Answer:
[tex]P(X<4)=P(\frac{X-\mu}{\sigma}<\frac{4-\mu}{\sigma})=0.2[/tex]
[tex]P(z<\frac{4-\mu}{\sigma})=0.2[/tex]
But we know which value of z satisfy the previous equation so then we can do this:
[tex]z=-0.842<\frac{4-\mu}{1.1}[/tex]
And if we solve for the mean we got
[tex]\mu =4 +0.842*1.1=4.926[/tex]
Step-by-step explanation:
Assuming this question "Suppose that the lifetimes of TV tubes are normally distributed with a standard deviation of 1.1 years. Suppose also that exactly 20% of the tubes die before 4 years. Find the mean lifetime of TV tubes. Carry your intermediate computations to at least four decimal places. Round your answer to at least one decimal place. ?
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Solution to the problem
Let X the random variable that represent the lifetimes of TV tubes of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(\mu,1.1)[/tex]
Where [tex]\mu[/tex] and [tex]\sigma=1.1[/tex]
And the best way to solve this problem is using the normal standard distribution and the z score given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
For this part we know the following condition:
[tex]P(X>4)=0.8[/tex] (a)
[tex]P(X<4)=0.2[/tex] (b)
Both conditions are equivalent on this case. We can use the z score again in order to find the value a.
As we can see on the figure attached the z value that satisfy the condition with 0.2 of the area on the left and 0.8 of the area on the right it's z=-0.842. On this case P(Z<-0.842)=0.2 and P(z>-0.842)=0.8
If we use condition (b) from previous we have this:
[tex]P(X<4)=P(\frac{X-\mu}{\sigma}<\frac{4-\mu}{\sigma})=0.2[/tex]
[tex]P(z<\frac{4-\mu}{\sigma})=0.2[/tex]
But we know which value of z satisfy the previous equation so then we can do this:
[tex]z=-0.842<\frac{4-\mu}{1.1}[/tex]
And if we solve for the mean we got
[tex]\mu =4 +0.842*1.1=4.926[/tex]
Mean lifetime is 4.9 years. Found using z-score of -0.8416 with 20% dying before 4 years, [tex]\(\sigma = 1.1\)[/tex].
To find the mean lifetime of TV tubes, we will use the properties of the normal distribution and the given information. Here's the step-by-step solution:
1. Identify the given values:
- Standard deviation [tex](\(\sigma\))[/tex]: 1.1 years
- Percentage of tubes that die before 4 years: 20% (or 0.20)
2. Find the z-score corresponding to the given percentage:
- Since the percentage is 20%, we need to find the z-score for which 20% of the area under the normal curve lies to the left.
- Using a standard normal distribution table or a calculator, the z-score for 0.20 is approximately -0.8416.
3. Set up the z-score formula:
The z-score formula is given by:
[tex]\[ z = \frac{X - \mu}{\sigma} \][/tex]
where [tex]\(X\)[/tex] is the value (4 years in this case), [tex]\(\mu\)[/tex] is the mean lifetime we want to find, and [tex]\(\sigma\)[/tex] is the standard deviation.
4. Substitute the known values into the z-score formula and solve for [tex]\(\mu\)[/tex]:
[tex]\[ -0.8416 = \frac{4 - \mu}{1.1} \][/tex]
Rearrange to solve for [tex]\(\mu\)[/tex]:
[tex]\[ -0.8416 \times 1.1 = 4 - \mu \][/tex]
[tex]\[ -0.92576 = 4 - \mu \][/tex]
[tex]\[ \mu = 4 + 0.92576 \][/tex]
[tex]\[ \mu \approx 4.9258 \][/tex]
5. Round the mean lifetime to one decimal place:
[tex]\[ \mu \approx 4.9 \][/tex]
So, the mean lifetime of the TV tubes is approximately 4.9 years.
A principal of $4700 was invested at 4.75% interest, compounded annually.
Lett be the number of years since the start of the investment. Let y be the value of the investment, in dollars.
Write an exponential function showing the relationship between y and t.
Answer:
Step-by-step explanation:
We would apply the formula for determining compound interest which is expressed as
y = P(1 + r/n)^nt
Where
y = the value of the investment at the end of t years
r represents the interest rate.
n represents the periodic interval at which it was compounded.
P represents the principal or initial amount invested
From the information given,
P = $4700
r = 4.75% = 4.75/100 = 0.0475
n = 1 because it was compounded once in a year.
Therefore, the exponential function showing the relationship between y and t is
y = 4700(1 + 0.0475/1)^1 × t
y = 4700(1.0475)^t
Elizabeth is returning to the United States from Canada. She changes the remaining 300 Canadian dollars she has to $250 US dollars. What was $1 dollar worth in Canadian dollars?
Answer:
$1 USD = $1.20 Canadian dollars
Step-by-step explanation:
Use proportions of Canadian dollars over US dollars so you get
300 CAD/250 USD = x CAD / 1 USD
solve for x to get 1.2
So $1 USD is equal to $1.20 Canadian dollars.
Answer:
$1.20
Step-by-step explanation:check work by any of these ways.
300/1.20=250,
300/250=1.20,
Or 1.20*250=300
A consumer’s preference relation ≿ is represented by the quasiconcave utility function: �(�/, �1) = �/ ?.D�1 ?.E She has $50 to spend and prices are �/ = $2 and �1 = $4. Compute the (a) compensating variation, (b) equivalent variation, and (c) the change in consumer surplus associated with an increase in the price of good 1 to $4. You can use any mathematical expression derived in lecture or in previous homework to answer this question.
Answer:
Step-by-step explanation:
You can find your answer in attached documents.
The student's question involves calculating compensating variation, equivalent variation, and change in consumer surplus due to a price increase in one of two goods, given a quasilinear utility function. These calculations require the application of microeconomic principles related to utility functions, indifference curves, and consumer choices, but cannot be completed without specific mathematical expressions and functions provided in the coursework.
Explanation:The question relates to a consumer's preference relation and utility maximization in the context of an increase in the price of one of the goods. The compensating variation, equivalent variation, and change in consumer surplus are economic measures used to quantify the impacts of price changes on consumer welfare. These concepts involve understanding indifference curves, marginal rates of substitution, and consumer budget constraints.
Computing the compensating and equivalent variations requires setting up and solving expenditure function problems before and after the price change, whereas the change in consumer surplus can be represented graphically and calculated as the area under the demand curve and above the price level. Unfortunately, without specific details or mathematical expressions for utility maximization and budget constraints, which are typically provided in academic coursework, it is not possible to provide explicit calculations for these measures.
Understanding the utility function's form and how it translates into demand can help derive these economic measures. For instance, the quasilinear utility function is notable for its constant elasticity of demand and specific responses to income and price changes which can be used to predict consumer behavior and welfare changes.
A standard medium pizza has a diameter of 12 inches and is cut into 8 slices.
What is the area and what is the one sector area of a slice of pizza.
Please show work. I've been stuck on this for a while and want to understand it.
Answer:
The answer to your question is Circle's area = 113.04 in² Slice' area = 14.13 in²
Step-by-step explanation:
Data
diameter = 12 in
8 slices
Area = ?
Area of a slice = ?
Process
1.- Calculate the area of the pizza
radius = 12/2
radius = 6 in
Area = πr²
-Substitution
Area = (3.14)(6)²
-Simplification
Area = 113.04 in²
2.- Calculate the area of the slice
We can calculate this area by two methods
a) Divide the area of the number of slices
Area of the slice = 113.04 / 8
= 14.13 in²
b) Using the area of a sector
-Find the angle of each slice
360 / 8 = 45°
-Convert 45° to rad
180° ------------------- πrad
45° ------------------- x
x = 45πrad/180
x = 1/4πrad
- Calculate the area of the sector
Area = 1/4(3.14)(6)²/2
Area = 113.04/8
Area = 14.13 in²
Answer:
Step-by-step explanation:
Assuming the pizza is circular, we would apply the formula for determining the area of a circle. It is expressed as
Area = πr²
Where
r represents the radius of the circle.
π is a constant whose value is 3.14
From the information given,
Diameter = 12 inches
Radius = 12/2 = 6 inches
Area = 3.14 × 6² = 113.04 inches²
The formula for determining the area of a sector is expressed as
Area of sector = θ/360 × πr²
Where θ represents the central angle.
The complete circle or pizza is 360°. Since it is divided into 8 equal parts, then the central angle formed by each sector is
360/8 = 45°
Therefore,
Area of sector = 45/360 × 3.14 × 6²
= 14.13 inches²
Erythromycin is a drug that has been proposed to possibly lower the risk of premature delivery. A related area of interest is its association with the incidence of side effects during pregnancy. Assume that 30% of all pregnant women complain of nausea between the 24th and 28th week of pregnancy. Furthermore, suppose that of 195 women who are taking erythromycin regularly during this period, 65 complain of nausea. Find the p-value for testing the hypothesis that incidence rate of nausea for the erythromycin group is greater than for a typical pregnant woman.
Answer:
[tex]z=\frac{0.333 -0.3}{\sqrt{\frac{0.3(1-0.3)}{195}}}=1.01[/tex]
[tex]p_v =P(z>1.01)=0.156[/tex]
So the p value obtained was a very low value and using the significance level asumed [tex]\alpha=0.05[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIl to reject the null hypothesis, and we can said that at 5% of significance the proportion of women who complain of nausea between the 24th and 28th week of pregnancy is not significantly higher than 0.3 or 30%
Step-by-step explanation:
Data given and notation
n=195 represent the random sample taken
X=65 represent the women who complain of nausea between the 24th and 28th week of pregnancy
[tex]\hat p=\frac{65}{195}=0.333[/tex] estimated proportion of women who complain of nausea between the 24th and 28th week of pregnancy
[tex]p_o=0.3[/tex] is the value that we want to test
[tex]\alpha[/tex] represent the significance level
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value (variable of interest)
Concepts and formulas to use
We need to conduct a hypothesis in order to test the claim that true proportion is higher than 0.3.:
Null hypothesis:[tex]p\leq 0.3[/tex]
Alternative hypothesis:[tex]p > 0.3[/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].
Calculate the statistic
Since we have all the info requires we can replace in formula (1) like this:
[tex]z=\frac{0.333 -0.3}{\sqrt{\frac{0.3(1-0.3)}{195}}}=1.01[/tex]
Statistical decision
It's important to refresh the p value method or p value approach . "This method 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 assumed is [tex]\alpha=0.05[/tex]. The next step would be calculate the p value for this test.
Since is a right tailed test the p value would be:
[tex]p_v =P(z>1.01)=0.156[/tex]
So the p value obtained was a very low value and using the significance level asumed [tex]\alpha=0.05[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIl to reject the null hypothesis, and we can said that at 5% of significance the proportion of women who complain of nausea between the 24th and 28th week of pregnancy is not significantly higher than 0.3 or 30%
To find the p-value for testing the hypothesis that the incidence rate of nausea for the erythromycin group is greater than for a typical pregnant woman, we can use the binomial probability formula.
Explanation:To find the p-value for testing the hypothesis that the incidence rate of nausea for the erythromycin group is greater than for a typical pregnant woman, we can use the binomial probability formula.
The formula is:
P(X ≥ k) = 1 - P(X < k)
where X is the number of women in the erythromycin group who complain of nausea, and k is the number of complaints of nausea in the control group. In this case, X = 65 and k = 0, because the question specifies that there are no complaints of nausea in a typical pregnant woman. By plugging these values into the formula, we can calculate the p-value.
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The department in which you work in your company has 24 employees: 10 women and 14 men. A team of 4 employees must be selected to represent the department at a companywide meeting in the headquarters of the company. The team must have two women and two men. How many different teams of 4 can be selected
Answer:
We conclude that 4095 different teams can be selected.
Step-by-step explanation:
We know that a company has 24 employees: 10 women and 14 men.
A team of 4 employees must be selected. The team must have two women and two men.
So from 10 women we choose 2 women:
[tex]C_2^{10}=\frac{10!}{2!(10-2)!}=45\\[/tex]
So from 14 men we choose 2 men:
[tex]C_2^{14}=\frac{14!}{2!(14-2)!}=91\\[/tex]
We get: 45 · 91 = 4095.
We conclude that 4095 different teams can be selected.
Final answer:
The total number of different teams that can be selected from the department to represent at the companywide meeting is 4095.
Explanation:
To select a team of 4 employees with 2 women and 2 men from a department with 10 women and 14 men, we can use the concept of combinations. We need to choose 2 women out of 10 and 2 men out of 14.
The number of ways to choose 2 women from 10 is given by 10C2 = 10! / (2! * (10-2)!) = 45.
The number of ways to choose 2 men from 14 is given by 14C2 = 14! / (2! * (14-2)!) = 91.
Therefore, the total number of different teams that can be selected is 45 * 91 = 4095.
A market researcher selects 500 drivers under 30 years of age and 500 drivers over 30 years of age. Identify the type of sampling used in this example.
Answer:
The type of sampling is stratified.
Step-by-step explanation:
Samples types are classified as:
Random: Basically, put all the options into a hat and drawn some of them.
Systematic: Every kth element is taken. For example, you want to survey something on the street, you interview every 5th person, for example.
Cluster: Divides population into groups, called clusters, and each element in the cluster is surveyed.
Stratified: Also divides the population into groups. However, then only some elements of the group are surveyed.
In this problem, we have that:
The drivers are divided into two groups according to their ages.
There are thousands and thousand of drivers both under and over 30 years of age, and 500 for each group(some elements of the group) are selected.
So the type of sampling is stratified.
The type of sampling used in the example is stratified sampling. The researcher divided the total population of drivers into two groups (under 30 and over 30), and selected 500 drivers from each group.
Explanation:In this example, the type of sampling used by the market researcher is called stratified sampling. In stratified sampling, the population is divided into different subgroups, or 'strata', and a sample is taken from each stratum. In this case, the researcher has divided the population of drivers into two strata based on their age: drivers under 30 and drivers over 30. 500 drivers are then selected from each of these strata.
Stratified sampling is a common method used in market research because it allows for a more accurate representation of the population, as each subgroup is adequately represented in the sample.
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The gypsy moth is a serious threat to oak and aspen trees. A state agriculture department places traps throughout the state to detect the moths. When traps are checked periodically, the mean number of moths per trap is only 1.2, but some traps have several moths. The distribution of moth counts in traps is strongly right skewed, with standard deviation 1.4. A random sample of 60 traps has x = 1 and s = 2.4.
Let X = the number of moths in a randomly selectd trap
(a) For the population distribution, what is the ...
...mean? =
...standard deviation? =
(b) For the distribution of the sample data, what is the ...
...mean? =
...standard deviation? =
(c) What shape does the distribution of the sample data probably have?
Exactly NormalApproximately Normal Right skewedLeft skewed
(d) For the sampling distribution of the sample mean with n = 60, what is the ...
...mean? =
...standard deviation? = (Use 3 decimal places)
(e) What is the shape of the sampling distribution of the sample mean?
Left skewedApproximately Normal Right skewedExactly Normal
(f) If instead of a sample size of 60, suppose the sample size were 10 instead. What is the shape of the sampling distribution of the sample mean for samples of size 10?
Approximately normalSomewhat left skewed Exactly NormalSomewhat right skewed
(g) Can we use the Z table to calculate the probability a randomly selected sample of 10 traps has a sample mean less than 1?
No, because the sampling distribution of the sample mean is somewhat right skewedYes, by the Central Limit Theorem, we know the sampling distribution of the sample mean is normal
Answer:
Step-by-step explanation:
Hello!
X: number of gypsy moths in a randomly selected trap.
This variable is strongly right-skewed. with a standard deviation of 1.4 moths/trap.
The mean number is 1.2 moths/trap, but several have more.
a.
The population is the number of moths found in traps places by the agriculture departments.
The population mean μ= 1.2 moths per trap
The population standard deviation δ= 1.4 moths per trap
b.
There was a random sample of 60 traps,
The sample mean obtained is X[bar]= 1
And the sample standard deviation is S= 2.4
c.
As the text says, this variable is strongly right-skewed, if it is so, then you would expect that the data obtained from the population will also be right-skewed.
d. and e.
Because you have a sample size of 60, you can apply the Central Limit Theorem and approximate the distribution of the sampling mean to normal:
X[bar]≈N(μ;σ²/n)
The mean of the distribution is μ= 1.2
And the standard deviation is σ/√n= 1.4/50= 0.028
f. and g.
Normally the distribution of the sample mean has the same shape of the distribution of the original study variable. If the sample size is large enough, as a rule, a sample of size greater than or equal to 30 is considered sufficient you can apply the theorem and approximate the distribution of the sample mean to normal.
You have a sample size of n=10 so it is most likely that the sample mean will have a right-skewed distribution as the study variable. The sample size is too small to use the Central Limit Theorem, that is why you cannot use the Z table to calculate the asked probability.
I hope it helps!
We are interested in conducting a study in order to determine the percentage of voters in a city who would vote for the incumbent mayor. What is the minimum sample size needed to estimate the population proportion with a margin of error not exceeding 4% at 95% confidence? where p is 50%.
Answer:
[tex]n=\frac{0.5(1-0.5)}{(\frac{0.04}{1.96})^2}=600.25[/tex]
And rounded up we have that n=601
Step-by-step explanation:
Previous concepts
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The population proportion have the following distribution
[tex]p \sim N(p,\sqrt{\frac{p(1-p)}{n}})[/tex]
Solution to the problem
In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2 =0.025[/tex]. And the critical value would be given by:
[tex]z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96[/tex]
The margin of error for the proportion interval is given by this formula:
[tex] ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex] (a)
And on this case we have that [tex]ME =\pm 0.04[/tex] and we are interested in order to find the value of n, if we solve n from equation (a) we got:
[tex]n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}[/tex] (b)
The estimated proportion is [tex] \hat p =0.5[/tex]. And replacing into equation (b) the values from part a we got:
[tex]n=\frac{0.5(1-0.5)}{(\frac{0.04}{1.96})^2}=600.25[/tex]
And rounded up we have that n=601
Suppose that combined verbal and math SAT scores follow a normal distribution with a mean 896 and standard deviation 174. Suppose further that Peter finds out he scored in the top 2.5 percentile (97.5% of students scored below him). Determine how high Peter's score must have been.
Let x* denote Peter's score. Then
P(X > x*) = 0.025
P((X - 896)/174 > (x* - 896)/174) = 0.025
P(Z > z*) = 1 - P(Z < z*) = 0.025
P(Z < z*) = 0.975
where Z follows the standard normal distribution (mean 0 and std dev 1).
Using the inverse CDF, we find
P(Z < z*) = 0.975 ==> z* = 1.96
Then solve for x*:
(x* - 896)/174 = 1.96 ==> x* = 1237.04
so Peter's score is roughly 1237.
The histogram to the right represents the weights (in pounds) of members of a certain high-school math team. How many team members are included in the histogram?
Answer:
Total member = a+b+c+d+e+f+g=5+4+2+0+5+0+4=20 member
Step-by-step explanation:
Let suppose the attached histogram
a) Total members whose weight 110 = 5
b) Total members whose weight 120 = 4
c) Total members whose weight 130 = 2
d) Total members whose weight 140 = 0
e) Total members whose weight 150 = 5
f) Total members whose weight 160 = 0
g) Total members whose weight 170 = 4
Total member = a+b+c+d+e+f+g=5+4+2+0+5+0+4=20 member
A histogram is a statistical graphical representation of data with the help of bars with different heights.
A population has mean 187 and standard deviation 32. If a random sample of 64 observations is selected at random from this population, what is the probability that the sample average will be less than 182
Answer:
11.51% probability that the sample average will be less than 182
Step-by-step explanation:
To solve this question, we have to understand the normal probability distribution and the central limit theorem.
Normal probability distribution:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
Central Limit theorem:
The Central Limit Theorem estabilishes that, for a random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], a large sample size can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex]
In this problem, we have that:
[tex]\mu = 187, \sigma = 32, n = 64, s = \frac{32}{\sqrt{64}} = 4[/tex]
What is the probability that the sample average will be less than 182
This is the pvalue of Z when X = 182. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{182 - 187}{4}[/tex]
[tex]Z = -1.2[/tex]
[tex]Z = -1.2[/tex] has a pvalue of 0.1151
11.51% probability that the sample average will be less than 182
find the horizontal asymptote of the graph of y= -2x^6+5x+8/8x^6+6x+5
a: y= -1/4
b: y=0
c= y=1
D= no horizontal asymptote
NEED HELP ASAP!!!!
Answer:
a) y = -¼
Step-by-step explanation:
(-2x^6+5x+8)/(8x^6+6x+5)
Just divide the leading coefficients
-2/8 = -¼
Horizontal asymptote:
y = -¼
Answer:a) y = -¼
Step-by-step explanation:
(-2x^6+5x+8)/(8x^6+6x+5)
Just divide the leading coefficients
Let X, the number of flaws on the surface of a randomly selected boiler of a certain type, have a Poisson distribution with parameter μ = 5. Use the cumulative Poisson probabilities from the Appendix Tables to compute the following probabilities. (Round your answers to three decimal places.)
(a) P(X ≤ 8)
(b) P(X = 8)
(c) P(9 ≤ X)
(d) P(5 ≤ X ≤ 8)
(e) P(5 < X < 8)
Answer:
[tex]a. \ P(X\leq 8)=0.932\\b. \ P(X=8)=0.065\\c. \ P(9\leq X)=0.068\\d. \ P(5\leq X \leq 8)=0.491\\[/tex]
e. P(5<X<8)=0.251
Step-by-step explanation:
Given that boiler follows a Poisson distribution,$\sim$
[tex]X $\sim$Poi(\mu=5)[/tex]
Poisson Distribution formula is given by the expression:-
[tex]p(x,\mu)=\frac{e^{-\mu}\mu^x}{x!}[/tex]
Our probabilities will be calculated as below:
a.
[tex]P(X=x)=\frac{5^xe^{-5}}{x!}\\P(X\leq 8)=P(X=0)+P(X=1)+...+P(X=8)\\=\frac{5^0e^{-5}}{1!}+...+\frac{5^8e^{-5}}{8!}\\=0.006738+...+0.065278\\=0.961[/tex]
b.
[tex]P(X=8)=\frac{5^8e^{-5}}{8!}\\=0.065[/tex]
c.
[tex]P(9\leq X)=1-P(X=0)-P(X=1)-...-P(X=8)\\=0.068[/tex]
d.
[tex]P(5\leq X\leq 8)=P(X=5)+P(X=6)+P(X=7)+P(X=8)\\=0.491[/tex]
e.P(5<X<8)
[tex]P(X=6)+P(X=7)\\=0.251[/tex]
Petroleum pollution in oceans stimulates the growth of certain bacteria. An assessment of this growth has been madew by counting the bacteria in each of 5 randomly chosen specimens of ocean water (of a fixed size). The 5 counts obtained were as follows.
41, 62, 45, 48, 69
Find the deviation of this sample of numbers. Round your answers to at least two decimal places.
Final answer:
The standard deviation of the sample of numbers is approximately 10.68, after calculating the mean, finding the deviations, squaring them, averaging the squared deviations to get the variance, and then taking the square root of the variance.
Explanation:
The student is asking how to calculate the standard deviation of a sample of numbers. The sample given consists of the counts of bacteria found in ocean water, which are 41, 62, 45, 48, 69. To calculate the standard deviation, follow these steps:
Calculate the mean (average) of the sample.Subtract the mean from each number in the sample to find their deviations.Square each deviation to get the squared deviations.Calculate the average of the squared deviations (this is the variance).Take the square root of the variance to get the standard deviation.Performing the calculations, you get:
Mean (average): (41 + 62 + 45 + 48 + 69) / 5 = 265 / 5 = 53Deviation for each number: -12, 9, -8, -5, 16Squared deviations: 144, 81, 64, 25, 256Variance: (144 + 81 + 64 + 25 + 256) / 5 = 570 / 5 = 114Standard deviation: √114 ≈ 10.68 (rounded to two decimal places)find the measure of exterior angle tuv below .
Given:
m∠T = 61°
m∠S = 5x - 15
m ext∠TUV = 8x - 8
To find:
The measure of exterior angle TUV
Solution:
STU is a triangle.
By exterior angle of triangle property:
The measure of an exterior angle is equal to the sum of the opposite interior angles.
m ext∠TUV = m∠T + m∠S
8x - 8 = 61 + 5x - 15
8x - 8 = 46 + 5x
Add 8 on both sides.
8x - 8 + 8 = 46 + 5x + 8
8x = 54 + 5x
Subtract 5x from both sides.
8x - 5x = 54 + 5x - 5x
3x = 54
Divide by 3 on both sides.
[tex]$\frac{3x}{3} =\frac{54}{3}[/tex]
x = 18
Substitute x = 18 in m ext∠TUV.
m ext∠TUV = 8(18) - 8
= 144 - 8
= 136
Therefore measure of exterior angle ∠TUV is 136°.
Avis Company is a car rental company that is located three miles from the Los Angeles airport (LAX). Avis is dispatching a bus from its offices to the airport every 2 minutes. The average traveling time (round-trip) is 10 minutes.
Answer:
Avis Company is a car rental company that is located three miles from the Los Angeles airport (LAX). Avis is dispatching a bus from its offices to the airport every 2 minutes. The average traveling time (round-trip) is 20 minutes.
a. How many Avis buses are traveling to and from the airport?
b. The branch manager wants to improve the service and suggests dispatching buses every 0.5 minutes. She argues that this will reduce the average traveling time from the airport to Avis offices to 2.5 minutes. Is she correct? If your answer is negative, then what will the average traveling time be?
The answers to the question are
a. 10 buses
b. Yes, increasing the fleet by a factor of 4 will reduce the apparent average travel time from the airport to Avid offices from 10 minutes to 2.5 minutes
Step-by-step explanation:
Number of buses dispatched every 20 minutes= 20÷2=10
Therefore 10 Avid buses are travelling to and from the Los Angeles airport
b. By dispatching every 0.5 minutes we have 20÷0.5 = 40 buses.
Since with 10 buses it takes 10 minutes to make the journey from the airport to Avid offices, then it would take 40 buses 10÷4 = 2.5 minutes time to deliver what it took 10 buses, 10 minutes to deliver.
The Avis Company has 10 buses traveling to and from LAX airport with an average round trip time of 20 minutes and a dispatch frequency of 2 minutes. Increasing the dispatch frequency to every 0.5 minutes does not reduce the average travel time; it remains at 20 minutes as it is dependent on the route's distance and traffic conditions, not the dispatch frequency.
Part A: Number of Avis Buses
To calculate the number of buses traveling to and from the airport, we use the average traveling time for a round trip and the dispatch frequency. With a round trip time of 20 minutes and buses leaving every 2 minutes, we divide the round trip time by the dispatch interval:
Total Buses = Round Trip Time / Dispatch Frequency
Total Buses = 20 minutes / 2 minutes
Total Buses = 10
Therefore, Avis has 10 buses traveling to and from the airport.
Part B: Average Traveling Time with Increased Dispatch Frequency
Increasing the dispatch frequency to every 0.5 minutes does not affect the average traveling time for a round trip. The travel time is determined by the distance between the Avis office and the airport and the traffic conditions, not merely the frequency of bus dispatches. If Avis dispatches buses every 0.5 minutes, there will be more buses in operation, but the average round trip time remains the same at 20 minutes, assuming no changes in traffic conditions or route speeds.
f(x) = x3 − 9x2 − 21x + 8
(a) Find the interval on which f is increasing.
Find the interval on which f is decreasing
(b) Find the local minimum and maximum values of f.
(c) Find the inflection point.
(d)Find the interval on which f is concave up.
(e)Find the interval on which f is concave down.
The function f(x) = x3 - 9x2 - 21x + 8 increases in the intervals (-∞, -1) and (7, ∞), and decreases in the interval (-1, 7). It achieves a local minimum at -190 and a local maximum at -23. The inflection point is at x = 3, with the function being concave up for x < 3 and concave down for x > 3.
Explanation:Given the function f(x) = x3 - 9x2 - 21x + 8, we need to find intervals where the function is increasing or decreasing, and determine its local minima/maxima, inflection points, and where it is concave up or down.
To find the intervals, we need to first take the derivative of f(x): f'(x) = 3x2 - 18x - 21 which becomes 3(x - 7)(x + 1). The critical points are then 7 and -1.
The function increases in the interval (-∞, -1) and decreases in the interval (-1, 7). The function again increases in the interval (7, ∞).
Local minima and maxima can be found by evaluating the function at the critical points. f(-1) = -23 and f(7) = -190, so the local maximum value of f is -23 and local minimum is -190.
To find the inflection point, we compute the second derivative f''(x) = 6x - 18, which simplifies to 6(x - 3), giving us an inflection point at x = 3.
The function is concave up for x < 3 and concave down for x > 3.
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a) f(x) is increasing for x in (- ∞, - 1) cup(7, infty)
f(x) is decreasing for x \in (- 1, 7)
b)Local minimum=-237 Local maximum=19
c)Point of inflection: (x, y) = (3, - 109)
d)Interval for concave upward curve: x \in (3, ∞)
e)Interval for concave upward curve: x \in (- ∞, 3)
The given function is [tex]f(x) = x ^ 3 - 9x ^ 2 - 21x + 8[/tex]
[tex]f' * (x) = 3x ^ 2 - 18x - 21[/tex]
f^ prime prime (x) = 6x - 18
a)f(x) is increasing for
f' * (x) > 0
[tex]3x ^ 2 - 18x - 21 > 0[/tex]
[tex]x ^ 2 - 6x - 7 > 0[/tex]
[tex]x ^ 2 - 7x + x - 7 > 0[/tex]
x(x - 7) + 1(x - 7) > 0
(x - 7)(x + 1) > 0
x > 7 and x > - 1or x < 7 and x < - 1
f(x) is decreasing for
f' * (x) < 0
3x ^ 2 - 18x - 21 < 0
x ^ 2 - 6x - 7 < 0
x ^ 2 - 7x + x - 7 < 0
x(x - 7) + 1(x - 7) < 0
(x - 7)(x + 1) < 0
x > 7 and x < - 1 or x < 7 and x > - 1
x=(-1,7)
b)for local extrema
f' * (x) = 0
3x ^ 2 - 18x - 21 = 0
x ^ 2 - 6x - 7 < 0
x ^ 2 - 7x + x - 7 = 0
x(x - 7) + 1(x - 7) = 0
(x - 7)(x + 1) = 0
x = 7 and x = - 1
At x=-1:
f(- 1) = (- 1) ^ 3 - 9 * (- 1) ^ 2 - 21(- 1) + 8 = 19
At x=7:
f(7) = (7) ^ 3 - 9 * (7) ^ 2 - 21(7) + 8 = - 237
Local minimum=-237
Local maximum=19
c) For point of inflection:
f^ prime prime (x) = 0
Rightarrow 6x - 18 = 0
Rightarrow x = 3
y = f(3) = (3) ^ 3 - 9 * (3) ^ 2 - 21(3) + 8 = - 109
(x, y) = (3, - 109)
d)Interval for concave upward curve:
f^ prime prime (x) > 0
6x - 18 > 0
x \in (3, ∞)
e)Interval for concave upward curve:
f^ prime prime (x) < 0
6x - 18 < 0
X€(−0,3)
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A corporate Web site contains errors on 50 of 1000 pages. If 100 pages are sampled randomly, without replacement, approximate the probability that at least 2 of the pages in error are in the sample. (Use normal approximation to the binomial distribution.)
To approximate the probability, we can use the normal approximation to the binomial distribution. We need to calculate the mean and standard deviation, and then use the normal distribution to find the cumulative probability for each number of errors. Finally, we add up the probabilities to get the probability of at least 2 errors.
Explanation:To approximate the probability, we can use the normal approximation to the binomial distribution. We know that there are 50 pages with errors out of 1000, so the probability of success is p = 50/1000 = 0.05. The probability of failure is q = 1 - p = 1 - 0.05 = 0.95.
The sample size is 100 pages. To find the probability that at least 2 of the pages in error are in the sample, we need to find the probability of getting 2 errors, 3 errors, 4 errors, and so on, up to 100 errors.
We can use the normal approximation to the binomial distribution by calculating the mean and standard deviation. The mean is given by np = 100 * 0.05 = 5, and the standard deviation is given by sqrt(npq) = sqrt(100 * 0.05 * 0.95) ≈ 2.179.
Next, we can use the normal distribution to find the probability. We need to calculate the z-scores for each number of errors, and then find the cumulative probability from the z-score table or using a calculator. Finally, we add up the probabilities for each number of errors to get the probability of at least 2 errors.
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Each group will submit a PowerPoint Presentation based on four different conflicts you have encountered. These conflicts can be work related or personal conflicts. The presentation will consist of 5 slides from each group member and must have at least 1 academic reference for each slide. Neither textbooks nor Wikipedia can be used as references. The cover slide and reference slide do not constitute part of the five slides per group member. The presentation will follow APA format in a number 12 font, and will be due midnight Friday of week 8.
In this question, we cannot provide most of the answer, as it requires the use of Powerpoint and the personal experiences of each participant. However, we are able to talk about some of the personal conflicts and work conflicts that students might have faced in their lives. Some examples that you could use in your text are:
Having a different perspective from your parents.Fighting with a friend over the best way to spend time together.Fighting with your partner about emotional matters.Believing that your boss is unfair in his/her requests to you.Believing that your professors are unfair in their assessment of you.A contractor is required by a county planning department to submit one, two, three, four, five, or six forms (depending on the nature of the project) in applying for a building permit. Let Y = the number of forms required of the next applicant. The probability that y forms are required is known to be proportional to y—that is, p(y) = ky for y = 1, , 6. (Enter your answers as fractions.) (a) What is the value of k? [Hint: 6 y = 1 p(y) = 1] k = 1/21 (b) What is the probability that at most four forms are required? 10/21 (c) What is the probability that between two and five forms (inclusive) are required?
Answer:
a)
[tex]k = \dfrac{1}{21}[/tex]
b) 0.476
c) 0.667
Step-by-step explanation:
We are given the following in the question:
Y = the number of forms required of the next applicant.
Y: 1, 2, 3, 4, 5, 6
The probability is given by:
[tex]P(y) = ky[/tex]
a) Property of discrete probability distribution:
[tex]\displaystyle\sum P(y_i) = 1\\\\\Rightarrow k(1+2+3+4+5+6) = 1\\\\\Rightarrow k(21) = 1\\\\\Rightarrow k = \dfrac{1}{21}[/tex]
b) at most four forms are required
[tex]P(y \leq 4) = \displaystyle\sum^{y=4}_{y=1}P(y_i)\\\\P(y \leq 4) = \dfrac{1}{21}(1+2+3+4) = \dfrac{10}{21} = 0.476[/tex]
c) probability that between two and five forms (inclusive) are required
[tex]P(2\leq y \leq 5) = \displaystyle\sum^{y=5}_{y=2}P(y_i)\\\\P(2\leq y \leq 5) = \dfrac{1}{21}(2+3+4+5) = \dfrac{14}{21} = 0.667[/tex]
The constant of proportionality (k) equals 1/21. The probability that at most four forms are required is 10/21. The probability that between two and five forms (inclusive) are required is 14/21 or 2/3 when simplified.
Explanation:The student has provided information regarding a probability distribution which indicates that the probability (p(y)) is proportional to the number of forms required (y), with y ranging from 1 to 6. Since we know probabilities must sum to 1, this allows us to find the constant of proportionality (k).
Part (a): Finding the Value of k
We can use the formula sum of all probabilities equals 1: 1 = k * (1 + 2 + 3 + 4 + 5 + 6). The sum of the numbers from 1 to 6 is 21, therefore 1 = 21k. Solving for k gives us k = 1/21.
Part (b): Probability of At Most Four Forms Required
To find the probability of at most four forms being required, we need to sum the probabilities for 1, 2, 3 and 4 forms: P(x ≤ 4) = k + 2k + 3k + 4k. Substituting the value of k, we get P(x ≤ 4) = (1/21)(1 + 2 + 3 + 4) = 10/21.
Part (c): Probability Between Two and Five Forms Required (Inclusive)
We calculate this by adding the probabilities for 2, 3, 4, and 5 forms: P(2 ≤ x ≤ 5) = 2k + 3k + 4k + 5k. With k = 1/21, this probability is P(2 ≤ x ≤ 5) = (1/21)(2 + 3 + 4 + 5) = 14/21 or 2/3 when simplified.
A Type I error is:
A. incorrectly specifying the null hypothesis.
B. rejecting the null hypothesis when it is true.
C. incorrectly specifying the alternative hypothesis.
D. accepting the null hypothesis when it is false.
Answer:
Type I error, also known as a “false positive” is the error of rejecting a null hypothesis when it is actually true. Can be interpreted as the error of no reject an alternative hypothesis when the results can be attributed not to the reality.
Type II error, also known as a "false negative" is the error of not rejecting a null hypothesis when the alternative hypothesis is the true. Can be interpreted as the error of failing to accept an alternative hypothesis when we don't have enough statistical power.
Solution to the problem
Based on the defnitions above we can conclude that the best answer for this case is:
B. rejecting the null hypothesis when it is true.
Step-by-step explanation:
Previous concepts
A hypothesis is defined as "a speculation or theory based on insufficient evidence that lends itself to further testing and experimentation. With further testing, a hypothesis can usually be proven true or false".
The null hypothesis is defined as "a hypothesis that says there is no statistical significance between the two variables in the hypothesis. It is the hypothesis that the researcher is trying to disprove".
The alternative hypothesis is "just the inverse, or opposite, of the null hypothesis. It is the hypothesis that researcher is trying to prove".
Type I error, also known as a “false positive” is the error of rejecting a null hypothesis when it is actually true. Can be interpreted as the error of no reject an alternative hypothesis when the results can be attributed not to the reality.
Type II error, also known as a "false negative" is the error of not rejecting a null hypothesis when the alternative hypothesis is the true. Can be interpreted as the error of failing to accept an alternative hypothesis when we don't have enough statistical power.
Solution to the problem
Based on the defnitions above we can conclude that the best answer for this case is:
B. rejecting the null hypothesis when it is true.
Football Strategies A particular football team is known to run 30% of its plays to the left and 70% to the right. A linebacker on an opposing team notes that the right guard shifts his stance most of the time (80%) when plays go to the right and that he uses a balanced stance the remainder of the time. When plays go to the left, the guard takes a balanced stance 90% of the time and the shift stance the remaining 10%. On a particular play, the linebacker notes that the guard takes a balanced stance.
a. What is the probability that the play will go to the left?
b. What is the probability that the play will go to the right?
c. If you were the linebacker, which direction would you prepare to defend if you saw the balanced stance?
Answer:
a) 0.659
b) 0.341
c) It'll be smarter to prepare to defend the left hand side as the probability of play going through that side when the right guard takes a balanced stance is almost double of the probability that the play would go through the right when the right guard takes a balanced stance.
Step-by-step explanation:
Considering the left first
Play goes through the left 30% of the time
- The right guard takes a balanced stance 90% of the time when play goes through the left, that is, 0.9 × 0.3 = 27% overall.
- He takes a shift stance 10% of the time that play goes through the left = 0.1 × 0.3 = 3% overall.
Play goes through the right, 70% of the time.
- He takes a balanced stance 20% of the time that play goes through the right, that is, 0.7 × 0.2 = 14% overall
- He takes a shift stance 80% of the time that play goes through the right, that is, 0.7 × 0.8 = 56% overall.
Then the right guard takes a balanced stance 27% + 14% of all the time = 41% overall.
a) Probability that play goes through the left with the right guard in a balanced stance = 27/41 = 0.659
b) Probability that play goes through the right with the right guard in a balanced stance = 14/41 = 0.341
c) It'll be smarter to prepare to defend the left hand side as the probability of play going through that side when the right guard takes a balanced stance is almost double of the probability that the play would go through the right when the right guard takes a balanced stance.
Using conditional probability, we can find the probabilities of the play going left or right. The probability of the play going left is 0.3, and the probability of the play going right is 0.7. If the linebacker sees a balanced stance, the probability of the play going left is approximately 0.658, so the linebacker should prepare to defend against a play going left.
Explanation:To find the probabilities of the play going left or right, we need to use conditional probability. Let's denote L as the event of the play going left, and R as the event of the play going right. We are given P(L) = 0.3 and P(R) = 0.7. We also have the following conditional probabilities: P(B|L) = 0.9 (balanced stance when play goes left) P(S|L) = 0.1 (shift stance when play goes left) P(B|R) = 0.2 (balanced stance when play goes right) P(S|R) = 0.8 (shift stance when play goes right)
a. To find the probability that the play will go left, we can use the Law of Total Probability. P(L) = P(L∩B) + P(L∩S) = P(L)P(B|L) + P(L)P(S|L) = 0.3 X 0.9 + 0.3 * 0.1 = 0.27 + 0.03 = 0.3.
b. Similarly, to find the probability that the play will go right, P(R) = P(R∩B) + P(R∩S) = P(R)P(B|R) + P(R)P(S|R) = 0.7 X 0.2 + 0.7 X 0.8 = 0.14 + 0.56 = 0.7.
c. If the linebacker sees a balanced stance, it means that P(B) = P(B|L)P(L) + P(B|R)P(R) = 0.9 X 0.3 + 0.2 X 0.7 = 0.27 + 0.14 = 0.41. Now, we can compare the conditional probability of the play going left given a balanced stance, P(L|B) = P(B|L)P(L)/P(B) = (0.9 X 0.3)/0.41 = 0.27/0.41 ≈ 0.658. Therefore, the linebacker should prepare to defend against a play going to the left.
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Seventy-five (75) customers at Crummy Burger today were entered in a drawing. Three names will be drawn, and each will win a prize. Determine the number of ways that the prizes can be distributed if first prize is a $25 gift card, second prize is a free Crummy Combo Meal, and third prize is a free Jumbo Crummy Burger.
Answer:
405150
Step-by-step explanation:
3 people are to be selected from 75 people. This could be done using the combination function, [tex]\binom{75}{3}[/tex]. But thus function only gives the number of ways of choosing any 3 out of the 75 customers. For any single selection of any 3 customers, there are 3! ways of sharing the three prizes among them. Hence the total number of ways of sharing the prizes is
[tex]\binom{75}{3}\times3! = 405150[/tex]
In fact, this is the permutation function given by
[tex]{}^{75}P_3 = \dfrac{75!}{(75-3)! = 405150}[/tex]
A standard deck of cards has 52 cards. The cards have one of two colors: 26 cards in the deck are red and 26 are black. The cards have one of four denominations: 13 cards are hearts (red), 13 cards are diamonds (red), 13 cards are clubs (black), and 13 cards are spades (black).
a. One card is selected at random and the denomination is recorded. What is the sample space S for the set of possible outcomes?
b. Two cards are selected at random and the color is recorded. What is the sample space S for the set of possible outcomes?
c. Two cards are selected at random and the denomination is recorded. The event H is defined as the event that the first card is hearts. What defines event H?
d. Two cards are selected at random and the denomination is recorded. The event D is defined as the event that the first card is diamonds and the second card is red. What defines event DC?
e. Two cards are selected at random. Event C is defined as the event that the first card is clubs, event R as the event that the first card is red, and event B as the event that the second card is black. Which events are disjoint?
Final answer:
The sample space S for a. is a set of all possible denominations of cards; the sample space S for b. is a set of all possible combinations of two cards with color recorded; event H is defined as the first card being hearts in c.; event DC is defined as the first card being diamonds and second card being red in d.; events C and B are disjoint in e.
Explanation:
a. One card is selected at random and the denomination is recorded. What is the sample space S for the set of possible outcomes?
The sample space S for this scenario is the set of all possible denominations of cards in a standard deck, which includes the numbers 1 through 10, as well as the face cards (J, Q, K) and the Ace (A).
b. Two cards are selected at random and the color is recorded. What is the sample space S for the set of possible outcomes?
The sample space S for this scenario is the set of all possible combinations of two cards from a standard deck, with the colors (red or black) of each card being recorded. Since there are 26 red cards and 26 black cards, the number of possible outcomes is 26 * 26 = 676.
c. Two cards are selected at random and the denomination is recorded. The event H is defined as the event that the first card is hearts. What defines event H?
Event H is defined as the event where the first card selected is a heart. In terms of the sample space, it can be defined as the subset of sample space S that includes all outcomes where the first card is a heart.
d. Two cards are selected at random and the denomination is recorded. The event D is defined as the event that the first card is diamonds and the second card is red. What defines event DC?
Event DC is defined as the event where the first card selected is a diamond and the second card selected is red. In terms of the sample space, it can be defined as the subset of sample space S that includes all outcomes where the first card is a diamond and the second card is red.
e. Two cards are selected at random. Event C is defined as the event that the first card is clubs, event R as the event that the first card is red, and event B as the event that the second card is black. Which events are disjoint?
Events C and B are disjoint because the first card cannot be both a club and black at the same time. However, events R and B are not disjoint because there are red clubs in the deck, which would satisfy both events.
The sample spaces for the following are as follows: a) Hearts, Diamonds Clubs, Spades. b) Red Red, Red Black, Black Red, Black Black.
c) Hearts - Hearts, Hearts - Diamonds, Hearts - Clubs, Hearts - Spades.
d) Diamond - Black, Black - Black, Black - Red, Hearts - Any.
e) The events C and R are disjoint.
A standard deck of cards has 52 cards, with 26 red and 26 black cards across four suits: hearts, diamonds, clubs, and spades, each containing 13 cards.
a. One card is selected at random and the denomination is recorded.
The sample space for the denomination of a single card is:
Hearts
Diamonds
Clubs
Spades
b. Two cards are selected at random and the color is recorded.
The sample space for the color of two cards selected can be:
Red, Red
Red, Black
Black, Red
Black, Black
c. Two cards are selected at random and the denomination is recorded. T
Event H is defined as drawing a heart for the first card. The possible outcomes for the second card can be any of the four denominations: hearts, diamonds, clubs, or spades. So, the event H sample space is:
Hearts - Hearts
Hearts - Diamonds
Hearts - Clubs
Hearts - Spades
d. Two cards are selected at random and the denomination is recorded.
Event D is defined as drawing a diamond for the first card and a red card for the second card. Since a red card can only be a diamond or heart, event DC (complement of event D) includes any draw combinations not fitting this pattern:
Diamond - Black
Black - Black
Black - Red
Hearts - Any
e. Two cards are selected at random.
Events C (first card is clubs) and R (first card is red) are disjoint since a card cannot simultaneously be clubs (black) and red. However, either of these events can be paired with event B (second card is black). Therefore, events C and R are disjoint.
The sample spaces for the following are as follows: a) Hearts, Diamonds Clubs, Spades. b) Red Red, Red Black, Black Red, Black Black.
c) Hearts - Hearts, Hearts - Diamonds, Hearts - Clubs, Hearts - Spades.
d) Diamond - Black, Black - Black, Black - Red, Hearts - Any.
e) The events C and R are disjoint.
For the Hawkins Company, the monthly percentages of all shipments received on time over the past 12 months are 80, 82, 84, 83, 83, 84, 85, 84, 82, 83, 84, and 83. Click on the datafile logo to reference the data. (a) Choose the correct time series plot. (i) (ii) (iii) (iv) What type of pattern exists in the data? (b) Compare a three-month moving average forecast with an exponential smoothing forecast for α = 0.2. Which provides the better forecasts using MSE as the measure of model accuracy? Do not round your interim computations and round your final answers to three decimal places. Three-month Moving Average Exponential smoothing MSE (c) What is the forecast for next month? If required, round your answer to two decimal places.
Answer:
Moving average MSE =1.235
Exponential smoothing MSE =3.555
forecast for next month =(83+84+83)/3=83.33
Step-by-step explanation:
a)
Select your answer - Plot (iii)
What type of pattern exists in the data? -Horizontal Pattern
b)
for Moving Average MSE =1.235
for Exponential smoothing MSE =3.555
Moving Average is better
c)
forecast for next month =(83+84+83)/3=83.33
It can be deduced from the graph that the type of pattern that exists in the data is a horizontal pattern.
How to interpret the graphFrom the complete question, when comparing the three-month moving average forecast with an exponential smoothing forecast, the forecast for moving average is 1.235 while that of exponential smoothing is 3.555. Therefore, moving average is better.
In conclusion, the forecast for next month will be:
= (83 + 84 + 83)/3
= 83.33
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7. At a city high school, past records indicate that the MSAT scores of students have a mean of 510 and a standard deviation of 90. One hundred students in the high school are to take the test. What is the probability that the sample mean score will be (a) within 10 of the population mean
Answer:
73.30% probability that the sample mean score will be within 10 of the population mean
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal probability distribution:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
When we are approximating a binomial distribution to a normal one, we have that [tex]\mu = E(X)[/tex], [tex]\sigma = \sqrt{V(X)}[/tex].
Central limit theorem:
The Central Limit Theorem estabilishes that, for a random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sample means with size n of at least 30 can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex]
In this problem, we have that:
[tex]\mu = 510, \sigma = 90, n = 100, s = \frac{90}{\sqrt{100}} = 9[/tex]
What is the probability that the sample mean score will be (a) within 10 of the population mean
This is the pvalue of Z when X = 510 + 10 = 520 subtracted by the pvalue of Z when X = 510 - 10 = 500.
X = 520
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{520 - 510}{9}[/tex]
[tex]Z = 1.11[/tex]
[tex]Z = 1.11[/tex] has a pvalue of 0.8665
X = 500
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{500 - 510}{9}[/tex]
[tex]Z = -1.11[/tex]
[tex]Z = -1.11[/tex] has a pvalue of 0.1335
0.8665 - 0.1335 = 0.7330
73.30% probability that the sample mean score will be within 10 of the population mean
The TurDuckEn restaurant serves 8 entr´ees of turkey, 12 of duck, and 10 of chicken. If customers select from these entr´ees randomly, what is the probability that exactly two of the next four customers order turkey entr´ees?
Answer:
The probability that exactly 2 of the next four customers order turkey entries is 6 * 0.0393359 = 0.2360
Step-by-step explanation:
There are a total of 8+12+10 = 30 entries, 8 of them being turkey. Lets compute the probability that the first 2 customers order turkey entries and the remaining 2 do not.
For the first customer, he has 8 turkey entries out of 30, so the probability for him to pick a turkey entry is 8/30. For the second one there will be 29 entries left, 7 of them being turkey. So the probability that he picks a turkey entry is 7/29. The third one has 28 entries left, 6 of them which are turkey and 22 that are not. The probability that he picks a non turkey entry is 22/28. And for the last one, the probability that he picks a non turkey entry is 21/27. So the probability of this specific event is
8/30*7/29*22/28*21/27 = 0.0393359
Any other event based on a permutation of this event will have equal probability (we are just swapping the order of the numerators when we permute). The total number of permutations is 6, and since each permutation has equal probability, then the probability that exactly 2 of the next four customers order turkey entries is 6 * 0.0393359 = 0.2360
Given the system of equations
(D+5)x+D2y=cost
D2x+(D+9)y=t2+3t
(a) Write the system as a differentialequation for x , in operator form, by filling thepolynomial operator in D (on the left-hand-sideof the equation) and the corresponding right-hand-side of theequation:
x=
(Write your expression for the right-hand-side as a function oft only---that is, with no derivativeoperators.)
(b) Note that we could equally well solve fory. Do this here, obtaining an equivalent differentialequation for y.
y=
(Write your expression for the right-hand-side as a function oft only---that is, with no derivativeoperators.)
The differential equations for x and y are:
x = (cost - D^2 (t2 + 3t)) / ((D+5) + D^2 (D+9))
y = (t2 + 3t - D2 cost) / (D2 / (D+5) + (D+9))
Here's how to solve the problem:
(a) Write the system as a differential equation for x:
Eliminate y: Substitute the second equation into the first equation to eliminate y.
(D+5)x + D2(t2 + 3t) = cost
Rearrange and define operator: Define P(D) = (D+5) + D^2 (D+9).
P(D) x = cost - D^2 (t2 + 3t)
Write as differential equation:
x = (cost - D^2 (t2 + 3t)) / P(D)
(b) Write the system as a differential equation for y:
Eliminate x: Substitute the first equation into the second equation to eliminate x.
D2((cost - D2y) / (D+5)) + (D+9)y = t2 + 3t
Rearrange and define operator: Define Q(D) = D2 / (D+5) + (D+9).
Q(D) y = t2 + 3t - D2 cost
Write as differential equation:
y = (t2 + 3t - D2 cost) / Q(D)
Therefore, the differential equations for x and y are:
x = (cost - D^2 (t2 + 3t)) / ((D+5) + D^2 (D+9))
y = (t2 + 3t - D2 cost) / (D2 / (D+5) + (D+9))
(a)[tex]\(x = \frac{1}{D^3 + 14D + 5}\cdot \left( \cos(t) \right)\)[/tex]
(b)[tex]\(y = \frac{1}{D^3 + 14D + 5}\cdot \left( t^2 + 3t \right)\)[/tex] by expressing (x) and (y) in terms of differential operators and given functions of (t), we derive their differential equations .
Explanation:The given system of equations can be transformed into differential equations using operator (D), representing differentiation with respect to a variable, often time.
(a) To derive the differential equation for (x), we isolate (x) in terms of the differential operator \(D\). Rearranging the first equation gives us [tex]\(x = \frac{\cos(t)}{D+5} - \frac{D^2y}{D+5}\).[/tex]Substituting the expression for (y) from the second equation into this yields[tex]\(x = \frac{\cos(t)}{D+5} - \frac{D^2}{D+5}\left(\frac{t^2 + 3t}{D+9}\right)\).[/tex] Simplifying further leads to [tex]\(x = \frac{1}{D^3 + 14D + 5}\cdot \left( \cos(t) \right)[/tex]\).
(b) Similarly, isolating \(y\) from the given equations leads to [tex]\(y = \frac{t^2 + 3t}{D+9} - \frac{D+5}{D+9}x\).[/tex]Substituting the expression for (x) from the first equation into this yields [tex]\(y = \frac{t^2 + 3t}{D+9} - \frac{D+5}{D+9}\left(\frac{\cos(t)}{D+5} - \frac{D^2y}{D+5}\right)\).[/tex]Further simplification results in [tex]\(y = \frac{1}{D^3 + 14D + 5}\cdot \left( t^2 + 3t \right)\).[/tex]
Therefore, by expressing (x) and (y) in terms of differential operators and given functions of (t), we derive their differential equations as specified in the final answer.
The rate constant for a certain reaction is kkk = 8.70×10−3 s−1s−1 . If the initial reactant concentration was 0.150 MM, what will the concentration be after 19.0 minutes?
The initial reactant concentration is 0.150 M. Using the first order kinetics equation, we converted the time to seconds and plugged in the given values for initial concentration, time and rate constant. The final concentration of the reactant after 19.0 minutes is 4.92 x 10^-5 M.
Explanation:The initial concentration of the reactant is 0.150 M and the rate constant, k = 8.70×10−3 s−1. We use the formula for first order kinetics, [A] = [A]0 * e^-kt, where [A] is the final concentration, [A]0 is the initial concentration, k is the rate constant and t is the time.
First, convert the time from minutes to seconds (since the rate constant is in s^-1). Therefore, t = 19 min * 60 s/min = 1140 s.
Then, substitute the given values into the equation: [A] = 0.150 M * e^-((8.70 x 10^-3 s^-1) * (1140 s)) = 0.150 M * e^-9.918 = 4.92 x 10^-5 M.
So, the concentration of the reactant after 19.0 minutes will be 4.92 x 10^-5 M.
Learn more about First Order Kinetics here:https://brainly.com/question/31201270
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Part A: Concentration ≈ 0.186 MM after 8.00 minutes.
Part B: Initial concentration ≈ 5.90 MM for zero-order reaction.
Part A:
Using the first-order integrated rate law equation:
[tex]\[ [A] = [A]_0 \times e^{-kt} \][/tex]
[tex]\[ [A] = 0.650 \times e^{-(3.70 \times 10^{-3} \times 8 \times 60)} \][/tex]
[tex]\[ [A] \approx 0.186 \, MM \][/tex]
Part B:
Using the zero-order integrated rate law equation:
[tex]\[ [A] = -kt + [A]_0 \][/tex]
[tex]\[ [A]_0 = -kt + [A] \][/tex]
[tex]\[ [A]_0 = -(3.00 \times 10^{-4} \times 80) + 3.50 \times 10^{-2} \][/tex]
[tex]\[ [A]_0 = 3.50 \times 10^{-2} + 2.40 \times 10^{-2} \][/tex]
[tex]\[ [A]_0 \approx 5.90 \, MM \][/tex]
The correct question is:
Part A
The rate constant for a certain reaction is [tex]$k k k=3.70 \times 10^{-3} \mathrm{~s}-1 \mathrm{~s}-1$[/tex]. If the initial reactant concentration was [tex]$0.650 \mathrm{MM}$[/tex], what will the concentration be after 8.00 minutes?
Part B
A zero-order reaction has a constant rate of [tex]$3.00 \times 10^{-4} \mathrm{M} / \mathrm{sM} / \mathrm{s}$[/tex]. If after 80.0 seconds the concentration has dropped to [tex]$3.50 \times 10^{-2} \mathrm{MM}$[/tex], what was the initial concentration?