Two balanced dice are rolled. Let X be the sum of the two dice. Obtain the probability distribution of X (i.e. what are the possible values for X and the probability for obtaining each value?). Check that the probabilities sum to one.


1) What is the probability for obtaining X \geq 8?

2) What is the average value of X?

Jack inherited a perpetuity-due, which he exchanged for a 25-year annuity-due. The present value of both are equal and based on different interest rates over the 25 years. The annual payment X of the annuity can be calculated using the formula for the present value of an annuity and the present value of the inherited perpetuity-due.

This problem is about financial mathematics, specifically involving perpetuities and annuities. In a perpetuity-due, the payments are made at the beginning of each period indefinitely. An annuity-due is similar, but the payments only last a specified number of years.

The present value (PV) of a perpetuity-due with annual payments P and annual interest rate r is calculated by PV = P / r. Given that P is $15,000 and r is 10%, the present value of the perpetuity-due that Jack inherits is $150,000.

When Jack swaps this for a 25-year annuity-due, with the first 10 years at 10% interest and the next 15 years at 8% interest, we calculate the annual payment X using the formula for the present value of an annuity. This formula involves dividing the total present value by the sum of the present value factors for each year at the respective interest rates.

Consequently, X can be calculated as follows: X = (PV of perpetuity-due) / ∑ (discount factors for each year). A detailed calculation will give the exact value of X.

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Answer:

a) [tex]P(X<80)=P(\frac{X-\mu}{\sigma}<\frac{80-\mu}{\sigma})=P(Z<\frac{80-72.6}{4.78})=P(z<1.548)[/tex]

And we can find this probability using the normal standard distirbution or excel and we got:

[tex]P(z<1.548)=0.939[/tex]

And that correspond to 93.9 %

b) [tex]P(60<X<80)=P(\frac{60-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{80-\mu}{\sigma})=P(\frac{60-72.6}{4.78}<Z<\frac{80-72.6}{4.78})=P(-2.636<z<1.548)[/tex]

And we can find this probability with this difference:

[tex]P(-2.636<z<1.548)=P(z<1.548)-P(z<-2.636)[/tex]

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

[tex]P(-2.636<z<1.548)=P(z<1.548)-P(z<-2.636)=0.939-0.0042=0.935 [/tex]

So we have approximately 93.5%

c) [tex]z=1.64<\frac{a-72.6}{4.78}[/tex]

And if we solve for a we got

[tex]a=72.6 +1.64*4.78=80.439[/tex]

So the value of velocity that separates the bottom 95% of data from the top 5% is 80.439.  

d) [tex]P(X>70)=P(\frac{X-\mu}{\sigma}>\frac{70-\mu}{\sigma})=P(Z>\frac{70-72.6}{4.78})=P(z>-0.544)[/tex]

And we can find this probability using the complement rule, normal standard distirbution or excel and we got:

[tex]P(z>-0.544)=1-P(z<-0.544) = 1-0.293=0.707 [/tex]

And that correspond to 70.7 %

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 vehicles speeds of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(72.6,4.78)[/tex]  

Where [tex]\mu=72.6[/tex] and [tex]\sigma=4.78[/tex]

We are interested on this probability

[tex]P(X<80)[/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(X<80)=P(\frac{X-\mu}{\sigma}<\frac{80-\mu}{\sigma})=P(Z<\frac{80-72.6}{4.78})=P(z<1.548)[/tex]

And we can find this probability using the normal standard distirbution or excel and we got:

[tex]P(z<1.548)=0.939[/tex]

And that correspond to 93.9 %

Part b

We want this probability

[tex]P(60<X<80)=P(\frac{60-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{80-\mu}{\sigma})=P(\frac{60-72.6}{4.78}<Z<\frac{80-72.6}{4.78})=P(-2.636<z<1.548)[/tex]

And we can find this probability with this difference:

[tex]P(-2.636<z<1.548)=P(z<1.548)-P(z<-2.636)[/tex]

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

[tex]P(-2.636<z<1.548)=P(z<1.548)-P(z<-2.636)=0.939-0.0042=0.935 [/tex]

So we have approximately 93.5%

Part c

For this part we want to find a value a, such that we satisfy this condition:

[tex]P(X>a)=0.05[/tex]   (a)

[tex]P(X<a)=0.95[/tex]   (b)

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

As we can see on the figure attached the z value that satisfy the condition with 0.95 of the area on the left and 0.05 of the area on the right it's z=1.64. On this case P(Z<1.64)=0.95 and P(z>1.64)=0.05

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

[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.95[/tex]  

[tex]P(z<\frac{a-\mu}{\sigma})=0.95[/tex]

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

[tex]z=1.64<\frac{a-72.6}{4.78}[/tex]

And if we solve for a we got

[tex]a=72.6 +1.64*4.78=80.439[/tex]

So the value of velocity that separates the bottom 95% of data from the top 5% is 80.439.  

Part d

[tex]P(X>70)[/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(X>70)=P(\frac{X-\mu}{\sigma}>\frac{70-\mu}{\sigma})=P(Z>\frac{70-72.6}{4.78})=P(z>-0.544)[/tex]

And we can find this probability using the complement rule, normal standard distirbution or excel and we got:

[tex]P(z>-0.544)=1-P(z<-0.544) = 1-0.293=0.707 [/tex]

And that correspond to 70.7 %

Answer:

71.08% probability that pˆ takes a value between 0.17 and 0.23.

Step-by-step explanation:

We use the binomial approxiation to the normal to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

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

The standard deviation of the binomial distribution is:

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

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that [tex]\mu = E(X)[/tex], [tex]\sigma = \sqrt{V(X)}[/tex].

In this problem, we have that:

[tex]p = 0.2, n = 200[/tex]. So

[tex]\mu = E(X) = np = 200*0.2 = 40[/tex]

[tex]\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{200*0.2*0.8} = 5.66[/tex]

In other words, find probability that pˆ takes a value between 0.17 and 0.23.

This probability is the pvalue of Z when X = 200*0.23 = 46 subtracted by the pvalue of Z when X = 200*0.17 = 34. So

X = 46

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

[tex]Z = \frac{46 - 40}{5.66}[/tex]

[tex]Z = 1.06[/tex]

[tex]Z = 1.06[/tex] has a pvalue of 0.8554

X = 34

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

[tex]Z = \frac{34 - 40}{5.66}[/tex]

[tex]Z = -1.06[/tex]

[tex]Z = -1.06[/tex] has a pvalue of 0.1446

0.8554 - 0.1446 = 0.7108

71.08% probability that pˆ takes a value between 0.17 and 0.23.

The probability that [tex]\( \hat{p} \)[/tex] takes a value between 0.17 and 0.23 is approximately 0.7108.

1. Given that 20% of adult women dye or highlight their hair, the true population proportion [tex]\( p \)[/tex] is 0.20.

2. We want to find the probability that a sample proportion [tex]\( \hat{p} \)[/tex] from a simple random sample (SRS) of size 200 falls within plus or minus 3 percentage points of this true value. In other words, we want to find [tex]\( P(0.17 < \hat{p} < 0.23) \)[/tex].

3. The standard error of [tex]\( \hat{p} \)[/tex] is given by:

[tex]\[ SE = \sqrt{\frac{p(1-p)}{n}} \][/tex]

where [tex]\( p = 0.20 \)[/tex] (the true population proportion) and [tex]\( n = 200 \)[/tex]  (the sample size).

4. Calculate the standard error:

[tex]\[ SE = \sqrt{\frac{0.20 \times (1-0.20)}{200}} = \sqrt{\frac{0.20 \times 0.80}{200}} \][/tex]

[tex]\[ SE = \sqrt{\frac{0.16}{200}} = \sqrt{0.0008} \approx 0.0283 \][/tex]

5. Next, we find the z-scores corresponding to the values 0.17 and 0.23 using the standard normal distribution table:

[tex]\[ z_{0.17} = \frac{0.17 - 0.20}{0.0283} \approx -1.0601 \][/tex]

[tex]\[ z_{0.23} = \frac{0.23 - 0.20}{0.0283} \approx 1.0601 \][/tex]

6. Using the z-scores, we find the corresponding probabilities from the standard normal distribution table:

[tex]\[ P(\hat{p} < 0.17) \approx P(Z < -1.0601) \approx 0.1446 \][/tex]

[tex]\[ P(\hat{p} < 0.23) \approx P(Z < 1.0601) \approx 0.8554 \][/tex]

7. Therefore, the probability that [tex]\( \hat{p} \)[/tex] takes a value between 0.17 and 0.23 is approximately:

[tex]\[ P(0.17 < \hat{p} < 0.23) = P(\hat{p} < 0.23) - P(\hat{p} < 0.17) \][/tex]

[tex]\[ \approx 0.8554 - 0.1446 = 0.7108 \][/tex]

8. Alternatively, we can find this probability directly using the cumulative distribution function (CDF) of the standard normal distribution:

[tex]\[ P(0.17 < \hat{p} < 0.23) = P(-1.0601 < Z < 1.0601) \][/tex]

[tex]\[ \approx \Phi(1.0601) - \Phi(-1.0601) \][/tex]

[tex]\[ \approx 0.8554 - 0.1446 = 0.7108 \][/tex]

9. Therefore, the probability that[tex]\( \hat{p} \)[/tex] takes a value between 0.17 and 0.23 is approximately 0.7108.

Answer:

[tex]x^{2} -10x+24[/tex]

Step-by-step explanation:

Answer:

The probability table is shown below.

A Poisson distribution can be used to approximate the model of the number of hurricanes each season.

Step-by-step explanation:

(a)

The formula to compute the probability of an event E is:

[tex]P(E)=\frac{Favorable\ no.\ of\ frequencies}{Total\ NO.\ of\ frequencies}[/tex]

Use this formula to compute the probabilities of 0 - 8 hurricanes each season.

The table for the probabilities is shown below.

(b)

Compute the mean number of hurricanes per season as follows:

[tex]E(X)=\frac{\sum x f_{x}}{\sum f_{x}}=\frac{176}{68}= 2.5882\approx2.59[/tex]

If the variable X follows a Poisson distribution with parameter λ = 7.56 then the probability function is:

[tex]P(X=x)=\frac{e^{-2.59}(2.59)^{x}}{x!} ;\ x=0, 1, 2,...[/tex]

Compute the probability of X = 0 as follows:

[tex]P(X=0)=\frac{e^{-2.59}(2.59)^{0}}{0!} =\frac{0.075\times1}{1}=0.075[/tex]

Compute the probability of X = 1 as follows:

[tex]\neq P(X=1)=\frac{e^{-2.59}(2.59)^{1}}{1!} =\frac{0.075\times7.56}{1}=0.1943[/tex]

Compute the probabilities for the rest of the values of X in the similar way.

The probabilities are shown in the table.

On comparing the two probability tables, it can be seen that the Poisson distribution can be used to approximate the distribution of the number of hurricanes each season. This is because for every value of X the Poisson probability is approximately equal to the empirical probability.

To address this question, probabilities from the given data are calculated first, after which Poisson distribution is applied to compute the probabilities using the mean number of hurricanes per season. A comparison of both results offers an insight into the accuracy of the Poisson distribution in modeling this phenomenon.

To answer this question, we first have to calculate probabilities based on the empirical data and then compare them to probabilities computed under the assumption of a Poisson distribution.

First, we count total seasons from 1940 through 2007, which is 68. Using these frequencies, we can calculate the probability of having 0-8 hurricanes each season as follows: the number of seasons with a certain number of hurricanes divided by the total number of seasons.

For the Poisson distribution, we first need to calculate the average (mean) number of hurricanes per season, which is the sum of the product of the number of hurricanes and its frequency divided by the total number of seasons. After finding the mean, we can compute the probability of experiencing 0-8 hurricanes using the Poisson formula: e^(-mean) * (mean^n) / n!. After that, we can compare these probabilities with the ones derived from the empirical data.

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Answer:

Since the balls are drawn with replacement, it means the individual probability  remain constant,

I have solved this problem on paper (Figures Attached).

Thanks.

Answer:

Since this problem related to the replacement problem thus

Pr(2 red)=(25/98)(25/98)=0.0651

Pr(5 green)=(19/98)(19/98)(19/98)(19/98)(19/98)

Pr(5 green)=2.73*10^-4

Pr(10 purple)=((30/98))^10=7.22*10^-6

Pr(5 blue)=((24/98))^5=8.8*10^-4

Answer:

0.0082 = 0.82% probability that he will pass

Step-by-step explanation:

For each question, there are only two possible outcomes. Either the students guesses the correct answer, or he guesses the wrong answer. The probability of guessing the correct answer for a question is independent of other questions. 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.

In this problem we have that:

[tex]n = 14, p = 0.3[/tex].

If the student makes knowledgeable guesses, what is the probability that he will pass?

He needs to guess at least 9 answers correctly. So

[tex]P(X \geq 9) = P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14)[/tex]

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

[tex]P(X = 9) = C_{14,9}.(0.3)^{9}.(0.7)^{5} = 0.0066[/tex]

[tex]P(X = 10) = C_{14,10}.(0.3)^{10}.(0.7)^{4} = 0.0014[/tex]

[tex]P(X = 11) = C_{14,11}.(0.3)^{11}.(0.7)^{3} = 0.0002[/tex]

[tex]P(X = 12) = C_{14,12}.(0.3)^{12}.(0.7)^{2} = 0.000024[/tex]

[tex]P(X = 13) = C_{14,13}.(0.3)^{13}.(0.7)^{1} = 0.000002[/tex]

[tex]P(X = 14) = C_{14,14}.(0.3)^{14}.(0.7)^{0} \cong 0 [/tex]

[tex]P(X \geq 9) = P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) = 0.0066 + 0.0014 + 0.0002 + 0.000024 + 0.000002 = 0.0082[/tex]

0.0082 = 0.82% probability that he will pass

Answer: area of fabric needed is 50.24 ft²

Step-by-step explanation:

The table in the shape of a circle has a diameter of 6 feet. This means that the diameter of the fabric that would just fit the table is 6 feet. Therefore, the diameter of the fabric needed to make a table cloth if it hangs 1 foot off the table would be 6 + 1 + 1 = 8 feet

The formula for determining the area of a circle is expressed as

Area = πr²

Where

r represents radius of the circle.

π is a constant whose value is 3.14

Radius = diameter/2. Therefore

r = 8/2 = 4

Area of fabric = 3.14 × 4²

= 50.24 ft²

Final answer:

To make a tablecloth for a circular table with a diameter of 6 feet, and an overhang of 1 foot, the student would need approximately 50.27 square feet of fabric.

Explanation:

The student is asking about finding the amount of fabric needed to create a tablecloth for a circular table with specific dimensions. Given that the table has a diameter of 6 feet, and the tablecloth needs to hang 1 foot off the table all the way around, we need to calculate the diameter of the fabric required.

To solve this, we need to add the overhang to the diameter of the table, considering that the overhang occurs on both sides:

Now, to find the area of fabric needed, we apply the formula for the area of a circle which is π × radius². First, we find the radius by halving the diameter:

Then, we calculate the area:

Finally, we can approximate π as 3.1416 to get an approximate area of:

Therefore, the student would need approximately 50.27 square feet of fabric to make the tablecloth.

Answer:

Below is the R code for the bootstrapping in exponential distribution. The result is attached below.

####################################

rm(list=ls(all=TRUE))

set.seed(12345)

N=c(10,100,500)

Rate=0.2

B=1000

MN=SE=rep()

for(i in 1:length(N))

{

n=N[i]

X=rexp(n,rate=Rate)

EST=1/mean(X)

ESTh=rep()

for(j in 1:B)

{

Xh=rexp(n,rate=EST)

ESTh[j]=1/mean(Xh)

}

MN[i]=mean(ESTh)

SE[i]=sd(ESTh)

}

cbind(N,Rate,MN,SE)

Step-by-step explanation:

Answer:

(-1,9)

Step-by-step explanation:

Answer:

Given p = 0.07 as the probability that someone is a universal donor

In case of Geometric Distribution, Probability of  getting the first success on nth trial is given by

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

where p is the probability of success on any one trial and (1-p) shows the probability of failure.

So the probability of the first subject to be a universal blood donor will be the seventh person is

[tex]P (X=7) = 0.07 (1-0.07) ^ {7-1} = 0.07 (0.93) ^ 6 = 0.07*0.647 = 0.0453[/tex]

So the final probability is 0.0453

Answer:

[tex]\frac{dy}{dx} = 2x ( 2cos2x) +sin2x[/tex]

[tex]\frac{dy}{dx} = 4x( cos2x)+sin2x[/tex]

Step-by-step explanation:

Given y = x sin2x .....(1)

Applying UV formula [tex]\frac{d(UV)}{dx} = u \frac{dv}{dx} + v\frac{du}{dx}[/tex]

Differentiating with respective to 'x' we get

[tex]\frac{dy}{dx} = x ( 2cos2x)\frac{d(2x)}{dx} +sin2x (1)[/tex]

[tex]\frac{dy}{dx} = x ( 2cos2x)(2) +sin2x (1)[/tex]

Final answer:-

[tex]\frac{dy}{dx} = 4x( cos2x)+sin2x[/tex]

Answer:

The set is a subspace

Step-by-step explanation:

We need to check 3 things: whether the 0 vector is in the set, whether the sum of 2 elements of the set is an element of the set and whether the product of an element of the set for a real scalar is an element of the set.

Yes: the 0 vector (0, 0, ..., 0) satysfies the set property: 0-0+0-0........-0 = 0.

Yes: Note that (v1+w1)-(v2+w2)+(v3+w3)- ..... - (vn+wn) = v1-v1+v3 - ... - vn + w1 - w2 + w3 - ... - wn = 0+0 = 0.

Yes: By taking r as common factor, we have rv1 - rv2 + rv3 - ... - rvn = r * (v1-v2+v3 - ... - vn) = r*0 = 0.

Thus, the described set is effectively a subspace.

The described set is a subspace of Rn (n even). It satisfies all three conditions of a subspace: containing the zero vector, closed under addition, and closed under scalar multiplication.

The set described is a subspace of ℝn (where n is even).

To determine if the set is a subspace, we need to check if it satisfies three conditions:

It contains the zero vector: The zero vector satisfies v1 - v2 + v3 - v4 + v5 - ... - vn = 0, so it is in the set.

Since the set satisfies all three conditions, it is a subspace of ℝn (where n is even).

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Answer:

a. A = {1, 2, 3, 4, 5, 6}

b. B = {1, 2, 3, 4, 5, 6}

c. C = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}

d. D = {-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5}

Step-by-step explanation:

a. the maximum value to appear in the two rolls

Since only the maximum value is computed, the variable can assume any integer from 1 to 6:

A = {1, 2, 3, 4, 5, 6}

b. the minimum value to appear in the two rolls;

Since only the minimum value is computed, the variable can assume any integer from 1 to 6:

B = {1, 2, 3, 4, 5, 6}

c. the sum of the two rolls;

The minimum value would be from rolling two ones (sum is 2) and the maximum value would be from rolling two sixes (sum is 12). Every integer in-between is possible:

C = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}

d. the value of the first roll minus the value of the second roll

The minimum value would be from rolling a one and a six (result is -5) and the maximum value would be from rolling a six and a one (result is 5). Every integer in-between is possible:

D = {-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5}

Final answer:

Explanation of possible values for maximum, minimum, sum, and difference in rolling two dice.

Explanation:

a. The possible values for the maximum value:

1, if the same number appears twice on both rolls

2 to 6, for individual numbers in different scenarios

b. The possible values for the minimum value:

1 to 5, as the minimum value will exclude the maximum value

c. The possible values for the sum:

2 to 12, representing all possible sums from rolling two dice

d. The possible values for the difference between two rolls:

-5 to 5, as the difference can range from -5 to 5

Answer: Length of arc DB is 14π feet.

Step-by-step explanation:

The sum of the angles on a straight line is 180 degrees. This means that

m∠DAB + m∠CAB = 180

m∠DAB + 40 = 180

m∠DAB = 180 - 40

m∠DAB = 140°

The formula for determining the length of an arc is expressed as

Length of arc = θ/360 × 2πr

Where

θ represents the central angle.

r represents the radius of the circle.

π is a constant whose value is 3.14

From the information given,

r = 18 feet

θ = 140°

Therefore,

Length of arc DB = 140/360 × 2 × π × 18

Length of arc DB = 14π feet

Answer:see the pictures attached

Step-by-step explanation:

Answer:

(1)

(a) Shown below.

(b) The probability that both impaneled jurors are women is 0.0667.

(2)

(a) Sampling 3 people and noting their gender.

(b) Shown below.

(c) The probability of each simple event is 0.125.

(d) The probability of selecting only one male is 0.375.

(e) The probability of selecting all 3 females is 0.125.

Step-by-step explanation:

(1)

Let the 4 men be denoted as: M₁, M₂, M₃ and M₄.

And the 2 women be denoted as: W₁ and W₂.

(a)

A jury of two is to be selected.

The simple events in this experiment are:

(M₁, M₂), (M₁, M₃), (M₁, M₄), (M₁, W₁), (M₁, W₂)

(M₂, M₃), (M₂, M₄), (M₂, W₁), (M₂, W₂)

(M₃, M₄), (M₃, W₁), (M₃, W₂)

(M₄, W₁), (M₄, W₂)

(W₁, W₂)

(b)

The total possible number of jury selections is, N = 15.

The possible combination such that both the jurors are woman is, n = 1.

Compute the probability of selecting  two women jurors as follows:

[tex]P(2\ juror\ are\ women)=\frac{n}{N} =\frac{1}{15}=0.0667[/tex]

Thus, the probability that both impaneled jurors are women is 0.0667.

(2)

(a)

The experiment consists of sampling 3 people from the voter registration and driving records and noting the gender of each person.

(b)

The simple events are:

S = {(M, M, M), (M, M, F), (M, F, M), (F, M, M), (M, F, F), (F, F, M), (F, M, F), (F, F, F)}

Total number of simple events = 8.

(c)

If the probability of selecting a male is same as the probability of selecting a female, i.e. P (M) = P (F) = [tex]\frac{1}{2}[/tex] then,

The probability of each simple event is:

[tex]\frac{1}{2} \times\frac{1}{2} \times\frac{1}{2}=\frac{1}{8}=0.125[/tex]

(d)

The number of simple events with only 1 male is n = 3.

Compute the probability that only one of the three is a man as follows:

[tex]P(1\ male)=\frac{n}{N} =\frac{3}{8} =0.375[/tex]

Thus, the probability of selecting only one male is 0.375.

(e)

The number of simple events with all 3 females is n = 1.

Compute the probability that all 3 females are selected as follows:

[tex]P(3\ female)=\frac{n}{N} =\frac{1}{8} =0.125[/tex]

Thus, the probability of selecting all 3 females is 0.125.

Answer:

0.0836

Step-by-step explanation:

Given that p1 be the proportion of successes in the first population and let p2 be the proportion of successes in the second population

[tex]H_0:p_1-p_2=0\\H_a:p_1-p_2\neq 0[/tex]

is the hypotheses created

This is two tailed

Test statistic Z = 1.73

corresponding p value

= 0.08363

=0.0836 (after rounding off to 4 decimals)

The p-value for this test is 0.0802.

The p-value for this two-tailed test can be calculated by doubling the probability of observing a test statistic at least as extreme as the observed z-score of 1.73, under the assumption that the null hypothesis is true.

Given that the z-score is 1.73, we need to find the area under the standard normal distribution curve that lies beyond this z-score. This area represents the probability of observing a value at least as extreme as 1.73 standard deviations from the mean under the null hypothesis.

Using a standard normal distribution table or a calculator, we can find the probability corresponding to a z-score of 1.73. Let's denote this probability as P(Z > 1.73).

P(Z > 1.73) = 1 - P(Z < 1.73)

Using a standard normal distribution table or calculator, we find that P(Z < 1.73) is approximately 0.9599. Therefore:

P(Z > 1.73) = 1 - 0.9599 = 0.0401

Since this is a two-tailed test, we need to consider the probability of observing a z-score at least as extreme in either direction (both tails of the distribution). Thus, we multiply the single-tail probability by 2:

p-value = 2 * P(Z > 1.73) = 2 * 0.0401 = 0.0802

Rounded to four decimal places, the p-value is 0.0802.

Therefore, the p-value for this test is 0.0802.

The answer is: 0.0802.

Answer:

a) Probability that a randomly selected commuter will spend more than 7 minutes waiting for GO-D.O.T = P(7 < x ≤ 20) = 0.65

b) Standard deviation of the uniform distribution = 5.77 minutes

c) Probability that a randomly selected commuter will spend longer than 10 minutes but no more than 17 minutes waiting for the GO-D.O.T = P(10 < x < 17) = 0.35

d) average waiting time for the uniform distribution = 10 minutes.

Step-by-step explanation:

This is a uniform distribution problem with lower limit of 0 minute and upper limit of 20 minutes.

a = 0, b = 20

Probability = f(x) = [1/(b-a)] ∫ dx (with the definite integral evaluated between the two intervals whose probability is required.

a) Probability that a randomly selected commuter will spend more than 7 minutes waiting for GO-D.O.T

P(7 < x ≤ 20) = f(x) = [1/(b-a)] ∫²⁰₇ dx

P(7 < x ≤ 20) = (20-7)/(20-0) = (13/20) = 0.65

b) Standard deviation of the uniform distribution

Standard deviation of a uniform distribution is given as

σ = √[(b-a)²/12]

σ = √[(20-0)²/12]

σ = √[20²/12]

σ = 5.77 minutes

c) Probability that a randomly selected commuter will spend longer than 10 minutes but no more than 17 minutes waiting for the GO-D.O.T = P(10 < x < 17)

P(10 < x < 17) = (17-10)/(20-0)

P(10 < x < 17) = (7/20) = 0.35

d) The average waiting time.

The average of a uniform distribution = (b+a)/2

Average waiting time = (20+0)/2

Average waiting time = 10 minutes

Hope this Helps!!!

Answer:

a) 81.85% of employees put between $9, 500 and $11,000 into the 401 k per year

b) 0.13% of employee put more than $11, 500 into the 401 k per year

c) 97.72% of employees put less than $11,000 into the 401k per year.

d) 97.72% of employees put more than $9,000 into the 401k per year

e) 2.15% of employees put between than $11,000 and $11, 500 into the 401k per year

f) An employee would need to put $10,640 into his or her 401 K to be in the upper 10% of investors

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

[tex]\mu = 10000, \sigma = 500[/tex]

a. what proportion of employees put between $9, 500 and $11,000 into the 401 k per year

This is the pvalue of Z when X = 11000 subtracted by the pvalue of Z when X = 9500. So

X = 11000

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

[tex]Z = \frac{11000 - 10000}{500}[/tex]

[tex]Z = 2[/tex]

[tex]Z = 2[/tex] has a pvalue of 0.9772.

X = 9500

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

[tex]Z = \frac{9500 - 10000}{500}[/tex]

[tex]Z = -1[/tex]

[tex]Z = -1[/tex] has a pvalue of 0.1587

0.9772 - 0.1587 = 0.8185

81.85% of employees put between $9, 500 and $11,000 into the 401 k per year

b. What proportion of employee put more than $11, 500 into the 401 k per year?

This is 1 subtracted by the pvalue of Z when X = 11500. So

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

[tex]Z = \frac{11500 - 10000}{500}[/tex]

[tex]Z = 3[/tex]

[tex]Z = 3[/tex] has a pvalue of 0.9987

1 - 0.9987 = 0.0013

0.13% of employee put more than $11, 500 into the 401 k per year

c. What proportional of employees put less than $11,000 into the 401k per year?

This is the pvalue of Z when X = 11000. So

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

[tex]Z = \frac{11000 - 10000}{500}[/tex]

[tex]Z = 2[/tex]

[tex]Z = 2[/tex] has a pvalue of 0.9772.

97.72% of employees put less than $11,000 into the 401k per year.

d. What proportional of employees put more than $9,000 into the 401k per year?

This is 1 subtracted by the pvalue of Z when X = 9000. So

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

[tex]Z = \frac{9000 - 10000}{500}[/tex]

[tex]Z = -2[/tex]

[tex]Z = -2[/tex] has a pvalue of 0.0228.

1 - 0.0228 = 0.9772

97.72% of employees put more than $9,000 into the 401k per year

e. What proportional of employees put between than $11,000 and $11, 500 into the 401k per year?

This is the pvalue of Z when X = 11500 subtracted by the pvalue of Z when X = 11000. So

X = 11500

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

[tex]Z = \frac{11500 - 10000}{500}[/tex]

[tex]Z = 3[/tex]

[tex]Z = 3[/tex] has a pvalue of 0.9987

X = 11000

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

[tex]Z = \frac{11000 - 10000}{500}[/tex]

[tex]Z = 2[/tex]

[tex]Z = 2[/tex] has a pvalue of 0.9772.

0.9987 - 0.0972 = 0.0215

2.15% of employees put between than $11,000 and $11, 500 into the 401k per year

f. How much would an employees need to put into his or her 401 K to be in the upper 10% of investors?

This is the value of Z when X has a pvalue of 1-0.1 = 0.9. So it is X when Z = 1.28.

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

[tex]1.28 = \frac{X - 10000}{500}[/tex]

[tex]X - 10000 = 500*1.28[/tex]

[tex]X = 10640[/tex]

An employee would need to put $10,640 into his or her 401 K to be in the upper 10% of investors

Answer:

Based on the graph below, what is the total number of solutions to the equation f(x)= g(x) will be 3.

Step-by-step explanation:

The intersection points of both graphs would be the total number of solutions to the equation f(x)= g(x).

From the given diagram, it is clear that both the graphs intersect at three locations points or intersection points. The approximations locations of The intersection points of  both graphs are

Therefore, based on the graph below, what is the total number of solutions to the equation f(x)= g(x) will be 3.

Answer: f(t) = 15 - 1.1t

Step-by-step explanation:

Let t represent the number of hours since the candle was lit.

A 15-inch candle is lit and burns at a constant rate of 1.1 inches per hour. This means that in t hours, the candle that would have burnt is 1.1t

The length of the candle that would be left after t hours is expressed as

15 - 1.1t

suppose f is a function such that f(t) represents the remaining length of the candle (in inches) t hours after it was lit, then

f(t) = 15 - 1.1t

Answer:

note:

solution is attached due to error in mathematical equation. please find the attachment

The conditional distribution of income level among single men should add up to 100%, since when calculating a conditional distribution we treat the subgroup as the total population. Thus, income levels among single men are all the possible outcomes for this group and should collectively correspond to 100% of cases.

The conditional distribution in this scenario is the distribution of income level among single men, being key to understanding he relative frequency of various income levels within this specific group. Since we are looking at a subset of the total population (in this case, single men), the percentages should indeed add up to 100%. This is because when we calculate the conditional distribution, we treat this subgroup as if it were the total population. Therefore, the total should be 100%, reflecting the entire subgroup, not compared to the entire population. An example could be: If 10% of single men are in low income, 30% are in middle income, and 60% are in high income, these percentages represent the income level distribution among single men, and add up to 100%.

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Answer:

Step-by-step explanation:

As per the given question, their current winning is $1 million.

The probability of the guessing to be true is 50% = [tex]\frac{50}{100} = \frac{1}{2}[/tex].

There is also a possibility of 50% to be wrong, which can reduced the winning amount to $500,000 that is the half of the current amount.

Hence, the contestant should not play.

Answer:

$4848.8

Step-by-step explanation:

(1 + 0.05)⁵ = 1.276

FV = 3800 × 1.276

= 4848.8

Final answer:

The future value of $3,800 invested for five years at a 5% interest rate, using the formula [tex]FV = PV[/tex] x [tex](1 + r)^t[/tex], is approximately $4,849.07.

Explanation:

You are asking how to calculate the future value of an amount of money when invested at a certain interest rate over a set period of time. Specifically, you want to know the future value of $3,800 invested for five years at a 5% interest rate.

To calculate the future value (FV) of money we use the formula:
[tex]FV = PV X (1 + r)^t[/tex]

Where:

PV is the present value or initial amount ($3,800)

r is the annual interest rate (5%, or as a decimal, 0.05)

t is the time in years the money is invested (5 years)

Using the formula, we get:
FV = $3,800 x (1 + 0.05)5

FV = $3,800 x (1.276281)

FV = $3,800 x 1.276 (rounded to three decimal places)

FV = $4,849.07 (rounded to two decimal places)

Hence, the future value of $3,800 invested for five years at a 5% interest rate would be approximately $4,849.07.

Based on the properties of the Normal Distribution and Standard Deviations, we find that about 0.3% of the items will weigh either less than 87 grams or more than 93 grams.

To solve this question, we need to understand and apply concepts of the Normal Distribution and Standard Deviations. A normal distribution is a common type of statistical distribution representing various types of data, characterized by a bell-shaped curve symmetric about its mean.

In this case, the item weights are normally distributed with a mean (average) of 90 grams and a standard deviation of 1 gram. Standard deviation basically tells us how much the data varies around the mean value.

The question now asks us to find out what percentage of weights are either less than 87 grams or more than 93 grams. We know that within 1 standard deviation of the mean (between 89 and 91 grams in this case), we have about 68.2% of the data. Similarly, within 2 standard deviations of the mean (between 88 and 92 grams), we have about 95.4% of the data. And within 3 standard deviations (between 87 and 93 grams), we have about 99.7% of the data.

So, since 87 grams and 93 grams are 3 standard deviations away from the mean, we know that about 99.7% of the weights lie between these two values. This means that the other 0.3%, or 0.15% on either side, is either below 87 grams or above 93 grams. Hence, 0.3% of the items will either weigh less than 87 grams or more than 93 grams.

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Answer:

A company has learned that the relationship between its advertising and sales shows diminishing marginal returns. That​ is, as it saturates consumers with​ ads, the benefits of increased advertising diminish. The company should expect to find a linear association between its advertising and sales - This statement is false

Step-by-step explanation:

According to the scenario given for the company, it was said that the marginal return diminished after a saturation point, therefore, the company should rather expect a non-linear pattern and not a linear pattern.

Therefore, the statement expressed in the question is false.

Final answer:

Mike's monthly car payment is approximately $489.99. The total cost of the car including interest over 60 months is $29,399.40, and the total interest Mike pays over the loan's lifetime is $4,399.40.

Explanation:

To determine Mike's monthly car payment, total cost of the car, and total interest paid over the life of the loan, we need to use the formula for the monthly payment on an installment loan, which can be found using the formula:

P = (Pv * r) / (1 - (1 + r)-n)

Where:

Lets calculate the monthly payment (P):

P = ($25,000 * 0.065/12) / (1 - (1+0.065/12)-60)
= $489.99 approximately

Mike's total cost of the car is the monthly payment multiplied by the number of payments:

$489.99 * 60 = $29,399.40

The total interest Mike pays is the total cost minus the loan amount:

$29,399.40 - $25,000 = $4,399.40

Thus, Mike's monthly payment is approximately $489.99, the total cost of his car will be $29,399.40, and the total interest paid over the life of the loan is $4,399.40.

Answer:

The Answer is "G. P"

Step-by-step explanation:

They ask for "Which of the following vectors must be a PARTICULAR solution to the system?" so the answer simply G. which is P

Answer:

The minimum score required for an A grade is 86.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

[tex]\mu = 72.1, \sigma = 9.5[/tex]

Find the minimum score required for an A grade.

Top 7%, which is the value of X when Z has a pvalue of 1-0.07 = 0.93. So it is X when Z = 1.475. So

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

[tex]1.475 = \frac{X - 72.1}{9.5}[/tex]

[tex]X - 72.1 = 1.475*9.5[/tex]

[tex]X = 86[/tex]

The minimum score required for an A grade is 86.