What is the inverse of -8+3i

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:

(a) Probability = 0.29947

Step-by-step explanation:

The probability of the hand containing exactly one ace would be:

Number of ways this can happen = 4C1 * 48C4      (using combinations)

Number of ways this can happen = 4 * 194580

Number of ways this can happen = 778,320

Total number possible hands = 2,598,960 (as stated in question)

Total probability of exactly one ace = Number of ways to get an ace / total number of ways

Total probability = 778320 / 2598960 = 0.29947

Thus, the probability of the hand containing exactly one ace will be 0.2994

Another way to solve this:

Probability of one ace and 5 other cards = [tex]\frac{4}{52}*\frac{48}{51}*\frac{47}{50}*\frac{46}{49}*\frac{45}{48}[/tex] = 0.059894

Number of ways to arrange 1 ace and 4 other cards = 5

Total probability = 0.0598 * 5 = 0.29947

The probability of getting exactly one ace in a five-card poker hand is 0.2556 (rounded to four decimal places).

To find the probability of getting exactly one ace in a five-card poker hand, we need to determine the number of favorable outcomes and divide it by the total number of possible outcomes. In a standard 52-card deck, there are 4 aces, and we need to choose 1 ace out of the 4. The remaining 4 cards in the hand can be chosen from the remaining 48 non-ace cards in the deck. Hence, the number of favorable outcomes is C(4,1) * C(48,4). The probability can be calculated as:

P(exactly one ace) = (C(4,1) * C(48,4)) / C(52,5)

Substituting the values and evaluating the expression, we get:

P(exactly one ace) = (4 * 171,230) / 2,598,960 = 0.2556 (rounded to four decimal places)

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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:

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:

(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:

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.

Answer:

We need to check the conditions in order to use the normal approximation.

[tex]np \geq 15[/tex]

[tex]n(1-p) \geq 15[/tex]

If we see that we satisfy the conditions and then we can apply the approximation.

If we appply the approximation the new mean and standard deviation are:

[tex]E(X)=np[/tex]

[tex]\sigma=\sqrt{np(1-p)}[/tex]

[tex] X \sim N (\mu = np, \sigma=\sqrt{np(1-p)}) [/tex]

So then the correct answer for this case would be:

E. Normal

Step-by-step explanation:

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Let X the random variable of interest, on this case we assume that:

[tex]X \sim Binom(n, p)[/tex]

The probability mass function for the Binomial distribution is given as:

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

Where (nCx) means combinatory and it's given by this formula:

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

We need to check the conditions in order to use the normal approximation.

[tex]np \geq 15[/tex]

[tex]n(1-p) \geq 15[/tex]

If we see that we satisfy the conditions and then we can apply the approximation.

If we appply the approximation the new mean and standard deviation are:

[tex]E(X)=np[/tex]

[tex]\sigma=\sqrt{np(1-p)}[/tex]

[tex] X \sim N (\mu = np, \sigma=\sqrt{np(1-p)}) [/tex]

So then the correct answer for this case would be:

E. Normal

The approximate shape of the sampling distribution of the sample will be normal.

Option E is correct.

It is a statistic that determines the probability of an event based on data from a small group within a large population.

Given that, [tex]np[/tex] is greater than or equal to 15 and [tex]n(1-p)[/tex] is greater than or equal to 15.

     [tex]np\geq 15[/tex]  and [tex]n(1-p)\geq 15[/tex]

So that the new mean and standard deviation will be,

           [tex]mean=\mu=np\\\\Deviation=\sigma=\sqrt{np(1-p)}[/tex]

Thus, the approximate shape of the sampling distribution of the sample will be normal.

Learn more about the sampling distribution here:

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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:

note:

please find the attachment

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: 5.5hours

Step-by-step explanation:

Total distance = 226miles

Total time = 7 hours

Let b represent total time spent while walking.

Distance (walking) = 4b

Distance ( riding) = 40(7-b)

Total distance 226 = 4b + 40(7-b)

226 = 4b + 280 - 40b

226 = 280-36b

b = 54/36

b = 1.5hours

Amount of time spent walking = 1.5hours

Amount of time spent riding = 7-1.5 = 5.5hours

Amount of time spent on bicycle = 5.5hours

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:

Step-by-step explanation:

The total number of lines, n(U) = 18

Let the number of lins with verb be n(V) = 11

Let the number of lines with adjectives be n(A) = 13

n(V n A) = 8

Find the number of lines that have a verb but no adjective, that is, n(V n A')

Mathematically, according to sets theory,

n(V) = n(V n A) + n(V n A')

So,

n(V n A') = n(V) - n(V n A) = 11 - 8 = 3.

Hence, only 3 lines have a verb but no adjectives.

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:

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:

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:

(a) No

(b) 0.21 or 21%

Step-by-step explanation:

(a) Since the outcome of the first game influences in the probability of winning the second game, the two games are not independent.

(b) The probability of losing both games is given by the product of the probability of losing the first game and the probability of losing the second game given that you have lost the first:

[tex]P = (1-0.7)*(1-0.3)\\P=0.21=21\%[/tex]

The probability you lose both games is 21%

Given Information:

Probability of wining 1st game = p₁ = 0.7

Probability of wining 2nd game given 1st game won = p₂|p₁ = 0.5

Probability of wining 2nd game given 1st game lost = p₂|q₁ = 0.3

Required Information:

(a) Are the two games independent = ?

(b) Probability of losing both games = ?

Answer:

(a) Are the two games independent = No

(b) Probability of losing both games = 0.21

Step-by-step explanation:

(a) Independent Events:

Two events are said to be independent when the success of one event is not affected by the success or failure of another event.

In this case, the probability of 2nd game depends on the success or failure of the 1st game, therefore, the two games are not independent.

(b) Probability of losing both games

The probability of losing the both games is the product of the probabilities of losing each game.

Probability of losing 1st game = 1 - Probability of wining 1st game

Probability of losing 1st game = 1 - 0.7 = 0.30

Probability of losing 2nd game = 1 - Probability of wining 2nd game given 1st game lost

Probability of losing 2nd game = 1 - 0.3 = 0.70

Please note that since we are finding the probability of losing both games that's why we used the condition of 1st game lost

Probability of losing both games = Probability of losing 1st game*Probability of losing 2nd game

Probability of losing both games = 0.30*0.70

Probability of losing both games = 0.21

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.

12x-5y=-20

y=x+4

12x-5(x+4)=-20

12x-5x-20=-20

7x=0,

So, we get: x=0 and y=4

Answer:

95% confidence interval for the mean amount of the active chemical in the drug = [ 90.088 , 94.312 ]

Step-by-step explanation:

We are given that a lab is testing the amount of a certain active chemical compound in a particular drug that has been recently developed. It is known that the standard deviation in the amount of the chemical is 6 mg.

A random sample of 31 batches of the new drug is tested and found to have a sample mean concentration of 92.2 mg of the active chemical i.e.;

Population standard deviation, [tex]\sigma[/tex] = 6 mg

Sample mean, [tex]Xbar[/tex] = 92.2 mg

Sample size, n = 31

Now, the pivotal quantity for 95% confidence interval is given by;

          [tex]\frac{Xbar -\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] ~ N(0,1)

So, 95% confidence interval for the mean amount of the active chemical in the drug is given by;

P(-1.96 < N(0,1) < 1.96) = 0.95

P(-1.96 < [tex]\frac{Xbar -\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] < 1.96) = 0.95

P(-1.96 * [tex]\frac{\sigma}{\sqrt{n} }[/tex] < [tex]Xbar - \mu[/tex] < 1.96 * [tex]\frac{\sigma}{\sqrt{n} }[/tex] ) = 0.95

P(Xbar - 1.96 * [tex]\frac{\sigma}{\sqrt{n} }[/tex] < [tex]\mu[/tex] < Xbar + 1.96 * [tex]\frac{\sigma}{\sqrt{n} }[/tex] ) = 0.95

95% confidence interval for [tex]\mu[/tex] = [Xbar - 1.96 * [tex]\frac{\sigma}{\sqrt{n} }[/tex] , Xbar + 1.96 * [tex]\frac{\sigma}{\sqrt{n} }[/tex] ]

                                                  = [ 92.2 - 1.96 * [tex]\frac{6}{\sqrt{31} }[/tex] , 92.2 + 1.96 * [tex]\frac{6}{\sqrt{31} }[/tex] ]

                                                  = [ 90.088 , 94.312 ]

To calculate the 95% confidence interval for the mean amount of the active chemical in the drug, substitute the sample mean concentration, standard deviation, and sample size into the formula: 95% Confidence Interval = Sample Mean ± (Z * (Standard Deviation / √Sample Size)).

To calculate the 95% confidence interval for the mean amount of the active chemical in the drug, we can use the formula:

95% Confidence Interval = Sample Mean ± (Z * (Standard Deviation / √Sample Size))

Given that the sample mean concentration is 92.2 mg, the standard deviation is 6 mg, and the sample size is 31 batches, we can substitute these values into the formula:

95% Confidence Interval = 92.2 mg ± (1.96 * (6 mg / √31))

Calculating this expression gives us a 95% confidence interval of approximately 90.855 mg to 93.545 mg.

Answer:

[tex]0.482 - 1.64 \sqrt{\frac{0.482(1-0.482)}{850}}=0.454[/tex]  

[tex]0.482 + 1.64 \sqrt{\frac{0.482(1-0.482)}{850}}=0.510[/tex]  

And the 90% confidence interval would be given (0.454;0.510).  

B. 0.4542 to 0.5105.

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".  

Solution to the problem

The estimated proportion of residents in the community that support the property tax levis is given by:

[tex] \hat p =\frac{x}{n}= \frac{410}{850}= 0.482[/tex]

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=1-0.9=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]  

And replacing into the confidence interval formula we got:  

[tex]0.482 - 1.64 \sqrt{\frac{0.482(1-0.482)}{850}}=0.4542[/tex]  

[tex]0.482 + 1.64 \sqrt{\frac{0.482(1-0.482)}{850}}=0.5105[/tex]  

And the 90% confidence interval would be given (0.4542;0.5105)

B. 0.4542 to 0.5105.

Answer:

90% confidence interval for p is [0.4542 , 0.5105] .

Step-by-step explanation:

We are given that a local board of education conducted a survey of residents in the community concerning a property tax levy on the coming local ballot. Of the 850 residents surveyed, 410 supported the property tax levy.

Let p = proportion of residents in the community that support the property tax levy

[tex]\hat p[/tex] = proportion of residents in the community that support the property tax levy in a survey of 850 residents = [tex]\frac{410}{850}[/tex] = [tex]\frac{41}{85}[/tex]

The pivotal quantity that will be used here population proportion p is;

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

So, 90% confidence interval for p is given by;

P(-1.6449 < N(0,1) < 1.6449) = 0.90 {At 10% significance level the z table give

                                                            critical value of 1.6449)

P(-1.6449 < [tex]\frac{\hat p - p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] < 1.6449) = 0.90

P( [tex]-1.6449 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }[/tex] < [tex]{\hat p - p}[/tex] < [tex]1.6449 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }[/tex] ) = 0.90

P( [tex]\hat p -1.6449 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }[/tex] < p < [tex]\hat p +1.6449 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }[/tex] ) = 0.90

90% confidence interval for p = [ [tex]\hat p -1.6449 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }[/tex] , [tex]\hat p +1.6449 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }[/tex] ]

                                       = [ [tex]\frac{41}{85} -1.6449 \times {\sqrt{\frac{\frac{41}{85} (1-\frac{41}{85} )}{850} }[/tex] , [tex]\frac{41}{85} +1.6449 \times {\sqrt{\frac{\frac{41}{85} (1-\frac{41}{85} )}{850} }[/tex] ]

                                       = [ 0.4542 , 0.5105 ]

Therefore, 90% confidence interval for p is [0.4542 , 0.5105] .

Answer:

A. Data

Step-by-step explanation: Data is a term used to describe facts, information or statistics that are collected together in order for it to be used as a reference or for analysis.

An effective data collection is one of the most important aspects in research,experiments or statistics as it helps to guarantee a reliable and effective outcome.

Data collection should be done in such a way that it helps to solve the problem which is being studied or handled.

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.

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:

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]

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.

https://brainly.com/question/33638776

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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).

Step-by-step explanation:

Hope it helps you in your learning process.