BJ's goal is to have $50,000 saved at the end of Year 5. At the end of Year 2, they can add $7,500 to their savings but they want to deposit the remainder they need to reach their goal today, Year 0, as a lump sum deposit. If they can earn 4.5 percent, how much must they deposit today

Answer:

The probability mass function of the number of modems in use at the given time is:

[tex]P (X=x)={1000\choose x}(0.01)^{x}(1-0.01)^{1000-x};\ x=0, 1, 2, ...49[/tex]

Step-by-step explanation:

Let the random variable X = number of modems in use.

The internet provider uses 50 modems.

The number of customers served by the internet user is, n = 1000.

The probability that a customer will require an internet connection is, p = 0.01.

It is provided that the customers are independent of each other.

The random variable X satisfies all the properties of a Binomial distribution with parameters n = 1000 and p = 0.01.

The probability mass function of a Binomial distribution is:

[tex]P (X=x)={n\choose x}(p)^{x}(1-p)^{n-x};\ x=0, 1, 2, ...49[/tex]

Then the probability mass function of the number of modems in use at the given time is:

[tex]P (X=x)={1000\choose x}(0.01)^{x}(1-0.01)^{1000-x};\ x=0, 1, 2, ...49[/tex]

The question is about calculating the probability mass function of a binomial distribution. Given a fixed number of independent trials (1000 customers), with a constant probability of success (0.01), the pmf is given by P(X = k) = C(n, k) * (p)^k * (1-p)^(n-k), where n=1000, p=0.01, and k varies from 0 to 50.

The situation described is a case of a binomial distribution. This is because there are a fixed number of independent trials (1000 customers attempting to connect), and each trial has two possible outcomes (the customer needs a connection or does not need a connection), with the probability of success (need a connection) being constant (0.01).

The probability mass function (pmf) of a binomial distribution is given by:

P(X = k) = C(n, k) * (p)^k * (1-p)^(n-k)

where:
- P(X = k) is the probability that exactly k trials are successful
- n is the total number of trials (here, 1000)
- k is the number of successful trials (modems in use)
- p is the probability of success on a single trial (here, 0.01)
- C(n, k) is the combination of n items taken k at a time.

To find the pmf of the number of modems in use, substitute the given values into the pmf formula for each possible value of k (0 to 50, because the ISP has only 50 modems). For each k, the result is the probability that k modems are in use at the given time.

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The 95% confidence interval for the population variance of the provided sample is between 2.34 and 24.87. This is calculated using the chi-square distribution, by finding the sample variance and then applying the chi-square formulas for each bound of the confidence interval.

The question is asking for a 95% confidence interval for the variance of a normal population given a random sample. We would first utilize the chi-square distribution, specifically the chi-square statistic (χ^2), which measures the discrepancy between the observed data and what we would expect if the null hypothesis were true. The chi-square statistic is calculated as: χ^2 = (n-1) * (sample variance) / (population variance).

In this question, we are being asked to find the confidence interval for the population variance, but we are given a sample. From our sample we calculate the sample variance and then use the chi-square distribution to find the confidence interval for the population variance.

To calculate the sample variance: Add up the squared deviation of each number from the mean and divide by n-1 for the sample variance. For this sample, the variance is 7.67.

For the chi-square distribution, the degrees of freedom (df) will be n-1, in this case, 7. The confidence interval for the variance is (df * s^2 / χ^2_right, df * s^2 / χ^2_left), where χ^2_right and χ^2_left represent the right and left critical values for the chi-square distribution respectively. For a 95% confidence interval with 7 degrees of freedom, the right critical value is 2.167 and the left critical value is 16.013. Substituting in our values, we get the confidence interval to be (7.67*7/16.013, 7.67*7/2.167). This simplifies to (2.34, 24.87).

Therefore, we estimate with 95 percent confidence that the true value of the population variance is between 2.34 and 24.87.

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The 95% confidence interval for the population variance of the given sample is calculated using Chi-Square distribution. The steps include finding the sample mean, sample variance, and then applying the Chi-Square values to get the interval. The final interval is approximately (1.179, 8.703).

To calculate the 95% confidence interval for the population variance from the given sample data (14, 10, 13, 16, 12, 18, 15, and 11), we need to follow these steps:

= 13.625

≈ 2.696

Lower limit = (df * s²) / χ² (0.975,7)

≈ (7 * 2.696) / 16.012

≈ 1.179
Upper limit = (df * s²) / χ² (0.025,7)

≈ (7 * 2.696) / 2.167

≈ 8.703

Thus, the 95% confidence interval for the population variance is approximately (1.179, 8.703).

Answer:

Step-by-step explanation:

sequence=seq(1,27, by=3)

new_seq=c()

for (i in sequence){

n_seq.append(log(i))

}

new_seq<- n_seq[-(2:5)]

seq_length=length(new_seq)

sorted_seq=sort(new_seq,decreasing=TRUE)

The average number of accidents in the intersection in a week, given by the expected value of the random variable, is calculated by multiplying each possible value by its corresponding probability and summing these products. The resulting average is 1.8 accidents per week.

This question is about calculating the expected value or mean of a random variable. For a discrete random variable, we calculate the expected value by summing the product of each possible value and its corresponding probability.

Here, you should multiply the number of accidents (x) by the corresponding probability (P(X = x)), then sum these products. So, the calculation becomes:

0*0.20 + 1*0.30 + 2*0.20 + 3*0.15 + 4*0.10 + 5*0.05 = 0 + 0.30 + 0.40 + 0.45 + 0.40 + 0.25 = 1.80.

Therefore, on average, there are 1.8 accidents in the intersection in a week, so the answer is (c).

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Answer:

the expected number of claims within a year is 10

Step-by-step explanation:

since the life expectancy of each man is independent from others, then the random variable X= number of man of 50 years who die within a year , follows a binomial distribution. Thus the expected value of X is given by

E(X) = n*p

where

p= probability that a man of age 50 dies within a year = 0.01

n = number of policies on men of age 50 = 1000

replacing values

E(X) = n*p = 0.01 * 1000 = 10

therefore the expected number of claims within a year is 10

Using the binomial distribution, it is found that the estimate of the number of claims that the company can expect from beneficiaries of these men within a year is of 10.

For each men, there are only two possible outcomes, either there is a claim(they die), or there is not a claim. The probability of a men dying is independent of any other men, hence, the binomial distribution is used to solve this question.

Binomial probability distribution

Probability of exactly x successes on n repeated trials, with p probability, and has expected value given by:

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

In this problem:

Then:

[tex]E(X) = 1000(0.1) = 10[/tex]

The expected number of claims is of 10.

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Answer:

Step-by-step explanation:

Hello!

The objective is to test if the new type of resin has a bonding time between 2 and 6 hs, less than 2 hs or more than 6 hs makes it undesirable.

The bonding time of resin is normally distributed with mean μ= 4.5 hs and standard deviation δ= 1.5 hs.

a. What is the probability that the bonding time will be less than 2 hours?

Symbolically:

P(X<2)

Using thestandard normal distribution you have to standardize this value to reach the corresponding probability using Z= (X-μ)/δ ~N(0;1)

P(Z<(2-4.5)/1.5)= P(Z<-1.67)= 0.047

b. What is the probability that the bonding time will be more than 6 hours?

P(X>6)= 1 - P(X≤6)

1 - P(Z≤(6-4.5)/1.5)= 1 - P(Z≤1)= 1 - 0.841= 0.159

c. How often would the performance be considered undesirable (in a value of probability)?

The performance will be considered undesirable when the bonding time is less than 2 hs or when the time is higher than 6 hs,

P(X<2) + P(X>6)= 0.047 + 0.159= 0.206

20.6% of the new resin will be considered undesirable.

d. Between what two times (equally before the mean and equally after the mean) accounts for the a drying time of 95%?

I'll start working with the standard normal distribution and then transform the values to X.

Considering that the mean if the standard normal distribution is zero and that the distribution is symmertical regarding it's mean.

You need to look for two values of Z equidistant from the mean that surround 1-α: 0.95 of the distribution, leaving the remaining α: 0.05 outside, equally distributed in two tails α/2: 0.025 (See attachment)

These two value would be the same but with different sign, symbolically:

P(-d≤Z≤d)= 0.95

We can  say that the accumulated probability until "d" is

P(Z≤d)= (1-α)+(α/2)

P(Z≤d)= 0.95+0.025= 0.975

And the accumulated  until "-d" is:

P(Z≤-d)= α/2

P(Z≤-d)= 0.025

Now you look in the table of the standard normal distribution for the corresponding Z values:

P(Z≤d)= 0.975 ⇒ d= 1.965

P(Z≤-d)= 0.025 ⇒ -d= -1.965

Now for each value you have to use the variable information to reverse the standardization and reach the corresponding bonds of the interval:

Upper bond:

Z= (d-μ)/δ

d= (Z*δ)+μ

d= (1.965*1.5)+4.5

d= 7.45

Lower bond

Z= (-d-μ)/δ

-d= (Z*δ)+μ

-d= (-1.965*1.5)+4.5

-d= 1.55

95% of the bonding time will be between 1.55 and 7.45 hs

I hope it helps!

Answer:

[tex]P(y = 2) = 0.294[/tex]

Step-by-step explanation:

If the times for service calls follow a normal distribution with mean μ = 60 minutes and standar deviation σ = 10 minutes, then we can calculate the probability that one call take more than 68.4 minutes, thus:

P(x>68.4) = 1 - P(X<68.4)

We standardize as follow

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

[tex]z = \frac{68.4 - 60}{10} = 0.84[/tex]

Since the table of the distribution normal, we obtain:

1-  P(z <0.84) = 1 - 0.7995  = 0.2005

Therefore:

P(x>68.4) = 0.2005

Now if you select random eight service calls, the number of calls takes more than 68.4 minutes is a binomial variable with

p = 0.2005

n = 8

And we need calculate this probability

[tex]P(y = 2) = (8C2)(0.2005)^{2}(1-2005)^{8-2}[/tex]

[tex]P(y = 2) = 0.294[/tex]

To find the probability that exactly two of the service calls take more than 68.4 minutes, we can use the properties of the normal distribution and the binomial distribution. First, we find the probability of a single service call taking more than 68.4 minutes using the z-score. Then, we use the binomial distribution to find the probability of exactly two service calls taking more than 68.4 minutes in a random sample of eight service calls.

To solve this problem, we will use the properties of the normal distribution. We know that the mean service time is 60 minutes and the standard deviation is 10 minutes. We need to find the probability that exactly two of the service calls take more than 68.4 minutes. First, we need to find the probability that a single service call takes more than 68.4 minutes. Using the z-score formula, we find that the z-score for 68.4 minutes is (68.4 - 60) / 10 = 0.84. Looking up this value in the standard normal distribution table, we find that the probability of a service call taking more than 68.4 minutes is approximately 0.2019. Since we have a random sample of eight service calls, we can use the binomial distribution to find the probability of exactly two service calls taking more than 68.4 minutes. The formula for this is P(X = k) = C(n, k) * p^k * (1-p)^(n-k), where n is the sample size, k is the number of successes, p is the probability of success, and C(n, k) is the binomial coefficient. Plugging in the values from our problem, we have P(X = 2) = C(8, 2) * 0.2019^2 * (1-0.2019)^(8-2). Evaluating this expression, we find that the probability of exactly two service calls taking more than 68.4 minutes is approximately 0.2262.

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Answer:

Therefore, the probability is P=1/4.

Step-by-step explanation:

We know that a super has a key ring with 12 keys. He has forgotten which one opens the apartment he needs to enter.

We conclude that each key has an equal probability of opening an apartment. Since there are 12 keys, it follows that the probability for each key is equal,

p = 1/12.

We calculate the probability he is able to enter before reaching the 4th key. So he will try three keys by then. We get:

[tex]P=p+p+p\\\\P=\frac{1}{12}+\frac{1}{12}+\frac{1}{12}\\\\P=\frac{3}{12}\\\\P=\frac{1}{4}[/tex]

Therefore, the probability is P=1/4.

Answer:

A = Total cost

B = Variable cost

C = Fixed cost

Step-by-step explanation:

A - Manufacturing cost includes all the cost directly and/or indirectly involved in the production process. It includes the variable component and fixed component, which represent the total cost. Hence, to arrive at the manufacturing margin, the total cost is subtracted from the net sales. The results us the manufacturing margin.

B - Contribution margin only includes and ultimately works with the individual variable element of cost, unlike the manufacturing margin. Hence, to arrive at the contribution margin, the variable cost element is subtracted from the manufacturing margin.

C - To arrive at the income from operations, the fixed cost element is subtracted from the contribution margin. With this, our income from operations can be derived.

Answer:

a) [tex] p = (3/4)^5 *(1/4) =0.0593[/tex]

b) [tex] P(X=6) = (6C6) (0.25)^6 (1-0.25)^{6-6}= 0.000244[/tex]

c) [tex] P(X \geq 1)[/tex]

And we can use the complement rule like this:

[tex]P(X \geq 1) = 1-P(X<1) = 1-P(X=0) [/tex]

[tex] P(X=0) = (6C0) (0.25)^0 (1-0.25)^{6-0}= 0.17798[/tex]

And replacing we have:

[tex]P(X \geq 1) = 1-P(X<1) = 1-P(X=0) =1-0.17798 = 0.822[/tex]

Step-by-step explanation:

Previous concepts

A Bernoulli trial is "a random experiment with exactly two possible outcomes, "success" and "failure", in which the probability of success is the same every time the experiment is conducted". And this experiment is a particular case of the binomial experiment.

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".

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]  

The complement rule is a theorem that provides a connection between the probability of an event and the probability of the complement of the event. Lat A the event of interest and A' the complement. The rule is defined by: [tex]P(A)+P(A') =1[/tex]

We can model the number of correct questions answered with a binomial distribution [tex] X \sim Binom(n = 6, p = 1/4=0.25)[/tex]

Solution to the problem

Assuming the following questions:

a) the first question she gets right is the 6th question?  

For this case we want the first 5 questions incorrect and the last one correct, assuming independence we have:

[tex] p = (3/4)^5 *(1/4) =0.0593[/tex]

(b) she gets all of the questions right?

For this case we want all the questions right so then we want this:

[tex] P(X=6) = (6C6) (0.25)^6 (1-0.25)^{6-6}= 0.000244[/tex]

(c) she gets at least one question right?

For this case we want this probability:

[tex] P(X \geq 1)[/tex]

And we can use the complement rule like this:

[tex]P(X \geq 1) = 1-P(X<1) = 1-P(X=0) [/tex]

[tex] P(X=0) = (6C0) (0.25)^0 (1-0.25)^{6-0}= 0.17798[/tex]

And replacing we have:

[tex]P(X \geq 1) = 1-P(X<1) = 1-P(X=0) =1-0.17798 = 0.822[/tex]

The probability of guessing all 6 answers correctly in a multiple choice exam is 1/4096.

In mathematics, especially in the field of probability, the question describes a situation in which the answer to each of the questions on the exam is guessed at random. To calculate the probability of correctly guessing the answer to an individual question with 4 choices, we need to remember that probability is calculated as a ratio of the favorable outcomes to the total number of outcomes. In this case, each question has 1 correct answer and 3 incorrect answers, so the probability of correctly guessing an answer is 1/4 or 0.25 for each question.

However, if we were to calculate the probability of correctly guessing all the answers, such a probability would be significantly smaller. Because the answers to the questions are guessed independently, their probabilities multiply. Thus, the probability of correctly guessing the answer to all 6 questions is (1/4)^6 = 0.000244140625.

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Final answer:

The correct statistical analysis for assessing the effects of two independent variables on a single outcome and understanding interactions between the variables is c) ANOVA 2-Way.

Explanation:

The appropriate statistical analysis for the researcher’s hypothesis, which involves understanding the effects of two independent variables (current employment status and prior cash welfare assistance) on a dependent variable (material well-being) is a two-way ANOVA. This statistical method is used to analyze the impact of two independent categorical variables on a continuous outcome and to interact between these variables.

A two-way ANOVA, also known as factorial ANOVA, is particularly useful in this case as the researcher is interested not just in the separate effect of each variable, but also in the potential interaction between current employment status and the experience of receiving cash welfare in the past. This interaction term allows the researcher to understand if the effect of employment status on material well-being differs based on whether the women had received cash welfare.

Therefore, the correct answer to the question is: c) ANOVA 2-Way

Answer:

(a) [tex]p=100x-14[/tex]

(b) [tex]p=\$46[/tex]

Step-by-step explanation:

Linear Modeling

It consists of finding an equation of a line that fits the conditions of a certain situation in real life. We'll use a linear model for the cookies of the students.

(a) We know that we have a total of n=100 cookies. If sold for $0.10 each, they lose $4. We have an initial condition (x,p) = (0.10,-4), where x is the price of each cookie and p(x) is the net profit from selling all the cookies. The second conditions are that when then the price is $0.50 each, there is a positive profit of $36, which is a second point (0.5,36). That is enough to build the linear function, that can be found by

[tex]\displaystyle p-p_1=\frac{p_2-p_1}{x_2-x_1}(x-x_1)[/tex]

[tex]\displaystyle p+4=\frac{36+4}{0.5-0.1}(x-0.1)[/tex]

Reducing

[tex]p=100x-14[/tex]

(b) If the students sell the cookies for x=0.60 each, the profit will be

[tex]p=100(0.6)-14=46[/tex]

[tex]p=\$46[/tex]

It's a reasonable answer because we have found that increasing the price, the profit will increase also. The model doesn't have any restriction for the price

Answer:

131200 Units

Step-by-step explanation:

Now, in the production process:

Units to be accounted for= Beginning Work in Process+Work Started

Units Accounted for=Ending Work in Process + Completed and Transferred out

Units to be accounted for = Units Accounted for

Beginning Work in Process =9900 Ending Work in Process = 49000 Total units to be accounted for= 180200

Since

Units to be accounted for = Units Accounted for

Then:

Units Accounted for=Ending Work in Process + Completed and Transferred out

180200=49000+Completed and Transferred out

Completed and Transferred out=180200-49000

=131200

131200 units were transferred to Department 2.

The number of units transferred out to Department 2 can be found by subtracting the Ending Work in Process from the Total units to be accounted for. In this case, that would be 180200 - 49000, which equals 131200 units.

In the scenario described, the number of units transferred out to Department 2 can be calculated by deducting the Ending Work in Process from the Total units to be accounted for. This is because the Total units to be accounted for includes all units at the start (Beginning Work in Process), all units processed during the period, and all units still in process at the end (Ending Work in Process). So, the units transferred to Department 2 were processed by Department 1 but are not part of Department 1's ending work in process.

Hence, using the provided numbers:

Units transferred out to Department 2 = Total units to be accounted for - Ending Work in Process = 180200 - 49000 = 131200

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Answer:

For bbff we have only 6.3% probability

Step-by-step explanation:

If the parents are heterozygous for both traits, them they are represented by:

BbFf × BbFf

Parent 1: BbFf

Parent 2: BbFf

We have to find the percentage of occurence of bb × ff, which is a child that has blue eyes and no freckles, with no dominant factor.

By distributing the possibilities in a Punnett square, vide picture. We have the following possibilities:

Genotype   Count     Percent  

bBfF  4     25

BBfF  2     12.5

bBFF  2     12.5

bBff          2     12.5

bbfF  2     12.5

BBFF  1     6.3

BBff      1            6.3

bbFF  1     6.3

bbff      1     6.3

For bbff we have only 6.3% probability

Answer:

E (13.755 ± 1.456)

Step-by-step explanation:

with 95% confidence interval and df=25

we can find t score on the t table : t*=2.06

(because the table has already provided us standard error , we DON'T have to calculate it by using  σsample=S/[tex]\sqrt{n}[/tex]) (standard error = standard deviation)

so the interval should be  13.755± 2.06(0.707)= 13.755± 1.456

The 95% confidence interval for [tex]\(\mu\)[/tex] is e. [tex]\(13.755 \pm 1.456\)[/tex], based on a sample mean of 13.755, standard error of 0.707, and 25 degrees of freedom.

To construct a 95% confidence interval for the mean [tex]\(\mu\)[/tex], we need to use the sample mean, the standard error, and the appropriate critical value for a t-distribution.

Given data:

- Sample mean [tex](\(\bar{x}\))[/tex]: 13.755

- Standard error (SE): 0.707

- Degrees of freedom (df): 25

For a 95% confidence interval with 25 degrees of freedom, we need the critical value [tex]\( t^* \).[/tex] We can find [tex]\( t^* \)[/tex] using a t-table or statistical software.

For 25 degrees of freedom, the critical value [tex]\( t^* \)[/tex] for a 95% confidence interval is approximately 2.064.

The formula for the confidence interval is:

[tex]\[\bar{x} \pm t^* \times \text{SE}\][/tex]

Substitute the values:

[tex]\[13.755 \pm 2.064 \times 0.707\][/tex]

Calculate the margin of error:

[tex]\[2.064 \times 0.707 \approx 1.459\][/tex]

Thus, the 95% confidence interval is:

[tex]\[13.755 \pm 1.459\][/tex]

So, the correct answer is:

e. [tex]\[\boxed{13.755 \pm 1.456}\][/tex]

Complete Question:

Answer:

Therefore logistic equation is

[tex]P(t) = \frac{4000}{1+ 24(\frac{36}{11})^t}[/tex]

Step-by-step explanation:

The logistic equation is

[tex]\frac{dP}{dt}= kP(1-\frac{P}{k})[/tex]

The analytic solution is

[tex]P(t)= \frac{K}{1+Ae^{-kt}}[/tex] ; [tex]A=\frac{K-P_0}{P_0}[/tex]

P(t) = the population at time t.

K = the carrying capacity = 4,000

k= constant proportionality

t= time

[tex]P_0[/tex] = 160.

The number of fish tripled in the first year.

P(1) = 3×160 =480

[tex]A= \frac{4000-160}{160}[/tex] =24

[tex]\therefore P(t) =\frac{4000}{1+24 e^{-kt}}[/tex] .....(1)

When t= 1 , P(1)=480

[tex]480=\frac{4000}{1+24 e^{-k}}[/tex]

[tex]\Rightarrow 1+24e^{-k} =\frac{4000}{480}[/tex]

[tex]\Rightarrow 24e^{-k} =\frac{4000}{480} -1[/tex]

[tex]\Rightarrow 24e^{-k}= \frac{25}{3}-1[/tex]

[tex]\Rightarrow 24e^{-k}=\frac{22}{3}[/tex]

[tex]\Rightarrow e^{-k} = \frac{22}{3\times 24}[/tex]

[tex]\Rightarrow e^{-k} = \frac{11}{36}[/tex]

taking ln both sides

[tex]\Rightarrow ln e^{-k} =ln \frac{11}{36}[/tex]

[tex]\Rightarrow {-k} = ln\frac{11}{36}[/tex]

[tex]\Rightarrow k = -ln\frac{11}{36}[/tex]

Putting the value of k in equation (1)

[tex]P(t)=\frac{4000}{1+24e^{ln\frac{36}{11}t}}[/tex]

[tex]\Rightarrow P(t) = \frac{4000}{1+24 (e^{ln\frac{36}{11}})^t}[/tex]

[tex]\Rightarrow P(t) = \frac{4000}{1+ 24(\frac{36}{11})^t}[/tex]

Therefore logistic equation is

[tex]P(t) = \frac{4000}{1+ 24(\frac{36}{11})^t}[/tex]

Answer:

Step-by-step explanation:

Each die has faces numbered from 1 to 6. Hence, the sums range from 2 to 12.

There are 6 × 6 = 36 possible combinations of the dice.

The sums, their possible combinations and their probabiities are as below:

2 = 1+1 1/36

3 = 1+2 = 2+1 2/36= 1/18

4 = 1+3 = 2+2 = 3+1 3/36= 1/12

5 = 1+4 = 2+3 = 3+2 = 4+1 4/36= 1/9

6 = 1+5 = 2+4 = 3+3 = 4+2 = 5+1 5/36

7 = 1+6 = 2+5 = 3+4 = 4+3 = 5+2 = 6+1 6/36= 1/6

8 = 2+6 = 3+5 = 4+4 = 5+3 = 6+2 5/36

9 = 3+6 = 4+5 = 5+4 = 6+3 4/36= 1/9

10 = 4+6 = 5+5 = 6+4 3/96= 1/12

11 = 5+6 = 6+5 2/36= 1/18

12 = 6+6 1/36

Sum of probabilities = 1/36 + 1/18 + 1/12 + 1/9 + 5/36 + 1/6 + 5/36 + 1/9 + 1/12 + 1/18 + 1/36 = 1

1) [tex]P(X\ge 8) = 5/36+1/9+1/12+1/18+1/36=15/36[/tex]

2) The average value of X = 1/36*(2+12) + 1/18*(3+11) + 1/12*(4+10)+ 1/9(5+9)+ 5/36(6+8)+ 1/6(7) = 14*(5/6)+7/6 = 77/6

Answer:

Option D) significantly different from 24

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 24

Sample mean, [tex]\bar{x}[/tex] = 25

Sample size, n = 16

Alpha, α = 0.05

Population standard deviation, σ = 2

First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu = 24\text{ years}\\H_A: \mu \neq 24\text{ years}[/tex]

We use Two-tailed z test to perform this hypothesis.

Formula:

[tex]z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }[/tex]

Putting all the values, we have

[tex]z_{stat} = \displaystyle\frac{25 - 24}{\frac{2}{\sqrt{16}} } = 2[/tex]

Now, [tex]z_{critical} \text{ at 0.05 level of significance } = \pm 1.96[/tex]

Since,  

The calculated z statistic does not lie in the acceptance region, we fail to accept the null hypothesis and reject it. We accept the alternate hypothesis.

Thus, it can be concluded that the mean age is

Option D) significantly different from 24

Answer:

Step-by-step explanation:

Hello!

The variable of interest is X: flexural strength of concrete beams.

Data:

7.3 6.8 8.6 9.7 11.6 9.7 6.8 7.7 11.8 7.4 8.1 8.7 6.3 7.9 7.0 7.9 7.8 6.5 6.3 7.0 7.5 7.7 7.2 11.3 9.0 10.7 5.1

a.

The point estimate of the mean value is the sample mean. You calculate it using the following formula:

X[bar]= ∑X/n= 219.40/27= 8.1259≅ 8.13

b)x

b.

The point estimate that divides the sample in 50%-50% is the measure of central tendency called Median(Me).

To reach the median you have to calculate it's position (PosMe)

For uneven samples PosMe= (n+1)/2= (27+1)/2= 14 ⇒ This means that the Median is the 14th observation of the data set.

So next is to order the data from lower to heighest and identify the value in the 14th position:

5,1 ; 6,3; 6,3; 6,5 ; 6,8; 6,8; 7; 7; 7,2; 7,3; 7,4 ; 7,5 ; 7,7 ; 7,7 ; 7,8 ; 7,9 ; 7,9 ; 8,1 ; 8,6 ; 8,7 ; 9 ; 9,7 ; 9,7 ; 10,7 ; 11,3 ; 11,6 ; 11,8

Me= 7.7

d) x tilde

c.

The point estimate of the population standard deviation is the sample standard deviation. The standard deviation is the square root of the variance, so the first step is to calculate the sample variance:

[tex]S^2= \frac{1}{n-1}*[sumX^2-\frac{(sumX)^2}{n} ][/tex]

∑X²= 1858.92

[tex]S^2= \frac{1}{26}*[1858.92-\frac{219.40^2}{27} ] = 2.926[/tex]

S= √2.926= 1.71

The standard deviation is a measurement of variability, it shows how to disperse the data is concerning the sample mean.

d. This estimate describes the spread of the data.

I hope it helps!

(a) Point estimate of mean strength: [tex]\( \bar{x} \) ≈ 8.118 MPa.[/tex] (d)

(b) Point estimate of separating strength: 7.7 MPa.

(d) (c) d) Spread of data.

(a) To calculate the point estimate of the mean value of strength, we'll use the sample mean estimator, denoted as [tex]\( \bar{x} \)[/tex].

Step 1:

Sum the data points:[tex]\( \sum{x_i} = 219.4 \)[/tex]

Step 2:

Divide the sum by the number of data points (n = 27): [tex]\( \bar{x} = \frac{\sum{x_i}}{n} = \frac{219.4}{27} \approx 8.118 \)[/tex] (rounded to three decimal places).

So, the point estimate of the mean strength is approximately 8.118 MPa.

(b) To find the strength value that separates the weakest 50% from the strongest 50%, we'll use the sample median estimator, denoted as [tex]\( \tilde{x} \).[/tex]

Step 1:

Arrange the data in ascending order.

Step 2:

Since there are 27 data points, the median will be the value of the 14th data point. Counting from the lowest value, the 14th value is 7.7.

So, the point estimate of the separating strength value is 7.7 MPa.

(c) To estimate the population standard deviation, we'll use the sample standard deviation estimator, denoted as (s).

Step 1:

Use the formula[tex]\( s = \sqrt{\frac{\sum{x_i^2} - \frac{(\sum{x_i})^2}{n}}{n-1}} \).[/tex]

Step 2:

Substitute the given values:[tex]\( s = \sqrt{\frac{1858.92 - \frac{(219.4)^2}{27}}{26}} \approx 1.624 \)[/tex](rounded to three decimal places).

This estimate describes the spread of the data.

So, the answers are:

(a) d)[tex]\( \bar{x} \)[/tex]

(b) d)[tex]\( \tilde{x} \)[/tex]

(c) d) This estimate describes the spread of the data.

Answer:

There are 73815 ways of selecting 4 of the lottery numbers.

Step-by-step explanation:

The total amount of ways to pick 4 numbers from a set of 38 ignoring the order is given by the combinatorial number of 38 with 4, denoted by [tex] {38 \choose 4} [/tex] , which is

[tex] {38 \choose 4} = \frac{38!}{4!(38-4)!} = 73815 [/tex]

Therefore, there are 73815 ways of picking 4 of the numbers.

Answer:

We conclude that have 73815 a different ways.

Step-by-step explanation:

We know that a certain lottery has 38 numbers. We assume that order of selection is not​ important. We calculate  how many different ways can 4 of the numbers be​ selected.

So out of 38 numbers we choose 4 numbers. We get:

[tex]C_4^{38}=\frac{38!}{4!(38-4)!}=73815[/tex]

We conclude that have 73815 a different ways.

The equation y = Ax + B represents a straight line. In the given example, 'A' is the slope of the line which is 3, and 'B' is the y-intercept which is 9. Hence, the sum of 'A' and 'B', or A + B, equals 12.

The student's question pertains to the equation of a straight line, which in slope-intercept form is y = mx + b, where 'm' represents the slope and 'b' represents the y-intercept. However, the equation mentioned in the question was written as y = Ax + B, where 'A' represents the slope and 'B' represents the y-intercept (equivalent to 'm' and 'b', respectively, in the more common form).

In the instance given, A (the slope) is 3 and B (the y-intercept) is 9. Therefore, when you add the values of A and B together as the question asked, the calculation is 3 + 9. So, A + B equals 12. In essence, the equation represents a line that, for every horizontal movement (x), the vertical movement (y) increases by 3 times the x value, and it intersects the y-axis at 9.

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The calculated value of A + B is -15/4 from the equation y = 3/4x - 9/2

From the question, we have the following parameters that can be used in our computation:

The graph

A linear equation is represented as

y = mx + c

Where

c = y when x = 0

m = slope

Using the above as a guide, we have the following:

m = (-6 + 3)/(-2 - 2)

m = 3/4

So, we have

y = 3/4x + c

Using the points, we have

3/4 * 6 + c = 0

Evaluate

c = -9/2

So, we have

y = 3/4x - 9/2

So, we have

A + B = 3/4 - 9/2

Evaluate

A + B = -15/4

Hence, the value of A + B is -15/4

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Answer:

C

Step-by-step explanation:

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Answer:

a) 1/8 b) 1/12 c) probability of obtaining sum of 9 in a single roll of die without having the same number repeated twice is less than than of sum of 10  

Step-by-step explanation:

Probability of any digit in a single roll = 1/6

a) sum of 11 will be obtained if each roll has the following result

1 and 4 and 6, 1 and 5 and 5, 1 and 6 and 4, 4 and 1 and 6, 4 and 6 and 1, 6 and 4 and 1, 6 and 1 and 4, 5 and 1 and 5, 5 and 5 and 1, 2 and 5 and 4, 2 and 4 and 5, 4 and 2 and 5, 4 and 5 and 2, 5 and 2 and 4, 5 and 4 and 2, 2 and 6 and 3, 2 and 3 and 6, 3 and 2 and 6, 3 and 6 and 2, 6 and 3 and 2, 6 and 2 and 3, 3 and 3 and 5, 3 and 5 and 3, 5 and 3 and 3, 3 and 4 and 4, 4 and 3 and 3, 3 and 4 and 3

27(1/6 × 1/6 × 1/6)= 1/8

b) 2 and 4 and 6, 2 and 5 and 5, 4 and 2 and 6, 4 and 6 and 2, 6 and 4 and 2, 6 and 2 and 4, 5 and 2 and 5, 5 and 5 and 2, 2 and 6 and 4, 3 and 3 and 6, 3 and 6 and 3, 6 and 3 and 3, 3 and 4 and 5, 3 and 5 and 4, 4 and 3 and 5, 4 and 5 and 3, 5 and 3 and 4, 5 and 4 and 3

18(1/6 × 1/6 × 1/6)= 1/12

c) six ways of obtaining 10:  3+1+6, 3+2+5, 3+3+4, 4+2+4,4+5+1, 6+2+2

  six ways of obtaining 9: 1+2+6, 1+3+5, 1+4+4, 2+2+5, 2+3+4,3+3+3

To get 10, there are only two ways of repeating a number out of 6 ways. To get 9, there are three ways of repeating a number out of 6 ways

The probability of sums from rolls of dice is based on the possible combinations of outcomes, where each side of a dice has a 1/6 chance. For three independent rolls, specific sums like 11 or 12 would require identifying all possible combinations that achieve these sums. Galileo's observation relates to the unequal probabilities of different number combinations despite having the same number of combinations for sums of 9 and 10.

The probabilities for the sums of dice rolls are calculated based on the number of ways those sums can be achieved using the possible outcomes of each individual roll. For a fair six-sided die, the probability of rolling any given number is 1/6.

(a) To find the probability that the sum of three rolls is 11, we would identify all the combinations of rolls that result in that sum and calculate the likelihood of each. Possible combinations for a sum of 11 include (5,5,1), (5,4,2), (6,4,1), etc. Each of the three dice has a 1/6 chance of landing on a specified number, and since each roll is independent, the probabilities are multiplied.

(b) Similarly, for a sum of 12, possible combinations include (4,4,4), (6,5,1), (6,4,2), etc.

(c) As for Galileo's observation about sums of 10 being more frequent than 9, it is important to note that while both sums have six distinct combinations, some combinations are more likely than others because rolls are independent and not all number combinations are equally probable. This leads to a higher overall probability for a sum of 10.

Answer:

⅛

Step-by-step explanation:

Let the total be X

4/8 = 1/2

X - ⅜X - ½X = ⅛X are orange

Fraction ⅛X ÷ X = ⅛

Answer:

The diameter of the circle is 9.7mm

Step-by-step explanation:

The area of a circle is given by the following formula:

[tex]A_{c} = \pi r^{2}[/tex]

In which r is the radius.

In this problem, we have that

[tex]A_{c} = 74 mm^{2}[/tex]

So

[tex]74 = \pi r^{2}[/tex]

[tex]r = \sqrt{\frac{74}{\pi}}[/tex]

[tex]r = 4.85mm[/tex]

The diameter is twice the radius:

So

[tex]d = 2r = 2*4.85 = 9.7[/tex]

The diameter of the circle is 9.7mm

Final answer:

To find the diameter of a circle with an area of 74 square millimeters, use the area formula A = πr², solve for r, then double the result to get the diameter. The estimated diameter is approximately 9.71 mm.

Explanation:

The question asks us to find the diameter of a circle with an area of 74 square millimeters. We use the formula for the area of a circle, which is A = πr² where A is the area and r is the radius. To find the diameter, we need to first solve for the radius, and then multiply by 2 since the diameter is twice the length of the radius.

Steps to Find the Diameter:

Assuming π is approximately 3.14, we can estimate:

A = 74 mm² = πr²

r = √(74/π) mm

r ≈ √(74/3.14) mm ≈ √(23.567) mm ≈ 4.855 mm

The radius is approximately 4.855 mm, so the diameter would be 2 × 4.855 mm ≈ 9.71 mm.

Answer:

95% confidence interval estimate for the standard deviation = [4.78 , 8.06]

Step-by-step explanation:

We are given that the manager of the service department of a local car dealership has noted that the service times of a sample of 30 new automobiles has a standard deviation of 6 minutes, i.e. n = 30 and s = 6.

The pivotal quantity for 95% confidence interval for the population variance is given by;

                   [tex]\frac{(n-1)s^{2} }{\sigma^{2} }[/tex]  ~ [tex]\chi^{2}__n_-_1[/tex]

95% confidence interval for the population variance is;

P(16.05 < [tex]\chi^{2}__2_9[/tex] < 45.72) = 0.95

P(16.05 < [tex]\frac{(n-1)s^{2} }{\sigma^{2} }[/tex] < 45.72) = 0.95

P( [tex]\frac{16.05}{(n-1)s^{2} }[/tex] < [tex]\frac{1 }{\sigma^{2} }[/tex] < [tex]\frac{45.72}{(n-1)s^{2} }[/tex] ) = 0.95

P( [tex]\frac{(n-1)s^{2}}{ 45.72}[/tex] < [tex]\sigma^{2}[/tex] <[tex]\frac{(n-1)s^{2}}{ 16.05}[/tex]  ) = 0.95

95% Confidence interval for [tex]\sigma^{2}[/tex] = [ [tex]\frac{(n-1)s^{2}}{ 45.72}[/tex] , [tex]\frac{(n-1)s^{2}}{ 16.05}[/tex] ]

                                                = [ [tex]\frac{(30-1)6^{2}}{ 45.72}[/tex] , [tex]\frac{(30-1)6^{2}}{ 16.05}[/tex] ]

                                                = [ 22.835 , 65.047 ]

So, 95% Confidence interval for [tex]\sigma[/tex] = [ [tex]\sqrt{ 22.835} ,\sqrt{ 65.047}[/tex] ] = [ 4.78 , 8.06 ]

Therefore, 95% confidence interval for the standard deviation of the service times for all their new automobiles is [ 4.78 , 8.06 ] .

The correct answer is b. 4.78 to 8.07.

The question asks us to determine the 95% confidence interval estimate for the standard deviation of service times of new automobiles at a dealership.

Given the sample size (n = 30) and the sample standard deviation (s = 6 minutes), we can calculate the confidence interval using the chi-squared (χ²) distribution.

1. Calculate the degrees of freedom (df):

df = n - 1 = 30 - 1 = 29

2. Find the critical values of chi-squared at 95% confidence level:

3. From chi-squared tables, χ²α/2, 29 = 45.722, and χ²1-α/2, 29 = 16.047

4. Use the critical values to compute the confidence interval:

Lower limit: √ ( (n - 1)s² / χ²α/2 ) = √ ((29)(6²) / 45.722 ) = √ 174 / 45.722 ) = 1.92 × 2.49 = 4.78

Upper limit: √ ( (n - 1)s² / χ²1-α/2 ) = √ ((29)(6²) / 16.047 ) = √ 174 / 16.047 ) = 1.92 × 5.14 = 8.07

Therefore, the 95% confidence interval for the standard deviation of service times for all new automobiles at the dealership is 4.78 to 8.07 minutes, so the correct answer is b. 4.78 to 8.07.

Complete question: The manager of the service department of a local car dealership has noted that the service times of a sample of 30 new automobiles has a standard deviation of 6 minutes. A 95% confidence interval estimate for the standard deviation of the service times for all their new automobiles is

a. 16.05 to 45.72

b. 4.78 to 8.07

c. 2.93 to 6.31

d. 22.83 to 65.06

Answer:

286 different treatment arrangements

Step-by-step explanation:

This is a combination problem.

We want to share 10 trees amongst 4 experimental groups.

13C3 = 286 different treatment arrangements.

There are 286 different treatment arrangements of trees possible.

The tree treatment includes the medical aid of trees in the growth and development such as fertilizers and various chemical that allows for their development.

As per the question the experiment was conducted to study the tree's growth and the treatments were used that included irrigation, fertilization. The row of 10 trees was used in the study and each tree was randomly taken.

This is a combination problem. About  10 trees are given that have 4 experimental groups. Thus the 13 C3 = 286 different treatment arrangements.

Find out more information about the treatments.

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Answer:

B

Step-by-step explanation:

we can plug 13 into each equation

A. 13+3 =10

16≠10, so A is wrong

B. 13-10=3

3=3, B is correct

C, is definitely wrong

D. 13+12=1

25≠1

Answer:

7.9

Step-by-step explanation:

Answer:

a) [tex]n=\frac{0.25(1-0.25)}{(\frac{0.06}{1.64})^2}=140.08[/tex]  

And rounded up we have that n=141

b) [tex]n=\frac{0.25(1-0.25)}{(\frac{0.04}{1.64})^2}=315.19[/tex]  

And rounded up we have that n=316

c) [tex]n=\frac{0.25(1-0.25)}{(\frac{0.03}{1.64})^2}=560.33[/tex]  

And rounded up we have that n=561

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".  

The population proportion have the following distribution

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

Solution to the problem

Part a

The estimated proportion for this case is [tex]\hat p =0.25[/tex]

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 90% of confidence, our significance level would be given by [tex]\alpha=1-0.90=0.1[/tex] and [tex]\alpha/2 =0.05[/tex]. And the critical value would be given by:

[tex]z_{\alpha/2}=-1.64, z_{1-\alpha/2}=1.64[/tex]

The margin of error for the proportion interval is given by this formula:  

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

And on this case we have that [tex]ME =\pm 0.06[/tex] and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

[tex]n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}[/tex]   (b)  

And replacing into equation (b) the values from part a we got:

[tex]n=\frac{0.25(1-0.25)}{(\frac{0.06}{1.64})^2}=140.08[/tex]  

And rounded up we have that n=141

Part b

[tex]n=\frac{0.25(1-0.25)}{(\frac{0.04}{1.64})^2}=315.19[/tex]  

And rounded up we have that n=316

Part c

[tex]n=\frac{0.25(1-0.25)}{(\frac{0.03}{1.64})^2}=560.33[/tex]  

And rounded up we have that n=561