You play two games against the same opponent. The probability you win the first game is 0.7. If you win the first​ game, the probability you also win the second is 0.5. If you lose the first​ game, the probability that you win the second is 0.3.(a) Are the two games independent?(b) What's the probability you lose both games?

We can calculate the 92% confidence interval for the mean assembly time to be approximately 15.572 to 16.828 minutes. The information provided, however, is insufficient to answer part b) of the question.

In order to construct a 92% confidence interval for the mean assembly time, we first have to find the z-score corresponding to this level of confidence. The z-score for a 92% confidence interval is approximately 1.75. This value is found using a standard normal distribution table (or z-table).

So, the 92% confidence interval for the mean assembly time of 16.2 minutes can be computed as follows: 16.2 ± 1.75*(3.6/sqrt(120)). This gives a confidence interval of approximately 15.572 to 16.828 minutes.

Regarding the second part of your question (b), it seems unfinished and lacks enough information for me to provide a complete answer. Typically, a desired margin of error or a certain level of confidence would need to be specified to determine the sample size.

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a) The 92% confidence interval for the mean assembly time is (15.75 minutes, 16.65 minutes).  b) About 993 workers are needed for the study to estimate the mean assembly time with a 0.2-minute margin of error.

To construct the confidence interval, use the formula for the confidence interval of the mean for a normal distribution: [tex]\[ \text{CI} = \bar{x} \pm z \times \frac{\sigma}{\sqrt{n}} \][/tex]

1. Calculate the standard error of the mean: [tex]\[ \text{SE} = \frac{\sigma}{\sqrt{n}} = \frac{3.6}{\sqrt{120}} \approx 0.329 \][/tex]

2. Find the critical z-value for a 92% confidence interval (using a z-table or calculator): [tex]\( z = 1.75 \) (approximately).[/tex]

3. Substitute the values into the formula: [tex]\[ \text{CI} = 16.2 \pm 1.75 \times 0.329 \][/tex]

4. Calculate the confidence interval: [tex]\[ \text{CI} = (16.2 - 0.575, 16.2 + 0.575) \][/tex]

[tex]\[ \text{CI} = (15.625, 16.775) \][/tex]

So, the 92% confidence interval for the mean assembly time is approximately (15.625 minutes, 16.775 minutes).

b) To estimate the mean assembly time with a desired margin of error, use the formula for the required sample size: [tex]\[ n = \left( \frac{z \times \sigma}{\text{ME}} \right)^2 \][/tex]

Given the desired margin of error (ME), let's say 0.2 minutes, and using \( z = 1.75 \) from the 92% confidence level:

[tex]\[ n = \left( \frac{1.75 \times 3.6}{0.2} \right)^2 \]\[ n = (31.5)^2 \]\[ n = 992.25 \][/tex]

So, approximately 993 workers should be involved in the study to estimate the mean assembly time with a margin of error of 0.2 minutes.

Answer:

Option a)  H0 : μ = 61 , Hα : μ > 61 .

Step-by-step explanation:

We are given that the time, in number of days, until maturity of a certain variety of tomato plant is Normally distributed, with mean μ and standard deviation s = 2.4.

You select a simple random sample of four plants of this variety and measure the time until maturity. The sample yields x bar = 65.

You read on the package of seeds that these tomatoes reach maturity, on average, in 61 days.

Now, since we know about;

sample size, n = 4

sample standard deviation, s = 2.4

sample, x bar = 65

The only line which represent information about the population mean is reading on the package of seeds that these tomatoes reach maturity, on average, in 61 days.

This shows that population mean , [tex]\mu[/tex] = 61 and in null and alternate hypothesis only population mean is tested.

So, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu[/tex] = 61

Alternate Hypothesis, [tex]H_1[/tex] : [tex]\mu[/tex] > 61 { we are asked for later than expected}

Answer:

a. H + K = {u + v}

b. See explanation below

Step-by-step explanation:

Given

H+K={w:w=u+v for some u in H and some v in K}

a.

From the (given) above,

We have that

0 = v1 + 0.u

v1 represents the zero vector in K

We also have that

0 = v + u1

u1 represents the zero vector in H

Up this point, we've shown that both H and K have vector factors, V

We'll use the available data to solve for the zero vector of H + K

This is given by

u1 + v1 + u2 + v2 ---- Rearrange

u1 + u2 + v1 + v2 ---- Let u3 = u1 + u2 and v3 = v1 + v2.

So, we have

u3 + v3 as the zero vector of H + K

H + K = {u + v}

So, H + K is a subspace of V

b

Since H, K are subspace of V, then they are

1. Both close under scalar multiplication

2. Both closed under vector addition

1 and 2 gives

c(u + v) where c is a constant

= c.u + c.v

We can then conclude that H is a subspace of H+K and K is a subspace of H+K because H + K is closed under scalar multiplication and vector addition

The sum of two subspaces H and K is a subspace of V as it contains the zero vector, is closed under addition, and under scalar multiplication. Additionally, H and K themselves also qualify as subspaces of this resultant sum subspace, H+K.

To show that H+K is a subspace of V, we need to prove that it satisfies three properties: it contains the zero vector, it is closed under addition, and it is closed under scalar multiplication.

1. Zero Vector: Since H and K are subspaces, they both contain the zero vector. So, 0 = 0 + 0 represents a vector in H+K (with the first 0 from H and the second from K).

2. Closure under Addition: If w1 and w2 are vectors in H+K, w1 can be written as h1 + k1 (h1 in H, k1 in K) and w2 as h2 + k2 (h2 in H, k2 in K). Adding w1 and w2 gives: (h1 + k1) + (h2 + k2) = (h1 + h2) + (k1 + k2), where h1+h2 is in H and k1+k2 is in K, hence, the result is in H+K.

3. Closure under Scalar Multiplication: Let w be in H+K and c be any scalar. Then w = h + k, so cw = c(h + k) = ch + ck, where ch is in H and ck is in K, hence, cw is in H+K.

To show that H is a subspace of H+K and K is a subspace of H+K, note that for any vector h in H, it can be written as h+0 (with 0 as the zero vector of K), and similarly for any vector k in K as 0+k. Thus, H and K are subspaces of H+K.

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Answer:

a) Let X the random variable that represent the diameters of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(2.514 cm,0.008 cm)[/tex]  

Where [tex]\mu=2.514[/tex] and [tex]\sigma=0.008[/tex]

b) For this case we can see the interval required on the figure attached,

c) [tex]P(2.49<X<2.51)=P(\frac{2.49-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{2.51-\mu}{\sigma})=P(\frac{2.49-2.514}{0.008}<Z<\frac{2.51-2.514}{0.008})=P(-3<z<-0.5)[/tex]

And we can find this probability with this difference:

[tex]P(-3<z<-0.5)=P(z<-0.5)-P(z<-3)[/tex]

And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.  

[tex]P(-3<z<-0.5)=P(z<-0.5)-P(z<-3)=0.309-0.00135=0.307[/tex]

Step-by-step explanation:

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".  

Part a

Let X the random variable that represent the diameters of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(2.514 cm,0.008 cm)[/tex]  

Where [tex]\mu=2.514[/tex] and [tex]\sigma=0.008[/tex]

Part b

For this case we can see the interval required on the figure attached,

Part c

We are interested on this probability

[tex]P(2.49<X<2.51)[/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]

If we apply this formula to our probability we got this:

[tex]P(2.49<X<2.51)=P(\frac{2.49-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{2.51-\mu}{\sigma})=P(\frac{2.49-2.514}{0.008}<Z<\frac{2.51-2.514}{0.008})=P(-3<z<-0.5)[/tex]

And we can find this probability with this difference:

[tex]P(-3<z<-0.5)=P(z<-0.5)-P(z<-3)[/tex]

And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.  

[tex]P(-3<z<-0.5)=P(z<-0.5)-P(z<-3)=0.309-0.00135=0.307[/tex]

Answer:

Part a

For this case after do the operations we got a value for the correlation coeffcient of:

[tex] r =0.506[/tex]

With this value we can find the determination coefficient:

[tex] r^2 = 0.506^2 = 0.256[/tex]

And with this value we can analyze the proportion of variance explained by one variable and the other. So we can conclude that 25.6% of the mother's BMI variation is explained by the daugther's BMI.

Part a

Since the BMI is a relation between height and weight, other possible variables that can explain the variability are (weight , height, age)  

Step-by-step explanation:

Previous concepts

The correlation coefficient is a "statistical measure that calculates the strength of the relationship between the relative movements of two variables". It's denoted by r and its always between -1 and 1.

And in order to calculate the correlation coefficient we can use this formula:  

[tex]r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^2 -(\sum x)^2][n\sum y^2 -(\sum y)^2]}}[/tex]  

Solution to the problem

Part a

For this case after do the operations we got a value for the correlation coeffcient of:

[tex] r =0.506[/tex]

With this value we can find the determination coefficient:

[tex] r^2 = 0.506^2 = 0.256[/tex]

And with this value we can analyze the proportion of variance explained by one variable and the other. So we can conclude that 25.6% of the mother's BMI variation is explained by the daugther's BMI.

Part a

Since the BMI is a relation between height and weight, other possible variables that can explain the variability are (weight , height, age)  

Final answer:

The correct mean time for the first three finishers is 5 minutes, after subtracting the 45-second error from the initially recorded mean time of 5.75 minutes.

Explanation:

The correct mean time for the first three finishers, after adjusting for the stopwatch error, can be calculated by subtracting the 45 seconds from the initially recorded meantime. Since the mean time was calculated to be 5.75 minutes (or 345 seconds), we must correct each time by subtracting 45 seconds and then recalculating the mean.

To find the correct mean time, we do the following steps:

The correct mean time for the first three finishers is therefore 5 minutes.

Answer:

Tuesday's Z-score is 7.56

Step-by-step explanation:

We are given that a department store, on average, has daily sales of 28,651.79 and the standard deviation of sales is $1000.

Also, it is given that on Tuesday, the store sold $36,211.08 worth of goods.

Let X = Daily sales of goods

So, X ~ N([tex]\mu = 28,651.79, \sigma^{2} =1000^{2}[/tex])

The z-score probability distribution is given by;

        Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ standard normal N(0,1)

Now, Tuesday,s Z-score is given by;

      Z = [tex]\frac{36,211.08-28,651.79}{1000}[/tex] = 7.56

Yes, Tuesday was an unusually good day as on this day more worth of sales takes place as compared to the average daily sales of $28,651.79 .

Answer:

The worth of the company in 2000 is $56,000.

Step-by-step explanation:

The growth rate of the company is:

[tex]f'(t)=2000-12t^{2}[/tex]

To determine the worth of the company in 2000, first compute the change in the net worth during the period 1990 (t = 0) to 2000 (t = 10) as follows:

[tex]\int\limits^{10}_{0} {2000-12t^{2}} \, dt =\int\limits^{10}_{0} {2000} \, dt-12\int\limits^{10}_{0} {t^{2}} \, dt=2000 |t|^{10}_{0}-12|\frac{t^{3}}{3}|^{10}_{0}\\=(2000\times10)-(4\times10^{3})\\=20000-4000\\=16000[/tex]

The increase in the company's net worth from 1990 to 2000 is $16,000.

If the company's worth was $40,000 in 1990 then the worth of the company in 2000 is:

Worth in 2000 = Worth in 1990 + Net increase in company's worth

                        [tex]=40000+16000\\=56000[/tex]

Thus, the worth of the company in 2000 is $56,000.

Answer:

[tex]\dfrac{d}{dx}f(x)=\dfrac{e^{2x}+e^{-2x}}{2}[/tex]

Step-by-step explanation:

It is given that

[tex]\sinh x=\dfrac{e^x-e^{-x}}{2}[/tex]

[tex]\cosh x=\dfrac{e^x+e^{-x}}{2}[/tex]

[tex]f(x)=\sinh x\cosh x=[/tex]

Using the given hyperbolic functions, we get

[tex]f(x)=\dfrac{e^x-e^{-x}}{2}\times \dfrac{e^x+e^{-x}}{2}[/tex]

[tex]f(x)=\dfrac{(e^x)^2-(e^{-x})^2}{4}[/tex]

[tex]f(x)=\dfrac{e^{2x}-e^{-2x}}{4}[/tex]

Differentiate both sides with respect to x.

[tex]\dfrac{d}{dx}f(x)=\dfrac{d}{dx}\left(\dfrac{e^{2x}-e^{-2x}}{4}\right )[/tex]

[tex]\dfrac{d}{dx}f(x)=\left(\dfrac{2e^{2x}-(-2)e^{-2x}}{4}\right )[/tex]

[tex]\dfrac{d}{dx}f(x)=\left(\dfrac{2(e^{2x}+e^{-2x})}{4}\right )[/tex]

[tex]\dfrac{d}{dx}f(x)=\dfrac{e^{2x}+e^{-2x}}{2}[/tex]

Hence, [tex]\dfrac{d}{dx}f(x)=\dfrac{e^{2x}+e^{-2x}}{2}[/tex].

Answer:

[tex]E(T)=E(X1 + X2 + X3 + X4 =E(x1) + E(x2) + E(x3) + E(x4) [/tex]

[tex]E(T) =900+900+900+900 =3600[/tex]

And the mean would be:

[tex] \bar X = \frac{T}{4} = \frac{3600}{4}= 900[/tex]

And the standard deviation of total time would be:

[tex]SD(T)=\sqrt(Var(T)) = sqrt(Var(X1) + Var(x2) + Var(x3) + Var(x4))[/tex]

[tex]\sigma=sqrt(30^2+30^2+30^2+30^2) =\sqrt(3600) =60[/tex]

Step-by-step explanation:

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 total life time of t four transistors T = X1 + X2 + X3 + X4 (where X1,X2,X3 and X4 are life time of individual transistors

For this case the mean length of time that four transistors will last

[tex]E(T)=E(X1 + X2 + X3 + X4 =E(x1) + E(x2) + E(x3) + E(x4) [/tex]

[tex]E(T) =900+900+900+900 =3600[/tex]

And the mean would be:

[tex] \bar X = \frac{T}{4} = \frac{3600}{4}= 900[/tex]

And the standard deviation of total time would be:

[tex]SD(T)=\sqrt(Var(T)) = sqrt(Var(X1) + Var(x2) + Var(x3) + Var(x4))[/tex]

[tex]\sigma=sqrt(30^2+30^2+30^2+30^2) =\sqrt(3600) =60[/tex]

The mean lifetime of four transistors is 3600 hours and the standard deviation of the lifetime is 60 hours.

The mean and standard deviation are statistical concepts used to describe a dataset. The mean, or average, is the sum of all data points divided by the number of data points. In this case, the mean lifetime of a single transistor is 900 hours. If you have four transistors, their total lifetime would be four times the mean of a single transistor. So, the mean lifetime for four transistors is 4 * 900 = 3600 hours.

The standard deviation measures the amount of variation in a set of values. When you are considering the lifetime of multiple transistors, you add the variances - not the standard deviations - and then take the square root. The variance is the square of the standard deviation. So, the standard deviation for the lifetime of four transistors is sqrt(4) * 30 = 60 hours.

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Answer:

Step-by-step explanation:

The number of observations in model 1 are more compared with model 2 and model 3. And, the coefficient values increased from model 1 to model 3. Moreover, the standard error values are increased. Hence, the test statistic value increases. Therefore, the variable “difference in match win probabilities” is significant.

State the null and alternative Hypotheses:

Null Hypothesis:

H0: The difference in match win probabilities is not significant.

Alternative Hypothesis:

H1: The difference in match win probabilities is significant.

From the given information (model 3), the coefficient of “The difference in match win probabilities” is -0.550 and the corresponding standard error is 0.240. The sample size is 1973.

The test statistic is -2.2917 and it is calculated below: on the picture attached.

Statistics homework question answer, step 1, image 1

The number of sample observations is n=1973 and consider the level of significance as 0.05.

The degree of freedom is, n – 1 = 1973–1 = 1972.

Answer:

a. (1.811, 1.929)

Step-by-step explanation:

We have to find find M as such

[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]

In which, z is Z(a/2), [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample. So

[tex]M = 2.58*\frac{0.125}{\sqrt{30}} = 0.059[/tex]

The lower end of the interval is the mean subtracted by M. So it is 1.87 - 0.059 = 1.811.

The upper end of the interval is the mean added to M. So it is 1.87 + 0.059 = 1.929

So the correct answer is:

a. (1.811, 1.929)

Answer:

15 sec

Step-by-step explanation:

Data:

let m = 600 kg

    b  = 50

the differential equation will be:

[tex]m\frac{dv}{dt} = mg - bv(t)[/tex]

The equation of motion of the object can be determined considering the forces acting on it, including the force due to gravity and the force due to air resistance. The equation can be rearranged and solved using the given values in Newton's method to approximate the time when the object strikes the ground.

The equation of motion of the object can be determined by considering the forces acting on it. The main force acting on the object is the force due to gravity, which is given by F = mg, where m is the mass of the object and g is the acceleration due to gravity. However, in this case, we also need to consider the force due to air resistance, which is proportional to the velocity of the object.

Let's denote the velocity of the object as v and the constant of proportionality for the force due to air resistance as b. The equation of motion can then be written as:

m(dv/dt) = mg - bv

This equation can be rearranged to get:

(dv/dt) + (b/m)v = g

Now, we can solve this differential equation to find the velocity as a function of time, and then integrate the velocity to get the position as a function of time. However, since the exponential term is too large to ignore, we can use Newton's method to approximate the time when the object strikes the ground. The equation of motion and the value of g are given, so we can substitute those values into Newton's method to solve for the time when the object strikes the ground.

Answer:

The 99% confidence interval for the population mean hours spent watching television per month is between 143.07 hours and 158.93 hours.

Step-by-step explanation:

We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:

[tex]\alpha = \frac{1-0.99}{2} = 0.005[/tex]

Now, we have to find z in the Ztable as such z has a pvalue of [tex]1-\alpha[/tex].

So it is z with a pvalue of [tex]1-0.005 = 0.995[/tex], so [tex]z = 2.575[/tex]

Now, find M as such

[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]

In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.

[tex]M = 2.575*\frac{32}{\sqrt{108}} = 7.93[/tex]

The lower end of the interval is the mean subtracted by M. So it is 151 - 7.93 = 143.07 hours

The upper end of the interval is the mean added to M. So it is 151 + 7.93 = 158.93 hours

The 99% confidence interval for the population mean hours spent watching television per month is between 143.07 hours and 158.93 hours.

Answer:

(B) 4 minutes after the rocket was launched, the rocket is 516 miles above the ground

Step-by-step explanation:

The function h(t) represents the height of the rocket above the launchpad after a time t minutes.

h(4) = 516 means that when t = 4 minutes, the height h is 516 miles above the launch pad. Note that the time t is measured from when the rocket is launched.

The first option indicates a rate of change, which is a velocity. This is not indicated in the question because velocity is a time derivative of the the height function.

Option C implies that the rocket is 516 miles currently but we do not know what time currently is from the time of launch.

The fourth option has reversed the roles of the variable by implying the time is 516 minutes while the height is 4 miles which is not what the function means.

Answer:

There is a 99% probability that \mu μ is between 3 and 9.

Step-by-step explanation:

For a confidence level of x% for a mean being found between a and b, it means that we are x% sure, that is, there is a x% probability that the true mean for the population is between a and b.

A 99% confidence interval for the mean μ of a population is computed from a random sample and found to be 6 ± 3.

99% probability that the true mean of the population is between 3 and 9.

So the correct answer is:

There is a 99% probability that \mu μ is between 3 and 9.

A 99% confidence interval of 6 ± 3 implies that we are 99% confident that the population mean is between 3 and 9. If further samples were taken, both the mean and margin of error could vary, depending on the specific data points and variability of each sample.

A 99% confidence interval refers to the range of values within which we are 99% confident that the population mean (μ) resides. In this case, the confidence interval is 6 ± 3, which implies that we are 99% confident that the mean is anywhere between 3 and 9. It does not mean that there is a 99% probability that the true mean is 6, or that the true margin of error is 3. The margin of error is a function of your data's variability and your sample size, not a fixed number.

If we take more random samples, the mean could be different because each sample may contain different individuals or data points with different distributions. However, if the sampling process is properly randomized and unbiased, on average, the mean of the samples should be close to the population mean.

While the margin of error would be expected to remain around 3 for similar sample sizes and variability, it is not guaranteed to always be 3, as it depends on the specific data in each sample.

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Answer:

0.16

Step-by-step explanation:

For a normal distribution, the 68-95-99.7 rule says that 68% of the distribution lies within 1 standard deviation from the mean, 95% within two standard deviations and 99.7% within three standard deviations.

From the question,

Mean μ = 298 ml

Standard deviation σ = 3 ml

A value of 295 ml is within one standard deviation = 68%.

Since it's on the lower side, it's within 68% ÷ 2 = 34%.

The lower half below the mean is 50% of the distribution. Hence, for a selection less than 1 standard deviation, the probability is

50% - 34% = 16% = 0.16

Final answer:

The probability that a randomly selected bottle contains less than 295 ml of cola is calculated using the normal distribution. With a z-score of -1, ths probability is approximately 15.87%.

Explanation:

To determine the probability that a randomly selected bottle contains less than 295 ml of cola, we can use the properties of the normal distribution.

Given that the mean is 298 ml and the standard deviation is 3 ml, we must first find the z-score associated with 295 ml.

The z-score formula is z = (X - mu) / SD where X is the value for which we want to find the probability.

Substituting the given values into the formula, we get:

z = (295 - 298) / 3 = -1.

A z-score of -1 tells us that 295 ml is one standard deviation below the mean.

To find the probability corresponding to this z-score, we refer to a standard normal distribution table or use relevant statistical software. If we look up the probability for z = -1 in the z-table, we find that the area to the left of z is approximately 0.1587.

Therefore, the probability that a randomly selected bottle contains less than 295 ml of cola is about 15.87%.

the answer is in the attachment

Answer: 63 defective widgets

Step-by-step explanation:

Given that the proportion should not exceed 5%, that is:

p< or = 5%.

So we take p = 5% = 0.05

q = 1 - 0.05 = 0.095

Where q is the proportion of non-defective

We need to calculate the standard error (standard deviation)

S = √pq/n

Where n = 1024

S = √(0.05 × 0.095)/1024

S = 0.00681

Since production is to maximize profit(profitable), we need to minimize the number of defective items. So we find the limit of defective product to make this possible using the Upper Class Limit.

UCL = p + Za/2(n-1) × S

Where a is alpha of confidence interval = 100 -90 = 10%

a/2 = 5% = 0.05

UCL = p + Z (0.05) × 0.00681

Z(0.05) is read on the t-distribution table at (n-1) degree of freedom, which is at infinity since 1023 = n-1 is large.

Z a/2 = 1.64

UCL = 0.05 + 1.64 × 0.00681

UCL = 0.0612

Since the UCL in this case is a measure of proportion of defective widgets

Maximum defective widgets = 0.0612 ×1024 = 63

Alternatively

UCL = p + 3√pq/n

= 0.05 + 3(0.00681)

= 0.05 + 0.02043 = 0.07043

UCL =0.07043

Max. Number of widgets = 0.07043 × 1024

= 72

The probability question from part (a) requires calculating the chance of getting all heads or all tails on multiple days in a year, which involves complex probability distributions. For part (b), using a Poisson distribution could be appropriate due to the rarity of the event and the high number of trials involved.

The question pertains to the field of probability theory and involves calculating the probability of specific outcomes when flipping a fair coin. For part (a), Jack flips a coin ten times each morning for a year, counting the days (X) when all flips are identical (all heads or all tails). The exact expression for P(X > 1), the probability of more than one such day, requires several steps. First, we find the probability of a single day having all heads or all tails, then use that to calculate the probability for multiple days within the year. For part (b), whether it is appropriate to approximate X by a Poisson distribution depends on the rarity of the event in question and the number of trials. A Poisson distribution is typically used for rare events over many trials, which may apply here.

For part (a), the probability on any given day is the sum of the probabilities of all heads or all tails: 2*(0.5^10). Over a year (365 days), we need to calculate the probability distribution for this outcome occurring on multiple days. To find P(X > 1), we would need to use the binomial distribution and subtract the probability of the event not occurring at all (P(X=0)) and occurring exactly once (P(X=1)) from 1. However, this calculation can become quite complex due to the large number of trials.

For part (b), given the low probability of the event (all heads or all tails) and the high number of trials (365), a Poisson distribution may be an appropriate approximation. The mean (λ) for the Poisson distribution would be the expected number of times the event occurs in a year. Since the probability of all heads or all tails is low, it can be considered a rare event, and the Poisson distribution is often used for modeling such scenarios.

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Answer:

[tex]A_h=150\sqrt{3}\ m^2[/tex]

Step-by-step explanation:

Regular Hexagon

For the explanation of the answer, please refer to the image below. Let's analyze the triangle shown inside of the hexagon. It's a right triangle with sides x,y, and z.

We know that x is half the length of the side length of the hexagon. Thus

[tex]x=5 m[/tex]

Note that this triangle repeats itself 12 times into the shape of the hexagon. The internal angle of the triangle is one-twelfth of the complete rotation angle, i.e.

[tex]\theta=360/12=30^o[/tex]

Now we have [tex]\theta[/tex], the height of the triangle y is easily found by

[tex]\displaystyle tan30^o=\frac{x}{y}[/tex]

Solving for y

[tex]\displaystyle y=\frac{x}{tan30^o}=\frac{5}{ \frac{1} {\sqrt{3} }}=5\sqrt{3}[/tex]

The value of z can be found by using

[tex]\displaystyle sin30^o=\frac{x}{z}[/tex]

[tex]\displaystyle z=\frac{x}{sin30^o}=\frac{5}{\frac{1}{2}}=10[/tex]

The area of the triangle is

[tex]\displaystyle A_t=\frac{xy}{2}=\frac{5\cdot 5\sqrt{3}}{2}=\frac{25\sqrt{3}}{2}[/tex]

The area of the hexagon is 12 times the area of the triangle, thus

[tex]\displaystyle A_h=12\cdot A_t=12\cdot \frac{25\sqrt{3}}{2}=150\sqrt{3}[/tex]

[tex]\boxed{A_h=150\sqrt{3}\ m^2}[/tex]

Answer:

T= A person selected in the poll travel to Europe

NT= A person selected in the poll NOT travel to Europe

For this case we have the following respondents for each event

n(T)= 68

n(NT) = 124

So then the total of people for the poll are:

[tex] n = n(T) + n(NT)= 68 +124= 192[/tex]

And we are interested on the probability of getting someone who has traveled to Europe, and we can use the empirical definition of probability given by:

[tex] p =\frac{Possible}{Total}[/tex]

And if we replace we got:

[tex] p = \frac{n(T)}{n}= \frac{68}{192}= 0.354[/tex]

So then the probability of getting someone who has traveled to Europe is 0.354

Step-by-step explanation:

For this case we define the following events:

T= A person selected in the poll travel to Europe

NT= A person selected in the poll NOT travel to Europe

For this case we have the following respondents for each event

n(T)= 68

n(NT) = 124

So then the total of people for the poll are:

[tex] n = n(T) + n(NT)= 68 +124= 192[/tex]

And we are interested on the probability of getting someone who has traveled to Europe, and we can use the empirical definition of probability given by:

[tex] p =\frac{Possible}{Total}[/tex]

And if we replace we got:

[tex] p = \frac{n(T)}{n}= \frac{68}{192}= 0.354[/tex]

So then the probability of getting someone who has traveled to Europe is 0.354

Answer: 60 gallons of the 20% solution and 120 gallons of the 50% solution should be mixed.

Step-by-step explanation:

Let x represent the number of gallons of 20% solution that should be mixed.

Let y represent the number of gallons of 50% solution that should be mixed.

A 20% solution of fertilizer is to be mixed with a 50% solution of fertilizer in order to get 180 gallons of a 40% solution. This means that

0.2x + 0.5y = 0.4×180

0.2x + 0.5y = 72- - - - - - - - - - - -1

Since the total number of gallons is 180, it means that

x + y = 180

Substituting x = 180 - y into equation 1, it becomes

0.2(180 - y) + 0.5y = 72

36 - 0.2y + 0.5y = 72

- 0.2y + 0.5y = 72 - 36

0.3y = 36

y = 36/0.3

y = 120

x = 180 - y = 180 - 120

x = 60

To get a 40% solution by mixing a 20% and 50% solution, you need 60 gallons of the 20% solution and 120 gallons of the 50% solution.

To solve this problem, we can use the concept of mixtures. Let's represent the amount of 20% solution as x gallons and the amount of 50% solution as y gallons. From the given information, we can set up the following equations:

x + y = 180 (equation 1)

0.2x + 0.5y = 0.4 * 180 (equation 2)

Simplifying equation 2 gives us 0.2x + 0.5y = 72. To eliminate decimals, we can multiply both sides by 10 to get 2x + 5y = 720.

Now we have a system of equations. We can solve it by substitution or elimination. Let's use elimination:

Multiplying equation 1 by 2 gives us 2x + 2y = 360. Subtracting this from equation 2 gives us 3y = 360. Dividing both sides by 3 gives us y = 120. Substituting this value into equation 1 gives us x + 120 = 180. Subtracting 120 from both sides gives us x = 60.

Therefore, we need 60 gallons of the 20% solution and 120 gallons of the 50% solution.

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Step-by-step explanation:

Find the area using a proportion.

A / (πr²) = θ / 360°

A / (π (20 ft)²) = 160° / 360°

A ≈ 558.5 ft²

Answer:

a) the probability is 0.12 (12%)

b) the probability is 0.61 (61%)

c) the probability is 0.476 (47.6%)

a) the probability is 0.538 (53.8%)

Step-by-step explanation:

a) denoting the event H= hits her target and G= a gust of wind appears hen

P(H∩G) = probability that a gust of wind appears * probability of hitting the target given that is windy = 0.3* 0.4 = 0.12 (12%)

b) for any given shot

P(H)= probability that a gust of wind appears*probability of hitting the target given that is windy + probability that a gust of wind does not appear*probability of hitting the target given that is not windy = 0.3*0.4+0.7*0.7 = 0.12+0.49 = 0.61 (61%)

c) denoting P₂  as the probability of hitting once in 2 shots  and since the archer can hit in the first shot or the second , then

P₂ = P(H)*(1-P(H))+ (1-P(H))*P(H) = 2*P(H) *(1-P(H)) = 2*0.61*0.39= 0.476 (47.6%)

d) for conditional probability we can use the theorem of Bayes , where

M= the archer misses the shot → P(M) = 1- P(H) = 0.39

S= it is not windy when the archer shots → P(S) = 1- P(G) = 0.7

then

P(S/M) = P(S∩M)/P(M) = 0.7*(1-0.7)/0.39 = 0.538 (53.8%)

where P(S/M)  is the probability that there was no wind when the archer missed the shot

Answer:

The answer is 1/n

Step-by-step explanation:

First number = Alice's number

Second Number = Bob's Number

There are already 2 good answers if the first number is AT LEAST 1 greater than the second.

However, what if the first number is EXACTLY 1 greater than the second?

I will assume the first number is removed from the array after being selected.

If the first number is 1, then it is impossible, because the second number cannot be less than 1.

If the first number is 2, then the second number could be 1, which has a probability of 1/(n-1)

If the first number is 3, then the second number could be 2, which has a probability of 1/(n-1)

etc…

There is a 1/n probability of each first number selected, so the answer would be:

0*1/n+1/n*1/(n−1)+1/n*1/(n−1)+⋯+1/n*1/(n−1)

There are n terms in that series.

That gives us:

0+1/n(n−1)+1/n(n−1)+⋯+1/n(n−1)

There are n-1 of those identical terms, because 1 of the terms is 0.

(n−1)/n(n−1)=1/n

So the answer is 1/n.

Answer:

a) P ( 3 ≤X≤ 5 ) = 0.02619

b) E(X) = 1

Step-by-step explanation:

Given:

- The CDF of a random variable X = { 0 , 1 , 2 , 3 , .... } is given as:

                    [tex]F(X) = P ( X =< x) = 1 - \frac{1}{(x+1)*(x+2)}[/tex]

Find:

a.Calculate the probability that 3 ≤X≤ 5

b) Find the expected value of X, E(X), using the fact that. (Hint: You will have to evaluate an infinite sum, but that will be easy to do if you notice that

Solution:

- The CDF gives the probability of (X < x) for any value of x. So to compute the P (  3 ≤X≤ 5 ) we will set the limits.

                   [tex]F(X) = P ( 3=<X =< 5) = [1 - \frac{1}{(x+1)*(x+2)}]\limits^5_3\\\\F(X) = P ( 3=<X =< 5) = [-\frac{1}{(5+1)*(5+2)} + \frac{1}{(3+1)*(3+2)}}\\\\F(X) = P ( 3=<X =< 5) = [-\frac{1}{(42)} + \frac{1}{(20)}}]\\\\F(X) = P ( 3=<X =< 5) = 0.02619[/tex]

- The Expected Value can be determined by sum to infinity of CDF:

                   E(X) = Σ ( 1 - F(X) )

                   [tex]E(X) = \frac{1}{(x+1)*(x+2)} = \frac{1}{(x+1)} - \frac{1}{(x+2)} \\\\= \frac{1}{(1)} - \frac{1}{(2)}\\\\= \frac{1}{(2)} - \frac{1}{(3)} \\\\= \frac{1}{(3)} - \frac{1}{(4)}\\\\= ............................................\\\\= \frac{1}{(n)} - \frac{1}{(n+1)}\\\\= \frac{1}{(n+1)} - \frac{1}{(n+ 2)}[/tex]

                   E(X) = Limit n->∞ [1 - 1 / ( n + 2 ) ]  

                   E(X) = 1

Answer:

Complete question for your reference is below.

A researcher interested in weight control wondered whether normal and overweight individuals differ in their reaction to the availability of food. Thus, normal and overweight participants were told to eat as many peanuts as they desired while working on a questionnaire. One manipulation was the proximity of the peanut dish (close or far from the participant); the second manipulation was whether the peanuts were shelled or unshelled. After filling out the questionnaire, the peanut dish was weighed to determine the amount of peanuts consumed.

1. Identify the design (e.g., 2 X 2 factorial).

2. Identify the total number of conditions.

3. Identify the manipulated variable(s).

4. Is this an IV X PV design? If so, identify the participant variable(s).

5. Is this a repeated measures design? If so, identify the repeated variable(s).

6. Identify the dependent variable(s).

Step-by-step explanation:

Please find attached file for complete answer solution and explanation.

The research design is a 2x2 factorial with the manipulated or independent variables being the proximity of the peanut dish and whether the peanuts are shelled, resulting in four conditions. It is not an IV x PV design, nor a repeated measures design. The dependent variable in this study is the amount of peanuts consumed.

The research design posed in this question appears to be a 2x2 factorial design. This is because there are two independent variables (the proximity of the peanut dish and whether the peanuts are shelled or not), each with two levels (close or far, and shelled or unshelled respectively).

Given this, there would be a total of four conditions in this study (close-shelled, close-unshelled, far-shelled, far-unshelled). The manipulated variables, also referred to as independent variables, in this study are the 'proximity of the peanut dish' and 'whether the peanuts are shelled or unshelled'.

This does not appear to be an IV X PV (Independent Variable x Participant Variable) design because there is no information about a participant variable being used here. It also does not seem to be a repeated measures design as individuals are not exposed to all conditions; rather they experience one specific condition. The dependent variable is the amount of peanuts consumed, as determined by weighing the peanut dish after participants have finished eating.

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Answer:

0.0064 is the probability that Isaac will find his first defective can among the first 50 cans.    

Step-by-step explanation:

W are given the following in the question:

Probability of defective can =

[tex]P(A) = \dfrac{1}{90}[/tex]

We have to find the probability that Isaac will find his first defective can among the first 50 cans.

Then the number of adults follows a geometric distribution, where the probability of success on each trial is p, then the probability that the kth trial (out of k trials) is the first success is

[tex]P(X=k) = (1-p)^{k-1}p[/tex]

We have to evaluate:

[tex]P(x = 50)\\= (1-\frac{1}{90})^{50-1}(\frac{1}{90})\\= 0.0064[/tex]

0.0064 is the probability that Isaac will find his first defective can among the first 50 cans.

The Main Answer for:

1. The value of the variable for which the trinomial [tex]a^2+7a+6[/tex] and the binomial a+1 have the same value is [tex]a=-1[/tex] or [tex]a=-5[/tex].

2. The value of the variable for which the trinomial [tex]3x^2-x+1[/tex] and the trinomial [tex]2x^2+5x-4[/tex] have the same value is [tex]x=1[/tex] or [tex]x=5[/tex].

To find the value of the variable for which two polynomials have the same value, we set them equal to each other and solve for the variable.

1. For the trinomial [tex]a^2+7a+6[/tex] and the binomial [tex]a+1[/tex]:

[tex]a^2+7a+6=a+1[/tex]

First, let's rewrite the equation in standard form:

[tex]a^2+7a+6-(a+1)=0\\a^2+7a+6-a-1=0\\a^2+6a+5=0[/tex]

Now, we can solve this quadratic equation by factoring:

[tex]a^2+6a+5=0\\(a+5)(a+1)=0[/tex]

Setting each factor equal to zero:

[tex]a+5=0[/tex]⇒[tex]a=-5[/tex]

[tex]a+1=0[/tex]⇒[tex]a=-1[/tex]

So, the value of the variable for which the trinomial [tex]a^2+7a+6[/tex] and the binomial a+1 have the same value is [tex]a=-1[/tex] or [tex]a=-5[/tex].

2. For the trinomial [tex]3x^2-x+1[/tex] and the trinomial [tex]2x^2+5x-4[/tex]:

[tex]3x^2-x+1=2x^2+5x-4[/tex]

Again, let's rewrite the equation in standard form:

[tex]3x^2-x+1-(2x^2+5x-4)=0\\3x^2-x+1-2x^2-5x+4=0\\X^2-6x+5=0[/tex]

Now, let's solve this quadratic equation by factoring:

[tex]X^2-6x+5=0\\(x-5)(x-1)=0[/tex]

Setting each factor equal to zero:

[tex]x-5=0[/tex]⇒[tex]x=5[/tex]

[tex]x-1=0[/tex]⇒[tex]x=1[/tex]

So, the value of the variable for which the trinomial [tex]3x^2-x+1[/tex] and the trinomial [tex]2x^2+5x-4[/tex] have the same value is [tex]x=1[/tex] or [tex]x=5[/tex].

COMPLETE QUESTION:

Find out for what value of the variable:

1. Do the trinomial [tex]a^2+7a+6[/tex] and the binomial [tex]a+1[/tex] have the same value?

2. Do the trinomial [tex]3x^2-x+1[/tex] and the trinomial [tex]2x^2+5x-4[/tex] have the same value?