The time, in number of days, until maturity of a certain variety of tomato plant is Normally distributed, with mean μ and standard deviation s = 2.4. You select a simple random sample of four plants of this variety and measure the time until maturity. The sample yields ¯x = 65. You read on the package of seeds that these tomatoes reach maturity, on average, in 61 days. You want to test to see if your seeds are reaching maturity later than expected, which might indicate that your package of seeds is too old.
The appropriate hypotheses are:

a) H0 : μ = 61 , Hα : μ > 61 .
b) H0 : μ = 65 , Hα : μ < 65 .
c) H0 : μ = 61 , Hα : μ < 61 .
d) H0 : μ = 65 , Hα : μ > 65 .

The equation for the least-squares regression line can be found by calculating the slope and y-intercept using the given data on knee height and overall height. The regression line is used for predicting the height from knee height.

To find the equation of the least-squares regression line, you first need to calculate the slope (b1) and y-intercept (b0) using the given data. The least-squares regression line is essentially a line of best fit that minimizes the sum of the squared residuals.

The formula for the slope (b1) of the regression line is: b1 = (∑xy - n * mean_x * mean_y) / (∑x^2 - n * mean_x^2) And the y-intercept (b0) is calculated as: b0 = mean_y - b1 * mean_x

Following these formulas and plugging in the given data (for knee height x and height y), we can find b1 and b0. Once we've done that, we can write the equation for the least-squares regression line in the form y = b0 + b1*x.

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The least-squares regression line equation is estimated by determining the slope (b1) and y-intercept (b0) of the line. These values are calculated from the given data sets for height and knee height using the formulas: b1 = [N(Σxy) - (Σx)(Σy)] / [N(Σx²) - (Σx)²] and b0 = (Σy - b1(Σx)) / N

The subject matter of this question is statistics, specifically, it's about finding the equation of a least-squares regression line. The least-squares regression line is a tool used in statistics to show the best possible mathematical relationship between two variables. In this case, the variables are height and knee height.

To calculate the least-squares regression line, we need to calculate the slope (b1) and y-intercept (b0) of the line. The formulas to calculate these are:

Where:

N = number of observations (5 in this case)

Σxy = sum of the product of x and y

Σx = sum of x

Σy = sum of y

Σx² = sum of squares of x

After calculating the values for b0 and b1, the equation for the least-squares regression line would be: y = b0 + b1*x. You would need to calculate these values using the provided datasets for height (x) and knee height (y).

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Answer:

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

[tex]np \geq 15[/tex]

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

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

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

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

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

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

So then the correct answer for this case would be:

E. Normal

Step-by-step explanation:

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

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

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

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

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

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

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

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

[tex]np \geq 15[/tex]

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

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

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

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

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

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

So then the correct answer for this case would be:

E. Normal

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

Option E is correct.

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

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

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

So that the new mean and standard deviation will be,

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

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

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Answer:

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

Step-by-step explanation:

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

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

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

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

Sample size, n = 31

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

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

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

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

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

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

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

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

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

                                                  = [ 90.088 , 94.312 ]

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

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

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

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

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

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

Answer:

56 different tests

Step-by-step explanation:

Given:

Number of wires available (n) = 8

Number of wires taken at a time for testing (r) = 5

In order to find the number of different tests required for every possible pairing of five wires, we need to find the combination rather than their permutation as order of wires doesn't disturb the testing.

So, finding the combination of 5 pairs of wires from a total of 8 wires is given as:

[tex]^nC_r=\frac{n!}{r!(n-r)!}[/tex]

Plug in the given values and solve. This gives,

[tex]^8C_5=\frac{8!}{5!(8-5)!}\\\\^8C_5=\frac{8\times 7\times 6\times 5!}{5!\times 3\times2\times1}\\\\^8C_5=56[/tex]

Therefore, 56 different tests are required for every possible pairing of five ​wires.

A total of 56 different tests are needed to check every possible combination of five wires out of eight in the cable.

The question is asking for the number of different tests required to test every possible pairing of five wires in an eight-wire cable. This is a combination problem, where we're looking for how many different ways we can combine five items from a group of eight. The formula for combinations is C(n, k) = n! / [k!(n - k)!], where n is the total number of items, k is the number of items to choose, and ! represents factorial. Plugging into this formula, we get C(8, 5) = 8! / [5!(8 - 5)!] = 56. Therefore, a total of 56 different tests are needed to check every possible combination of five wires out of eight.

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Answer:

The probability that a randomly selected student is female or a physics major is 0.575.

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

Desired outcomes:

Students that are female or physics majors.

17 female

10 physics majors, of which 4 are female.

This means that there are 10 total physics majors and 17-4 = 13 non physics majors female. So

[tex]D = 13 + 10 = 23[/tex]

Total outcomes:

The class has 40 students, so [tex]T = 40[/tex]

Probability

[tex]P = \frac{23}{40} = 0.575[/tex]

The probability that a randomly selected student is female or a physics major is 0.575.

Answer:

Step-by-step explanation:

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

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

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

n(V n A) = 8

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

Mathematically, according to sets theory,

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

So,

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

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

Answer:

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

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

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

B. 0.4542 to 0.5105.

Step-by-step explanation:

Previous concepts  

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

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

Solution to the problem

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

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

The confidence interval for a proportion is given by this formula  

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

For the 90% confidence interval the value of [tex]\alpha=1-0.9=0.1[/tex] and [tex]\alpha/2=0.05[/tex], with that value we can find the quantile required for the interval in the normal standard distribution.  

[tex]z_{\alpha/2}=1.64[/tex]  

And replacing into the confidence interval formula we got:  

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

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

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

B. 0.4542 to 0.5105.

Answer:

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

Step-by-step explanation:

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

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

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

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

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

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

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

                                                            critical value of 1.6449)

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

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

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

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

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

                                       = [ 0.4542 , 0.5105 ]

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

Answer: The addition of the new point alters the previous standard deviation greatly

Step-by-step explanation:

Let the initial five points be : 2 3 4 5 and 6. In order to calculate the standard deviation for this data, we will need to calculate the mean first.

Mean = summation of scores/number of scores.

The mean is therefore: (2+3+4+5+6)/5 = 20/5 = 4.

We'll also need the sum of the squares of the deviations of the mean from all the scores.

Since mean = 4, deviation of the mean from the score "2" = score(2) - mean (4)

For score 3, it is -1

For 4, it's 0

For 5 it's 1

For 6 it's 2.

The squares for -2, -1, 0, 1, and 2 respectively will be 4, 1 , 0, 1, 4. Summing them up we have 10 i.e (4+1+0+1+4=10).

Calculating the standard deviation, we apply the formula:

√(summation of (x - deviation of mean)^2)/N

Where N means the number of scores.

The standard deviation = √(10/5) = 1.4142

If we add another score or point that is far away from the original points, say 40, what happens to the standard deviation. Let's calculate to find out.

i.e we now have scores: 2, 3, 4, 5, 6 and 40

We calculate by undergoing same steps.

Firstly mean. The new mean = (2+3+4+5+6+40)/6 = 60/6 = 10.

The mean deviations for the scores : 2, 3, 4, 5, 6 and 40 are -8, -7, -6, -5, -4 and 30 respectively. The squares of these deviations are also 64, 49, 36, 25, 16 and 900 respectively as well. Their sum will then be 1090. i.e. (64+49+36+25+16+900 = 1090).

The new standard deviation is then=

√(1090/6)

= √181.67

= 13.478.

It's clear that the addition of a point that's far away from the original points greatly alters the size of the standard deviation as seen /witnessed in this particular instance where the standard deviation rises from 1.412 to 13.478

Answer:

X is a discrete random variable.

X can take values from 0 to 12:

[tex]X\in[0,1,2,3,4,5,6,7,8,9,10,11,12][/tex]

Step-by-step explanation:

(a) X = the number of unbroken eggs in a randomly chosen standard egg carton

X is a discrete random variable.

The minimum amount of eggs broken is 0 and the maximum amount of eggs broken is 12 (assuming a dozen egg carton).

Then, X can take values from 0 to 12:

[tex]X\in[0,1,2,3,4,5,6,7,8,9,10,11,12][/tex]

About the probability ditribution nothing can be said, because there is no information about it (it can be a binomial, uniform or non-standard distribution).

The set of possible values for the random variable X, representing the number of unbroken eggs in an egg carton, is {0, 1, 2, 3, ..., 12}. X is a discrete random variable because it takes on distinct, countable values with no gaps in between.

The random variable X represents the number of unbroken eggs in a randomly chosen standard egg carton. To describe the set of possible values for this variable, we can consider the number of eggs in a standard carton, which is typically 12. Therefore, the set of possible values for X would be {0, 1, 2, 3, ..., 12}, as it encompasses all possible outcomes, ranging from having no unbroken eggs (0) to having all 12 unbroken eggs.

Now, let's determine whether the variable X is discrete or not. A discrete random variable is one that can take on a countable number of distinct values with gaps in between.

In this case, X is indeed a discrete random variable because it can only take on integer values from 0 to 12, and there are no intermediate values between these whole numbers. Each value represents a distinct and countable outcome based on the number of unbroken eggs, making it a discrete random variable.

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Answer:

D) $110 and $190

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 = 150

Standard deviation = 20

95% of the monthly food expenditures are between what two amounts?

By the Empirical Rule, within 2 standard deviations of the mean

150 - 2*20 = $110

150 + 2*20 = $190

So the correct answer is:

D) $110 and $190

Answer: D) $110 and $190

Step-by-step explanation:

The Empirical rule states that for a normal distribution, nearly all of the data will fall within three standard deviations of the mean . The empirical rule is further illustrated below

68% of data falls within the first standard deviation from the mean.

95% fall within two standard deviations.

99.7% fall within three standard deviations.

From the information given, the mean is $150 and the standard deviation is $20.

2 standard deviations = 2 × 20 = 40

150 - 40 = $110

150 + 40 = 190

Therefore, about 95% of the monthly food expenditures are between $110 and $190

The area of the circular fire pit is 4096 square inches.

Explanation:

Given that the rectangular fire pit is 7 feet by 6 feet.

We need to determine the area of the largest circular fire that can be made in this fire pit.

The diameter of the circular fire is 6 feet

The radius is given by

[tex]r=\frac{6}{2} =3[/tex]

Radius is 3 feet.

The area of the largest circular fire pit can be determined using the formula,

[tex]Area=\pi r^2[/tex]

Substituting the values in the formula, we have,

[tex]Area = (3.14)(3)^2[/tex]

        [tex]=(3.14)(9)[/tex]

[tex]Area= 28.26 \ ft^2[/tex]

We need to convert feet to inches by multiplying by 12, we get,

[tex]Area = 28.26\times (12)^2[/tex]

[tex]Area = 4096.44 \ in^2[/tex]

Rounding off to the nearest square inch, we get,

[tex]Area= 4096 \ in^2[/tex]

Thus, the area of the circular fire pit is 4096 square inches.

Answer:

-1

Step-by-step explanation:

We evaluate [tex]\sin(2\pi+3\pi/2)[/tex]

In [tex]2\pi+3\pi/2[/tex], [tex]2\pi[/tex] is a complete revolution and is the same as 0. So we have

[tex]\sin3\pi/2 = \sin(\pi+\pi/2)[/tex]

One [tex]\pi[/tex] is a half revolution, putting the point at (-1, 0). [tex]\pi/2[/tex] is a quarter of a revolution. A quarter circle from (-1, 0) anticlockwise is (0, -1).

The sine is the y-coordinate of a point along a unit circle.

Hence, [tex]\sin(2\pi+3\pi/2)=-1[/tex]

Using the cyclical nature of the unit circle and the sine function, the mathematical expression sin (2π + 3π/2) can be simplified to sin (3π/2). travelling 3π/2 around the unit circle takes us to the point where sin is -1.

The question requires the evaluation of the mathematical expression sin (2π + 3π/2). This can be solved by utilizing the cyclical nature of the unit circle and the sine function. Since the distance around the unit circle is 2π, adding or subtracting multiples of 2π from the angle doesn't change the result of the sin function.

So, we can simplify the function sin (2π + 3π/2) to sin (3π/2). Because every π/2 around the unit circle the sine function repeats, sin (π/2) is 1, sin (2π/2) or sin (π) is 0, sin (3π/2) is -1, and sin (4π/2) or sin (2π) is 0. So sin (3π/2) equals -1.

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Answer:

12.0 (3 sf)

Step-by-step explanation:

E(X) = 0.08(1)+0.2(3)+0.32(5)+0.4(10)

E(X) = 6.28

E(X²) = .08(1²)+.2(3²)+.32(5²)+.4(10²)

E(X²) = 49.88

Var(X) = E(X²) - [E(X)]²

= 49.88 - 6.28² = 10.4416

Var(1.07X) = 1.07² Var(X)

= 1.1449×10.4416 = 11.95458784

12.0 (3 sf)

The variance of the amount paid on the bill (Y) is $10.62.

To find the variance of Y, we need to calculate the expected value of Y first, and then use that to compute the variance.

Step 1: Calculate the expected value of Y (E(Y)).

E(Y) = Σ [P(X) * Y]

where P(X) is the probability of each price category.

E(Y) = (0.08 * $1) + (0.20 * $3) + (0.32 * $5) + (0.40 * $10)

E(Y) = $0.08 + $0.60 + $1.60 + $4.00

E(Y) = $6.28

Step 2: Find the variance of Y.

Variance of Y (Var(Y)) = Σ [P(X) * (Y - E(Y))²]

Var(Y) = (0.08 * ($1 - $6.28)²) + (0.20 * ($3 - $6.28)²) + (0.32 * ($5 - $6.28)²) + (0.40 * ($10 - $6.28)²)

Var(Y) = (0.08 * $27.92) + (0.20 * $10.22) + (0.32 * $1.58) + (0.40 * $14.58)

Var(Y) = $2.24 + $2.04 + $0.51 + $5.83

Var(Y) = $10.62

The variance of Y is $10.62.

The variance measures the spread or dispersion of the values around the expected value, which, in this case, is $6.28.

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Answer: 5.5hours

Step-by-step explanation:

Total distance = 226miles

Total time = 7 hours

Let b represent total time spent while walking.

Distance (walking) = 4b

Distance ( riding) = 40(7-b)

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

226 = 4b + 280 - 40b

226 = 280-36b

b = 54/36

b = 1.5hours

Amount of time spent walking = 1.5hours

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

Amount of time spent on bicycle = 5.5hours

Answer:

53

Step-by-step explanation:

47+6=53

Answer:

53

Step-by-step explanation:

Answer:

Linear Relationship

Step-by-step explanation:

The size is steadily progressing and so is the height.

Answer:

Step-by-step explanation:

linear relationship

Answer:

Percentage of students who scored greater than 700 = 97.72%

Step-by-step explanation:

We are given that the College Board originally scaled SAT scores so that the scores for each section were approximately normally distributed with a mean of 500 and a standard deviation of 100.

Let X = percentage of students who scored greater than 700.

Since, X ~ N([tex]\mu, \sigma^{2}[/tex])

The z probability is given by;

          Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)    where, [tex]\mu[/tex] = 500  and  [tex]\sigma[/tex] = 100

So, P(percentage of students who scored greater than 700) = P(X > 700)

   P(X > 700) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{700-500}{100}[/tex] ) = P(Z < 2) = 0.97725 or 97.72% Therefore, percentage of students who scored greater than 700 is 97.72%.

The question is wrong since it is not possible to have 111 defective graphite rackets when the total number of graphite racket is 100.

Question:

A sample of 100 wood and 100 graphite tennis rackets are taken from the warehouse. Assuming that If 88 wood and 90 graphite are defective and one racket is randomly selected from the​ sample, find the probability that the racket is wood or defective.

Given Information:  

Total wood = 100

Total graphite = 100

Defective wood = 88

Non-defective wood = 12

Defective graphite = 90

Non-defective graphite = 10

Required Information:  

Probability of racket being selected is wood or defective = ?

Answer:

P(wood or defective) = 0.95

Step-by-step explanation:

The probability of selecting a wood racket is

P(wood) = number of wood rackets/total number of rackets

P(wood) = 100/200 = 1/2

The probability of selecting a defective racket is

P(defective) = number of defective rackets/total number of rackets

P(defective) = 88+90/200 = 178/200 = 89/100

There is double counting of wood so we have to subtract the probability of wood and defective

P(wood and defective) = 88/200 = 11/25

P(wood or defective) = P(wood)  + P(defective) - P(wood and defective)

P(wood or defective) = 1/2 + 89/100 - 11/25

P(wood or defective) = 0.95

Alternatively:

P(defective) = number of defective rackets/total number of rackets

P(defective) = 88+90/200 = 178/200 = 89/100

P(wood and non-defective) = 12/200 = 3/50

There is no double counting here so we dont have to subtract anything

P(wood or defective) = P(wood)  + P(wood and non-defective)

P(wood or defective) = 89/100 + 3/50

P(wood or defective) = 0.95

Answer:

[tex]t=\frac{201-190}{\frac{10}{\sqrt{25}}}=5.5[/tex]    

[tex]p_v =P(t_{(24)}>5.5)=0.00000589[/tex]  

If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that the true mean is higher than 190

Step-by-step explanation:

Data given and notation  

[tex]\bar X=201[/tex] represent the mean

[tex]s=10[/tex] represent the sample standard deviation for the sample  

[tex]n=25[/tex] sample size  

[tex]\mu_o =190[/tex] represent the value that we want to test

[tex]\alpha=0.05[/tex] represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

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

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is higher than 190, the system of hypothesis would be:  

Null hypothesis:[tex]\mu \leq 190[/tex]  

Alternative hypothesis:[tex]\mu > 190[/tex]  

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex]  (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic

We can replace in formula (1) the info given like this:  

[tex]t=\frac{201-190}{\frac{10}{\sqrt{25}}}=5.5[/tex]    

P-value

The first step is calculate the degrees of freedom, on this case:  

[tex]df=n-1=25-1=24[/tex]  

Since is a one side test the p value would be:  

[tex]p_v =P(t_{(24)}>5.5)=0.00000589[/tex]  

Conclusion  

If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that the true mean is higher than 190

Answer:

(-2, 6)

Step-by-step explanation:

f(x) = x² + 4x + 10

f(x) = x² + 4x + 4 + 6

f(x) = (x + 2)² + 6

The vertex is at (-2, 6).

Answer: it is cheaper to buy for sweatshirts at $9.69 each.

Step-by-step explanation:

The regular price for one sweatshirt is $12.95. if a customer bought three shirts at the regular price of $12.95, a fourth sweatshirt was free

It means that the cost of buying 4 shirts is

12.95 × 3 = $38.85

Due to a special, the price of one sweatshirt was $9.69. It means that the cost of buying 4 shirts at this price is

9.69 × 4 = $38.76

Therefore, it is cheaper to buy for sweatshirts at $9.69 each than to buy at $12.95 each and get a free shirt

Answer: it is cheaper to buy for sweatshirts at $9.69 each.

Answer: 95 servings of applesauce

Step-by-step explanation:

It takes approximately 4 medium apples to make 3 servings of homemade apple sauce. It means that the number of servings of homemade apple sauce that can be made from 1 medium apple is

3/4 = 0.75 servings

Approximately 126 medium apples were purchased. Therefore, the number of servings of applesauce that they can make with the bushel is

126 × 0.75 = 95 servings of applesauce rounded to the nearest whole number.

Answer:

(a) Probability = 0.29947

Step-by-step explanation:

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

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

Number of ways this can happen = 4 * 194580

Number of ways this can happen = 778,320

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

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

Total probability = 778320 / 2598960 = 0.29947

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

Another way to solve this:

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

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

Total probability = 0.0598 * 5 = 0.29947

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

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

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

Substituting the values and evaluating the expression, we get:

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

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Step-by-step explanation:

Hope it helps you in your learning process.

Answer:

A. Data

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

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

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

Answer:

(a) No

(b) 0.21 or 21%

Step-by-step explanation:

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

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

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

The probability you lose both games is 21%

Given Information:

Probability of wining 1st game = p₁ = 0.7

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

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

Required Information:

(a) Are the two games independent = ?

(b) Probability of losing both games = ?

Answer:

(a) Are the two games independent = No

(b) Probability of losing both games = 0.21

Step-by-step explanation:

(a) Independent Events:

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

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

(b) Probability of losing both games

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

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

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

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

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

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

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

Probability of losing both games = 0.30*0.70

Probability of losing both games = 0.21

Answer:

121+11y+y^2 not perfect square trinomial because the second member should be twice the value of the products of the first and second monomers.

Step-by-step explanation:

(A+B) ^2=A^2 +2*A*B+B^2

Answer:

The probability that at most 10 of the next 150 items produced are unacceptable is 0.8315.

Step-by-step explanation:

Let X = number of items with unacceptable quality.

The probability of an item being unacceptable is, P (X) = p = 0.05.

The sample of items selected is of size, n = 150.

The random variable X follows a Binomial distribution with parameters n = 150 and p = 0.05.

According to the Central limit theorem, if a sample of large size (n > 30) is selected from an unknown population then the sampling distribution of sample mean can be approximated by the Normal distribution.

The mean of this sampling distribution is: [tex]\mu_{\hat p}= p=0.05[/tex]

The standard deviation of this sampling distribution is: [tex]\sigma_{\hat p}=\sqrt{\frac{ p(1-p)}{n}}=\sqrt{\frac{0.05(1-.0.05)}{150} }=0.0178[/tex]

If 10 of the 150 items produced are unacceptable then the probability of this event is:

[tex]\hat p=\frac{10}{150}=0.067[/tex]

Compute the value of [tex]P(\hat p\leq 0.067)[/tex] as follows:

[tex]P(\hat p\leq 0.067)=P(\frac{\hat p-\mu_{p}}{\sigma_{p}} \leq\frac{0.067-0.05}{0.0178})=P(Z\leq 0.96)=0.8315[/tex]

*Use a z-table for the probability.

Thus, the probability that at most 10 of the next 150 items produced are unacceptable is 0.8315.

Final answer:

Using the normal approximation to the binomial distribution, the probability that at most 10 of the next 150 items produced are unacceptable is approximately 86.43%.

Explanation:

Approximating the Probability of Defective Items:

To approximate the probability that at most 10 of the next 150 items produced are unacceptable when each item is of acceptable quality independently with probability 0.95, we use the binomial probability formula or normal approximation. However, since the number of trials is large (n = 150), we can use the normal approximation to the binomial distribution to simplify the calculation.

First, we find the mean (μ) and standard deviation (σ) of the binomial distribution:

Next, we convert the binomial problem to a normal distribution problem by finding the z-score for 10.5 (since we are looking for "at most" 10, we use 10 + 0.5 for continuity correction).

The z-score is calculated as follows:

Finally, we look up the z-score in a standard normal distribution table, or use a calculator to find the cumulative probability for Z ≤ 1.10, which is approximately 0.8643. Therefore, the probability that at most 10 of the next 150 items are unacceptable is roughly 86.43%.

Answer:

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

[tex]np=13*0.4=5.2 \geq 5[/tex]

[tex]n(1-p)=13*(1-0.4)=7.8 \geq 5[/tex]

Assuming that each trial is independent and we have a sample obtained from a random sampling method.

Then we can conclude that we can use the normal approximation since all the conditions are satisfied.

Step-by-step explanation:

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

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

[tex]X \sim Binom(n=13, p=0.4)[/tex]

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

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

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

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

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

[tex]np=13*0.4=5.2 \geq 5[/tex]

[tex]n(1-p)=13*(1-0.4)=7.8 \geq 5[/tex]

Assuming that each trial is independent and we have a sample obtained from a random sampling method.

Then we can conclude that we can use the normal approximation since all the conditions are satisfied.

12x-5y=-20

y=x+4

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

12x-5x-20=-20

7x=0,

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