The states of Ohio, Iowa, and Idaho are often confused, probably because the names sound so similar. Each year, the State Tourism Directors of these three states drive to a meeting in one of the state capitals to discuss strategies for attracting tourists to their states so that the states will become better known. The location of the meeting is selected at random from the three state capitals. The shortest highway distance from Boise, Idaho to Columbus, Ohio passes through Des Moines, Iowa. The highway distance from Boise to Des Moines is 1350 miles, and the distance from Des Moines to Columbus is 650 miles. Let d1 represent the driving distance from Columbus to the meeting, with d2 and d3 representing the distances from Des Moines and Boise, respectively

a. Find the probability distribution of d1 and display it in a table.
b. What is the expected value of d1?
c. What is the value of the standard deviation of d1?
d. Consider the probability distributions of d2 and d3. Is either probability distribution the same as the probability distribution of d1? Justify your answer.
e. Define a new random variable t = d1 + d2. Find the probability distribution of t.

The covariance between two investments such as the S&P index fund and Core Bonds fund is calculated using the standard deviations of each investment and their correlation coefficient. This measure indicates if their returns move together. However, key data is not provided in the question.

In finance, the covariance between two investments, such as the S&P index fund and a Core Bonds fund, is used as a measure of how their returns move together. The formula for covariance is standard deviation of instrument one multiplied by standard deviation of instrument two, multiplied by their correlation coefficient. However, there is missing data in your question: the standard deviations and correlation coefficient were not provided. Once you have those, use the formula: Covariance = (Standard Deviation of S&P) * (Standard Deviation of Core Bonds) * (Correlation coefficient of the S&P and Core Bonds). Now, if you find that the outcome is a large number, it means that the returns move a lot in unison, for both good and bad outcomes. A low positive or negative number means that any movements are not strongly linked.

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Answer:

Reject  There is sufficient evidence to support the claim that true mean is actually higher than the claim amount $1800.

Step-by-step explanation:

Based on the decision rule, the test statistic is lies in the rejection region. So reject the null hypothesis at 5% level of significance.

There is sufficient evidence to support the claim that the true mean is actually higher than the claim amount $1800.

Solution is attached below

Answer:

We conclude that the true mean life of a biomedical device is greater than 5200 hours.

Step-by-step explanation:

We are given that the life in hours of a biomedical device under development in the laboratory is known to be approximately normally distributed. For this a random sample of 15 devices is selected and found to have an average life of 5323.8 hours and a sample standard deviation of 220.9 hours.

We have to test that the true mean life of a biomedical device is greater than 5200 or not.

Let, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu \leq[/tex] 5200 {means that the true mean life of a biomedical device is less than or equal to 5200 hours}

Alternate Hypothesis, [tex]H_1[/tex] : [tex]\mu[/tex] > 5200 {means that the true mean life of a biomedical device is greater than 5200 hours}

The test statistics that will be used here is;

        T.S. = [tex]\frac{Xbar-\mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]

where, Xbar = sample average life = 5323.8 hours

               s = sample standard deviation = 220.9 hours

               n = sample devices = 15

So, test statistics = [tex]\frac{5323.8-5200}{\frac{220.9}{\sqrt{15} } }[/tex] ~ [tex]t_1_4[/tex]

                            = 2.171

Since, we are not given with the significance level, so we assume it to be 5%, now the critical value of t at 14 degree of freedom in t table is given as 1.761. Since our test statistics is more than the critical value of t which means our test statistics will lie in the rejection region. So, we have sufficient evidence to reject our null hypothesis.

Therefore, we conclude that the true mean life of a biomedical device is greater than 5200 hours.

Answer:

The question has some details missing. The remaining part of the question says ;

Determine

a) The stagnation temperature Tox in K

b) The stagnation pressure Pox in bar

c) The pressure Px in bar

d) The pressure py in bar

e) The stagnation pressure Poy in bar

f) The stagnation temperature Toy in K

g) If the throat area is 7.6 x 10^-4m2, and the exit plane pressure is 2.4bar, determine the mass flow rate in kg/s and the exit area in m2

Step-by-step explanation:

The detailed step by step calculation and appropriate substitution is carefully shown in the attached files

Answer:

(a) The highest 20% weight correspond to the weight 434.32 grams.

(b) The middle 60% weight correspond to the weights 434.32 grams and 395.68 grams.

(c) The highest 80% weight correspond to the weight 395.68 grams.

(d) The highest 80% weight correspond to the weight 391.08 grams.

Step-by-step explanation:

Let X = weight of a small Starbucks coffee.

It is provided: [tex]X\sim N(\mu = 415\ grams, \sigma=23\ grams)[/tex]

(a)

Compute the value of x fro P (X > x) = 0.20 as follows:

[tex]P (X>x)=0.20\\P(\frac{X-\mu}{\sigma}>\frac{x-415}{23} )=0.20\\P (Z>z)=0.20\\1-P(Z<z)=0.20\\P(Z<z)=0.80[/tex]

Use a standard normal table.

The value of z is 0.84.

The value of x is:

[tex]0.84=\frac{x-415}{23}\\0.84\times23=x-415\\x=415+19.32\\=434.32[/tex]

Thus, the highest 20% weight correspond to the weight 434.32 grams.

(b)

Compute the value of x fro P (x₁ < X < x₂) = 0.60 as follows:

[tex]P(x_{1}<X<x_{2})=0.60\\P(\frac{x_{1}-415}{23}<\frac{X-\mu}{\sigma}< \frac{x_{2}-415}{23})=0.60\\P(-z<Z<z)=0.60\\P(Z<z)-P(Z<-z)=0.60\\P(Z<z)-[1-P(Z<z)]=0.60\\2P(Z<z)=1.60\\P(Z<z)=0.80[/tex]

Use a standard normal table.

The value of z is 0.84.

The value of x₁ and x₂ are:

[tex]z=\frac{x_{1}-415}{23}\\0.84=\frac{x_{1}-415}{23}\\x_{1}=415+(0.84\times23)\\=434.32[/tex]

[tex]-z=\frac{x_{2}-415}{23}\\0.84=\frac{x_{2}-415}{23}\\x_{1}=415-(0.84\times23)\\=395.68[/tex]

Thus, the middle 60% weight correspond to the weights 434.32 grams and 395.68 grams.

(c)

Compute the value of x fro P (X > x) = 0.80 as follows:

[tex]P (X>x)=0.80\\P(\frac{X-\mu}{\sigma}>\frac{x-415}{23} )=0.80\\P (Z>z)=0.80\\1-P(Z<z)=0.80\\P(Z<z)=0.20[/tex]

Use a standard normal table.

The value of z is -0.84.

The value of x is:

[tex]-0.84=\frac{x-415}{23}\\-0.84\times23=x-415\\x=415-19.32\\=395.68[/tex]

Thus, the highest 80% weight correspond to the weight 395.68 grams.

(d)

Compute the value of x fro P (X < x) = 0.15 as follows:

[tex]P (X<x)=0.15\\P(\frac{X-\mu}{\sigma}<\frac{x-415}{23} )=0.15\\P (Z<z)=0.15[/tex]

Use a standard normal table.

The value of z is -1.04.

The value of x is:

[tex]-1.04=\frac{x-415}{23}\\-1.04\times23=x-415\\x=415-23.92\\=391.08[/tex]

Thus, the highest 80% weight correspond to the weight 391.08 grams.

Final answer:

The question pertains to finding the weight that corresponds to a specific event for a normally distributed random variable, using the mean and standard deviation provided for small Starbucks coffee weights.

Explanation:

The weight of a small Starbucks coffee is a normally distributed random variable with a mean of 415 grams and a standard deviation of 23 grams. To find the weight that corresponds to a specific event, you need to use the formula for the z-score: Z = (X - μ) / σ, where Z is the z-score, X is the weight in grams, μ is the mean weight, and σ is the standard deviation.

Given that the mean (μ) is 415 grams and the standard deviation (σ) is 23 grams, you can calculate the z-score for any specific weight (X) to see how it relates to the distribution. Without a specific event or weight provided in this instance, we can discuss the process rather than a specific outcome. To calculate the percentile or probability for a given weight, the found z-score can then be used with a standard normal distribution table.

Answer: it will take her 20 minutes to send all 8 texts.

Step-by-step explanation:

If she sends 8 texts averaging 80 words each, it means that the number of words in the 8 texts would be

8 × 80 = 640 words

Katy can text 32 words per minute on her phone. Therefore, the time it will take her to send 8 texts or 640 words would be

640/32 = 20 minutes.

Answer:

(a) The 90% confidence interval is: (0.60, 0.63) Correct interpretation is (3).

(b) The 95% confidence interval is: (0.42, 0.46) Correct interpretation is (1).

(c) First, the confidence level in part(b) is more than the confidence level in part(a) is, so the critical value of part(b) is more than the critical value of part(a), Second the margin of error in part(b) is more than than in part(a).

Step-by-step explanation:

(a)

Let X = number of dog owners who take more pictures of their dog than of their significant others or friends.

Given:

X = 610

n = 1000

Confidence level = 90%

The (1 - α)% confidence interval for population proportion is:

[tex]CI=\hat p\pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]

The sample proportion is:

[tex]\hat p=\frac{X}{n}=\frac{610}{1000}=0.61[/tex]

The critical value of z for a 90% confidence level is:

[tex]z_{\alpha/2}=z_{0.10/2}=z_[0.05}=1.645[/tex]

Construct a 90% confidence interval for the population proportion of dog owners who take more pictures of their dog than their significant others or friends as follows:

[tex]CI=\hat p\pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}\\=0.61\pm 1.645\sqrt{\frac{0.61(1-0.61}{1000}}\\=0.61\pm 0.015\\=(0.595, 0.625)\\\approx(0.60, 0.63)[/tex]

Thus, the 90% confidence interval for the population proportion of dog owners who take more pictures of their dog than their significant others or friends is (0.60, 0.63).

Interpretation:

There is a 90% chance that the true proportion of dog owners who take more pictures of their dog than of their significant others or friends falls within the interval (0.60, 0.63).

Correct option is (3).

(b)

Let X = number of dog owners who are more likely to complain to their dog than to a friend.

Given:

X = 440

n = 1000

Confidence level = 95%

The (1 - α)% confidence interval for population proportion is:

[tex]CI=\hat p\pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]

The sample proportion is:

[tex]\hat p=\frac{X}{n}=\frac{440}{1000}=0.44[/tex]

The critical value of z for a 95% confidence level is:

[tex]z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96[/tex]

Construct a 95% confidence interval for the population proportion of dog owners who are more likely to complain to their dog than to a friend as follows:

[tex]CI=\hat p\pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}\\=0.44\pm 1.96\sqrt{\frac{0.44(1-0.44}{1000}}\\=0.44\pm 0.016\\=(0.424, 0.456)\\\approx(0.42, 0.46)[/tex]

Thus, the 95% confidence interval for the population proportion of dog owners who are more likely to complain to their dog than to a friend  is (0.42, 0.46).

Interpretation:

There is a 95% chance that the true proportion of dog owners who are more likely to complain to their dog than to a friend falls within the interval (0.42, 0.46).

Correct option is (1).

(c)

The confidence interval in part (b) is wider than the confidence interval in part (a).

The width of the interval is affected by:

The confidence level in part (b) is more than that in part (a).

Because of this the critical value of z in part (b) is more than that in part (a).

Also the margin of error in part (b) is 0.016 which is more than the margin of error in part (a), 0.015.

First, the confidence level in part(b) is more than the confidence level in part(a) is, so the critical value of part(b) is more than the critical value of part(a), Second the margin of error in part(b) is more than than in part(a).

a. (a) With a 90% level of confidence, this is (0.60, 0.63) (3) is the correct interpretation.

b. (0.42, 0.46) represents the 95% confidence interval. (1) is the correct interpretation.

c. First, the confidence level in part(b) is more than the confidence level in part(a) is, so the critical value of part(b) is more than the critical value of part(a), Second the margin of error in part(b) is more than in part(a).

a)

The total number of dog owners is: n = 1,000

The number of people said that they take more pictures of their dog than their significant others is: x = 610

Calculate the sample proportion of people said that they take more pictures of their dog than their significant others.

[tex]\bar p = \frac{x}{n}[/tex]

[tex]\bar p = \frac{610}{1000}[/tex]

= 0.61

From the Z-tabulated values, the Z-critical value at the 90% confidence level is: 1.645

Calculate the 90% confidence interval for the proportion of dog owners who take more pictures of their dog than of their significant others or friends.

90% C.I. = [tex]\bar p +Z_{critical}\sqrt{\frac{\bar p (1-\bar p)}{n }[/tex]

90% C.I. = 0.61 ± 1.645 [tex]\sqrt{\frac{0.61(1-0.61)}{1000} }[/tex]

90% C.I. = 0.64 ± 0.015

90% C.I. = (0.595 ; 0.625) ≈ (0.60, 0.63)

Hence, the required 90% confidence interval is: (0.60 ∠ p ∠ 0.63)

We are 90% confident that the true proportion of dog owners who take more pictures of their dog than of their significant others or friends falls within this interval.

b)

The total number of dog owners is: n = 1,000

The number of people said that they take more pictures of their dog than their significant others is: x = 440

Calculate the sample proportion of people said that they take more pictures of their dog than their significant others.

[tex]\bar p = \frac{x}{n}[/tex]

[tex]\bar p = \frac{440}{1000}[/tex]

= 0.44

From the Z-tabulated values, the Z-critical value at the 95% confidence level is: 1.96

Calculate the 90% confidence interval for the proportion of dog owners who take more pictures of their dog than of their significant others or friends.

90% C.I. = [tex]\bar p +Z_{critical}\sqrt{\frac{\bar p (1-\bar p)}{n }[/tex]

90% C.I. = 0.44 ± 1.96 [tex]\sqrt{\frac{0.44(1-0.44)}{1000} }[/tex]

90% C.I. = 0.44 ± 0.016

90% C.I. = (0.424 ; 0.456) ≈ (0.42, 0.46)

Hence, the required 95% confidence interval is: (0.42 ∠ p ∠ 0.46)

We are 95% confident that the true proportion of dog owners who are more likely to complain to their dog than to a friend falls within this interval.

C)

Compared to component (a), part (b)'s confidence interval is more expansive.

Several factors influence the interval's width:

Part (b) has a higher degree of confidence than part (a).

As a result, component (b)'s critical value of z is greater than part (a)'s.

Additionally, part (b) has a margin of error of 0.016, which is greater than part (a)'s margin of error of 0.015.

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Answer:

In the Step-by-step explanation.

Step-by-step explanation:

If a new sample is taken out of the same population,

a. The population proportion, p will not change. It is an unknown value that is estimated with the statistics of the samples.

b. The sample proportion, p^ is expected to change, because it is a new sample that has its own statistic value that may or may not be equal to the first sample.

c. The standard deviation of p^ is expected to change, because it depends on the sample and its size.

d. The standard error of p^ will change, because it depends on the sample.

e. The sampling distribution of p^, including its shape, mean, and standard deviation, will not change, because it is estimated with the data of the previous sample and it is supposed to be a property of the population and the sample size. Although the new information can be used to review the sample mean and standard deviation.

Answer:

1. D

2. 6 7/12

3. 3 1/4

Answer: 95 servings of applesauce

Step-by-step explanation:

It takes approximately 4 medium apples to make 3 servings of homemade apple sauce. It means that the number of servings of homemade apple sauce that can be made from 1 medium apple is

3/4 = 0.75 servings

Approximately 126 medium apples were purchased. Therefore, the number of servings of applesauce that they can make with the bushel is

126 × 0.75 = 95 servings of applesauce rounded to the nearest whole number.

Answer:

Ted will need [tex]3\frac{3}{4}[/tex] cups of pretzels to make 15 cups of trail mix.

Step-by-step explanation:

Ted is making trail mix for a party. He mixes 1 1/2 cups of nuts, 1/4 cup of raisins, and 1/4 cup of pretzels.

So, 1/4 cup of pretzels to make 1 trail mix

       x cups of pretzels to make 15 trail mix

Using the ratio and proportional

∴ x = (1/4) * 15 = 3.75 = [tex]3\frac{3}{4}[/tex] cups.

Ted will need [tex]3\frac{3}{4}[/tex] cups of pretzels to make 15 cups of trail mix.

Answer:

Correct option: (a)

Step-by-step explanation:

A confidence interval is an interval estimate of the parameter value.

A (1 - α)% confidence interval implies that the confidence interval has a (1 - α)% probability of consisting the true parameter value.

OR

If 100 such confidence intervals are made then (1 - α) of these intervals would consist the true parameter value.

The 92% confidence interval for the mean annual phone charge of all Vopstra customers is:

[tex]\$470\pm \$65=(\$405, \$535)[/tex]

This confidence interval implies that true mean annual phone charge of all Vopstra customers is contained in the interval ($405, $535) with 0.92 probability.

Thus, the correct option is (a).

Answer:

56 different tests

Step-by-step explanation:

Given:

Number of wires available (n) = 8

Number of wires taken at a time for testing (r) = 5

In order to find the number of different tests required for every possible pairing of five wires, we need to find the combination rather than their permutation as order of wires doesn't disturb the testing.

So, finding the combination of 5 pairs of wires from a total of 8 wires is given as:

[tex]^nC_r=\frac{n!}{r!(n-r)!}[/tex]

Plug in the given values and solve. This gives,

[tex]^8C_5=\frac{8!}{5!(8-5)!}\\\\^8C_5=\frac{8\times 7\times 6\times 5!}{5!\times 3\times2\times1}\\\\^8C_5=56[/tex]

Therefore, 56 different tests are required for every possible pairing of five ​wires.

A total of 56 different tests are needed to check every possible combination of five wires out of eight in the cable.

The question is asking for the number of different tests required to test every possible pairing of five wires in an eight-wire cable. This is a combination problem, where we're looking for how many different ways we can combine five items from a group of eight. The formula for combinations is C(n, k) = n! / [k!(n - k)!], where n is the total number of items, k is the number of items to choose, and ! represents factorial. Plugging into this formula, we get C(8, 5) = 8! / [5!(8 - 5)!] = 56. Therefore, a total of 56 different tests are needed to check every possible combination of five wires out of eight.

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Answer/ Explanation:

Since X is exponentially distributed, its expected value is given by E[X]=1/λ=2.

Therefore,  E[Y]=E[1−2X]=E[1]+E[−2X]=E[1]−2E[X]=1−2E[X]=1−2⋅2=−3.

Hence,

We define the moment-generating function of Y as MY(t). It is given by

MY(t)=E[etY]=E[et(1−2X)]=E[ete−2tX]=E[et]E[e−2tX].

If I give you the hint that E[g(Y)]=∫∞0g(y)fY(y)dy, where fY(y) is the probability density function of Y, can you also solve for the moment generating function of Y?

We have E[X2]=2/λ2=2/(0.5)2=8. Thus,

E[Y2]=E[(1−2X)2]=E[1−4X+4X2]=E[1]−4E[X]+4E[X2]=1−4⋅2+4⋅8=25.

So,

Var(Y)=E[Y2]−E[Y]2=25−(−3)2=16.

Continuing for the moment-generating function:

MY(t)=E[et]E[e−2tX]=etE[e−2tX]=et∫∞x=0e−2txfX(x)dx,

where fX(x) is the probability density function of X and thus satisfies fX(x)=λe−λx. Substituting yields

MY(t)=et∫∞x=0e−2txλe−λxdx=λet∫∞x=0e−x(2t+λ)dx=λet2t+λ.

It is also good to note that

If you are after expectation, variance or moment generating function of Y then it is not needed to find the PDF of Y (see the answer of Ritz).

This is not an answer on the question in the title, but one on the question in the body.

FY(y)=P(Y≤y)=P(1−2X≤y)=P(X≥0.5−0.5y)=1−FX(0.5−0.5y)

Note that the last equality demands that FX is continuous.

Differentating on both sides gives fY on LHS and an expression in fX on RHS.

Answer:

After use the formula we got the following result for the 90% confidence interval (0.13 <p<0.34)

And the conclusion for this case would be:

e. With 90% confidence, the proportion of all students who take notes is between 0.13 and 0.34.

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

Description in words of the parameter p

[tex]p[/tex] represent the real population proportion of students who take notes

[tex]\hat p[/tex] represent the estimated proportion of students who take notes

n is the sample size required  

[tex]z_{\alpha/2}[/tex] represent the critical value for the margin of error  

The population proportion have the following distribution  

[tex]p \sim N(p,\sqrt{\frac{p(1-p)}{n}})[/tex]

Numerical estimate for p

In order to estimate a proportion we use this formula:

[tex]\hat p =\frac{X}{n}[/tex] where X represent the number of people with a characteristic and n the total sample size selected.

Confidence interval

The confidence interval for a proportion is given by this formula  

[tex]\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]  

For the 90% confidence interval the value of [tex]\alpha=1-0.90=0.1[/tex] and [tex]\alpha/2=0.05[/tex], with that value we can find the quantile required for the interval in the normal standard distribution.  

[tex]z_{\alpha/2}=1.64[/tex]  

After use the formula we got the following result for the 90% confidence interval (0.13 <p<0.34)

And the conclusion for this case would be:

e. With 90% confidence, the proportion of all students who take notes is between 0.13 and 0.34.

Answer: between 20.3 and 20.9

Step-by-step explanation:

Answer:

195

Step-by-step explanation:

The relationship between the covariance (cov_AB), and the correlation coefficient (ρ_AB = 0.65), and the standard deviations (σ_A = 12 and σ_B = 25) for the securities A and B is :

[tex]cov_{A,B} = \rho_{A,B}*\sigma_A*\sigma_B[/tex]

Applying the given data:

[tex]cov_{A,B} = 0.65*12*25\\cov_{A,B} = 195[/tex]

The covariance between these two securities is 195.

The sample space includes all possible arrangements of n parking spaces. The probability (P(A)) that Mary and Tom select spaces at most 2 apart is [tex]\(\frac{3}{n-1}\)[/tex], assuming equal likelihood of choice.

The sample space consists of all possible arrangements of cars in the parking lot, given that there are n parking spaces. Each arrangement is equally likely.

For Mary and Tom to select parking spaces that are at most 2 spaces apart, we consider the following scenarios:

1. Mary selects any space, and Tom selects the same space or an adjacent space.

2. Mary selects any space, and Tom selects any space 1 space away.

3. Mary selects any space, and Tom selects any space 2 spaces away.

The number of favorable outcomes is 3n, and the total number of possible outcomes is [tex]\(n \times (n-1)\)[/tex] (since Mary and Tom can choose from n spaces each). Therefore, the probability P(A) is given by:

[tex]\[ P(A) = \frac{3n}{n \times (n-1)} \][/tex]

Simplify the expression:

[tex]\[ P(A) = \frac{3}{n-1} \][/tex]

Answer:

0.6604

Step-by-step explanation:

Given that a market  research firm knows from historical data that telephone surveys have a 36% response rate.

Sample size of random sample = 280

We know for samples randomly drawn of large size sample proportion follows a normal distribution with mean= sample proportin and std error

= [tex]\sqrt{\frac{pq}{n} }[/tex]

Substitute p = 0.36 and q = 1-0.36= 0.64

p follows N with mean = 0.36 and std dev = [tex]\sqrt{\frac{0.36*0.64}{\sqrt{280} } } \\=0.0287[/tex]

Using normal distribution values we can find\

[tex]P(33.5p.c. < p < 39pc)\\= P(0.335<p<0.39)\\= F(0.39)-F(0.335)\\= 0.852183-0.191735\\=0.660448[/tex]

Answer:

Probability that the response rate will be between 33.5% and 39% = 0.66176 .

Step-by-step explanation:

We are given that a market research firm knows from historical data that telephone surveys have a 36% response rate.

The probability criterion we will use here is;

           [tex]\frac{\hat p- p}{\sqrt{\frac{\hat p (1-\hat p)}{n} } }[/tex] ~ N(0,1)

Here, p = 0.36 and n = sample size = 280

Let [tex]\hat p[/tex] = response rate

So, P(0.335 <= [tex]\hat p[/tex] <= 0.39) = P([tex]\hat p[/tex] <= 0.39) - P([tex]\hat p[/tex] < 0.335)

P([tex]\hat p[/tex] <= 0.39) = P( [tex]\frac{\hat p- p}{\sqrt{\frac{\hat p (1-\hat p)}{n} } }[/tex] <= [tex]\frac{0.39- 0.36}{\sqrt{\frac{0.39 (1-0.39)}{280} } }[/tex] ) = P(Z <= 1.03) = 0.84849

P([tex]\hat p[/tex] < 0.335) = P( [tex]\frac{\hat p- p}{\sqrt{\frac{\hat p (1-\hat p)}{n} } }[/tex] < [tex]\frac{0.335- 0.36}{\sqrt{\frac{0.335 (1-0.335)}{280} } }[/tex] ) = P(Z < -0.89) = 1 - P(Z <= 0.89)

                                                                  = 1 - 0.81327 = 0.18673

Therefore, P(0.335 <= [tex]\hat p[/tex] <= 0.39) = 0.84849 - 0.18673 = 0.66176

Hence, probability that the response rate will be between 33.5% and 39% is 0.66176 or 66.18 % .

Answer:

note:

please find the attachment

Answer:

0.99827 = 99.827% probability that when a theft occurs, at least one of the 3 systems will detect it.

Step-by-step explanation:

For each system, there are only two possible outcomes. Either it detects the theft, or it does not. The probability of a system detecting a theft is independent of other systems. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

And p is the probability of X happening.

Each of the 3 systems detects theft with a probability of 0.88 independently of the others.

This means that [tex]n = 3, p = 0.88[/tex]

What is the probability that when a theft occurs, at least oneof the 3 systems will detect it?

[tex]P(X \geq 1) = P(X = 1) + P(X = 2) + P(X = 3)[/tex]

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 1) = C_{3,1}.(0.88)^{1}.(0.12)^{2} = 0.03802[/tex]

[tex]P(X = 2) = C_{3,2}.(0.88)^{2}.(0.12)^{1} = 0.27878[/tex]

[tex]P(X = 3) = C_{3,3}.(0.88)^{3}.(0.12)^{0} = 0.68147[/tex]

[tex]P(X \geq 1) = P(X = 1) + P(X = 2) + P(X = 3) = 0.03802 + 0.27878 + 0.68147 = 0.99827[/tex]

0.99827 = 99.827% probability that when a theft occurs, at least one of the 3 systems will detect it.

The probability that at least one of three independent alarm systems, each with a detection probability of 0.88, will detect a theft is approximately 0.99827.

The problem you're asking about falls under the subject of probability, an area of mathematics that measures the likelihood an event will occur. The question asks for the probability that at least one of three independent alarm systems will detect a theft. These systems each have a detection probability of 0.88.

To solve this, it's easier to calculate the probability that none of the systems detect the theft and then subtract that from 1. The likelihood that a system will not detect a theft is 1 - 0.88, which equals 0.12. Since the systems are independent, the probabilities multiply, so: (0.12)^3 = 0.001728. But we want the opposite of this, so we subtract it from 1: 1 - 0.001728 = 0.998272 which is approximately 0.99827 when rounded to five decimal places. That's the probability that at least one alarm system will detect a theft.

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Step-by-step explanation:

Below is an attachment containing the solution.

Answer:

19.53125 cm/s

Step-by-step explanation:

h : s

15 : 8

s/h = 8/15

s = 8h/15

V = ⅓×s²×h

V = ⅓(8h/15)²×h = 64h³/675

dV/dh = 64h²/225

At h=3,

dV/dh = 64(3²)/225 = 64/25

dh/dV = 25/64

dV/dt = 50

dh/dt = dh/dV × dV/dt

= 25/64 × 50

= 19.53125 cm/s

Both male and female customers show a preference for the standard-engine Honda Civic, but the data suggests that women are slightly more likely to purchase this model than men are.

According to this data, out of the total 311 customers, 194 are male and 117 are female. Among the male customers, 77 purchased a Hybrid Honda Civic and 117 purchased a Standard-engine Honda Civic. While among the female customers, 34 purchased a Hybrid Honda Civic and 83 purchased a Standard-engine Honda Civic. If we compare the proportions, we can see that approximately 60% (117 out of 194) of male customers and approximately 70% (83 out of 117) of female customers purchased a Honda Civic with a standard engine. Hence, we can conclude that both women and men have shown a preference for the standard-engine Honda Civic, but the data suggest that women are somewhat more likely to purchase a standard engine Honda Civic than men are.

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The data suggests that women are more likely to purchase a standard-engine Honda Civic compared to men, and men are more likely to choose a hybrid engine compared to women.

The data from the local Honda dealership regarding customer choices between Hybrid Honda Civic and Standard-engine Honda Civic suggests a relationship between gender and engine type choice. To determine this, we look at the proportions of each gender choosing each type of engine relative to their total gender group. Specifically, out of 194 males, 77 chose the hybrid (which is approximately 39.7%), and 117 chose the standard (approximately 60.3%). For females, 34 out of 117 chose the hybrid (approximately 29.1%), and 83 chose the standard (approximately 70.9%).

By comparing these percentages, it is clear that women are less likely to purchase a Hybrid Honda Civic than men because a smaller percentage of women chose the hybrid compared to men. Conversely, a larger percentage of women chose the standard engine compared to men, suggesting that women are more likely to purchase a Honda Civic with a standard engine than men.

Answer:

The correct optiion is C

Step-by-step explanation:

Beer =$1 and amount sold is x

so $1×x= $x which is the profit on beer

Wine=$2 and amount sold is y

so, $2×y= $2y which is the profit on wine

so an algebraic formulation of the profit function will be,

the sum off both the profit of the beer and wine which is

x+2y= max(x+2y)

Answer:

The new mean of processed transactions is 8

Step-by-step explanation:

The teller averages 5 transactions every 15 minutes.

Taking this to a single transaction basis gives; The teller averages one transaction every 3 minutes.

So, when the time changes to 25 minutes, there is the need to find the number of 3-minutes obtainable from a 25-minute interval

New mean of processed transaction = [tex]\frac{25 minutes}{3 minutes}[/tex] = 8.333

The new mean of transaction every 25 minutes is about 8 transactions

Answer: 8.33/ 8 transactions on the average

Step-by-step explanation:

Since the teller is capable of processing five (5) different transactions on the average every fifteen minutes (15).

This means that he is capable of processing a transaction in three minutes:

15minutes/3transactions

= 3 minutes per transaction.

Judging by this speed ( 3 minutes per transaction); we can deduce the number of transactions the teller is capable of processing on the average in 25 minutes.

Since 3 minutes --------- 1 transaction, in 25 minutes the teller will process :

(25/3) × (1/1)

= 8.333 transactions

approximately 8 transactions.

Answer:

                                          [tex]C = 1/18[/tex]

Step-by-step explanation:

Remember that for a probability density function

                                     [tex]\int_{-\infty}^{\infty } f(x) dx = 1[/tex]

Since [tex]f(x) = 0[/tex]   outside [tex][0,2][/tex]   we would have that

                                       [tex]\int_{0}^{2} C(10-x) dx = 1[/tex]

Therefore  

                                              [tex]18C = 1 \\C = 1/18[/tex]

Answer:

a. P(AnB)

b. P(B|A)

c. [tex]P(A^I|B)[/tex]

d. P(A or B)

e. [tex]P(B^I or A)[/tex]

Step-by-step explanation:

Since  A= driver is under 25 years old (1)

B = driver has received a speeding ticket (2)

a.The probability the driver is under 25 years old and has recieved a speeding ticket.

this simple means the intersection of both set, which can be written as

P(AnB)

b. The probability a driver who is under 25 years old has received a speeding ticket.

This is a conditional probability, probability that B will occur given that A as occur.

P(B|A)

c. the probability a driver who has received a speeding ticket is 25 years or older.

[tex]P(A^I|B)[/tex]

d. The probability the driver is under 25 years old or has received a speeding ticket.

P(A or B)

e. The probability the driver is under 25 years old or has not received a speeding ticket.

[tex]P(B^I or A)[/tex]

Answer: 0.9013

Step-by-step explanation:

Given mean, u = 10, standard deviation =8

P(X) =P(Z= X - u /S)

We are to find P(X> or =12)

P(X> or = 12) = P(Z> 12-10/8)

P(Z>=2/8) = P(Z >=0.25)

P(Z) = 1 - P(Z<= 0.25)

We read off Z= 0.25 from the normal distribution table

P(Z) = 1 - 0.0987 = 0.9013

Therefore P(X> or=12) = 0.9013

Note the question was given as an incomplete question the correct and complete question had to be searched online via Google. So the data used are those gotten from the online the Googled question.

Answer:

Correlation between x and y is 0.9508          

Step-by-step explanation:

We are given the following in the question:

Distance(x):   1     4       6     6       6       7

      Days(y):  2    20    29    38     44     50

We have to find the correlation between x and y.

[tex]\sum y = 183\\\sum x=30\\\bar{x} = \displaystyle\frac{\sum x}{n} = \frac{30}{6} = 5\\\\\bar{y} = \displaystyle\frac{\sum y}{n} = \frac{183}{6} = 30.5\\\\(x-\bar{x}) = -4,-1,+1,+1,+1,+2\\(y-\bar{y}) = -28.5,-10.5,-1.5,7.5,13.5,19.5\\\sum (x-\bar{x}) (y-\bar{y}) = 183\\\sum(x-\bar{x})^2 = 24\\\sum(y-\bar{y})^2= 1543.5\\[/tex]

Formula:

[tex]r = \dfrac{\sum(x-\bar{x})(y - \bar{y})}{\sqrt{\sum(x-\bar{x})^2\sum(y-\bar{y})^2}}[/tex]

Putting values, we get,

[tex]r = \dfrac{183}{\sqrt{24\times 1543.5}} = 0.9508[/tex]

Thus, correlation between x and y is 0.9508

To find the correlation between the distance and the number of days until deaths began, we can use a statistical measure called the correlation coefficient (r). The formula for calculating r is given, and we can use the provided data to calculate the correlation coefficient, which is approximately -0.994, indicating a strong negative correlation.

To find the correlation between the distance and the number of days until deaths began, we can use a statistical measure called the correlation coefficient (r). The formula for calculating r is:

r = (n(Σxy) - (Σx)(Σy)) / √((n(Σx^2) - (Σx)^2)(n(Σy^2) - (Σy)^2))

Using the provided data, we can calculate the correlation coefficient:

After performing the calculations, we find that the correlation coefficient (r) is approximately -0.994, indicating a strong negative correlation between distance and the number of days until deaths began.

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Answer:

1. Assuming with replacement, the probability is 7.91%; or

2. Assuming without replacement, the probability is 8.64%

Step-by-step explanation:

Total number of golf balls = 24

Let Pr denotes probability, P denotes ProFlights, D denotes DistMax.

The probability of selecting 5 balls can be with or without replacement. Since the question is silent on this, the answers to the methods are provided as follows:

1. Assuming with replacement

Pr(4 P and 1 D)  = (18/24) × (18/24) × (18/24) × (18/24) × (6/24)

                         = 0.75 × 0.75 × 0.75 × 0.75 × 0.25

                         = 0.0791  = 7.91%

2. Assuming without replacement

Here, we assume that 4 ProFlights are selected first before 1 DistMax is selected, and the probability is as follows:

Pr(4 P and 1 D) = (18/24) × (17/23) × (16/22) × (15/21) × (6/20)

                        = 0.7500 × 0.7391  × 0.7273  × 0.7143  × 0.3000  

                        =  0.0864  = 8.64%

Therefore, the probability that he reaches in his bag and randomly selects 5 golf balls and 4 of them are ProFlights and the other 1 is DistMax is 7.91% assuming with replacement or 8.64% assuming without replacement.

Final answer:

To find the probability, we use the concept of combinations.

Explanation:

To find the probability that the golfer randomly selects 5 golf balls and 4 of them are ProFlights and the other 1 is DistMax, we need to use the concept of combinations. The total number of ways to select 5 golf balls from a bag containing 24 balls is C(24,5). The number of ways to select 4 ProFlight balls and 1 DistMax ball from their respective totals is C(18,4) * C(6,1). Therefore, the probability is given by:

P = (C(18,4) * C(6,1)) / C(24,5)

Answer:

[tex]P(X<200)=P(\frac{X-\mu}{\sigma}<\frac{200-\mu}{\sigma})=0.1[/tex]  

[tex]P(z<\frac{200-\mu}{\sigma})=0.1[/tex]

But we know which value of z satisfy the previous equation so then we can do this:

[tex]z=-1.28<\frac{200-\mu}{7.8}[/tex]

And if we solve for [tex]\mu[/tex] we got

[tex]\mu=200 +1.28*7.8=209.984[/tex]

Step-by-step explanation:

Assuming this question "The weight, in grams, of beans in a tin is normally distributed with mean m and standard deviation 7.8 grams. Given that 10% of tins contain less than 200 grams, Find the mean m. explain why. ?

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

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

[tex] X \sim N (\mu =m, 7.8)[/tex]

Where [tex]\mu =m[/tex]  and [tex] \sigma =7.8[/tex]

And the best way to solve this problem is using the normal standard distribution and the z score given by:

[tex]z=\frac{x-\mu}{\sigma}[/tex]

For this part we know the following condition:

[tex] P(X>200) =0.9[/tex]   (a)

[tex] P(X<200) = 0.1[/tex]  (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value m.  

As we can see on the figure attached the z value that satisfy the condition with 0.1 of the area on the left and 0.9 of the area on the right it's z=-1.28.. On this case P(Z<-1.28)=0.1 and P(z>-1.28)=0.9

If we use condition (b) from previous we have this:

[tex]P(X<200)=P(\frac{X-\mu}{\sigma}<\frac{200-\mu}{\sigma})=0.1[/tex]  

[tex]P(z<\frac{200-\mu}{\sigma})=0.1[/tex]

But we know which value of z satisfy the previous equation so then we can do this:

[tex]z=-1.28<\frac{200-\mu}{7.8}[/tex]

And if we solve for [tex]\mu[/tex] we got

[tex]\mu=200 +1.28*7.8=209.984[/tex]

The mean weight of the beans in a tin m  must be greater than 200 grams.

 The problem states that the weight of beans in a tin is normally distributed with a mean m and a standard deviation of 7.8 grams.

It is also given that 10% of the tins contain less than 200 grams.

Since the distribution is normal, we can use the properties of the normal distribution to solve for the mean weight m

 In a normal distribution, approximately 68% of the data falls within one standard deviation of the mean, 95% within two standard deviations, and 99.7% within three standard deviations.

The fact that 10% of the tins weigh less than 200 grams indicates that 200 grams is less than one standard deviation below the mean.

 The 10% corresponds to the bottom 10% of the distribution, which is less than the mean by a certain number of standard deviations.

The standard normal distribution table (or Z-table) tells us that a Z-score of approximately -1.28 corresponds to the 10th percentile (since 10% of the distribution is to the left of this Z-score).

Using the Z-score formula:

[tex]\[ Z = \frac{X - \mu}{\sigma} \][/tex]

[tex]\[ \mu = X - Z \times \sigma \][/tex]

[tex]\[ \mu = 200 - (-1.28) \times 7.8 \][/tex]

[tex]\[ \mu = 200 + 10.064 \][/tex]

mu = 210.064

 Therefore, the mean weight m  (denoted by mu in our calculations) must be greater than 200 grams, specifically approximately 210.064 grams, to ensure that only 10% of the tins weigh less than 200 grams.