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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The least squares regression line minimizes the sum of the:
(A) Differences between actual and predicted Y values.
(B) Absolute deviations between actual and predicted Y values.
(C) Absolute deviations between actual and predicted X values
(D) Squared differences between actual and predicted Y values
(E) Squared differences between actual and predicted X values.
Answer:
d) Squared differences between actual and predicted Y values.
Step-by-step explanation:
Regression is called "least squares" regression line. The line takes the form = a + b*X where a and b are both constants. Value of Y and X is specific value of independent variable.Such formula could be used to generate values of given value X.
For example,
suppose a = 10 and b = 7. If X is 10, then predicted value for Y of 45 (from 10 + 5*7). It turns out that with any two variables X and Y. In other words, there exists one formula that will produce the best, or most accurate predictions for Y given X. Any other equation would not fit as well and would predict Y with more error. That equation is called the least squares regression equation.
It minimize the squared difference between actual and predicted value.
The least squares regression line minimizes the sum of the squared differences between the actual and predicted Y values, making option (D) the correct answer.
Explanation:The least squares regression line is a method used in statistics to find the line that best fits a set of data points. This method minimizes the sum of squared differences between the actual Y values and the predicted Y values on the regression line. In other words, the objective of the least squares regression line is to find the coefficients that minimize the SSE (sum of squared errors). So, the correct answer is:
(D) Squared differences between actual and predicted Y values.
Using this method allows for the most accurate prediction within the given set of data, but it is important to note that it may not be suitable for predicting values outside that set.
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For the function given below, find a formula for the Riemann sum obtained by dividing the interval [0,3] into n equal subintervals and using the right-hand endpoint for each c[Subscript]k. Then take a limit of this sum as n approaches infinity to calculate the area under the curve over [0,3].
f(x)=2x^2
We divide [0, 3] into [tex]n[/tex] subintervals,
[tex]\left[0,\dfrac3n\right]\cup\left[\dfrac3n,\dfrac6n\right]\cup\left[\dfrac6n,\dfrac9n\right]\cup\cdots\cup\left[\dfrac{3(n-1)}n,3\right][/tex]
so that the right endpoint of each subinterval is given according to the arithmetic sequence,
[tex]r_k=\dfrac{3k}n[/tex]
for [tex]1\le k\le n[/tex].
The Riemann sum is then
[tex]\displaystyle\sum_{k=1}^nf(r_k)\Delta x_k[/tex]
where
[tex]\Delta x_k=r_k-r_{k-1}=\dfrac{3k}n-\dfrac{3(k-1)}n=\dfrac3n[/tex]
With [tex]f(x)=2x^2[/tex], we have
[tex]\displaystyle\frac3n\sum_{k=1}^n2\left(\frac{3k}n\right)^2=\frac{54}{n^3}\sum_{k=1}^nk^2[/tex]
Recall that
[tex]\displaystyle\sum_{k=1}^nk^2=\frac{n(n+1)(2n+1)}6[/tex]
The area under the curve [tex]f(x)[/tex] over the interval [0, 3] is then
[tex]\displaystyle\int_0^32x^2\,\mathrm dx=\lim_{n\to\infty}\frac{54n(n+1)(2n+1)}{6n^3}=\lim_{n\to\infty}9\left(2+\frac3n+\frac1{n^2}\right)=\boxed{18}[/tex]
During a recent drought, a water utility in a certain town sampled 100 residential water bills and found out that 73 of the residences had reduced their water consumption over that of the previous year. Find a 95% confidence interval for the proportion of residences that reduced their water
Answer:
(0.6430, 0.8170)
Step-by-step explanation:
Given that during a recent drought a water utility in a certain town sampled 100 residential water bills and found out that 73 of the residences had reduced their water consumption over that of the previous year.
Sample size n = 100
Sample proportion p = [tex]\frac{73}{100} =0.73[/tex]
q = 1-p = 0.23
Std error of proportion = [tex]\sqrt{\frac{pq}{n} } \\=\sqrt{\frac{0.73*0.27}{100} } \\=0.04440[/tex]
95% Z critical value = 1.96
Margin of error = [tex]1.96*0.0444 = 0.0870[/tex]
Confidence interval = sample proportion ±margin of error
0.642983946
0.817016054
(0.6430, 0.8170)
The 95% confidence interval for the proportion of residences that reduced their water is (0.6430, 0.8170).
What is Confidence interval?
This is used to determine the measure of how much uncertainty there is with any particular statistic.
Sample size (n) = 100
Sample proportion(p) = 73/100 = 0.73
q = 1-p = 0.23
√pq/n = √0.73×0.27/100 = 0.04440
95% Z critical value = 1.96
Margin of error = 1.96×0.04440 = 0.0870
Confidence interval = sample proportion ± margin of error
= (0.6430, 0.8170)
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What is the value of c so that -9 and 9 are both solutions of x^2 + c = 103?
Answer:
c = 22
Step-by-step explanation:
The given equation can be written as ...
x^2 = 103 -c
In order for -9 and +9 to be solutions, the equation needs to be ...
x^2 = 81
So, we must have ...
103 -c = 81
c = 103 -81 = 22
The value of c must be 22.
Last year, during an investigation of the time spent reading e-mails on a daily basis, researchers found that on Monday the average time was 50 minutes. Office workers claim that with the increased spam and junk mail, this time has now increased. To conduct a test, a sample of 25 employees is selected with the following results: sample mean = 51.05 minutes and sample standard deviation = 14.2 minutes. What is the critical value for ? = .05 to test the hypotheses. Assume the time spent reading e-mails is normally distributed.
Set up the problem.
Calculate the test statistic.
State your conclusion. Reject/Fail?
Based on your decision about the hypothesis testing what kind error is possible? Type 1 or Type 2
Answer:
We conclude that with the increased spam and junk mail, the time spent reading e-mails on a daily basis has not increased.
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = 50 minutes
Sample mean, [tex]\bar{x}[/tex] = 51.05 minutes
Sample size, n = 25
Alpha, α = 0.05
Sample standard deviation, σ = 14.2 minutes
First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu = 50\text{ minutes}\\H_A: \mu > 50\text{ minutes}[/tex]
We use One-tailed t test to perform this hypothesis.
Formula:
[tex]t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }[/tex]
Putting all the values, we have
[tex]t_{stat} = \displaystyle\frac{51.05 - 50}{\frac{14.2}{\sqrt{25}} } = 0.3697[/tex]
Now, [tex]t_{critical} \text{ at 0.05 level of significance, 24 degree of freedom } = 1.710882[/tex]
Since,
[tex]t_{stat} < t_{critical}[/tex]
We fail to accept the alternate hypothesis and reject the alternate hypothesis. We accept the null hypothesis. Thus, we conclude that with the increased spam and junk mail, the time spent reading e-mails on a daily basis has not increased.
There maybe a chance to commit type II error, defined as fail to reject the null hypothesis when it is false.
If n is a positive integer, how many 5-tuples of integers from 1 through n can be formed in which the elements of the 5-tuple are written in decreasing order but are not necessarily distinct? In other words, how many 5-tuples of integers (h, i, j, k, m) are there with n ≥ h ≥ i ≥ j ≥ k ≥ m ≥ 1?
Answer:
[tex] n + 4 {n \choose 2} + 6 {n \choose 3} + 4 {n \choose 4} + {n \choose 5} [/tex]
Step-by-step explanation:
Lets divide it in cases, then sum everything
Case (1): All 5 numbers are different
In this case, the problem is reduced to count the number of subsets of cardinality 5 from a set of cardinality n. The order doesnt matter because once we have two different sets, we can order them descendently, and we obtain two different 5-tuples in decreasing order.
The total cardinality of this case therefore is the Combinatorial number of n with 5, in other words, the total amount of possibilities to pick 5 elements from a set of n.
[tex]{n \choose 5 } = \frac{n!}{5!(n-5)!}[/tex]
Case (2): 4 numbers are different
We start this case similarly to the previous one, we count how many subsets of 4 elements we can form from a set of n elements. The answer is the combinatorial number of n with 4 [tex] {n \choose 4} . [/tex]
We still have to localize the other element, that forcibly, is one of the four chosen. Therefore, the total amount of possibilities for this case is multiplied by those 4 options.
The total cardinality of this case is [tex] 4 * {n \choose 4} . [/tex]
Case (3): 3 numbers are different
As we did before, we pick 3 elements from a set of n. The amount of possibilities is [tex] {n \choose 3} . [/tex]
Then, we need to define the other 2 numbers. They can be the same number, in which case we have 3 possibilities, or they can be 2 different ones, in which case we have [tex] {3 \choose 2 } = 3 [/tex] possibilities. Therefore, we have a total of 6 possibilities to define the other 2 numbers. That multiplies by 6 the total of cases for this part, giving a total of [tex] 6 * {n \choose 3} [/tex]
Case (4): 2 numbers are different
We pick 2 numbers from a set of n, with a total of [tex] {n \choose 2} [/tex] possibilities. We have 4 options to define the other 3 numbers, they can all three of them be equal to the biggest number, there can be 2 equal to the biggest number and 1 to the smallest one, there can be 1 equal to the biggest number and 2 to the smallest one, and they can all three of them be equal to the smallest number.
The total amount of possibilities for this case is
[tex] 4 * {n \choose 2} [/tex]
Case (5): All numbers are the same
This is easy, he have as many possibilities as numbers the set has. In other words, n
Conclussion
By summing over all 5 cases, the total amount of possibilities to form 5-tuples of integers from 1 through n is
[tex] n + 4 {n \choose 2} + 6 {n \choose 3} + 4 {n \choose 4} + {n \choose 5} [/tex]
I hope that works for you!
Consider the force field and circle defined below.
F(x, y) = x2 i + xy j
x2 + y2 = 9
(a) Find the work done by the force field on a particle that moves once around the circle oriented in the clockwise direction.
By Green's theorem,
[tex]\displaystyle\int_{x^2+y^2=9}\vec F(x,y)\cdot\mathrm d\vec r=\iint_D\left(\frac{\partial(xy)}{\partial x}-\frac{\partial(x^2)}{\partial y}\right)\,\mathrm dx\,\mathrm dy=\iint_Dy\,\mathrm dx\,\mathrm dy[/tex]
where [tex]C[/tex] is the circle [tex]x^2+y^2=9[/tex] and [tex]D[/tex] is the interior of [tex]C[/tex], or the disk [tex]x^2+y^2\le1[/tex].
Convert to polar coordinates, taking
[tex]\begin{cases}x=r\cos\theta\\y=r\sin\theta\end{cases}\implies\mathrm dx\,\mathrm dy=r\,\mathrm dr\,\mathrm d\theta[/tex]
Then the work done by [tex]\vec F[/tex] on the particle is
[tex]\displaystyle\iint_Dy\,\mathrm dx\,\mathrm dy=\int_0^{2\pi}\int_0^3(r\sin\theta)r\,\mathrm dr\,\mathrm d\theta=\left(\int_0^{2\pi}\sin\theta\,\mathrm d\theta\right)\left(\int_0^3r^2\,\mathrm dr\right)=\boxed0[/tex]
The work done in a force field on a particle moving around a circle is found by calculating and integrating the line integral of the force field over the path defined by the circle. It involves substituting the parametric representation of the circle into the force field equation and incorporating the directional aspect of the line integral.
Explanation:This problem is related to calculating work done in a force field and involves concepts from vector calculus. The work done is calculated based on the line integral of the force field F over a path C, defined by the parametric representation of the circle. More specifically, we'll need to find the line integral of F along the path C, and then integrate that from 0 to 2π (since the particle moves around the circle once).
The parametric representation of the circle x² + y² = 9 is x = 3cosθ, y = 3sinθ, where -π ≤ θ ≤ π. Substitute these into the force field equation you'll get F(3cosθ, 3sinθ) = 9cos²θ i + 9cosθsinθ j.
To find the work done, we'll compute the line integral of F over the path C, which in this case is the circle. Since the movement is in the clockwise direction and when it comes to line integrals, the direction matters, we'll need to use -θ instead of θ to represent our parameter. So we're integrating F along C from 0 to 2π.
The exact calculation of the integral might require a bit of time and effort, but you should end up with the work done by the force field on the particle that moves once around the circle oriented in the clockwise direction.
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In the past the average age of employees of a large corporation has been 40 years. Recently, the company has been hiring older individuals. In order to determine whether there has been an increase in the average age of all the employees, a sample of 25 employees was selected. The average age in the sample was 45 years with a standard deviation of 5 years. Assume the distribution of the population is normal let α= 0.5 A, state the null and the alterative hypothesis B, test to determine whether or not the mean age of all the employees is significantly more than 40 years.
Answer:
a) Null hypothesis:[tex]\mu \leq 40[/tex]
Alternative hypothesis:[tex]\mu > 40[/tex]
b) [tex]t=\frac{45-40}{\frac{5}{\sqrt{25}}}=5[/tex]
[tex]p_v =P(t_{(24)}>5)=2.08x10^{-5}[/tex]
If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that mean age of all employees is significantly more than 40 years at 5% of signficance.
Step-by-step explanation:
1) Data given and notation
[tex]\bar X=45[/tex] represent the sample mean
[tex]s=15[/tex] represent the sample standard deviation
[tex]n=25[/tex] sample size
[tex]\mu_o =40[/tex] represent the value that we want to test
[tex]\alpha=0.05[/tex] represent the significance level for the hypothesis test.
t would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value for the test (variable of interest)
2) State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the mean age of all the employees is significantly more than 40 years, the system of hypothesis are :
Null hypothesis:[tex]\mu \leq 40[/tex]
Alternative hypothesis:[tex]\mu > 40[/tex]
Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".
3) Calculate the statistic
We can replace in formula (1) the info given like this:
[tex]t=\frac{45-40}{\frac{5}{\sqrt{25}}}=5[/tex]
4) P-value
First we need to calculate the degrees of freedom given by:
[tex]df=n-1=25-1=24[/tex]
Since is a one-side upper test the p value would be:
[tex]p_v =P(t_{(24)}>5)=2.08x10^{-5}[/tex]
5) Conclusion
If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the mean age of all employees it's significantly more than 40 years at 5% of signficance.
To test whether the mean age of employees has increased, use a one-sample t-test with a significance level of 0.05.
Explanation:To test whether the mean age of all the employees is significantly more than 40 years, we can use a hypothesis test.
The null hypothesis (H0) states that the mean age is equal to or less than 40 years, while the alternative hypothesis (Ha) states that the mean age is greater than 40 years.
We can conduct a one-sample t-test to compare the sample mean of 45 years to the population mean of 40 years. We will use a significance level (α) of 0.05. If the p-value of the test is less than α, we reject the null hypothesis and conclude that there is evidence of a significant increase in the average age of all the employees.
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Confidence Interval and Hypothesis Test on the Proportion 1. A composites manufacturer is having serious problems with porosity in their parts. A Quality Engineer samples 300 parts and finds 58 defective. (a) Test the hypothesis that defective rate (proportion defective) exceeds 15%. Test at α = 0.05. What is the parameter of interest? What assumptions are made? Show mathematical evidence to support assumption.
Answer:
Parameter of interest= proportion of parts that present serious problems with porosity (defective)
z= 5.08
Based on the p value obtained and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of parts defective (problems with porosity) NOT exceeds 0.15 or 15% .
Step-by-step explanation:
1) Data given and notation n
n=300 represent the random sample taken
X=58 represent the parts that present serious problems with porosity in the sample
[tex]\hat p=\frac{58}{300}=0.193[/tex] estimated proportion of parts that present serious problems with porosity in the sample
[tex]p_o=0.15[/tex] is the value that we want to test
[tex]\alpha=0.05[/tex] represent the significance level
Confidence=95% or 0.95
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value (variable of interest)
2) Concepts and formulas to use
We need to conduct a hypothesis in order to test the claim that defective rate (proportion defective) exceeds 15%. :
Null hypothesis:[tex]p\leq 0.15[/tex]
Alternative hypothesis:[tex]p>0.15[/tex]
We assume that the proportion follows a normal distribution.
When we conduct a proportion test we need to use the z statistic, 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,higher or less) from a hypothesized value [tex]p_o[/tex].
3) Calculate the statistic
Since we have all the info requires we can replace in formula (1) like this:
[tex]z=\frac{0.193 -0.15}{\sqrt{\frac{0.15(1-0.15)}{300}}}=2.086[/tex]
4) 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 provided [tex]\alpha=0.05[/tex]. The next step would be calculate the p value for this test.
Since is a one side test the p value would be:
[tex]p_v =P(z>2.086)=0.0185[/tex]
Based on the p value obtained and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of parts defective (problems with porosity) NOT exceeds 0.15 or 15% .
Consider the following binomial experiment. The probability that a green jelly bean is chosen at random from a large package of jelly beans is 2/5. If Sally chooses 13 jelly beans, what is the probability that at most two will be green jelly beans?
a) 0.0349
b) 0.3100
c) 0.0579
d) 0.9421
e) 0.5000
To answer this question, we use the binomial probability formula to calculate the probability of picking 0, 1, and 2 green jelly beans out of 13. Summing up these probabilities gives us our answer. The closest option to our calculated value is (b) 0.3100.
Explanation:The subject of this question is the calculation of probabilities that is handled in Mathematics, specifically in Statistics. To solve this, we can use the binomial probability formula P(X=k) = C(n, k) * (p^k) * (1 - p)^(n - k), where k is the number of successful trials, n is the total number of trials, and p is the probability of success in one trial.
In this case, we want to find the probability that at most 2 jelly beans will be green if Sally chooses 13. We need to calculate the probability of getting 0, 1, and 2 green jelly beans, then sum those probabilities. Therefore, P(X≤2) = P(X=0) + P(X=1) + P(X=2).
Using the above binomial formula, we will not get an exact value from the options provided. Only option (b) 0.3100, is closest to our calculated value and would be correct if we were to pick from these options.
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In March 2007, Business Week reported that at the top 50 business schools, students studied an average of 14.6 hours. You wonder whether the amount UMSL students study is different from this 14.6 hour benchmark. Set up the hypotheses used in this situation.
Answer:
The hypotheses used in this situation
[tex]H_0:\mu = 14.6[/tex]
[tex]H_a:\mu \neq 14.6[/tex]
Step-by-step explanation:
We are given that Business Week reported that at the top 50 business schools, students studied an average of 14.6 hours.
Mean = [tex]\mu = 14.6[/tex]
Claim : The amount UMSL students study is different from this 14.6 hour benchmark.
The hypotheses used in this situation
[tex]H_0:\mu = 14.6[/tex]
[tex]H_a:\mu \neq 14.6[/tex]
A Snapshot in USA Today indicates that 51% of Americans say the average person is not very considerate of others when talking on a cellphone. Suppose that 100 Americans are randomly selected. Find the approximate probability that 62 or more Americans would indicate that the average person is not very considerate of others when talking on a cellphone. (Use the normal approximation. Round your answer to four decimal places.)
Answer:
There is an 1.39% probability that 62 or more Americans would indicate that the average person is not very considerate of others when talking on a cellphone.
Step-by-step explanation:
For each American, there are only two possible outcomes. Either they indicate that the average person is not very considerate of others when talking on a cellphone, of they indicate that the average person is considerate. This means that we use the binomial probability distribution to solve this problem.
This distribution can be approximated to a normal distribution, using the expected value and the standard deviation.
The expected value of the binomial distribution is:
[tex]E(X) = np[/tex]
The standard deviation of the binomial distribution is:
[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]
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), \sigma = \sqrt{V(X)}[/tex]
In this problem, we have that:
There are 100 americans, so [tex]n = 100[/tex]
51% of Americans say the average person is not very considerate of others when talking on a cellphone. This means that [tex]p = 0.51[/tex].
Find the approximate probability that 62 or more Americans would indicate that the average person is not very considerate of others when talking on a cellphone.
This is 1 subtracted by the pvalue of Z when [tex]X = 62[/tex].
We have that
[tex]\mu = E(X) = np = 100*0.51 = 51[/tex]
[tex]\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{100*0.51*0.49} = 5[/tex]
So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{62 - 51}{5}[/tex]
[tex]Z = 2.2[/tex]
[tex]Z = 2.2[/tex] has a pvalue of 0.9861.
This means that there is a 1-0.9861 = 0.0139 = 1.39% probability that 62 or more Americans would indicate that the average person is not very considerate of others when talking on a cellphone.
You are given 5 to 4 odds against tossing three heads heads with three coins, meaning you win $5 if you succeed and you lose $4 if you fail. Find the expected value (to you) of the game. Would you expect to win or lose money in 1 game? In 100 games? Explain. Find the expected value (to you) for the game.- $___ (Type an integer or a decimal rounded to the nearest hundredth as needed.)
Answer:
$-2.875 in 1 game
$-287.5 in 100 games
Step-by-step explanation:
As the probability of getting head in tossing a fair coin is 0.5, the probability of getting 3 heads with 3 coins is 0.5*0.5*0.5 = 0.125
So you have a 0.125 chance of winning and 0.875 chance of losing a game
The expected value for 1 game would be
0.125*5 - 0.875*4 = -2.875
On average you would be losing 2.875 per game. In 100 games you should expect to lose 287.5 dollar
A function y(t) satisfies the differential equation dy dt = y 4 − 6y 3 + 5y 2 . (a) What are the constant solutions of the equation? (Recall that these have the form y = C for some constant, C.) (b) For what values of y is y increasing? (c) For what values of y is y decreasing?
Answer:
Hence increasing in (-\infty,0) U (1,5)
c) Decreasing in (0,1)
Step-by-step explanation:
Given that y(t) satisfies the differential equation
[tex]\frac{dy}{dt} =y^4-6y^2+5y^2\\=y^2(y^2-6y+5)\\=y^2(y-1)(y-5)[/tex]
Separate the variables to have
[tex]\frac{dy}{y^2(y-1)(y-5)} =dt[/tex]
Left side we can resolve into partial fractions
Let [tex]\frac{1}{y^2(y-1)(y-5)} =\frac{A}{y} +\frac{B}{y^2}+\frac{C}{y-1} \frac{D}{y-5}[/tex]
Taking LCD we get
[tex]1= Ay(y-1)(Y-5) +B(y-1)(y-5)+Cy^2 (y-5)+Dy^2 (y-1)\\Put y =1\\1 = -4C\\Put y =5\\ 1 = 25(4)D\\Put y =0\\1=5B\\[/tex]
By equating coeff of y^3 we have
A+C+D=0
[tex]C=\frac{-1}{4} \\D=\frac{1}{100} \\B =\frac{1}{5} \\A = -C-D = \frac{6}{25}[/tex]
Hence left side =
[tex]\frac{6}{25y} +\frac{1}{5y^2}+\frac{-1}{4(y-1)}+ \frac{1}{100(y-5)}=dt\\\frac{6}{25}ln y -\frac{1}{5y}-\frac{1}{4}ln|(y-1)| +\frac{1}{100}ln|y-5| = t+C[/tex]
b) y is increasing whenever dy/dt>0
dy/dt =0 at points y =0, 1 and 5
dy/dt >0 in (-\infty,0) U (1,5)
Hence increasing in (-\infty,0) U (1,5)
c) Decreasing in (0,1)
Answer:
a) y = 0 , 5,1
b) y ⊂ (- ∞,0) ∪ (0,1)∪(5,∞)
Step-by-step explanation:
Given data:
differential equation is given as
[tex]\frac{dy}[dt} = y^4 -6y^3+ 5y^2[/tex]
a) constant solution
[tex] y^4 -6y^3+ 5y^2 = 0 [/tex]
taking y^2 from all part
[tex]y^2(y^2 - 6y -5) = 0[/tex]
solution of above equation is
y = 0 , 5,1
b) for which value y is increasing
[tex]\frac{dy}{dt} > 0[/tex]
y^2(y - 5) (y -1) > 0
y ⊂ (- ∞,0) ∪ (0,1)∪(5,∞)
Given below are the number of successes and sample size for a simple random sample from a population. xequals6, nequals50, 90% level a. Determine the sample proportion. b. Decide whether using the one-proportion z-interval procedure is appropriate. c. If appropriate, use the one-proportion z-interval procedure to find the confidence interval at the specified confidence level. d. If appropriate, find the margin of error for the estimate of p and express the confidence interval in terms of the sample proportion and the margin of error.
Answer:
a. Sample proportion ^p= 0.12
b. It is appropiate.
c. [0.0447;0.1953]
d. [^p ± d]
Step-by-step explanation:
Hello!
Given the information I'll assume that the variable of study has a binomial distribution:
X~Bi(n;ρ)
The sample data:
n= 50
"Success" x= 6
Sample proportion ^p= x/n = 6/50 = 0.12
Now, your study variable has a binomial distribution, but remember that the Central Limit Theorem states that given a big enough sample size (usually n≥ 30) you can approximate the sample proportion distribution to normal.
Since the sample is 50 you can apply the approximation, your sample proportion will have the following distribution:
^p≈ N( p; [p(1 - p)]/n)
With E(^p)= p and V(^p)= [p(1 - p)]/n.
This allows you to estimate the population proportion per Confidence Interval using the Z-distribution:
[^p±[tex]Z_{1-\alpha /2}[/tex]*√(^p(1 - ^p)/n)]
Since you are estimating the value of p, you'll use the estimated standard deviation (i.e. with the sample proportion instead of the population proportion)
to calculate the interval.
At level 90% the interval is:
[0.12±1.64*√([0.12(1 - 0.12)]/50)]
[0.0447;0.1953]
The margin of error (d) of an interval is half its amplitude (a)
if a= Upper bond - Low bond
then d= (Upper bond - Low bond)/2
d= (0.1953-0.0447)/2
d= 0.0753
And since the interval structure is "estimator" -/+ "margin of error" you can write it as:
[^p ± d]
I hope you have a SUPER day!
Studies show that gasoline use for compact cars sold in the United States is normally distributed, with mean of 25.5 miles per gallon (mpg) and a standard deviation of 4.5 mpg. If a manufacturer wishes to develop a compact car that outperforms 95% of the current compacts in fuel economy, what must the gasoline use rate for the new car be?
Answer:
32.9 mpg
Step-by-step explanation:
Population mean (μ) = 25.5 mpg
Standard deviation (σ) = 4.5 mpg
Assuming a normal distribution for gasoline use, the manufacturer wants his car to be at the 95-th percentile of the distribution. The 95-th percentile has a corresponding z-score of 1.645. The expression for the z-score for a given gasoline use rate 'X' is:
[tex]z=\frac{X-\mu}{\sigma} \\1.645=\frac{X-25.5}{4.5} \\X=32.9\ mpg[/tex]
The gasoline use rate for the new car must be at least 32.9 mpg
To outperform 95% of the current compacts in fuel economy, the gasoline use rate for the new car must be at least 33.7775 miles per gallon.
Explanation:To develop a compact car that outperforms 95% of the current compacts in fuel economy, the gasoline use rate for the new car must be at least as good as the top 5% of the current compacts. To find this, we use the z-score formula: z = (x - mean) / standard deviation. Since we want to find the value for x that corresponds to the top 5% (or 0.05) of the distribution, we can find the z-score by using the inverse normal distribution table. Once we have the z-score, we can use the formula z = (x - mean) / standard deviation to solve for x.
Using the inverse normal distribution table, we find that the z-score corresponding to the top 5% is approximately 1.645. Plugging this value into the formula and rearranging to solve for x, we get:
x = mean + (z * standard deviation)
Substituting in the given values, we have:
x = 25.5 + (1.645 * 4.5) = 33.7775
Therefore, the gasoline use rate for the new car must be at least 33.7775 miles per gallon to outperform 95% of the current compacts in fuel economy.
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What are the quotient and remainder when a) 19 is divided by 7? b) −111 is divided by 11? c) 789 is divided by 23? d) 1001 is divided by 13? e) 0 is divided by 19? f ) 3 is divided by 5? g) −1 is divided by 3? h) 4 is divided by 1?
Answer:
a) quotient = 2, remainder = 5
b) quotient = -10, remainder = -1
c) quotient = 34, remainder = 7
d) quotient = 77, remainder = 0
e) quotient = 0, remainder = 0
f) quotient = 0, remainder = 3
g) quotient = 0, remainder = -1
h) quotient = 4, remainder = 0
Step-by-step explanation:
The relationship between the quotient, remainder and divisor is that the product of the quotient and divisor added to the reminder gives the dividend.
Based on this, when
a)19 is divided by 7 = 19/7
= 2[tex]\frac{5}{7}[/tex]
Hence the quotient is 2, the remainder is 5
Applying the same principle, when
b) −111 is divided by 11
= -10[tex]\frac{1}{11}[/tex]
Hence the quotient is -10, the remainder is -1
c) 789 is divided by 23
=34[tex]\frac{7}{23}[/tex]
Hence the quotient is 34, the remainder is 7
d) 1001 is divided by 13
= 77
Hence the quotient = 77, the remainder is 0
e) 0/19 = 0
Hence the quotient is 0, the remainder is 0
f) 3/5
Hence the quotient is 0, the remainder is 3
g) -1/3
Hence the quotient is 0, the remainder is -1
h) 4/1
Hence the quotient is 4, the remainder is 0
The division operations result in the following quotients: 2, -10, 34, 77, 0, 0, 0, 4, and remainders: 5, 0, 3, 0, 0, 3, -1, 0 respectively.
Explanation:Here are the results of the given divisions: a) When 19 is divided by 7, the quotient is 2 and the remainder is 5. b) When −111 is divided by 11, the quotient is −10 and there is no remainder. c) When 789 is divided by 23, the quotient is 34 and the remainder is 3. d) When 1001 is divided by 13, the quotient is 77 and there is no remainder. e) When 0 is divided by 19, the quotient and the remainder are both 0. f) When 3 is divided by 5, the quotient is 0 and the remainder is 3. g) When −1 is divided by 3, the quotient is 0 and the remainder is -1. h) When 4 is divided by 1, the quotient is 4 and there is no remainder.
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On a standardized test, Paul answered the first 22 questions in 5 minutes. There are 77 questions on the test.
If he continues to answer questions at the same rate, how long will it take him to complete the test
from start to finish?
Answer: time it will take him to 17.5 minutes to complete the test from start to finish
Step-by-step explanation:
During the standardized test, Paul answered the first 22 questions in 5 minutes.
If he answers 22 questions 5 minutes
He would answer one question in 5/22 = 0.227 minutes
He continues to answer questions at the same rate. This means that his unit rate of answering 1 question in 0.227 minutes is constant throughout the test.
Total number of questions on the test is 77. The time it will take him to complete the test from start to finish will be
Unit rate of answering questions × total number of questions
= 0.227 × 77 = 17.5 minutes
Razorback Realtors (RR) has apartment complexes in Fayetteville that primarily rent to University of Arkansas students. Historically, 85% of tenants pay their rent on time. RR wants to be paid on time but also wants to help students improve their credit score. Since late payments negatively impact a tenant’s credit score, RR would like to decrease late payments in the future. A new program was implemented where if a tenant was not late with their rent during a six-month period, then RR would be give them a $50 Wal-Mart gift card. After six months of the program, 44 out of 50 tenants paid their rent on time. What is the p-value?
Answer:
p-value= 0.27760
Step-by-step explanation:
Hello!
The RR apartment complex wants to decrease the proportion of tenants that pay their rent late. For this, they implemented the 6month pay on time and receive a Wal-Mart $50 gift card program.
If the historical proportion of tenants that pay on time is 85%, then the proportion of tenants that are late with their rent is 15%
If they want to test whether the proportion of the tenants that pay late decreased with the new program, the hypothesis is:
H₀: ρ ≥ 0.15
H₁: ρ < 0.15
α:?
The statistic is:
Z= ^ρ - ρ ≈ N(0;1)
√[(ρ(1 - ρ))/n]
If 44 out of 50 paid on time, this means that 6 paid late. Since for this hypothesis we are testing the proportion of people that paid late, these 6 tenants will be our number of "success"
The sample proportion is: ^ρ= 06/50 = 0.12
Z= 0.12 - 0.15 = -0.59
√[(0.15*0.85)/50]
The p-value is one-tailed (left) as the test:
P(Z ≤ -0.59) = 0.27760
I hope you have a SUPER day!
The head of institutional research at a university believed that the mean age of full-time students was declining. In 1995，the mean age of a full-time student was known to be 27.4 years. After looking at the enrollment records of all 4934 full-time students in the current semester, he found that the mean age was 27.1 years, with a standard deviation of 7.3 years. He conducted a hypothesis of H0: μ = 27.4 years versus H1: μ < 27.4 years and obtained a P-value of 0.0020. He concluded that the mean age of full-time students did decline. Is there anything wrong with his research?A. No, the hypothesis test was conducted correctly, and the correct conclusion was made.B. Yes, a P-value only indicates the likelihood of getting a result as extreme or more extreme as the one found, the head of institutional research needs to include a confidence level.C. Yes, the head of institutional research has access to the entire population, inference is unnecessary. He can say with 100% confidence that the mean age has decreased.D. Yes, the head of institutional research stated the hypotheses incorrectly; a left-tailed hypothesis test was conducted instead of a two-tailed test.
Answer:
A. No, the hypothesis test was conducted correctly, and the correct conclusion was made.
Step-by-step explanation:
Given that the head of institutional research at a university believed that the mean age of full-time students was declining. In 1995，the mean age of a full-time student was known to be 27.4 years.
For this he has recorded all 4934 students and got average as 27.1 years
This is very much perfect because the question is about the mean age of a full time student. when all full time students in the university are taken into account and mean arrived this is a reliable procedure.
The hypothesis also is right as this is left tailed test.
Since p =0.002<0.05 he rejected null hypothesis.
Conclusion is the mean age did decline. Everything is right.
A. No, the hypothesis test was conducted correctly, and the correct conclusion was made.
As a general rule, the sampling distribution of the sample proportions can be approximated by a normal probability distribution whenever
a)np greater than or equal to 5
b) n(1-p) greater than or equal to 5
c) n greater than or equal to 30
d) both a and b are true
A continuous random variable is uniformly distributed between a and b. The probability density function between a and b is
a) zero.
b) (a - b).
c) (b - a).
d) 1/(b - a).
A population has a mean of 84 and a standard deviation of 12. A sample of 36 observations will be taken. The probability that the sample mean will be between 80.54 and 88.9 is
a) 0.0347.
b) 0.7200.
c) 0.9511.
d) None of the alternative answers is correct.
In probability, the sampling distribution of the sample proportions can be approximated by a normal probability distribution when D. Both a and b are true.
How to calculate the probability?The sampling distribution of the sample proportions can be approximated by a normal probability distribution whenever np is greater than or equal to 5 and when n(1-p) is greater than or equal to 5.
When the continuous random variable is uniformly distributed between a and b, the probability density function between a and b is 1/(b - a).
The probability that the sample mean will be between 80.54 and 88.9 will be:
= P[Z = 88.9 - 84)/(12/✓36)] - P(Z = 80.54 - 84)/(12/✓36)]
= P(Z = 2.45) - P(Z = -1.73)
= 0.9929 - 0.0418
= 0.9511
Therefore, the probability is 0.9511.
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A researcher wishes to estimate, with 95% confidence, the population proportion of adults who think the president of their country can control the price of gasoline. Her estimate must be accurate within 5% of the true proportion.
(a) No preliminary estimate is available. Find the minimum sample size needed.
(b) Find the minimum sample size needed, using a prior study that found that 40% of the respondents said they think their president can control the price of gasoline.
(c) Compare the results from parts (a) and (b).
To estimate the required sample size to estimate the population proportion, we can use the formula n = (Z^2 * p * (1 - p)) / E^2. In part (a), assume p=0.5, while in part (b), use a known value of p=0.4 from a prior study. Compare the results in part (c).
Explanation:To find the minimum sample size needed to estimate the population proportion of adults who think the president of their country can control the price of gasoline with a 95% confidence level and an accuracy within 5%, we can use the formula:
n = (Z^2 * p * (1 - p)) / E^2
Where:
n = required sample size
Z = Z-value for the desired confidence level (for 95%, Z-value is approximately 1.96)
p = estimated proportion from a prior study or 0.5 if no prior study is available
E = desired margin of error (5% or 0.05)
Using this formula, we can plug in the values and calculate the minimum sample size. In part (a), p is assumed to be 0.5, while in part (b), the p-value is given as 0.4 from a prior study. Finally, in part (c), you can compare the results obtained from the different approaches.
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To estimate a population proportion with 95% confidence and an error of 5%, a sample size of approximately 385 is needed without a preliminary estimate. If a prior estimate of 40% is used, the sample size required is about 370. Using an estimate closer to 0.5 generally results in a larger, more conservative sample size.
A researcher wishes to estimate the population proportion of adults who think the president of their country can control the price of gasoline with 95% confidence. The estimate must be accurate within 5% of the true proportion.
No preliminary estimate available: When you do not have a preliminary estimate, the most conservative approach is to use 0.5 as the estimated population proportion (p). The formula for the minimum sample size (n) is:
n = (Z² × p × (1 - p)) / E²For a 95% confidence level, Z is 1.96, and the margin of error (E) is 0.05.n = (1.96² × 0.5 × 0.5) / 0.05² = 384.16Therefore, the minimum sample size needed is approximately 385.
Using a prior study where 40% of respondents think the president can control gasoline prices: Here, p is 0.4.
n = (Z² × p × (1 - p)) / E²n = (1.96² × 0.4 × 0.6) / 0.05² = 369.6Thus, the minimum sample size needed is approximately 370.
Comparison: The sample size required using no preliminary estimate (385) is slightly larger than the sample size required using a prior study (370). Using an estimated proportion closer to 0.5 increases the required sample size, ensuring a more conservative (more confident) estimate.A television set marked $350 is sold at a discount of 15%.?
Answer:
$297.50
Step-by-step explanation:
Do 15% × $350 = $52.50
Subtract that from the original
350 - 52.50 =$297.5
Answer:
$297.50
Step-by-step explanation:
15% of 350 = 0.15 x 350
0.15 x 350 = 52.50
350 - 52.50 = 297.5
Hope This Helps! :D
Katie is biking to the beach. The graph gives her distance from the beach x hours after she began riding.
1. What does the x-intercept mean?
2. What does the y-intercept mean?
(Examine the labels on the horizontal and vertical axes).
3. Find the slope of the line using the slope formula. Show all work. What does the slope represent in this problem? Use a complete sentence to answer.
Answer:
1. Distance from Katie's House to the beach = 75 km
2. Hours that it takes to get to the beach = 6 hrs
3. -12.5 which tells us the rate at which katie traveled which was 12.5 km per hour
Steps to 3.
Rise/Run = -75/6 = -12.5 = slope
Answer:
Step-by-step explanation:
The graph is a plot of distance in kilometers on the vertical axis against time, in hours on the horizontal axis.
1) The x-intercept is the point on the horizontal axis at which y = 0. This means that at this point, the distance from the beach to her house is zero. From the graph, the point is 6 hours.
2) The y-intercept is the point on the vertical axis at which x = 0. This means that this point is the point at which she started to ride. At this point,time = 0 hours and the distance from the beach to her house is 75 kilometers.
3) slope is expressed as
Change in y / change in x
We will pick 2 points on the y axis and two corresponding points on the x axis.
Change in y = y2 - y1
Change in x = x2 - x1
Slope = (0-75) / (6 - 0) = -75/6
Slope = - 12.5 kilometers per hour
The slope represents the rate at which her distance from the beach changes with time. It represents her velocity. It is negative because her distance from the beach is decreasing with time
You wish to test the claim that p > 33 at a level of significance of a = 0.05 and are given sample 19) statistics n = 5O x = 33.3. Assume the population standard deviation is 12. Compute the value of the standardized test statistic. Round your answer to two decimal places.
Answer:
test statistic is 0.176
Step-by-step explanation:
Given Data
p>33
a=0.05
n=50
x=33.3
d(population deviation)=12
Test statistics=?
Solution
Test statistic z=(p-x)\(d/sqrt(50))
z=(33.3-30)\(12\sqrt(50))
z=0.176
A study was conducted to identify the relationship between the hours of practice put in by the University at Buffalo football team and the success they had in games won. The Pearson's correlation coefficient was found to be .78. What type of relationship does this represent- positive or direct/negative or inverse? What is the strength of this relationship- weak, moderate, or strong?
A. Direct, strong
B. Direct, moderate
C. Inverse, strong
D. Inverse moderate.
Answer:
A. Direct, strong
Step-by-step explanation:
If the Pearson's correlation coefficient is positive the relationship coefficient is direct. If it is negative, it is inverse.
If is considerate to be strong if it is larger than 0.7
In this problem, we have that:
The Pearson's correlation coefficient was found to be .78. It is positive, and larger than 0.7. So the correct answer is:
A. Direct, strong
Which equation has the solutions x = -3 ± √3i/2 ?
2x2 + 6x + 9 = 0
x2 + 3x + 12 = 0
x2 + 3x + 3 = 0
2x2 + 6x + 3 = 0
Answer:Answer is option C : [[tex]x^{2}[/tex] + 3x + 3 ] =0
Note: None of options matches with given question.
instead of "-3" , there should be "-[tex]\frac{3}{2}[/tex]".
Step-by-step explanation:
Note: None of options matches with given question.
instead of "-3" , there should be "[tex]\frac{3}{2}[/tex]".
Here, First thing you have to observe the nature of roots.
∴ x = -[tex]\frac{3}{2}[/tex]+[tex]\frac{\sqrt{3}}{2}[/tex]i and x = -[tex]\frac{3}{2}[/tex]-[tex]\frac{\sqrt{3}}{2}[/tex]
∴ [ x+([tex]\frac{3}{2}[/tex]-[tex]\frac{\sqrt{3}}{2}[/tex]i) ][ x+([tex]\frac{3}{2}[/tex]+[tex]\frac{\sqrt{3}}{2}[/tex]i) ]=0
∴ [ [tex]x^{2}[/tex] + x([tex]\frac{3}{2}[/tex]+[tex]\frac{\sqrt{3}}{2}[/tex]i)+ x([tex]\frac{3}{2}[/tex]-[tex]\frac{\sqrt{3}}{2}[/tex]i) + ([tex]\frac{3}{2}[/tex]-[tex]\frac{\sqrt{3}}{2}[/tex]i)([tex]\frac{3}{2}[/tex]+[tex]\frac{\sqrt{3}}{2}[/tex]i) ]=0
∴ [[tex]x^{2}[/tex] + [tex]\frac{3}{2}[/tex]x + [tex]\frac{\sqrt{3}}{2}[/tex]ix + [tex]\frac{3}{2}[/tex]x - [tex]\frac{\sqrt{3}}{2}[/tex]ix + (3-[tex]\frac{\sqrt{3}}{2}[/tex]i)(3+[tex]\frac{\sqrt{3}}{2}[/tex]i) ] =0
∴ [[tex]x^{2}[/tex] + 3x + ([tex]\frac{3}{2}[/tex]-[tex]\frac{\sqrt{3}}{2}[/tex]i)([tex]\frac{3}{2}[/tex]+[tex]\frac{\sqrt{3}}{2}[/tex]i) ] =0
∴ [[tex]x^{2}[/tex] + 3x + [tex]\frac{9}{4}[/tex] - ([tex]\frac{\sqrt{3}}{2}[/tex]i)([tex]\frac{\sqrt{3}}{2}[/tex]i) ] =0
∴ [[tex]x^{2}[/tex] + 3x + [tex]\frac{9}{4}[/tex] - ([tex]\frac{3}{4}[/tex]) [tex]i^{2}[/tex] ] =0
∴ [[tex]x^{2}[/tex] + 3x + [tex]\frac{9}{4}[/tex] + ([tex]\frac{3}{4}[/tex]) ] =0
∴ [[tex]x^{2}[/tex] + 3x + [tex]\frac{12}{4}[/tex] ] =0
∴ [[tex]x^{2}[/tex] + 3x + 3 ] =0
Thus, Answer is option C : [[tex]x^{2}[/tex] + 3x + 3 ] =0
Answer:
C
Step-by-step explanation:
The amount of filling in a Doughiest Donut Boston cream donut follows a Normal distribution, with a mean of 3 ounces and a standard deviation of 0.4 ounce. A random sample of 36 donuts is selected every day and measured. What is the probability the mean weight will exceed 3.1 ounces?
Answer:
The probability the mean weight will exceed 3.1 ounces is 0.0668
Step-by-step explanation:
We have a random sample of size n = 36 measures which comes from a normal distribution with a mean of 3 ounces and a standard deviation of 0.4 ounces. Then, we know that the mean weight is also normally distributed with the same mean of 3 ounces and a standard deviation of [tex]0.4/\sqrt{36} = 0.4/6[/tex]. The z-score associated to 3.1 is (3.1-3)/(0.4/6) = 1.5. We are looking for P(Z > 1.5) = 0.0668, i.e., the probability the mean weight will exceed 3.1 ounces is 0.0668
g Twenty percent of drivers driving between 11 PM and 3 AM are drunken drivers. Using the binomial probability formula, find the probability that in a random sample of 12 drivers driving between ll PM and 3 AM, two to four will be drunken drivers. (Round to 4 digits, ex. 0.1234)
Answer: p(2 lesser than or equal to x lesser than or equal to 4) = 0.6526
Step-by-step explanation:
20% of drivers driving between 11 PM and 3 AM are drunken drivers.
We want to use the binomial distribution to determine the probability that in a random sample of 12 drivers driving between 11 PM and 3 AM, two to four will be drunken drivers.
The formula for binomial distribution is
P( x = r) = nCr × q^n-r × p^r
x = number of drivers
p = probability that the drivers that are drunken.
q= 1-p = probability that the drivers are not drunken.
n = number of sampled drivers.
From the information given,
p = 20/100 = 0.2
q = 1 - p = 1 - 0.2 = 0.8
n = 12
We want to determine
p(2 lesser than or equal to x lesser than or equal to 4)
It is equal to p(x=2) + p(x= 3) + p(x=4)
p(x=2) = 12C2 × 0.8^10 × 0.2^2 = 0.2835
p(x=3) = 12C3 × 0.8^9 × 0.2^3 = 0.2362
p(x=4) = 12C4 × 0.8^8 × 0.2^4 = 0.1329
p(2 lesser than or equal to x lesser than or equal to 4) = 0.2835 + 0.2362 + 0.1329 = 0.6526
The popularity of computer, video, online, and virtual reality games has raised concerns about their ability to negatively impact youth. The data in this exercise are based on a recent survey of 14 ‑ to 18 ‑year‑olds in Connecticut high schools. Assume the table displays the grade distributions of boys who have and have not played video games.
Grade Average
A's and B's C's D's and F's
Played games 730 444 190
Never played games 214 137 87
Give the conditional distribution of the grades of those who have played games. (Enter your answers rounded to two decimal places.)
Answer:
P(The grade of the boy is A| He has played video games)
is, [tex]\simeq 0.54[/tex]
P(The grade of the boy is B| He has played video games)
is [tex]\simeq 0.33[/tex]
P(The grade of the boy is C| He has played video games)
is [tex]\simeq 0.14[/tex]
Step-by-step explanation:
The total no. of boys who have played video games,
= (730 + 444 + 190)
=1360
Now, from the given data,
P(The grade of the boy is A| He has played video games)
= [tex]\frac {730}{1360}[/tex]
[tex]\simeq 0.54[/tex]
P(The grade of the boy is B| He has played video games)
= [tex]\frac {444}{1360}[/tex]
[tex]\simeq 0.33[/tex]
P(The grade of the boy is C| He has played video games)
= [tex]\frac {190}{1360}[/tex]
[tex]\simeq 0.14[/tex]
The conditional distribution of grades for boys who have played games shows that 53% received A's and B's, 33% received C's, and 14% received D's and F's. This is calculated by dividing each grade category by the total number of boys who played games and rounding to two decimal places.
To find the conditional distribution of grades for boys who have played games, we need to calculate the proportion of each grade category relative to the total number who played.
Steps to Calculate Conditional Distribution
Calculate the total number of boys who played games:Therefore, the conditional distribution of grades for boys who have played games is 0.53 for A's and B's, 0.33 for C's, and 0.14 for D's and F's.