Mary got 85% correct on her math test. Mary missed 6 questions. How many total questions were on her math final?

When calculating A×B, the number of columns of A = number of rows of B.

Thus, only last case is possible.

Answer:

(35,0)

Step-by-step explanation:

Consider the diagram below, the starting point is given as A and the finish point given as C.

Using similar right-angle triangle, we have that:

[tex]\frac{|AM|}{|BM|}= \frac{|BM|}{|MC|}\\\frac{7}{14}= \frac{14}{x}\\7x=14 X 14\\x=196/7=28[/tex]

Therefore to find the point where Derek stops at C, we first determine the distance |AC|

|AC|=7+28=35

The Coordinates at C where Derek stops is (35,0)

Derek walks along a road described by the equation y = 2x. At the point (7,14), he turns right and walks perpendicularly to his original path until he reaches the x-axis. Upon reaching the x-axis, his stopping point is at (35,0).

When Derek reached the point (7,14), he turned right and started walking perpendicular to the original road. Given that this road is represented by the linear equation y = 2x, a perpendicular path would have a negative reciprocal slope. Therefore, the path he took after turning is represented by y = -1/2x + b. As he turned at the point (7, 14), substituting these coordinates into the equation provides b = 17.5. Since he stopped on the x-axis where y = 0, putting this into the equation gives x = 35. So, Derek stopped at (35,0).

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Answer:

The 99​% confidence interval for his expected proportion of the​ vote is (0.5516, 0.6336). Since the interval for the proportion is above 50%, he would be confident of winning based on the pool.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.

[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

In which

z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].

For this problem, we have that:

[tex]n = 950, p = \frac{563}{950} = 0.5926[/tex]

99% confidence level

So [tex]\alpha = 0.01[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.01}{2} = 0.995[/tex], so [tex]Z = 2.575[/tex].

The lower limit of this interval is:

[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.5926 - 2.575\sqrt{\frac{0.5926*0.4074}{950}} = 0.5516[/tex]

The upper limit of this interval is:

[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.5926 + 2.575\sqrt{\frac{0.5926*0.4074}{950}} = 0.6336[/tex]

The 99​% confidence interval for his expected proportion of the​ vote is (0.5516, 0.6336). Since the interval for the proportion is above 50%, he would be confident of winning based on the pool.

Answer:

10⁵

Step-by-step explanation:

10×10×10×10×10

= 10⁵ or 100,000

Answer:

So, the probability is P=0.9953.

Step-by-step explanation:

We know  that a market research firm conducts telephone surveys with a 44% historical response rate.

We get that:

[tex]p=44\%=0.44=\mu_{\hat{x}}\\\\n=400\\\\\hat{p}=\frac{150}{400}=0.375\\\\[/tex]

We calculate the standar deviation:

[tex]\sigma_{\hat{p}}=\sqrt{\frac{0.44(1-0.44)}{400}}\\\\\sigma_{\hat{p}}=0.025[/tex]

So, we get

[tex]z=\frac{\hat{p}-\mu_{\hat{p}}}{\sigma{\hat{p}}}\\\\z=\frac{0.375-0.44}{0.025}}\\\\z=-2.6[/tex]

We use a probability table to calculate it

[tex]P(\hat{p}>0.375)=P(z>-2.6)=1-P(z<-2.6)=1-0.0047=0.9953[/tex]

So, the probability is P=0.9953.

To find the probability of at least 150 of 400 individuals responding given a 44% response rate, calculate the mean and standard deviation and then find the z-score for 150 responses. The probability of getting at least 150 responses is the area to the right of the z-score in the standard normal distribution.

The question asks for the probability that in a new sample of 400 telephone numbers, at least 150 individuals will respond, given a historical response rate of 44%. To determine this probability, we can approximate the binomial distribution to a normal distribution because the sample size is large (n=400).

First, we calculate the mean (μ) and the standard deviation (σ) for the number of responses. The mean is given by μ = np = 400 × 0.44 = 176. The standard deviation is σ = √(np(1-p)) = √(400 × 0.44 × 0.56) ≈ 9.92.

To calculate the probability of getting at least 150 responses, we would find the z-score for 150, which is z = (X - μ)/σ = (150 - 176)/9.92 ≈ -2.62. We then look up this z-score in a standard normal distribution table or use a calculator to determine the area to the right of this z-score, which represents the probability of getting more than 150 responses.

The question is related to market research and involves using statistical methods to calculate probabilities, which requires an understanding of binomial distributions and normal approximations.

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Answer:

P(T∩E) = 0.017

Step-by-step explanation:

Since every fourth CD is tested. Thus if T is the event that represents 4 disks being tested,

P(T) = 1/4 = 0.25

Let Fi represent event of failure rate. So from the question,

P(F1) = 0.01 ; P(F2) = 0.03 ; P(F3) =0.02 ; P(F4) = 0.01

Also Let F'i represent event of success rate. And we have;

P(F'1) = 1 - 0.01 = 0.99 ; P(F'2) = 1 - 0.03 = 0.97; P(F'3) = 1 - 0.02 = 0.98; P(F'4) = 1 - 0.01 = 0.99

Since all programs run independently, the probability that all programs will run successfully is;

P(All programs to run successfully) =

P(F'1) x P(F'2) x P(F'3) x P(F'4) =

0.99 x 0.97 x 0.98 x 0.97 = 0.932

Now, that all 4 programs failed will be = 1 - 0.932 = 0.068

Let E be denote that the CD fails the test. Thus P(E) = 0.068

Now, since testing and CD's defection are independent events, the probability that one CD was tested and failed will be =P(T∩E) = P(T) x P(E)= 0.25 x 0.068 = 0.017

Final answer:

The probability that a CD fails any of the four independent tests, given individual failure rates of 0.01, 0.03, 0.02, and 0.01, is approximately 6.88%.

Explanation:

To calculate the probability that a CD fails any test, we should first understand that the probability of failing any particular test is the same as 1 minus the probability of passing that test. Given the failure rates of 0.01, 0.03, 0.02, and 0.01 for the four independent tests, the probabilities of a CD passing each test are 0.99, 0.97, 0.98, and 0.99, respectively.

The probability that the CD passes all four tests is the product of the individual probabilities of passing each test (since the tests are independent):

Subtracting this from 1 gives the probability that a CD fails at least one test:

Let's perform the calculation:

P(pass all tests) = 0.99 * 0.97 * 0.98 * 0.99 ≈ 0.9312

P(fail any test) = 1 - 0.9312 ≈ 0.0688

Therefore, the probability that a CD fails any test is approximately 0.0688 or 6.88%.

Answer:

Step-by-step explanation:

Hello!

You have two events.

A: The employee is bilingual.

The probability of the employee being bilingual is P(A)= 30/85= 0.35

And

B: The employee has a graduate degree.

Additionally, you know that the probability of an employee having a graduate degree given that he is bilingual is:

P(B/A)= 0.37

You need to calculate the probability of the employee being bilingual and having a graduate degree. This is the intersection between the two events, symbolically:

P(A∩B)

The events A and B are not independent, which means that the occurrence of A modifies the probability of occurrence of B.

Applying the definition of conditional probability you have that:

P(B/A)= [P(A∩B)]/P(A)

From this definition, you can clear the probability of the intersection between A and B

P(A∩B)= P(B/A)* P(A)= 0.37*0.35= 0.1295≅ 0.13

I hope it helps!

Final answer:

The probability that a randomly selected employee from the division is bilingual and has a graduate degree is approximately 12.9%.

Explanation:

The question asks for the probability that a randomly selected employee from a division with 85 employees is bilingual and has a graduate degree. We know that 30 employees are bilingual, and 37% of these bilingual employees have a graduate degree. To find this probability, we will perform a two-step calculation:

First, calculate the number of bilingual employees with a graduate degree by taking 37% of 30: number of bilingual employees with graduate degrees = 0.37 × 30.

Then, divide this number by the total number of employees to get the probability: probability that an employee is bilingual with a graduate degree = (number of bilingual employees with graduate degrees) / 85.

Now let's calculate the values:

0.37 × 30 = 11.1. Since we cannot have a fraction of a person, we'll round down to 11 bilingual employees with a graduate degree.

(11 / 85) = approximately 0.129, or 12.9%.

So, the probability that a randomly selected employee is bilingual and has a graduate degree is roughly 12.9%.

Answer:

Please find attached file for complete answer solution and explanation of same question

Step-by-step explanation:

Answer:

$374,484

Step-by-step explanation:

Amount payable as income tax = 0.39 X $255,600 = $99,684

The amount Marcus receives if he deferred income tax that year = $(274,800 + 99,684) = $374,484

Answer:

Probability of having two hits in the same region = 0.178

mu: average number of hits per region

x: number of hits

e: mathematical constant approximately equal to 2.71828.

Step-by-step explanation:

We can describe the probability of k events with the Poisson distribution, expressed as:

[tex]P(x=k)=\frac{\mu^ke^{-\mu}}{k!}[/tex]

Being μ the expected rate of events.

If 535 bombs hit 553 regions, the expected rate of bombs per region (the events for this question) is:

[tex]\mu=\frac{\#bombs}{\#regions} =\frac{535}{553}= 0.9674[/tex]

For a region to being hit by two bombs, it has a probability of:

[tex]P(x=2)=\frac{\mu^2e^{-\mu}}{2!}=\frac{0.9674^2e^{-0.9674}}{2!}=\frac{0.9359*0.38}{2}=0.178[/tex]

Answer:

37.5 revolutions

Step-by-step explanation:

The average rotation speed for the first second and for the last two seconds is:

[tex]V_1 = \frac{0+500}{2}\\ V_1 = 250\ rpm[/tex]

For the next 3.0 seconds, the rotation speed is V = 500 rpm.

The total number of revolutions, converting rpm to rps, is given by:

[tex]n=\frac{1*V_1+3*V+2*V_1}{60}\\n=\frac{1*250+3*500*2*250}{60}\\n=37.5\ revolutions[/tex]

The DVD makes 37.5 revolutions.

Answer:

$0.09 tax per every dollar spent.

Step-by-step explanation:

45.78-42 which gives you the amount of money added on by the tax. which = $3.78 then divide that buy how much the game cost before tax was added to get the amount of tax per each $$$

Answer:

$0.09 tax per every dollar spent.

Step-by-step explanation:

You take the $45.78, and subtract that by $42 (the original cost) and then get $3.78. Then divide by every dollar spent, and you get the final rate, $0.09 tax rate.

Answer:

[tex]5314.4-1.675\frac{116.68}{\sqrt{52}}=5287.30[/tex]  

[tex]5314.4+1.675\frac{116.68}{\sqrt{52}}=5341.5[/tex]  

So on this case the 90% confidence interval would be given by (5287.3;5341.5)

And the best intrpretation is:

 5) We are 90% confident that the average number of plucks to failure for all 'high E' strings is between 5,287.3 and 5,341.5

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".  

[tex]\bar X=5314.4[/tex] represent the sample mean  

[tex]\mu[/tex] population mean (variable of interest)  

[tex]s=116.68[/tex] represent the sample standard deviation  

n=52 represent the sample size  

90% confidence interval  

The confidence interval for the mean is given by the following formula:  

[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)  

The degrees of freedom are given by:

[tex] df = n-1= 52-1=51[/tex]

Since the Confidence is 0.90 or 90%, the value of [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.05[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,51)".And we see that [tex]t_{\alpha/2}=1.675[/tex]  

Now we have everything in order to replace into formula (1):  

[tex]5314.4-1.675\frac{116.68}{\sqrt{52}}=5287.30[/tex]  

[tex]5314.4+1.675\frac{116.68}{\sqrt{52}}=5341.5[/tex]  

So on this case the 90% confidence interval would be given by (5287.3;5341.5)

And the best intrpretation is:

 5) We are 90% confident that the average number of plucks to failure for all 'high E' strings is between 5,287.3 and 5,341.5

We are not given the value of

Mount Everest

Elevation

h'(x)

So we are going to based our parameters needed to solve this question o assumptions. The main thing is to understand the process in solving this question.

So here is it!.

A climber on Mount Everest is 8000 meters from the start of his trail and at elevation  10000 meters above sea level. At (x) meters from the start, the elevation of the trail is h(x) meters above sea level. If  h' (x) = 0.5 for near , what is the approximate elevation another 8 meters along the trail

Answer:

10004 meters

Step-by-step explanation:

The rate of change of elevation, h' (x) = 0.5 near 8000 meters.

Thus; we can say that the elevation increases by 0.5 for each meter traveled at a given distance(x)

So, we need to determine the new elevation after 8 meters traveled from ( 8000 to 8008).Then the elevation change can now be written as:

=  (0.5 × 8)

= 4 meters.

And also the new elevation will be:

10000 + 4 meters

= 10004 meters

Answer:

0.8894 is the probability that the test result comes back negative if the disease is present .

Step-by-step explanation:

We are given the following in the question:

P(Disco Fever) = P( Disease) =

[tex]P(D) = \dfrac{1}{2}\times 1\% = 0.5 \times 0.01 = 0.005[/tex]

Thus, we can write:

P(No  Disease) =

[tex]P(ND) =1 - P(D)= 0.995[/tex]

P(Test Positive given the presence of the disease) = 0.99

[tex]P(TP | D ) = 0.99[/tex]

P( false-positive) = 4%

[tex]P( TP | ND) = 0.04[/tex]

We have to evaluate the probability that the test result comes back negative if the disease is present, that is

P(test result comes back negative if the disease is present)

By Bayes's theorem, we can write:

[tex]P(ND|TP) = \dfrac{P(ND)P(TP|ND)}{P(ND)P(TP|ND) + P(D)P(TP|D)}\\\\P(ND|TP) = \dfrac{0.995(0.04)}{0.995(0.04) + 0.005(0.99)} = 0.8894[/tex]

0.8894 is the probability that the test result comes back negative if the disease is present .

Answer:

a.) 0.0822

b.) 1

c.) 0.3659

Step-by-step explanation:

Probability distribution formula is often denoted by :

P(X=r) = nCr × p^r × q^n-r

Where n = total number of samples

r = number of successful outcome of sample

p = probability of success

q = probability of failure.

If we take 4 samples,

3 of this 4 samples are successful

the success rate =S= 31% = 0.31

Failure rate = F= 0.69

a.) Then probability distribution of X becomes:

P(X=3) = 4C3 × 0.31³ × 0.69¹

P(X=3) = 0.0822 (4d.p),

b. Most likely value of X = expected value = np

= 4 × 0.31

= 1.24 ≈ 1

c.) probability that at least 2 out of the 4 have insurance = Probability that 2 have insurance) + probability that 3 have insurance + probability that 4 have insurance.

P(X=2) = 4C2 × 0.31² × 0.69² = 0.2745

P(X=3), as calculated earlier = 0.0822

P(X=4) = 4C4 × 0.31^4 × 0.69^0 = 0.0092

Total probability of having at least 2 out of those 4 insured = 0.2745 + 0.0822 + 0.0092 =0.3659.

Answer:

The rate of study is 5 items per hour.

Step-by-step explanation:

Number of items a person can learn after t hours of instruction, w(t) is given by:

[tex]w(t)=15\sqrt[3]{t^{2}}[/tex]

We want to determine the rate of learning at any time t. The rate is the derivative of w(t) with respect to time.

[tex]\frac{dw(t)}{dt} =\frac{d}{dt} 15\sqrt[3]{t^{2}}[/tex]

[tex]\frac{dw(t)}{dt} =15\frac{d}{dt} {t^{2/3}}[/tex]

[tex]\frac{dw(t)}{dt} =15X\frac{2}{3} {t^{2/3-1}}[/tex]

[tex]\frac{dw(t)}{dt} =10 {t^{-\frac{1}{3} }}=\frac{10}{t^\frac{1}{3}}[/tex]

Therefore, the rate of learning at any time t

[tex]\frac{dw(t)}{dt} =\frac{10}{t^\frac{1}{3}}[/tex]

At the end of 8 hours, t=8

[tex]\frac{dw(t)}{dt} =\frac{10}{8^\frac{1}{3}}[/tex]

[tex]\frac{dw(t)}{dt} =\frac{10}{2}[/tex]=5

The rate of study is 5 items per hour.

Answer:

a) 0.8413

b) 30

c) A. The population is symmetrically distributed, such that the Central Limit Theorem will likely hold for samples of size 25.

d) 29.63

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

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.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex]

In this problem, we have that:

[tex]\mu = 29, \sigma = 5[/tex]

a. If you select a random sample of 25 full-time college students, what is the probability that the mean time spent on academic activities is at least 28 hours per week?

25 students, so [tex]n = 25, s = \frac{5}{\sqrt{25}} = 1[/tex]

This is 1 subtracted by the pvalue of Z when X = 28. So

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

By the Central Limit Theorem

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

[tex]Z = \frac{28 - 29}{1}[/tex]

[tex]Z = -1[/tex]

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

1 - 0.1587 = 0.8413

0.8413 is the answer.

b. If you select a random sample of 25 full-time college students, there is an 84 % chance that the sample mean is less than how many hours per week?

Value of X when Z has a pvalue of 0.84. So X when Z = 1.

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

[tex]1 = \frac{X - 29}{1}[/tex]

[tex]X - 29 = 1[/tex]

[tex]X = 30[/tex]

c. What assumption must you make in order to solve (a) and (b)? (choose between A through D)

Central limit theorem works if the population is normally distributed, or if the sample means are of size at least 30

So the correct answer is:

A. The population is symmetrically distributed, such that the Central Limit Theorem will likely hold for samples of size 25.

d. If you select a random sample of 64 full-time college students, there is an 84 % chance that the sample mean is less than how many hours per week?

Now we have n = 64, so [tex]s = \frac{5}{\sqrt{64}} = 0.63[/tex]

Value of X when Z has a pvalue of 0.84. So X when Z = 1.

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

[tex]0.63 = \frac{X - 29}{1}[/tex]

[tex]X - 29 = 0.63[/tex]

[tex]X = 29.63[/tex]

In this statistics problem related to full-time college students' time spent on academic activities, Z scores and the Central Limit Theorem are used to find probabilities and mean values for different sample sizes. Assumptions about the symmetry of population distribution are also discussed.

This question is about understanding and applying concepts of probability, sample means, and Central Limit Theorem in statistics.

a. The probability that the mean time is at least 28 hours can be found by calculating a Z score. The formula for Z score is (X - μ) ⸫ (σ ⸫ √n), where X is the value we are testing (28 hours in this case), μ is the population mean (29 hours), σ is the standard deviation (5 hours), and n is the sample size (25 students). After calculating the Z score, use a standard normal distribution table to find the probability.

b. To solve this, we will again use the Z score formula but in a different way. Use the given % chance and look up the corresponding Z score on a standard normal distribution table. Then use this Z score in the Z score formula to find the X value, which is the number of hours.

c. The assumption we have to make here is option A. For the Central Limit Theorem to hold, the population distribution does not have to be normal but it should be symmetric. Moreover, for sample sizes of 30 or greater, Central Limit Theorem holds regardless of the shape of the population distribution.

d. This is similar to part b but with a larger sample size (64 students). Use the same procedures to find the answer.

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Answer:

The code is as given below to be copied in a new matlab script m file. The screenshots are attached.

Step-by-step explanation:

As the question is not complete, the complete question is attached herewith.

The code for the problem is as follows:

%Defining the given matrices:

%P is the matrix showing the percentage of changes in voterbase

P = [ 0.8100 0.0800 0.1600 0.1000;

0.0900 0.8400 0.0500 0.0800;

0.0600 0.0400 0.7400 0.0400;

0.0400 0.0400 0.0500 0.7800];

%x0 is the vector representing the current voterbase

x0 = [0.5106; 0.4720; 0.0075; 0.0099];

%In MATLAB, the power(exponent) operator is defined by ^

%After 3 elections..

x3 = P^3 * x0;

disp("The voterbase after 3 elections is:");

disp(x3);

%After 6 elections..

x3 = P^6 * x0;

disp("The voterbase after 6 elections is:");

disp(x3);

%After 10 elections..

x10 = P^10 * x0;

disp("The voterbase after 10 elections is:");

disp(x10);

%After 30 elections..

x30 = P^30 * x0;

disp("The voterbase after 30 elections is:");

disp(x30);

%After 60 elections..

x60 = P^60 * x0;

disp("The voterbase after 60 elections is:");

disp(x60);

%After 100 elections..

x100 = P^100 * x0;

disp("The voterbase after 100 elections is:");

disp(x100);

The output is as well as the code in the matlab is as attached.

Answer:

The voter-base after 3 elections is:

0.392565, 0.400734, 0.109855, 0.096846

The voter-base after 6 elections is:

0.36168, 0.36294, 0.14176, 0.13362

The voter-base after 10 elections is:

0.35405, 0.34074, 0.15342, 0.15178

Step-by-step explanation:

This question is incomplete. I will proceed to give the complete question. Then I will add a screenshot of my code solution to this question. After which I will give the expected outputs.

Let's use the results of the 2012 presidential election as our x0. Looking up the popular vote totals, we find that our initial distribution vector should be (0.5106, 0.4720, 0.0075, 0.0099)T. Enter the matrix P and this vector x0 in MATLAB:

P = [ 0.8100 0.0800 0.1600 0.1000;

0.0900 0.8400 0.0500 0.0800;

0.0600 0.0400 0.7400 0.0400;

0.0400 0.0400 0.0500 0.7800];

x0 = [0.5106; 0.4720; 0.0075; 0.0099];

According to our model, what should the party distribution vector be after three, six and ten elections?

Please find the code solution in the images attached to this question.

The voter-base after 3 elections is therefore:

0.392565, 0.400734, 0.109855, 0.096846

The voter-base after 6 elections is therefore:

0.36168, 0.36294, 0.14176, 0.13362

The voter-base after 10 elections is therefore:

0.35405, 0.34074, 0.15342, 0.15178

Answer:

The sample mean is 13.

The sample variance is 0.2286.

The sample standard deviation is 0.4781.

Step-by-step explanation:

The sample is:

S = {12.6, 12.9, 13.4, 12.3, 13.6, 13.5, 12.6, 13.1}

The sample is of size n = 8.

The formula to compute the sample mean, sample variance and sample standard deviation are:

[tex]\bar x=\frac{1}{n} \sum x[/tex]

[tex]s^{2}=\frac{1}{n}\sum (x-\bar x)^{2} \\s=\sqrt{\frac{1}{n}\sum (x-\bar x)^{2} }\\[/tex]

Compute the sample mean as follows:

[tex]\bar x=\frac{1}{n} \sum x\\=\frac{1}{8}(12.6+ 12.9+ 13.4+ 12.3+ 13.6+ 13.5+ 12.6+ 13.1)\\=\frac{104}{8}\\ =13[/tex]

The sample mean is 13.

Compute the sample variance as follows:

[tex]s^{2}=\frac{1}{n-1}\sum (x-\bar x)^{2} \\=\frac{1}{8-1}[(12.6-13)^{2}+(12.9-13)^{2}+(13.4-13)^{2}+...+(13.1-13)^{2}] \\=\frac{1}{7}\times1.6\\=0.2286[/tex]

The sample variance is 0.2286.

Compute the sample standard deviation as follows:

[tex]s=\sqrt{s^{2}}\\=\sqrt{0.2286}\\=0.4781[/tex]

The sample standard deviation is 0.4781.

Answer:

The sample mean is 13.

The sample variance is 0.2286.

The sample standard deviation is 0.4781.

Step-by-step explanation:

Good luck

Answer:

i) [tex] P(9.5 < X<11)[/tex]

And we can solve this problem using 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(9.5<X<11)=P(\frac{9.5-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{11-\mu}{\sigma})=P(\frac{9.5-10}{1.414}<Z<\frac{11-10}{1.414})=P(-0.354<z<0.707)[/tex]

And we can find this probability with this difference:

[tex]P(-0.354<z<0.707)=P(z<0.707)-P(z<-0.354)[/tex]

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

[tex]P(-0.354<z<0.707)=P(z<0.707)-P(z<-0.354)=0.7602-0.3617=0.3985[/tex]

ii) The z scores for this case are:

[tex] z_1 = \frac{9.5-10}{1.414}= -0.354[/tex]

[tex] z_2 = \frac{11-10}{1.414}= 0.707[/tex]

And we can check the answer with the following excel code:

=NORM.DIST(0.707,0,1,TRUE)-NORM.DIST(-0.354,0,1,TRUE)

iii) [tex]P(X>a)=0.975[/tex]   (a)

[tex]P(X<a)=0.025[/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 0.025 of the area on the left and 0.975 of the area on the right it's z=-1.96. On this case P(Z<-1.96)=0.025 and P(z>-1.96)=0.975

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

[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.975[/tex]  

[tex]P(z<\frac{a-\mu}{\sigma})=0.975[/tex]

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

[tex]z=-1.96<\frac{a-10}{1.414}[/tex]

And if we solve for a we got

[tex]a=10 -1.96*1.414=7.228[/tex]

So the value of height that separates the bottom 2.5% of data from the top 97.5% is 7.228.

iv) [tex] z = \frac{7.228-10}{1.414}= -1.96[/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

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

[tex]X \sim N(10,\sqrt{2})[/tex]  

Where [tex]\mu=10[/tex] and [tex]\sigma=1.414[/tex]

Part i

For this case we want this probability:

[tex] P(9.5 < X<11)[/tex]

And we can solve this problem using 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(9.5<X<11)=P(\frac{9.5-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{11-\mu}{\sigma})=P(\frac{9.5-10}{1.414}<Z<\frac{11-10}{1.414})=P(-0.354<z<0.707)[/tex]

And we can find this probability with this difference:

[tex]P(-0.354<z<0.707)=P(z<0.707)-P(z<-0.354)[/tex]

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

[tex]P(-0.354<z<0.707)=P(z<0.707)-P(z<-0.354)=0.7602-0.3617=0.3985[/tex]

Part ii

The z scores for this case are:

[tex] z_1 = \frac{9.5-10}{1.414}= -0.354[/tex]

[tex] z_2 = \frac{11-10}{1.414}= 0.707[/tex]

And we can check the answer with the following excel code:

=NORM.DIST(0.707,0,1,TRUE)-NORM.DIST(-0.354,0,1,TRUE)

Part iii

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

[tex]P(X>a)=0.975[/tex]   (a)

[tex]P(X<a)=0.025[/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 0.025 of the area on the left and 0.975 of the area on the right it's z=-1.96. On this case P(Z<-1.96)=0.025 and P(z>-1.96)=0.975

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

[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.975[/tex]  

[tex]P(z<\frac{a-\mu}{\sigma})=0.975[/tex]

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

[tex]z=-1.96<\frac{a-10}{1.414}[/tex]

And if we solve for a we got

[tex]a=10 -1.96*1.414=7.228[/tex]

So the value of height that separates the bottom 2.5% of data from the top 97.5% is 7.228.

Part iv

The z score for this value is given by:

[tex] z = \frac{7.228-10}{1.414}= -1.96[/tex]

Final answer:

The question involves calculating probabilities and finding values within a normal distribution regarding the length of grass blades. It covers finding specific probabilities, standardizing values, and locating a value below which a certain percentage of data lies, all based on a given mean and variance.

Explanation:

A meticulous gardener is interested in the length of blades of grass on his lawn. He believes that blade length X follows a normal distribution centered on 10 mm with a variance of 2 mm.

Find the probability that a blade of grass is between 9.5 and 11 mm long.

What are the standardized values of 9.5 and 11 in the context of this distribution? Using the standardized values, confirm that you can obtain the same probability you found in (i) with the standard normal density.

Below which value is the shortest 2.5 percent of blade lengths found?

Standardize your answer from (iii).

The standard deviation (sqrt(variance)) is sqrt(2) mm. The standardized value, or z-score, is computed as Z = (X - μ)/σ, where X is the value, μ is the mean (10 mm), and σ is the standard deviation.

For X = 9.5, Z = (9.5 - 10) / sqrt(2) = -0.3536.

For X = 11, Z = (11 - 10) / sqrt(2) = 0.7071.

To find the probability between 9.5 and 11 mm, we look up these z-scores in the standard normal distribution table or use a calculator.

To find the value below which the shortest 2.5 percent of blade lengths are found, we look up the z-score that corresponds to the cumulative area of 0.025 in the standard normal distribution table. Then, we use the z-score formula in reverse to find the original value in mm.

Answer:

So, the sample mean is 31.3.

So, the sample standard deviation is 6.98.

Step-by-step explanation:

We have a list of data for the speeds (in miles per hour) of the 20 vehicles. So, N=20.

We calculate the sample mean :

[tex]\mu=\frac{19 +19 +22 +24 +25 +27 +28+ 37 +35 +30+ 37+ 36+ 39+ 40+ 43+ 30+ 31+ 36+ 33+ 35}{20}\\\\\mu=\frac{626}{20}\\\\\mu=31.3[/tex]

So, the sample mean is 31.3.

We use the formula for a sample standard deviation:

[tex]\sigma=\sqrt{\frac{1}{N-1}\sum_{i=1}^{N}(x_i-\mu)^2}[/tex]

Now, we calculate the sum

[tex]\sum_{i=1}^{20}(x_i-31.3)^2=(19-31.3)^2+(19-31.3)^2+(22-31.3)^2+(24-31.3)^2+(25-31.3)^2+(27-31.3)^2+(28-31.3)^2+(37-31.3)^2+(35-31.3)^2+(30-31.3)^2+(37-31.3)^2+(36-31.3)^2+(39-31.3)^2+(40-31.3)^2+(43-31.3)^2+(30-31.3)^2+(31-31.3)^2+(36-31.3)^2+(33-31.3)^2+(35-31.3)^2\\\\\sum_{i=1}^{20}(x_i-31.3})^2=926.2\\[/tex]

Therefore, we get

[tex]\sigma=\sqrt{\frac{1}{N-1}\sum_{i=1}^{N}(x_i-\mu)^2}\\\\\sigma=\sqrt{\frac{1}{19}\cdot926.2}\\\\\sigma=6.98[/tex]

So, the sample standard deviation is 6.98.

Answer:

[tex]P(X>10.983)=P(\frac{X-\mu}{\sigma}>\frac{10.983-\mu}{\sigma})=P(Z>\frac{10.983-10.5}{0.3})=P(z>1.61)[/tex]

And we can find this probability using the complement rule and with excel or the normal standard table:

[tex]P(z>1.61)=1-P(z<1.61)=1-0.946=0.054[/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

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

[tex]X \sim N(10.5,0.3)[/tex]  

Where [tex]\mu=10.5[/tex] and [tex]\sigma=0.3[/tex]

We are interested on this probability

[tex]P(X>10.983)[/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>10.983)=P(\frac{X-\mu}{\sigma}>\frac{10.983-\mu}{\sigma})=P(Z>\frac{10.983-10.5}{0.3})=P(z>1.61)[/tex]

And we can find this probability using the complement rule and with excel or the normal standard table:

[tex]P(z>1.61)=1-P(z<1.61)=1-0.946=0.054[/tex]

Answer:

a) [tex] \bar X = \frac{\sum_{i=1}^n X_i}{n}[/tex]

And replacing we got:

[tex] \bar X = 27.2[/tex]

b) For this case we have n =48 observations and we can calculate the median with the average between the 24th and 25th values on the dataset ordered.

20.4 20.4 20.5 21.7 22.3 23.3 23.6 23.7 23.7 23.7  23.9 23.9 24.1 24.3 24.7 24.9 25.1 25.1 25.1 25.2  25.3 25.6 26.1 26.1 26.4 26.5 26.7 27.1 27.1 27.4  27.6 28.4 28.6 28.6 28.7 28.8 28.8 29.6 31.0 32.0  32.0 32.4 32.5 32.9 33.1 34.5 38.4 44.1

For this case the median would be:

[tex] Median = \frac{26.1+26.4}{2}=26.25 \approx 26.3[/tex]

c) [tex] Mode= 23.2, 25.1[/tex]

And both with a frequency of 3 so then we have a bimodal distribution for this case

Step-by-step explanation:

For this case we have the following dataset:

23.6, 26.5, 28.6, 28.6, 23.7, 25.3, 24.9, 28.7, 26.7, 32.4, 20.4, 23.9, 32.0, 32.5, 23.7, 26.1, 21.7, 26.1, 38.4, 24.1, 20.5, 25.2, 31, 26.4, 27.1 ,25.1, 25.1, 23.7, 23.9, 32.9, 28.8, 25.6, 28.8, 28.4, 32, 27.6, 29.6, 44.1, 27.1, 24.7, 22.3, 24.3, 23.3, 27.4, 20.4, 25.1, 34.5, 33.1

Part a

We can calculate the mean with the following formula:

[tex] \bar X = \frac{\sum_{i=1}^n X_i}{n}[/tex]

And replacing we got:

[tex] \bar X = 27.2[/tex]

Part b

For this case we have n =48 observations and we can calculate the median with the average between the 24th and 25th values on the dataset ordered.

20.4 20.4 20.5 21.7 22.3 23.3 23.6 23.7 23.7 23.7  23.9 23.9 24.1 24.3 24.7 24.9 25.1 25.1 25.1 25.2  25.3 25.6 26.1 26.1 26.4 26.5 26.7 27.1 27.1 27.4  27.6 28.4 28.6 28.6 28.7 28.8 28.8 29.6 31.0 32.0  32.0 32.4 32.5 32.9 33.1 34.5 38.4 44.1

For this case the median would be:

[tex] Median = \frac{26.1+26.4}{2}=26.25 \approx 26.3[/tex]

Part c

For this case the mode would be:

[tex] Mode= 23.2, 25.1[/tex]

And both with a frequency of 3 so then we have a bimodal distribution for this case

The mean commute time for the 48 cities is 25.6 minutes, the median commute time is 26.5 minutes and the mode of the commute times is 28.6 minutes.

To address the student's question about commute times, let's go through the calculations step by step:

(a) Mean Commute Time

The mean (average) is calculated by summing all the commute times and dividing by the number of cities (48).

Sum of all commute times = 23.6 + 26.5 + 28.6 + 28.6 + 23.7 + 25.3 + 24.9 + 28.7 + 26.7 + 32.4 + 20.4 + 23.9 + 32.0 + 32.5 + 23.7 + 26.1 + 21.7 + 26.1 + 38.4 + 24.1 + 20.5 + 25.2 + 31.0 + 26.4 + 27.1 + 25.1 + 25.1 + 23.7 + 23.9 + 32.9 + 28.8 + 25.6 + 28.8 + 28.4 + 32.0 + 27.6 + 29.6 + 44.1 + 27.1 + 24.7 + 22.3 + 24.3 + 23.3 + 27.4 + 20.4 + 25.1 + 34.5 + 33.1 = 1229.3 minutes

Mean = Sum / Number of cities = 1229.3 / 48 ≈ 25.6 minutes

(b) Median Commute Time

To find the median, list the commute times in numerical order and find the middle value. Since we have an even number of cities (48), the median will be the average of the 24th and 25th values.

Ordered list: 20.4, 20.4, 20.5, 21.7, 22.3, 23.3, 23.6, 23.7, 23.7, 23.9, 23.9, 24.1, 24.3, 24.7, 24.9, 25.1, 25.1, 25.1, 25.2, 25.3, 25.6, 26.1, 26.1, 26.4, 26.5, 26.7, 27.1, 27.1, 27.4, 27.6, 28.4, 28.6, 28.6, 28.7, 28.8, 28.8, 29.6, 31.0, 32.0, 32.0, 32.4, 32.5, 32.9, 33.1, 34.5, 38.4, 44.1

The 24th and 25th values are 26.4 and 26.5.

Median = (26.4 + 26.5) / 2 = 26.45 ≈ 26.5 minutes

(c) Mode(s) Commute Time

The mode is the value that appears most frequently in the list.

From the list, we see that the most frequently occurring value is 28.6, which occurs 2 times.

Therefore, the mode is 28.6 minutes.

So, (a) the mean commute time is 25.6 minutes, (b) the median commute is 26.5 minutes and (c) the mode of the commute times is 28.6 minutes.

Answer:

Step-by-step explanation:

Lets take 1000 output in total.

From these 1000 outputs, 700 are from F1 and 300 are from F2.

Defective from F1 is [tex]\frac{700\times3}{100} = 21[/tex] and defective from F2 is [tex]\frac{300\times2}{100} = 6[/tex].

a.

Total received part is 1000.

Total defectives are (21+6) = 27.

The probability of a received part being defective is [tex]\frac{27}{1000}[/tex].

b.

The probability of the randomly chosen defective part from F1 is [tex]\frac{21}{27} = \frac{7}{9}[/tex].

The probability that a received part is defective is 2.7%, while the probability that a randomly chosen defective part came from factory F1 is 77.78%.

To find the probability that a received part is defective, we need to consider the probabilities of receiving a defective part from each factory. Let's assume that the VCR manufacturer receives 100 parts. From factory F1, 70% of the parts are received, which means 70 parts. The probability of a part from F1 being defective is 3%, so the number of defective parts from F1 is 70 * (3/100) = 2.1. From factory F2, which accounts for the remaining 30% of the parts (30 parts), the probability of a part being defective is 2%, resulting in 30 * (2/100) = 0.6 defective parts. Therefore, the total number of defective parts is 2.1 + 0.6 = 2.7. Since there are 100 parts in total, the probability that a received part is defective is 2.7/100 = 0.027, or 2.7%.

To find the probability that a randomly chosen defective part came from factory F1, we can use the concept of conditional probability. The probability of a part coming from F1 given that it is defective can be found using the formula P(F1|Defective) = P(F1 and Defective) / P(Defective). We already know that P(F1 and Defective) = 2.1/100 and P(Defective) = 2.7/100. Substituting these values, we get P(F1|Defective) = (2.1/100) / (2.7/100) = 2.1/2.7 = 0.7778, or 77.78%.

Answer:

Step-by-step explanation:

Let us assume that the test scores of the students were normally distributed. we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ

Where

x = test scores of students.

µ = mean test scores

σ = standard deviation

From the information given,

µ = 510

σ = 100

We want to find the probability that a student scored below 300. It is expressed as

P(x ≤ 300)

For x = 300

z = (300 - 510)/100 = - 2.1

Looking at the normal distribution table, the probability corresponding to the z score is 0.018

Therefore, the percentage of students that scored below 300 is

0.018 × 100 = 1.8%

Answer:

We need a sample size of at least 23 for a 99% confidence interval, with a tolerable interval width of 0.4.

We need a sample size of at least 9 for a 95% confidence interval with a tolerable width of 0.5,

Step-by-step explanation:

How large a sample size is required if want a 99% confidence interval, with a tolerable interval width of 0.4?

We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:

[tex]\alpha = \frac{1-0.99}{2} = 0.005[/tex]

Now, we have to find z in the Ztable as such z has a pvalue of [tex]1-\alpha[/tex].

So it is z with a pvalue of [tex]1-0.005 = 0.995[/tex], so [tex]z = 2.575[/tex]

Now, find the margin of error(width) as such

[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]

In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.

For this item, we have:

[tex]M = 0.4, \sigma = 0.75[/tex]. So

[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]

[tex]0.4 = 2.575*\frac{0.75}{\sqrt{n}}[/tex]

[tex]0.4\sqrt{n} = 1.93125[/tex]

[tex]\sqrt{n} = \frac{1.93125}{0.4}[/tex]

[tex]\sqrt{n} = 4.828125[/tex]

[tex]\sqrt{n}^{2} = (4.828125)^{2}[/tex]

[tex]n = 23[/tex]

We need a sample size of at least 23 for a 99% confidence interval, with a tolerable interval width of 0.4.

How large a sample would we need if were interested in a 95% confidence interval with a tolerable width of 0.5?

We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:

[tex]\alpha = \frac{1-0.95}{2} = 0.025[/tex]

Now, we have to find z in the Ztable as such z has a pvalue of [tex]1-\alpha[/tex].

So it is z with a pvalue of [tex]1-0.025 = 0.975[/tex], so [tex]z = 1.96[/tex]

Now, find the margin of error(width) as such

[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]

In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.

For this item, we have:

[tex]M = 0.5, \sigma = 0.75[/tex]. So

[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]

[tex]0.5 = 1.96*\frac{0.75}{\sqrt{n}}[/tex]

[tex]0.5\sqrt{n} = 1.47[/tex]

[tex]\sqrt{n} = \frac{1.47}{0.5}[/tex]

[tex]\sqrt{n} = 2.94[/tex]

[tex]\sqrt{n}^{2} = (2.94)^{2}[/tex]

[tex]n \cong 9[/tex]

We need a sample size of at least 9 for a 95% confidence interval with a tolerable width of 0.5,

Answer:

The number of digits in a randomly selected row until a 1 is found.

Final answer:

The random variable describing 'The number of digits in a randomly selected row until a 1 is found' is geometric, as it fits the criteria of a geometric distribution, which includes the number of trials needed for the first success, with constant success probability and independent trials.

Explanation:

The student has asked which of the given random variables is geometric. Among the options provided, the one describing a geometric random variable is "The number of digits in a randomly selected row until a 1 is found." A geometric distribution is concerned with the number of Bernoulli trials required to get the first success. In this case, obtaining a '1' when randomly selecting digits can be considered a success, and all trials are independent with the probability of success (getting a '1') remaining constant with each trial.

Other options such as the number of 1s in a row of 50 random digits or the number of tails when a coin is tossed 60 times describe binomial random variables, where we are interested in the number of successes within a fixed number of trials, not the trial number of the first success.

Answer:for FRR method the difference equation is given by:

u(k+1)=0.8125u(k)+(-24.68)e(k)+25e (k+1)

Step-by-step explanation:see the pictures attached

Answer:

11/50

Step-by-step explanation:

39 of 50 states are under court order, so 11 of 50 states are not under court order.

Final answer:

The probability that a randomly selected state is not under a court order to alleviate overcrowding and poor conditions in its prisons is 11/50.

Explanation:

To find the probability that a randomly selected state is not currently under court order to alleviate overcrowding and poor conditions in one or more of their prisons, we can use the complement rule in probability. This rule states that the probability of an event not happening is equal to one minus the probability of the event happening. Given that 39 states are under such a court order, the probability that a state is under court order is 39/50. Therefore, the probability that a state is not under court order is 1 - (39/50).

Calculating this, we get:

P(Not under court order) = 1 - (39/50) = (50/50) - (39/50) = 11/50

The probability that a randomly selected state is not currently under a court order to alleviate overcrowding and poor conditions in its prisons is 11/50, when expressed as a reduced fraction.

Answer:

The range of GPA of 95% students is from 1.9 to 3.18.

Step-by-step explanation:

The Empirical Rule states that, for a normally distributed random variable:

68% of the measures are within 1 standard deviation of the mean.

95% of the measures are within 2 standard deviation of the mean.

99.7% of the measures are within 3 standard deviations of the mean.

In this problem, we have that:

Mean = 3.04

Standard deviation = 0.57

Assuming that students' GAPs are normally distributed, what is the range of GPA of 95% students

Within 2 standard deviations of the mean

3.04 - 2*0.57 = 1.9

3.04 + 2*0.57 = 3.18

The range of GPA of 95% students is from 1.9 to 3.18.