In a start-up company which has 20 computers, some of the computers are infected with virus. The probability that a computer is infected with the virus is 0.4, independently of other computers. A technician tests the computers, one after another, to see if they are infected.1. What is the probability that she has to test at least 5 computers to find the first (if any) defective one?2. Find the probability that on this day at least 5 computers are infected.

Answer:

a) 1/2 (50%)

b) 1/2 (50%)

c) 3/4 (75%)

d) 3/4 (75%)

Step-by-step explanation:

for a uniform distribution

P(X=x)= x-A/(B-A) , for A≤x≤B

then

P(X=x)= (x+8)/16 , for -8≤x≤8

a) P(X<0)= interval chosen / total interval = (0-(-8))/(8-(-8)) = 1/2 (50%)

b)  P(-4<X<4)= interval chosen / total interval =(4-(-4))/(8-(-8)) = 1/2 (50%)

c)  P(-5<X<7)= interval chosen / total interval =(7-(-5))/(8-(-8)) = 3/4 (75%)

d) for −8 < k and  k + 4 < 8 → k < 4 , then -8<k<4 , thus k

P(X=k)= P(-8<X<4) =  interval chosen / total interval =(4-(-8))/(8-(-8)) = 3/4 (75%)

Answer:

1) Null hypothesis:[tex]p=0.53[/tex]  

2)Alternative hypothesis:[tex]p \neq 0.53[/tex]  

3) [tex]z=\frac{0.4 -0.53}{\sqrt{\frac{0.53(1-0.53)}{100}}}=-2.605[/tex]  

4) We assume that [tex]\alpha=0.05[/tex]

[tex]p_v =2*P(z<-2.605)=0.0092[/tex]  

So the p value obtained was a very low value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of people who would vote for the incumbent is different from 0.53.  

Step-by-step explanation:

Data given and notation  

n=100 represent the random sample taken

[tex]\hat p=0.4[/tex] estimated proportion of people who would vote for the incumbent

[tex]p_o=0.53[/tex] is the value that we want to test

[tex]\alpha=0.05[/tex] represent the significance level  (assumed)

Confidence=95% or 0.95  (Assumed)

z would represent the statistic (variable of interest)

[tex]p_v[/tex] represent the p value (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion is 0.53 or not.:  

1) Null hypothesis:[tex]p=0.53[/tex]  

2)Alternative hypothesis:[tex]p \neq 0.53[/tex]  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

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

The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].

3) Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

[tex]z=\frac{0.4 -0.53}{\sqrt{\frac{0.53(1-0.53)}{100}}}=-2.605[/tex]  

4) Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level assumed [tex]\alpha=0.05[/tex]. The next step would be calculate the p value for this test.  

Since is a bilateral test the p value would be:  

[tex]p_v =2*P(z<-2.605)=0.0092[/tex]  

So the p value obtained was a very low value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of people who would vote for the incumbent is different from 0.53 .  

Answer:

The shape of the distribution of the sample mean would be bell-shaped(normally distributed).

Step-by-step explanation:

The Central Limit Theorem answers this question.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], a large sample size, of at least 30, can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\frac{\sigma}{\sqrt{n}}[/tex].

In this problem, we have that

The population distribution is right-skewed.

Sampling distribution of the sample mean with size 40.

The Central Limit Theorem estabilishes that the sampling distribution of the sample mean will be normally distributed, even if the distribution of the population is not.

So the shape of the distribution of the sample mean would be bell-shaped(normally distributed).

Answer:

3) One-sided,compare two populations

Step-by-step explanation:

A manufacturer wants to investigate the whether there is an increase in the mean gas mileage of mid-sized sedan having new type of tire compared to mean gas mileage of mid-sized sedan not having new type of tire.

So, there are two populations :

population 1 : mid-sized sedan having new type of tire.

population 2 : mid-sized sedan not having new type of tire.

So, two populations are being compared.

Also, A manufacturer is investigating whether there is an increase in mean gas mileage.The alternative hypothesis will be right tailed and so, this corresponds to one sided test.

Final answer:

To determine if the new tires increase fuel efficiency, the manufacturer should use a one-sided hypothesis test to analyze changes in a single population's mean gas mileage.

Explanation:

The automotive manufacturer should perform a one-sided hypothesis test to analyze a change in a single population. This is because they are specifically interested in testing if the mean gas mileage for mid-sized sedans using the new tires is greater than 32 miles per gallon, not whether it is different (either less or greater). A two-sided test would be used if they were interested in any difference from the historic mean, rather than specifically an increase.

Answer:

PV=$9,143.88

Step-by-step explanation:

Compound Interest

When a principal amount (also called present value PV) is saved at some rate of interest r for some time t, the future value FV that includes the original investment plus the interests is computed as

[tex]FV=PV\left(1+r\right)^t[/tex]

If the investment is compounded other than annually, then r and t must be scaled to the proper time units.

If we already know the future value, then the present value is computed by solving the above equation

[tex]PV=FV\left(1+r\right)^{-t}[/tex]

The Annual Percentage Rate (APR) is 6.8% compounded weekly, so the value of r is (assuming 52 weeks per year)

[tex]\displaystyle r=\frac{6.8}{100\times 52}=0.00131[/tex]

And the time is computed in weeks

t=4*52=208

The present value of the investment Trevor needs to save now is

[tex]PV=12,000\left(1+0.00131\right)^{-208}[/tex]

[tex]\boxed{PV=\$9,143.88}[/tex]

Trevor will need to invest $9,143.88 into the account to have $12,000 after 4 years

Answer:

Step-by-step explanation: From the above question, the possible outcome are:

a) Admitted unconditionally √

b) Awarded a degree √

c) Not admitted √

d) Granted a visa ×

e) Admitted conditionally √

Answer:

The answer is :- ( -4 , 3 )

Answer: 2/3 hours

Step-by-step explanation: 1 1/3 will equal 4/3 as you have a whole number equaling 3/1 so then you divide by two and get 2/3

Answer:

a) A. Because the population diameter of​ Ping-Pong balls is approximately normally​ distributed, the sampling distribution of samples of 4 will also be approximately normal.

b) [tex]P(\bar X <1.28)=P(Z<\frac{1.28-1.32}{\frac{0.08}{\sqrt{4}}}=-1)[/tex]

And using a calculator, excel or the normal standard table we have that:

[tex]P(Z<-1)=0.159[/tex]

c) [tex]P(1.28< \bar X <1.34)=P(\frac{1.28-1.32}{\frac{0.08}{\sqrt{4}}}<Z<\frac{1.34-1.32}{\frac{0.08}{\sqrt{4}}})= P(-1< Z<0.5)[/tex]

And using a calculator, excel or the normal standard table we have that:

[tex]P(-1<Z<0.5)=P(Z<0.5)-P(Z<-1) = 0.691-0.159=0.532 [/tex]

Step-by-step explanation:

Previous concepts

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

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

Solution to the problem

Part a

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

[tex]X \sim N(1.32,0.08)[/tex]  

Where [tex]\mu=1.32[/tex] and [tex]\sigma=0.08[/tex]

And we select a sample of size n =4 and we want to find the distribution for [tex]\bar X[/tex]

Since the distribution for X is normal then the distribution for the sample mean [tex]\bar X[/tex] is normal and given by:

[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]

So the best option for this case would be:

A. Because the population diameter of​ Ping-Pong balls is approximately normally​ distributed, the sampling distribution of samples of 4 will also be approximately normal.

Part b

[tex] P(\bar X<1.28)[/tex]

We can use the z score given by:

[tex]z= \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

[tex]P(\bar X <1.28)=P(Z<\frac{1.28-1.32}{\frac{0.08}{\sqrt{4}}}=-1)[/tex]

And using a calculator, excel or the normal standard table we have that:

[tex]P(Z<-1)=0.159[/tex]

Part c

[tex] P(1.28< \bar X<1.34)[/tex]

[tex]P(1.28< \bar X <1.34)=P(\frac{1.28-1.32}{\frac{0.08}{\sqrt{4}}}<Z<\frac{1.34-1.32}{\frac{0.08}{\sqrt{4}}})= P(-1< Z<0.5)[/tex]

And using a calculator, excel or the normal standard table we have that:

[tex]P(-1<Z<0.5)=P(Z<0.5)-P(Z<-1) = 0.691-0.159=0.532 [/tex]

The sampling distribution of the mean diameter of the ping-pong balls, given that the population diameter is normally distributed, will also be normally distributed. The expected mean is 1.32 inches and the standard deviation will be 0.04 inches.

The correct answer to this question should be A. Because the population diameter of​ Ping-Pong balls is approximately normally distributed, the sampling distribution of samples of 4 will also be approximately normal. This is grounded on the Central Limit Theorem which states that the distribution of sample means tends to be normal regardless of the shape of the population distribution, especially when the sample size is large.

In this case, though our sample size (4) is fairly small due, because the distribution of the population is specified as normal, the sampling distribution will remain normal. The important numbers to know would be the expected value (mean which is the same as population mean i.e., 1.32 inches) and the standard deviation (which is the population standard deviation divided by square root of sample size i.e., 0.08/sqrt(4)).

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Answer:

%age of multifactor productivity reduction = 8.33%

Step-by-step explanation:

Total cost for chemicals = $10

Total cost of labor = $40

Total cost of misc = $5

Total initial cost = 10+40+5 = $55

Increase in use of chemical = 50%

Increase in cost of chemical = 10+ (0.5)(10)  =15

[Note (0.5)(10) is 50% of initial chemical cost]

Total increase in cost= $15+ $40 + $5 =%60

[tex]\%age\,\, of\,\, increase\,\, in \,\,cost =\frac{initial\,\, cost}{incresed\,\,cost}\times 100\%\\\\=\frac{55}{60}\times 100\%\\\\\\= 91.67\%[/tex]

%age of multifactor productivity reduction = 100 - 91.67 = 8.33%

Final answer:

The multifactor productivity of the cleaning company has fallen by approximately 9.09% after increasing the use of chemicals by 50%.

Explanation:

The student is asking about how the increase in chemical use affects the multifactor productivity (MFP) of a cleaning company. The original costs of cleaning one house are $10 for chemicals, $40 for labor, and $5 for miscellaneous expenses. With a 50% increase in chemical use, the new cost for chemicals becomes $15.

Firstly, we need to calculate the total initial cost: $10 (chemicals) + $40 (labor) + $5 (misc) = $55. After the increase in chemical use, the new total cost is $15 (chemicals) + $40 (labor) + $5 (misc) = $60. To find the percentage decrease in MFP, we compare the change in input cost to the original input cost.

The productivity decrease is calculated as: (($60 - $55) / $55) * 100 = (5 / 55) * 100 \\approx 9.09%. Thus, the multifactor productivity has fallen by approximately 9.09%.

Answer:

Step-by-step explanation:

For the disease having force of mortality  µ.

we apply exponential distribution law

probability of dying within 20 years

p( x <20) = [tex]1-e^{-\mu}[/tex] = .1

[tex]e^{-\mu}[/tex] = .9

For the disease having force of mortality  2µ.

we apply exponential distribution law

probability of dying within 20 years

p( x <20) = 1 - [tex]e^{-2\mu}[/tex]

= 1 - ( .9)²

= 1 - .81

= .19

or 19%

Answer:

a) There is a 15.87% probability that a part is defective.

b) The expected number of parts defective in such a sample is 0.7935.

Step-by-step explanation:

To solve this question, we use concepts of the normal probability distribution and the binomial probability distribution.

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.

Binomial probability distribution

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

The expected value of the binomial distribution is:

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

(a) what is the probability that a part is defective?

The tensile strength of a metal part is normally distributed with mean 40 pounds and standard deviation

5 pounds. A part is considered defective if the tensile strength is less than 35 pounds.

Here we have [tex]\mu = 40, \sigma = 5[/tex]

This probability is the pvalue of Z when X = 35.

So

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

[tex]Z = \frac{35 - 40}{5}[/tex]

[tex]Z = -1[/tex]

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

There is a 15.87% probability that a part is defective.

(b) If a testing sample consists of 5 parts, what is the expected number of parts defective in such a sample? Assume that each part is independent of the others.

This is the expected value of a binomial distribution when [tex]n = 5, p = 0.1587[/tex].

So

[tex]E(X) = np = 5*0.1587 = 0.7935[/tex]

The expected number of parts defective in such a sample is 0.7935.

Answer: Interaction

Step-by-step explanation: Interaction in statistics occurs mostly during factorial experiments and analyzes of regression. When two variables interact, it means the

relationship between them and some other variable (dependent), depends on the value of the other interacting variable.Simply put, the relationship between the dependent variable and one independent variable is different at different values of a second independent variable.

Finding interacting variables involve having a factorial design. Here, independent variables are "crossed" with one another so that there are observations at every levels of combination of the independent variables.

Interaction effect refers to the scenario in which the relationship between the dependent variable and one independent variable is different at different values of a second independent variable.

Interaction effect refers to the scenario in which the relationship between the dependent variable and one independent variable is different at different values of a second independent variable.

For example, let's say we are studying the effects of studying time and caffeine consumption on test scores. We find that the interaction effect between studying time and caffeine consumption is significant, indicating that the relationship between studying time and test scores depends on the level of caffeine consumption.

This concept is commonly explored in statistics and regression analysis.

hence, the final answer is Interaction effect .

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Answer:

Sample Space = {SSS,SSF,SFF,SFS,FSS,FSF,FFS,FFF}

Step-by-step explanation:

Given a sale representative makes presentations about a product in three homes per day.

In each home, Sale is denoted by S and No Sale is denoted by F.

Now the cases for required sample space will be :

So our Required sample space for the experiment

    = {SSS,SSF,SFF,SFS,FSS,FSF,FFS,FFF}.

Answer:

(a) 0.0492

(b) 0.0984

Step-by-step explanation:

The probability distribution of blood type in the US is:

[tex]\begin{array}{cc}Type&Probability\\O&0.46\\A&0.41\\B&0.12\\AB&0.01\end{array}[/tex]

(a) The probability that the wife has type A and the husband has type B is:

[tex]P(w=A\ and\ h=B) =0.41*0.12\\P(w=A\ and\ h=B) =0.0492[/tex]

(b) The probability that one of the couple has type A blood and the other has type B is given by the probability that the wife has type A and the husband has type B added to the probability that the wife has type B and the husband has type A.

[tex]P= P(w=A\ and\ h=B) + P(w=B\ and\ h=A)\\P=0.41*0.12+0.12*0.41=0.0984[/tex]

Final answer:

The probability that the wife has type A blood and the husband has type B blood is 4.92%. The probability that one of the couple has type A blood and the other has type B blood is 9.84%.

Explanation:

The student is being asked to calculate probabilities related to the blood types of a randomly chosen married couple in the United States using the given distribution of ABO blood types. The probabilities for each blood type are: type O at 0.46, type A at 0.41, type B at 0.12, and type AB at 0.01. To solve these questions, we use the principle that the probability of independent events occurring together is the product of their respective probabilities.

Answer:

49.32% probability that the proportion in our sample of green candies will be more than 25%.

Step-by-step explanation:

For each candy, there are only two possible outcomes. Either it is green, or it is not. The probabilities of each candy being green are independent from each other. 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:

According to the manufacturer of the candy Skittles, 25% of the candy produced are green. This means that [tex]p = 0.25[/tex]

If we take a random sample of 10 bags of Skittles, what is the probability that the proportion in our sample of green candies will be more than 25%?

10 bags, so [tex]n = 10[/tex]

This is [tex]P(X > 0.25*10) = P(X > 2.5)[/tex]

The number of bags is a discrete number, so [tex]P(X > 2.5) = P(X \geq 3)[/tex].

Either the number of green bags is lower than 3, or it is greater or equal than 3. The sum of these probabilities is decimal 1. Mathematically, this means that:

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

We want [tex]P(X \geq 3)[/tex]. So

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

In which

[tex]P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)[/tex]

So

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

[tex]P(X = 0) = C_{10,0}.(0.25)^{0}.(0.75)^{10} = 0.0563[/tex]

[tex]P(X = 1) = C_{10,1}.(0.25)^{1}.(0.75)^{9} = 0.1689[/tex]

[tex]P(X = 2) = C_{10,2}.(0.25)^{2}.(0.75)^{8} = 0.2816[/tex]

[tex]P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0563 + 0.1689 + 0.2816 = 0.5068[/tex]

Then

[tex]P(X \geq 3) = 1 - P(X \leq 2) = 1 - 0.5068 = 0.4932[/tex]

49.32% probability that the proportion in our sample of green candies will be more than 25%.

Final answer:

The subject matter revolves around the concept of probability in mathematics, particularly involving calculations using the binomial and Poisson distributions, and statistical methods for analyzing sample data.

Explanation:

The question posed relates to the concept of probability in mathematics, specifically the calculation of the probability of obtaining a certain number of items of a particular color in a random sample from a given population. This field of study typically involves understanding and applying theories of probability distributions, such as the binomial distribution and the Poisson distribution, as well as using statistical methods to estimate population parameters and test hypotheses.

For example, to find the probability of drawing exactly 3 green marbles from a bag containing 50 marbles of which 15 are green, one would need to use the binomial distribution formula P(X = k) = C(n, k) p^k (1-p)^(n-k), where 'C(n, k)' is the combination of 'n' items taken 'k' at a time, 'p' is the probability of success on a single trial, and 'X' is the random variable representing the number of successes in 'n' trials. Similar concepts are applicable to the drawing of candies from a bag and calculating the proportion of colors in a sample to match or compare with theoretical distributions.

Exercises that involve calculating mean and standard deviations for repeated sampling, such as in the exercise with M&M's, are instances of applying statistical methods to analyze data. This process of analysis provides a deeper understanding of the underlying distributions and the behavior of random variables.

Answer:

(a) P(all of the next three vehicles inspected pass) = 0.512 .

(b) P(at least one of the next three inspected fails) = 0.488 .

(c) P(exactly one of the next three inspected passes) = 0.096 .

(d) P(at most one of the next three vehicles inspected passes) = 0.104 .

(e) Probability that all three pass given that at least one of the next three vehicles passes inspection = 0.516 .

Step-by-step explanation:

We are given the Probability of all vehicles examined at a certain emissions inspection station pass the inspection to be 80%.

So, Probability that the next vehicle examined fails the inspection is 20%.

Also, it is given that successive vehicles pass or fail independently of one another.

(a) P(all of the next three vehicles inspected pass) = Probability that first vehicle, second vehicle and third vehicle also pass the inspection

              = 0.8 * 0.8 * 0.8 = 0.512

(b) P(at least one of the next three inspected fails) =

        1 - P(none of the next three inspected fails) = 1 - P(all next three passes)

    = 1 - (0.8 * 0.8 * 0.8) = 1 - 0.512 = 0.488 .

(c) P(exactly one of the next three inspected passes) is given by ;

Hence, P(exactly one of the next three inspected passes) = Add all above cases ;

             (0.8 * 0.2 * 0.2) + (0.2 * 0.8 * 0.2) + (0.2 * 0.2 * 0.8) = 0.096 .

(d) P(at most one of the next three vehicles inspected passes) = P(that none

     of the next three vehicle passes) + P(Only one of the next three passes)

We have calculated the P(Only one of the next three passes) in the above part of this question;

 Hence, P(at most one of the next three vehicles inspected passes) =

              (0.2 * 0.2 * 0.2) + (0.8 * 0.2 * 0.2) + (0.2 * 0.8 * 0.2) + (0.2 * 0.2 * 0.8)

             = 0.008 + 0.096 = 0.104 .

(e) Probability that all three pass given that at least one of the next three vehicles passes inspection is given by;

P(All three passes/At least one of the next three vehicles passes inspection)

= P( All next three passes [tex]\bigcap[/tex] At least one of next three passes) /

      P(At least one of next three passes)

=  P( All next three passes ) / P(At least one of next three passes)

= P( All next three passes ) / 1 - P(none of the next three passes)

 =  [tex]\frac{0.8*0.8*0.8}{1-(0.2*0.2*0.2)}[/tex] = 0.516 .

Therefore, Probability that all three pass given that at least one of the next three vehicles passes inspection = 0.516 .

a) P(all of the next three vehicles pass) = 0.512

(b) P(at least one of the next three inspected fails) = 0.488

(c) P(exactly one of the next three inspected passes) = 0.096

(d) P(at most one of the next three vehicles inspected passes) = 0.104

(e) Given that at least one of the next three vehicles passes inspection, the probability that all three pass is approximately 1.049 (rounded to three decimal places).

Let's calculate the probabilities step by step:

(a) To find the probability that all of the next three vehicles pass, we can use the probability of a single vehicle passing (0.80) raised to the power of 3 (because the events are independent).

P(all pass) = (0.80)^3 = 0.512

(b) To find the probability that at least one of the next three vehicles fails, we can use the complement rule. It's easier to find the probability that none of them fail and then subtract that from 1.

P(at least one fails) = 1 - P(all pass) = 1 - 0.512 = 0.488

(c) To find the probability that exactly one of the next three vehicles passes, we need to consider three cases: Pass-Fail-Fail, Fail-Pass-Fail, and Fail-Fail-Pass. Each of these cases has a probability of (0.80) * (0.20) * (0.20).

P(exactly one passes) = 3 * (0.80) * (0.20) * (0.20) = 0.096

(d) To find the probability that at most one of the next three vehicles passes, we sum the probabilities of exactly one passing and none passing.

P(at most one passes) = P(none pass) + P(exactly one passes) = (0.20)^3 + 0.096 = 0.104

(e) To calculate the conditional probability that all three pass given that at least one passes, we use the formula for conditional probability:

P(all three pass | at least one passes) = P(all three pass and at least one passes) / P(at least one passes)

P(all three pass and at least one passes) = P(all pass) = 0.512 (from part a)

So, P(all three pass | at least one passes) = 0.512 / 0.488 ≈ 1.049 (rounded to three decimal places).

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Answer:

(1) Correct option is (A)

(2) P (H or G) = 0.80

(3) Correct option is (C)

Step-by-step explanation:

Mutually exclusive events are those events that cannot occur at same time.

If events A and B are mutually exclusive then, P (A and B) = 0.

Given: P (G) = 0.50 and P (H) = 0.30

(1)

The statement is: P (H|G) = 0.40.

The conditional probability of event B given event A is:

[tex]P(B|A)=\frac{P(A\cap B)}{P(A)}[/tex]

The probability statement P (H|G) can be written as:

[tex]P(H|G)=\frac{P(H\cap G)}{P(G)}[/tex]

But as H and G are mutually exclusive, P (H ∩ G) = 0.

Hence, P (H|G) = 0.

Thus, the provided statement is false because events H and G are mutually exclusive, which makes P(H ∩ G) = 0.

Option (A) is correct.

(2)

The addition rule of probability states that:

[tex]P(A\cup B)=P(A)+P(B)-P(A\cap B)[/tex]

Compute the value of P (H or G) as follows:

[tex]P(H\cup G)=P(H)+P(G)-P(H\cap G\\=0.30+0.50-0\\=0.80[/tex]

Thus, the value of P (H or G) is 0.80.

(3)

Independent events are those events that are not affected by the occurrence of other events.

Events H and G are mutually exclusive events, i.e. occurrence of one affects the occurrence of other.

Thus, events H and G are dependent events.

Option (C) is correct.

Mutually exclusive events have a probability of 0 when they occur together. To find the probability of G OR H, use P(H) + P(G) - P(H AND G). G and H are independent events because they are mutually exclusive.

G and H are mutually exclusive events, which means they cannot occur at the same time. The probability of two mutually exclusive events happening together, P(H AND G), is always 0.

Therefore, the statement P(H | G) = 0.4 must be false. The correct answer is (A.) The events are mutually exclusive, which makes P(H AND G) = 0; therefore, P(H | G) = 0.

To find P(H OR G), you can use the formula P(H OR G) = P(H) + P(G) - P(H AND G). Plugging in the given values, P(H OR G) = 0.3 + 0.5 - 0 = 0.8.

G and H are mutually exclusive events, which means that the occurrence of one event does not affect the probability of the other event happening.

Therefore, G and H are independent events.

The correct answer is (B.) G and H are independent events because they are mutually exclusive.

Final answer:

The structure of a fugue includes elements such as the subject, counter-subject, expositions, and episodes. Bach used techniques like inversion, retrograde, augmentation, and diminution to develop his fugues.

Explanation:

The structure of a fugue traditionally includes a subject, which is the main theme introduced at the beginning, a counter-subject, which complements the subject, expositions where the subject is introduced in different voices, and episodes, which are free sections providing contrast. Johann Sebastian Bach used various techniques to create musical development within his fugues, some of which include:

Inversion, where the subject is mirrored vertically.

Retrograde, where the subject is played backwards.

Augmentation, where the subject is presented with longer note values, thus slower.

Diminution, where the subject is presented with shorter note values, making it faster.

These techniques helped to create complexity and interest, ensuring that the fugue evolves musically throughout the piece.

Answer:

[tex]P(3.9<X<5)=P(\frac{3.9-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{5-\mu}{\sigma})=P(\frac{3.9-4.8}{0.3}<Z<\frac{5-4.8}{0.3})=P(-3<z<0.67)[/tex]

[tex]P(-3<z<0.67)=P(z<0.67)-P(z<-3)[/tex]

[tex]P(-3<z<0.67)=P(z<0.67)-P(z<-3)=0.748571-0.00135=0.74722[/tex]

d. .7462

Step-by-step explanation:

Previous concepts

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

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

Solution to the problem

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

[tex]X \sim N(4.8,0.3)[/tex]  

Where [tex]\mu=4.8´[/tex] and [tex]\sigma=0.3[/tex]

We are interested on this probability

[tex]P(3.9<X<5.0)[/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(3.9<X<5)=P(\frac{3.9-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{5-\mu}{\sigma})=P(\frac{3.9-4.8}{0.3}<Z<\frac{5-4.8}{0.3})=P(-3<z<0.67)[/tex]

And we can find this probability with this difference:

[tex]P(-3<z<0.67)=P(z<0.67)-P(z<-3)[/tex]

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

[tex]P(-3<z<0.67)=P(z<0.67)-P(z<-3)=0.748571-0.00135=0.74722[/tex]

And on this case the most accurate answer would be:

d. .7462

Answer: option d is the correct answer.

Step-by-step explanation:

For healthy females, x has a approximately normal distribution. We would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ

Where

x = red blood counts.

µ = mean count

σ = standard deviation

From the information given,

µ = 4.8

σ = 0.3

We want to find the that a healthy female has a red blood count between 3.9 and 5.0. It is expressed as

P(3.9 ≤ x ≤ 5) = P(x ≤ 5) - P(x ≤ 3.9)

For x = 3.9,

z = (3.9 - 4.8)/0.3 = - 3

Looking at the normal distribution table, the probability corresponding to the z score is 0.00135

For x = 5,

z = (5 - 4.8)/0.3 = 0.67

Looking at the normal distribution table, the probability corresponding to the z score is 0.7486

P(3.9 ≤ x ≤ 5) = = 0.7486 - 0.00135

= 0.747

Answer: the polynomial representing the shaded region is

2x² - 6x - 1

Step-by-step explanation:

The area of the square is expressed as 4x² - 2x - 6

The area of the triangle is expressed as 2x² + 4x - 5. The area of the shaded region would be the area of the square - the area of the area of the triangle.

The polynomial that represents the area of the shaded region would be

4x² - 2x - 6 - (2x² + 4x - 5)

Collecting like terms, it becomes

= 4x² - 2x² - 2x - 4x - 6 + 5

= 2x² - 6x - 1

Final answer:

The area of the shaded region is represented by the polynomial [tex]2x^2 - 6x - 1[/tex] square inches, which is the difference between the area of the square and the area of the triangle.

Explanation:

The student asked for the polynomial that represents the area of the shaded region given the area of the square and the area of the triangle within it. To find this area, we must subtract the area of the triangle from the area of the square. The given area of the square is [tex]4x^2 - 2x - 6[/tex] square inches, and the area of the triangle is [tex]2x^2 + 4x - 5[/tex] square inches. Therefore, the polynomial representing the area of the shaded region is found by the following subtraction:

Area of the square - Area of the triangle = [tex](4x^2 - 2x - 6) - (2x^2 + 4x - 5) = 2x^2 - 6x - 1[/tex]

This simplifies to 2x^2 (from the square's area) minus 6x (the change in linear dimensions caused by the triangle) minus 1 (the constant term from completing the square). Thus, the polynomial representing the shaded region is [tex]2x^2 - 6x - 1[/tex] square inches.

Answer:

[tex]P(X>55.4)=P(\frac{X-\mu}{\sigma}>\frac{55.4-\mu}{\sigma})=P(Z>\frac{55.4-55}{0.5})=P(Z>0.8)[/tex]

[tex]P(Z>0.8)=1-P(Z\leq 0.8)[/tex]

[tex]P(Z>0.8)=1-0.788=0.212[/tex]

Step-by-step explanation:

Assuming this previous info : "Trucks carry barrels of crude oil from a port in Texas to a distributer in New Mexico on long  trailers. Filling equipment is used to fill the barrels with the oil. When functioning properly,  the actual volume of oil loaded into each barrel by the filling equipment at the port is  approximately normally distributed with a mean of 55 gallons and a standard deviation of 0.5  gallons. If the mean is greater than 55.4 gallons, the filling mechanism is overfilling."

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

[tex]X \sim N(55,0.5)[/tex]  

Where [tex]\mu=55[/tex] and [tex]\sigma=0.5[/tex]

We are interested on this probability

[tex]P(X>55.4)[/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>55.4)=P(\frac{X-\mu}{\sigma}>\frac{55.4-\mu}{\sigma})=P(Z>\frac{55.4-55}{0.5})=P(Z>0.8)[/tex]

And we can find this probability using the complement rule:

[tex]P(Z>0.8)=1-P(Z\leq 0.8)[/tex]

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

[tex]P(Z>0.8)=1-0.788=0.212[/tex]

Final answer:

The probability of a randomly selected barrel containing 55.4 gallons or more is calculated using normal distribution and z-scores. Since the value of 55.4 gallons is significantly higher than the mean with a large positive z-score, the probability is very small and close to 0.

Explanation:

The question involves finding the probability that the volume of oil in a randomly selected barrel will be 55.4 gallons or more, given that the normal operating conditions of a tank are 45 gallons with a 3-gallon standard deviation. This is a problem of probability involving normal distribution, where we are looking for the area under the normal curve to the right of 55.4 gallons.

To find this probability, we first calculate the z-score, which is the number of standard deviations away from the mean the value of 55.4 gallons is. The z-score formula is given by Z = (X - μ) / σ, where X is the value of interest (55.4 gallons), μ is the mean (45 gallons), and σ is the standard deviation (3 gallons). Plugging in the numbers, we get Z = (55.4 - 45) / 3 = 3.467.

Next, we use z-score tables or a calculator to find the probability corresponding to the calculated z-score. The probability represents the area to the left of the z-score under the normal curve. Since we are interested in the probability of the volume being more than 55.4 gallons, we need to subtract this value from 1 to find the area to the right of the z-score.

If using a standard normal distribution table, which gives the area to the left, we find the value corresponding to Z = 3.467. If such a precise value is not available in the table (which is likely because 3.467 is quite high for typical z-score tables), we would take the value for the closest z-score provided or use software or a calculator with statistical functions. Most standard normal tables do not extend beyond a z-score of 3.49, which is associated with a probability very close to 1. Therefore, the probability of the barrel having 55.4 gallons or more is very small and could be approximated as P(Z > 3.467) ≈ 0.

Answer:

The most that a bag can weigh and not need to be repackaged is 15.355 ounces.

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 = 13.50, \sigma = 1.06[/tex]

Bags in the upper 4% are too heavy and must be repackaged. What is the most that a bag can weigh and not need to be repackaged?

Bags in the upper 4% have a pvalue of 1-0.04 = 0.96. So the most that a bag can weight and not need to be repackaged is the value of X when Z has a pvalue of 0.9599. So this is X when [tex]Z = 1.75[/tex]

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

[tex]1.75 = \frac{X - 13.50}{1.06}[/tex]

[tex]X - 13.50 = 1.75*1.06[/tex]

[tex]X = 15.355[/tex]

The most that a bag can weigh and not need to be repackaged is 15.355 ounces.

Answer:

66.76% probability that the levee will NEVER fail in the next 20 years.

Step-by-step explanation:

For each year, there are only two possible outcomes. Either a levee fails during the year, or no levees fail. In each year, the probabilities of levees failing are independent from each other. So we use the binomial probability distribution to solve this problem.

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:

A levee was designed to protect against floods with an annual exceedance probability of 0.02. This means that [tex]p = 0.02[/tex]

What is the risk that the levee will NEVER fail in the next 20 years?

This is [tex]P(X = 0)[/tex] when [tex]n = 20[/tex]. So

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

[tex]P(X = 0) = C_{20,0}*(0.02)^{0}*.(0.98)^{20} = 0.6676[/tex]

66.76% probability that the levee will NEVER fail in the next 20 years.

The risk that a levee, designed to withstand floods with an annual exceedance probability of 0.02, will never fail in the next 20 years is approximately 67%, assuming each year's flood risk is independent and identical.

The question asks about a levee's failure probability over the next 20 years. If a levee is designed to protect against floods with an annual exceedance probability of 0.02, this means that there's a 2% chance each year that the levee will fail due to a flood.

To determine the risk that the levee will never fail in 20 years, we must first understand the probability that it won't fail in any given year, which is 1 - 0.02 = 0.98 (or 98%). The probability that the levee will not fail in 20 consecutive years is therefore 0.98^(20), or approximately 0.67 (or 67%), assuming each year's flood risk is independent and identical.

This calculation, while seemingly straightforward, doesn't account for variables like changes in climate, improvements in the levee's design and maintenance, or other unforeseen factors that might affect the levee's performance.

https://brainly.com/question/32117953

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Answer: t= 460.52 minutes

Step-by-step explanation:Q'=Q/100

Q'= rate in and out of water

Finding the differential equation

Let Q'(t)= The quantity of dye in the tank for t time

But rate in=0 Q/200 ×2=Q'

Q'/Q=-1/100

Dividing by Q gives

Ln/Q/ + c = -1/100 + c1

Integrating both sides gives

Ln/Q/ = -(1/100)t + c2

But c+c1=C2= A constant

Q=C2e(-t/100)

200e-(t/100)

t= ln200

t=460.52minutes

Final answer:

To find the time it takes for the dye concentration in the tank to reach 1% of its original value, we can use the concept of dilution. By applying the dilution formula, we can calculate that it would take approximately 166.67 hours for the concentration to reduce to 1% of its original value.

Explanation:

To find the time that will elapse before the concentration of dye in the tank reaches 1% of its original value, we can use the concept of dilution. The tank is being rinsed with fresh water flowing in at a rate of 2 L/min, and the well-stirred solution is flowing out at the same rate.

The concentration of dye in the tank after a certain amount of time can be calculated using the formula:

C1V1 = C2V2

Where:

Rearranging the formula to solve for V2:

V2 = (C1V1) / C2 = (1 g/L × 200 L) / (0.01 g/L) = 20000 L

So, it would take 20000 L / 2 L/min = 10000 min = 166.67 hours for the concentration of dye in the tank to reach 1% of its original value.

Answer:

a) 2.2 × 10⁸ beats

b) 2.20 × 10⁸ beats

c) 2.207 × 10⁸ beats

Step-by-step explanation:

Data provided in the question:

Average heart rate of a typical person = 70.0 beats/min

Now,

In the given cases, the significance is on the significant figures after the decimal

Therefore,

the answer is will be provided accordingly

Now,

a) Time = 6.0 years

[since 1 significant figure after decimal. answer will be give in  1 significant figure after decimal ]

time in minutes = 6.0 × 365 × 24 × 60

= 3.1 × 10⁶ minutes

Total beats = Average heart rate × Time

= 70 × 3.1 × 10⁶

= 2.2 × 10⁸ beats

b)  Time = 6.00 years

[since 2 significant figure after decimal. answer will be give in 2 significant figure after decimal ]

time in minutes = 6.00 × 365 × 24 × 60

= 3.15 × 10⁶ minutes

Total beats = Average heart rate × Time

= 70 × 3.15 × 10⁶

= 2.20 × 10⁸ beats

c) Time = 6.000 years

[since 3 significant figure after decimal. answer will be give in 3 significant figure after decimal ]

time in minutes = 6.000 × 365 × 24 × 60

= 3.154 × 10⁶ minutes

Total beats = Average heart rate × Time

= 70 × 3.154 × 10⁶

= 2.207 × 10⁸ beats

Answer:

standard deviation of these expected returns = 0.0295 or 2.95%

Step-by-step explanation:

The detailed step is shown in the attachment

To calculate standard deviation of a portfolio's returns, one must compute the average return, find deviations, square them, average those, and take the square root. However, without weights for each stock in the portfolio, the calculation cannot be completed with the provided information.

The student asked what the standard deviation of the expected returns for a portfolio consisting of four stocks is, with returns of minus 9%, 11%, 13%, and 17% respectively over the next four years. To calculate standard deviation, we need to follow several steps:

However, the information given does not include all the necessary data, such as the proportion of each stock within the portfolio. Without individual stock weights, we cannot compute the exact number for the portfolio's standard deviation. Typically, such a calculation would also require historical return data or expected return data for each stock to compute the portfolio variance and standard deviation.

Answer:  MUb / Pb ∠ MUd / Pd

Step-by-step explanation:

First let us define the parameters given;

we have that :

Marginal utility of from book MUb = 44

Marginal utility of from DVD MUd = 1212

Cost of DVD Pd = $ 66

Cost of Book Pb = $ 33

Using,

⇒ MUb / Pb = 44 / 33 = 1.33

⇒ MUd / Pd = 1212 / 66 = 18.36

From this we can see that;

MUb / Pb ∠ MUd / Pd

The question asked is if he is maximizing his utility?

From the law of equi-marginal utility, a consumer will maximize utility only when the utility derived from every unit of spending on goods and services is the same. That is to say that for this to happen,

MUb / Pb = MUd /Pd.

But from the question solved, this is not true, so from the answer gotten The consumer is not maximizing his utility because;

MUb / Pb ∠ MUd / Pd

cheers i hope this helps.

Answer: G and F are mutually exclusive because they cannot occur together

Step-by-step explanation:

According to the definition of mutually exclusive events,

The events which can not occur together and probability of them occurring together is 0 are known as mutually exclusive events.

The first statement gives an implication that if one happens then other happens meaning they could both still happen so it is not true.

The second statement contradict the question about being mutually exclusive events.

The third statement also is a implication that if one event occurs then other does or does not occur.

The last statement is correct one that conforms with the question and obeys the definition of mutually exclusive events.

Answer:

The probability that of the 2 students selected 1st is taking the class for an elective and 2nd for major is 0.121.

Step-by-step explanation:

The number of students taking the class because it is a major requirement is, n (M) = 38.

The number of students taking the class because it as an elective is, n (E) = 6.

The total number of students is, N = 44.

Two students are selected at random.

Assume that the selection is without replacement.

Compute the  probability that the 1st student selected is taking the class as an elective and the 2nd is taking it because it is a major requirement as follows:

P (1st Elective ∩ 2nd Major) = P (1st Elective) × P (2nd Major)

                                              [tex]=\frac{6}{44}\times \frac{38}{43}\\ =0.120507\\\approx0.121[/tex]

Thus, the probability that of the 2 students selected 1st is taking the class for an elective and 2nd for major is 0.121.