection region.
d. Calculate test statistic.
e. Make decision
f. Write conclusion.
SHOW ALL YOUR WORK
Answer:
We accept the alternate hypothesis. We conclude that the mean lifetime of tires is is less than 33,000 miles.
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = 33,000 miles
Sample mean, [tex]\bar{x}[/tex] = 32, 450 miles
Sample size, n = 18
Alpha, α = 0.05
Sample standard deviation, s = 1200 miles
a) First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu = 33000\text{ miles}\\H_A: \mu < 33000\text{ miles}[/tex]
b) Level of significance:
[tex]\alpha = 0.05[/tex]
c) We use One-tailed t test to perform this hypothesis.
d) 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{32450 - 33000}{\frac{1200}{\sqrt{18}} } = -1.9445[/tex]
Now, [tex]t_{critical} \text{ at 0.05 level of significance, 17 degree of freedom } = -1.7396[/tex]
Rejection area:
[tex]t < -1.7396[/tex]
Since,
[tex]t_{stat} < t_{critical}[/tex]
e) We fail to accept the null hypothesis and reject it as the calculated value of t lies in the rejection area.
f) We accept the alternate hypothesis. We conclude that the mean lifetime of tires is is less than 33,000 miles.
If 4 items are chosen at random without replacement from 7 items, in how many ways can the 4 items be arranged, treating each arrangement as a different event (i.e., if order is important)?
A. 35
B. 840
C. 5040
D. 24
Answer:
840
Step-by-step explanation:
A manufacturer of tires wants to advertise a mileage interval that ex-cludes no more than 10% of the mileage on tires he sells. All he knowsis that, for a large number of tires tested, the mean mileage was 25,000miles, and the standard deviation was 4000 miles. What interval wouldyou suggest?
The suggested mileage interval, excluding no more than 10% of the mileage on tires, is approximately 18,420 to 31,580 miles.
To determine the mileage interval that excludes no more than 10% of the mileage on tires, we can use the standard normal distribution and the properties of the normal curve. The mileage data is normally distributed with a mean [tex](\(\mu\))[/tex] of 25,000 miles and a standard deviation [tex](\(\sigma\))[/tex] of 4,000 miles.
To find the interval, we can use the Z-score formula:
[tex]\[ Z = \frac{X - \mu}{\sigma} \][/tex]
To exclude no more than 10% of the mileage, we need to find the Z-score corresponding to the 5th percentile on each side of the mean, as 10% is split between the lower and upper tails of the distribution.
Using a standard normal distribution table or a calculator, the Z-score for the 5th percentile is approximately -1.645 (negative due to being in the lower tail).
Now, we can use the Z-score formula to find the values of [tex]\(X\)[/tex] (mileage) corresponding to these Z-scores:
[tex]\[ X_{\text{lower}} = \mu + Z \times \sigma \][/tex]
[tex]\[ X_{\text{upper}} = \mu - Z \times \sigma \][/tex]
Substitute the values:
[tex]\[ X_{\text{lower}} = 25,000 - (-1.645) \times 4,000 \][/tex]
[tex]\[ X_{\text{upper}} = 25,000 + (-1.645) \times 4,000 \][/tex]
Calculating these values:
[tex]\[ X_{\text{lower}} \approx 31,580 \][/tex]
[tex]\[ X_{\text{upper}} \approx 18,420 \][/tex]
Therefore, the suggested mileage interval is approximately 18,420 miles to 31,580 miles to exclude no more than 10% of the mileage on the tires.
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:
False
Step-by-step explanation:
Confidence intervals provide a range for a population parameter at a given significance level. The parameter can be mean, standard deviation etc.
In this example population is the prices of the rents of all the unfurnished one-bedroom apartments in the Boston area
significance level is 95%. Thus, the chance being the true population parameter in the given interval is 95%.
But, "This interval describes the price of 95% of the rents of all the unfurnished one-bedroom apartments in the Boston area." statement is false because the population parameter is missing. Confidence interval may describe population mean for example but it does not describe the whole population.
Read the severe weather warning and answer the question. Warning Below freezing temperatures will be accompanied by strong winds. Heavy snowfall is expected to last for several days. What type of severe weather is being described? Blizzard Drought Hurricane Tornado
Answer:
Hi! The answer is A, Blizzard.
Recently did this test on FLVS
Hope this helped!
Have a terrific Tuesday!!
~Lola
Answer:
Tornado
Step-by-step explanation:
Its not blizzard or hurricane.
The probability of a successful optical alignment in the assembly of an optical data storage product is p = 0.6. Assume the trials are independent. Round your answers to four decimal places (e.g. 98.7654). (a) What is the probability that the 1st successful alignment requires exactly 4 trials? (b) What is the probability that the 1st successful alignment requires at most 4 trials? (c) What is the probability that the 1st successful alignment requires at least 4 trials?
Answer:
a) [tex]P(X=4)=(1-0.6)^{4-1} 0.6 = 0.0384[/tex]
b) [tex]P(X\leq 4)=0.6+0.24+0.096+0.0384=0.9744[/tex]
c) [tex]P(X\geq 4)=1-P(X<4)=1-P(X\leq 3)=1-[0.6+0.24+0.096]=0.064[/tex]
Step-by-step explanation:
The geometric distribution represents "the number of failures before you get a success in a series of Bernoulli trials. This discrete probability distribution is represented by the probability density function:"
[tex]P(X=x)=(1-p)^{x-1} p[/tex]
Let X the random variable that measures the number os trials until the first success, we know that X follows this distribution:
[tex]X\sim Geo (1-p)[/tex]
Part a
For this case we want this probability
[tex]P(X=4)=(1-0.6)^{4-1} 0.6 = 0.0384[/tex]
Part b
For this case we want this probability:
[tex]P(X\leq 4)=P(X=1)+P(X=2)+P(X=3)+P(X=4)[/tex]
If we find the individual probabilities we got:
[tex]P(X=1)=(1-0.6)^{1-1} 0.6 = 0.6[/tex]
[tex]P(X=2)=(1-0.6)^{2-1} 0.6 = 0.24[/tex]
[tex]P(X=3)=(1-0.6)^{3-1} 0.6 = 0.096[/tex]
[tex]P(X=4)=(1-0.6)^{4-1} 0.6 = 0.0384[/tex]
And replacing we have:
[tex]P(X\leq 4)=0.6+0.24+0.096+0.0384=0.9744[/tex]
Part c
For this case at least 4 trials means that the random variable X needs to be 4 or more
[tex]P(X\geq 4)=1-P(X<4)=1-P(X\leq 3)=1-[P(X=1)+P(X=2)+P(X=3)][/tex]
And we found already the probabilities P(X=1),P(X=2) and P(X=3) so we just need to replace:
[tex]P(X\geq 4)=1-P(X<4)=1-P(X\leq 3)=1-[0.6+0.24+0.096]=0.064[/tex]
The three scenarios are calculated using a combination of geometric and binomial probability distribution. For the first success to happen on the 4th trial is 0.0384. For a success to happen within the first four trials is 0.8847, and for the first success to need at least four trials is 0.3600.
Explanation:This question has to do with the concept of probability distribution in math, and specifically with the geometric distribution and binomial probability distribution. Given the probability of a successful optical alignment p = 0.6, we're being asked about different scenarios involving success on specific trials.
(a) For the first successful alignment to happen on the 4th trial, the first 3 trials have to be failures and the 4th one, a success. The probability would be (0.4)^3 * 0.6 = 0.0384.
(b) This means we want a successful alignment on the 1st, 2nd, 3rd or 4th trials. For this we sum up the probabilities of each case: (0.6) + (0.4*0.6) + (0.4)^2*0.6 + (0.4)^3*0.6 = 0.8847.
(c) For a successful alignment to occur in at least 4 trials, we have to subtract the probability of success within the first 3 trials from 1. Therefore, 1 - ((0.6) + (0.4*0.6) + (0.4)^2*0.6) = 0.3600.
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A random sample of n1 = 49 measurements from a population with population standard deviation σ1 = 3 had a sample mean of x1 = 12. An independent random sample of n2 = 64 measurements from a second population with population standard deviation σ2 = 4 had a sample mean of x2 = 14. Test the claim that the population means are different. Use level of significance 0.01.What distribution does the sample test statistic follow? Explain.
Answer:
We reject the null hypothesis that the population means are equal and accept the alternative hypothesis that the population means are different.
Step-by-step explanation:
We have large sample sizes [tex]n_{1} = 49[/tex] and [tex]n_{2} = 64[/tex], the unbiased point estimate for [tex]\mu_{1}-\mu_{2}[/tex] is [tex]\bar{x}_{1} - \bar{x}_{2}[/tex], i.e., 12-14 = -2.
The standard error is given by [tex]\sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}[/tex], i.e.,
[tex]\sqrt{\frac{(3)^{2}}{49}+\frac{(4)^{2}}{64}}[/tex] = 0.6585.
We want to test [tex]H_{0}: \mu_{1}-\mu_{2} = 0[/tex] vs [tex]H_{1}: \mu_{1}-\mu_{2} \neq 0[/tex] (two-tailed alternative). The rejection region is given by RR = {z | z < -2.5758 or z > 2.5758} where -2.5758 and 2.5758 are the 0.5th and 99.5th quantiles of the standard normal distribution respectively. The test statistic is [tex]Z = \frac{\bar{x}_{1} - \bar{x}_{2}-0}{\sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}}[/tex] and the observed value is [tex]z_{0} = \frac{-2}{0.6585} = -3.0372[/tex]. Because -3.0372 fall inside RR, we reject the null hypothesis.
The test statistic follow a standard normal distribution because we are dealing with large sample sizes.
In this scenario of comparing two independent samples and given that the sample sizes are large, the sample test statistic follows the Standard Normal distribution or Z-distribution. The Z-test statistic representing the difference in sample means (in units of standard error) is compared with critical values for a two-tailed test at 0.01 significance level to determine if there's sufficient evidence to reject the null hypothesis that the two population means are equal.
Explanation:The test in your question pertains to a hypothesis testing scenario featuring two independent samples. This scenario typically involves two population means given that population standard deviations are known. The distribution followed by the sample test statistic in such cases is the Standard Normal distribution or Z-distribution, as the sample sizes (n1 = 49, n2 = 64) are sufficiently large. To test the claim that population means are different (at a significance level of 0.01), you'd typically construct a Z-test statistic that represents the difference in sample means (x1 - x2) in units of its standard error. The Z-test statistic is calculated as follows:
[tex]Z = \frac{x_1 - x_2}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}}[/tex]Here, x1 and x2 are the sample means, σ1 and σ2 are the population standard deviations and n1 and n2 are the samples sizes. The resulting Z-score can be compared with critical Z-scores for a two-tailed test at the given level of significance (0.01) to determine whether or not the null hypothesis (two population means are equal) can be rejected.
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The average exam score of students of a large class is 70 with a standard deviation of 10. A sample of 36 students is selected, and the mean score of these students is computed. The sampling distribution of the sample mean has approximately a normal distribution because of
(A) the 68.3-95.4-99.7 rule.
(B) the law of large number.
(C) the central limit theorem.
Answer:
The sampling distribution of the sample mean has approximately a normal distribution because of
c) the central limit theorem.
Step-by-step explanation:
Given that the average exam score of students of a large class is 70 with a standard deviation of 10.
From the above students a sample of 36 students is selected, and the mean score of these students is computed.
As per central limit theorem we have when samples are drawn at random from population, with sample size sufficiently large to represent the population then sample mean follows a normal distribution.
Here population size N = 70 and sample size n =36
we can say sample size is greater than 30 and sufficiently large to represent the population. Also we can assume that these are randomly drawn.
So the answer would be
The sampling distribution of the sample mean has approximately a normal distribution because of
c) the central limit theorem.
The 3 × 3 matrix P satisfies the matrix equation P^2 = P.
(a) What are the possibilities for the determinant of P?
(b) Explain why there are no other possibilities.
(c) For each possible determinant, give an example of P with that determinant.
Answer: The answers are given below.
Step-by-step explanation: Given that a 3 × 3 matrix P satisfies the matrix equation P² = P.
We are to
(a) find the possibilities for the determinant of P.
(b) explain the reason behind there are no other possibilities.
(c) give an example of P, for each possible determinant.
(a) According to the given information, we have
[tex]P^2=P\\\\\Rightarrow P^2-P=0\\\\\Rightarrow P(P-I)=0\\\\\Rightarrow P=0,~~~P=I.[/tex]
So, P can be either a zero matrix of order 3 or an identity matrix of order 3.
If P = 0, then det(P) = 0 and if P = I, then det(P) = 1.
Therefore the possible determinants of P are 0 and 1.
(b) There can be any other determinant other than 0 and 1, because if so, then the given equation P² = P will not be satisfied.
(c) If |P| = 0, then the matrix P can be can be of the form as follows:
[tex]P=\left[\begin{array}{ccc}0&0&0\\0&0&0\\0&0&0\end{array}\right] .[/tex]
If |P| = 1, the the matrix P can be of the form as follows :
[tex]P=\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right] .[/tex]
Thus, all the parts are answered.
1. A researcher is interested in knowing about the number of hours UCF students sleep per night. In a survey of 400 UCF students, the average number of hours slept per night was 6.5 with a population standard deviation of 2 hours. a. Create a 95% confidence interval for the true number of hours slept by UCF students?
Answer: (6.304, 6.696)
Step-by-step explanation:
The confidence interval for population mean is given by :-
[tex]\overline{x}\pm z^*\dfrac{\sigma}{\sqrt{n}}[/tex]
, where [tex]\sigma[/tex] = Population standard deviation.
n= sample size
[tex]\overline{x}[/tex] = Sample mean
z* = Critical z-value .
Let x denotes the number of hours slept by UCF students.
Given : [tex]\sigma=2\ hours[/tex]
n= 400
[tex]\overline{x}= 6.5\ hours[/tex]
Two-tailed critical value for 95% confidence interval = [tex]z^*=1.96[/tex]
Then, the 95%confidence interval for the true number of hours slept by UCF students will be :-
[tex]6.5\pm(1.96)\dfrac{2}{\sqrt{400}}\\\\=6.5\pm(1.96)\dfrac{2}{20}\\\\=6.5\pm0.196=(6.5-0.196,\ 6.5+0.196)=(6.304,\ 6.696)[/tex]
Hence, the 95% confidence interval for the true number of hours slept by UCF students : (6.304, 6.696)
A machine that paints traffic stripes on roads is mounted on a truck and set to a width of 4 inches. Road crews adjust the mount to ensure the width is correct. A road inspector checks the width of 45 random stripes to see if the machine has slipped out of adjustment. The mean diameter for this sample is x = 3.87 inches with a standard deviation of s = 0.5 inches. Does this indicate that the machine has slipped out of adjustment and the average width of stripes is no longer μ = 4 inches? Use a 5% level of significance.
Let [tex]\mu[/tex] denotes the average width of stripes .
As per given , we have
[tex]H_0:\mu=4\\H_a:\mu\neq4[/tex]
, since [tex]H_a[/tex] is two-tailed , so the test is a two-tailed test.
Also, population standard deviation is unknown , so we perform two-tailed t-test.
For Sample size : n= 45
Sample mean : [tex]\overline{x}=3.87[/tex]
Sample standard deviation : s= 0.5 inches
Test statistic : [tex]t=\dfrac{\overline{x}-\mu}{\dfrac{s}{\sqrt{n}}}[/tex]
i.e. [tex]t=\dfrac{3.87-4}{\dfrac{0.5}{\sqrt{45}}}\approx-1.74[/tex]
Significance level = [tex]\alpha=0.05[/tex]
By using t-value table,
Two-tailed critical t-value = [tex]t_{\alpha/2,df}=t_{0.025,\ 44}=\pm2.0154[/tex] [df = n-1]
Decision : Since the test statistic value (-1.74) lies with in the interval (-2.0154, 2.0154) , it means we are failed to reject the null hypothesis .
Conclusion: We have sufficient evidence to support the claim that the machine has slipped out of adjustment and the average width of stripes is no longer μ = 4 inches.
After calculating the z-score for the provided data, the result (-1.962) lies within the critical values for a 5% level of significance. Therefore, we cannot reject the null hypothesis, hence there's no sufficient evidence to state that the machine is out of adjustment.
Explanation:To determine if the machine has slipped out of adjustment, we should conduct a hypothesis test. We can set the null hypothesis (H0) as μ = 4 (the machine is correctly adjusted), and the alternative hypothesis (Ha) as μ ≠ 4 (the machine has slipped). The sample size is large enough (>30) to use the z-score.
The z-score can be calculated using the formula: z = (x - μ)/(s/√n), where x is the sample mean, μ is the population mean, s is the standard deviation, and n is the sample size.
So, z = (3.87 - 4) / (0.5/√45) = -1.962. The critical z-value for a 5% level of significance (two-tailed test) is approximately +/- 1.96. Our calculated z-value falls within this range so we cannot reject the null hypothesis. Therefore, we don't have sufficient evidence to state that the machine has slipped out of adjustment.
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in a study of red/green color blindness, 600 men and 2150 women are randomly selected and tested. Among the men, 56 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. (Note: Type ‘‘p_m′′ for the symbol pm , for example p_mnot
Answer:
z=13.36 (Statistic)
[tex]p_v =P(Z>13.36)\approx 0[/tex]
The p value is a very low value and using any significance level for example [tex]\alpha=0.05, 0,1,0.15[/tex] always [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can say the the proportion of men with red/green color blindness is significant higher than the proportion of female with red/green color blindness .
Step-by-step explanation:
1) Data given and notation
[tex]X_{MCB}=56[/tex] represent the number of men with red/green color blindness
[tex]X_{WCB}=5[/tex] represent the number of women with red/green color blindness
[tex]n_{MCB}=600[/tex] sample of male selected
[tex]n_{WCB}=600[/tex] sample of demale selected
[tex]p_{MCB}=\frac{56}{600}=0.093[/tex] represent the proportion of men with red/green color blindness
[tex]p_{WCB}=\frac{5}{2150}=0.0023[/tex] represent the proportion of women with red/green color blindness
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the value for the test (variable of interest)
2) Concepts and formulas to use
We need to conduct a hypothesis in order to check if the proportion for men with red/green color blindness is a higher than the rate for women , the system of hypothesis would be:
Null hypothesis:[tex]p_{MCB} \leq p_{WCB}[/tex]
Alternative hypothesis:[tex]p_{MCB} > \mu_{WCB}[/tex]
We need to apply a z test to compare proportions, and the statistic is given by:
[tex]t=\frac{p_{MCB}-p_{WCB}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{MCB}}+\frac{1}{n_{WCB}})}}[/tex] (1)
Where [tex]\hat p=\frac{X_{MCB}+X_{WCB}}{n_{MCB}+n_{WCB}}=\frac{56+5}{600+2150}=0.0221[/tex]
t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.
3) Calculate the statistic
Replacing in formula (1) the values obtained we got this:
[tex]z=\frac{0.093-0.0023}{\sqrt{0.0221(1-0.0221)(\frac{1}{600}+\frac{1}{2150})}}=13.36[/tex]
4) Statistical decision
For this case we don't have a significance level provided [tex]\alpha[/tex], but we can calculate the p value for this test.
Since is a one side test the p value would be:
[tex]p_v =P(Z>13.36)\approx 0[/tex]
So the p value is a very low value and using any significance level for example [tex]\alpha=0.05, 0,1,0.15[/tex] always [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can say the the proportion of men with red/green color blindness is significant higher than the proportion of female with red/green color blindness .
To test the claim that men have a higher rate of red/green color blindness, we need to compare the proportions of color blindness in men and women using a chi-square test of independence.
Explanation:To test the claim that men have a higher rate of red/green color blindness, we need to compare the proportions of color blindness in men and women.
Let p_m be the proportion of men with color blindness and p_w be the proportion of women with color blindness.
Null hypothesis: p_m = p_w
Alternative hypothesis: p_m > p_w
To test this hypothesis, we can use a chi-square test of independence. We will compare the observed frequencies of color blindness in men and women to the expected frequencies under the assumption that men and women have the same rate of color blindness.
The chi-square test statistic is calculated as follows:
X^2 = (O_m - E_m)^2/E_m + (O_w - E_w)^2/E_w
where O_m and O_w are the observed frequencies of color blindness in men and women, and E_m and E_w are the expected frequencies of color blindness in men and women.
If the chi-square test statistic is large enough, we reject the null hypothesis and conclude that men have a higher rate of color blindness than women.
(1 point) The matrix A=⎡⎣⎢−4−4−40−8−4084⎤⎦⎥A=[−400−4−88−4−44] has two real eigenvalues, one of multiplicity 11 and one of multiplicity 22. Find the eigenvalues and a basis of each eigenspace. λ1λ1 = equation editorEquation Editor has multiplicity 11, with a basis of equation editorEquation Editor . λ2λ2 = equation editorEquation Editor has multiplicity 22, with a basis of equation editorEquation Editor .
Answer:
We have the matrix [tex]A=\left[\begin{array}{ccc}-4&-4&-4\\0&-8&-4\\0&8&4\end{array}\right][/tex]
To find the eigenvalues of A we need find the zeros of the polynomial characteristic [tex]p(\lambda)=det(A-\lambda I_3)[/tex]
Then
[tex]p(\lambda)=det(\left[\begin{array}{ccc}-4-\lambda&-4&-4\\0&-8-\lambda&-4\\0&8&4-\lambda\end{array}\right] )\\=(-4-\lambda)det(\left[\begin{array}{cc}-8-\lambda&-4\\8&4-\lambda\end{array}\right] )\\=(-4-\lambda)((-8-\lambda)(4-\lambda)+32)\\=-\lambda^3-8\lambda^2-16\lambda[/tex]
Now, we fin the zeros of [tex]p(\lambda)[/tex].
[tex]p(\lambda)=-\lambda^3-8\lambda^2-16\lambda=0\\\lambda(-\lambda^2-8\lambda-16)=0\\\lambda_{1}=0\; o \; \lambda_{2,3}=\frac{8\pm\sqrt{8^2-4(-1)(-16)}}{-2}=\frac{8}{-2}=-4[/tex]
Then, the eigenvalues of A are [tex]\lambda_{1}=0[/tex] of multiplicity 1 and [tex]\lambda{2}=-4[/tex] of multiplicity 2.
Let's find the eigenspaces of A. For [tex]\lambda_{1}=0[/tex]: [tex]E_0 = Null(A- 0I_3)=Null(A)[/tex].Then, we use row operations to find the echelon form of the matrix
[tex]A=\left[\begin{array}{ccc}-4&-4&-4\\0&-8&-4\\0&8&4\end{array}\right]\rightarrow\left[\begin{array}{ccc}-4&-4&-4\\0&-8&-4\\0&0&0\end{array}\right][/tex]
We use backward substitution and we obtain
1.
[tex]-8y-4z=0\\y=\frac{-1}{2}z[/tex]
2.
[tex]-4x-4y-4z=0\\-4x-4(\frac{-1}{2}z)-4z=0\\x=\frac{-1}{2}z[/tex]
Therefore,
[tex]E_0=\{(x,y,z): (x,y,z)=(-\frac{1}{2}t,-\frac{1}{2}t,t)\}=gen((-\frac{1}{2},-\frac{1}{2},1))[/tex]
For [tex]\lambda_{2}=-4[/tex]: [tex]E_{-4} = Null(A- (-4)I_3)=Null(A+4I_3)[/tex].Then, we use row operations to find the echelon form of the matrix
[tex]A+4I_3=\left[\begin{array}{ccc}0&-4&-4\\0&-4&-4\\0&8&8\end{array}\right] \rightarrow\left[\begin{array}{ccc}0&-4&-4\\0&0&0\\0&0&0\end{array}\right][/tex]
We use backward substitution and we obtain
1.
[tex]-4y-4z=0\\y=-z[/tex]
Then,
[tex]E_{-4}=\{(x,y,z): (x,y,z)=(x,z,z)\}=gen((1,0,0),(0,1,1))[/tex]