Consider a value to be significantly low if its z score less than or equal to minus 2 or consider a value to be significantly high if its z score is greater than or equal to 2. A test is used to assess readiness for college. In a recent​ year, the mean test score was 21.6 and the standard deviation was 5.4. Identify the test scores that are significantly low or significantly high. What test scores are significantly​ low?

Answer:

(a) The probability that there are no dandelions in a randomly selected area of 1 square meter in this region is 0.1353.

(b) The probability that there are at least one dandelion in a randomly selected area of 1 square meter in this region is 0.8647.

Step-by-step explanation:

Let X = number of dandelions per square meter.

The average number of dandelions per square meter is, λ = 2.

The random variable X follows a Poisson distribution with parameter λ = 2.

The probability mass function of X is:

[tex]P(X=x)=\frac{e^{-2}2^{x}}{x!};\ x=0,1,2,3...[/tex]

(a)

Compute the value of P (X = 0) as follows:

[tex]P(X=0)=\frac{e^{-2}2^{0}}{0!}=\frac{0.13534\times 1}{1}=0.1353[/tex]

Thus, the probability that there are no dandelions in a randomly selected area of 1 square meter in this region is 0.1353.

(b)

Compute the value of P (X ≥ 1) as follows:

P (X ≥ 1) = 1 - P (X < 1)

             = 1 - P (X = 0)

             [tex]=1-0.1353\\=0.8647[/tex]

Thus, the probability that there are at least one dandelion in a randomly selected area of 1 square meter in this region is 0.8647.

To find the probability of no dandelions, use the Poisson distribution formula with the mean value of 2. To find the probability of at least one dandelion, use the complement rule.

To find the probability of no dandelions in a randomly selected area of 1 square meter in this region, we need to use the Poisson distribution. The mean number of dandelions per square meter is 2. The probability of no dandelions is given by the formula: P(X=0) = e^(-2) * (2^0 / 0!). Calculating this gives us approximately 0.1353.

To find the probability of at least one dandelion, we can use the complement rule. The probability of at least one dandelion is equal to 1 minus the probability of no dandelions. So, P(X >= 1) = 1 - P(X=0). Calculating this gives us approximately 0.8647.

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Answer:

68% of the sample can be expected to fall between 28 and 32 cm

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 = 30

Standard deviation = 2

What proportion of the sample can be expected to fall between 28 and 32 cm

28 = 30 - 2

28 is one standard deviation below the mean

32 = 30 + 2

32 is one standard deviation above the mean.

By the Empirical Rule, 68% of the sample can be expected to fall between 28 and 32 cm

Answer:

Probability a team from national football conference wins the same year will be equal to 0.53

Step-by-step explanation:

Probability that a National League team wins baseball world series = 0.44

Probability that a National Football conference team wins the           Superbowl = 0.53

The two events are independent, since the sports played are different (baseball and football). This is a logical assumption to make.

Since these events are independent, the occurrence of one event will not change the probability of the other event. This means that it does not matter whether the national league team won or lost the world series. The probability of the football team winning the Superbowl will remain the same, which is 0.53.

Question 4: The value of x is 4.5.

Question 5: The value of x is 15.6.

Solution:

Question 4:

The given triangle is a right triangle.

Using trigonometric formulas for a right triangle.

[tex]$\sin \theta=\frac{\text { opposite }}{\text { hypotenuse }}[/tex]

[tex]$\sin C=\frac{AB}{BC}[/tex]

[tex]$\sin 45^\circ=\frac{x}{6.4}[/tex]

[tex]$\frac{1}{\sqrt{2} } =\frac{x}{6.4}[/tex]

Multiply 6.4 on both sides, we get

[tex]$\frac{6.4}{\sqrt{2} }=x[/tex]

x = 4.5

The value of x is 4.5.

Question 5:

The given triangle is a right triangle.

Using trigonometric formulas for a right triangle.

[tex]$\sin \theta=\frac{\text { opposite }}{\text { hypotenuse }}[/tex]

[tex]$\sin E=\frac{EF}{DF}[/tex]

[tex]$\sin 60^\circ=\frac{x}{18}[/tex]

[tex]$\frac{\sqrt3}{2} =\frac{x}{18}[/tex]

Multiply 18 on both sides, we get

x = 15.6

The value of x is 15.6.

Answer:

the probability that his mother would return on time to catch Gabe stealing is given by:[tex]\frac{15}{60}=0.25[/tex].

Step-by-step explanation:

probability is given by; the required outcome over the number of possible outcome.

Step1; we have to subtract the first initial 30 secs that Gabe spent wondering if he will be caught. i.e 90 sec-30 secs= 60 Sec.

Step2: we divide the number of seconds that he can finish a cookie without being caught over the number of seconds left before his mother returns. that is why we have [tex]\frac{15}{60} = 0.25[/tex]

The probability that his mother returns in time to catch Gabe stealing a cookie is 0.25.

Judy could return at any time in the next 90 seconds with equal probability. For the first 30 seconds, Gabe sheepishly wonders if he will get caught trying to grab a nearby cookie. And, it takes Gabe 15 seconds to grab a cookie from the jar.

Now the probability is

[tex]= 15 \div (90 - 30)\\\\= 15 \div 60[/tex]

= 0.25

Hence, we can conclude that The probability that his mother returns in time to catch Gabe stealing a cookie is 0.25.

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Answer:

Step-by-step explanation:

Given Data

Suppose That [tex]y_{1}[/tex] is a solution of L[tex]y_{1}[/tex] = F(x)

and [tex]y_{2}[/tex] is a solution  of [tex]Ly_{2}[/tex] = g(x)

L is liner operator

∴ [tex]L(y_{1}+y_{2} )[/tex] = L[tex]y_{1}[/tex] +[tex]Ly_{2}[/tex]

L(y) = F(x) + g(x)

[tex]y_{1}+y_{2}[/tex] is the solution to  L(y) = F(x) + g(x)

E.g the liner operator be L = [tex]\frac{d}{dx}[/tex]

[tex]\frac{d}{dx}[/tex] [tex]y_{1}[/tex]  = f(x)

[tex]\frac{d}{dx}[/tex] [tex]y_{2}[/tex] = g(x)

[tex]\frac{d}{dx}[/tex] [tex](y_{1}+y_{2} )[/tex] = [tex]\frac{d}{dx}[/tex] [tex]y_{1}[/tex] + [tex]\frac{d}{dx}[/tex] [tex]y_{2}[/tex] = f(x) + g(x)

Final Answer:

y = y1 + y2 solves the equation Ly = f(x) + g(x).

Explanation:

Certainly! To prove the superposition principle for nonhomogeneous linear differential equations, we will utilize the properties of linear operators and the given solutions for the differential equations.
Let's define L as a linear differential operator. Being linear implies that for any functions u(x) and v(x), and any constants a and b, the operator satisfies the following properties:
1. L(u + v) = L(u) + L(v)  (additivity)
2. L(au) = aL(u)           (homogeneity)

Given there are two functions y1(x) and y2(x) that are solutions to the nonhomogeneous linear differential equations:
Ly1 = f(x)
Ly2 = g(x)

We need to prove that if you take a linear combination of y1 and y2, the result y = y1 + y2 will also be a solution to the combined nonhomogeneous equation Ly = f(x) + g(x).

Here's how we do it:
Consider a linear combination of y1 and y2, denoted as y = y1 + y2. Apply the linear operator L to both sides of this equation:
L(y) = L(y1 + y2)

Since L is a linear operator, we can apply the additivity property:
L(y) = L(y1) + L(y2)

Now, we know that y1 and y2 are solutions to their respective nonhomogeneous equations, so we can substitute f(x) for L(y1) and g(x) for L(y2):
L(y) = f(x) + g(x)

Thus, we have shown that y = y1 + y2 solves the equation Ly = f(x) + g(x).

This is the proof of the superposition principle for nonhomogeneous linear differential equations. It shows that solutions to such equations can be added together to obtain a new solution corresponding to the sum of the nonhomogeneous parts.

Answer:

Probability = 0.23572 .

Step-by-step explanation:

We are given that the length of time needed to complete a certain test is normally distributed with mean 35 minutes and standard deviation 15 minutes.

Let X = length of time needed to complete a certain test

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] = 35  and  [tex]\sigma[/tex] = 15

So, P(31 < X < 40) = P(X < 40) - P(X <= 31)

P(X < 40) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{40-35}{15}[/tex] ) = P(Z < 0.33) = 0.62930

P(X <= 31) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{31-35}{15}[/tex] ) = P(Z < -0.27) = 1 - P(Z <= 0.27)

                                               = 1 - 0.60642 = 0.39358

Therefore, P(31 < X < 40) = 0.62930 - 0.39358 = 0.23572 .  

To find the probability that it takes between 31 and 40 minutes to complete the test when the test completion times are normally distributed with a mean (μ) of 35 minutes and a standard deviation (σ) of 15 minutes, we can use the properties of the normal distribution.

Firstly, we need to standardize our times (31 and 40 minutes) to find the corresponding z-scores. The z-score is a measure of how many standard deviations an element is from the mean. We use the following formula to calculate the z-score:

\[ z = \fraction{(X - \mu)}{\sigma} \]

where:
- \( X \) is the value for which we are finding the z-score
- \( \mu \) is the mean
- \( σ \) is the standard deviation

We will calculate the z-scores for both 31 minutes and 40 minutes.

For \( X = 31 \):
\[ z_{31} = \fraction{(31 - 35)}{15} = \fraction{-4}{15} \approx -0.267 \]

For \( X = 40 \):
\[ z_{40} = \fraction{(40 - 35)}{15} = \fraction{5}{15} \approx 0.333 \]

Next, we look up these z-scores in the standard normal distribution (z-distribution) table, which will give us the area to the left of each z-score.

If we don't have the z-distribution table available, we might use statistical software or a calculator with a normal distribution function. But let's proceed as if we are using the z-table.

Let's say our z-table gives us the following areas:

Area to the left of \( z_{31} \approx -0.267 \): Approximately 0.3944 (please note that actual values will depend on the specific z-table or calculator you are using).

Area to the left of \( z_{40} \approx 0.333 \): Approximately 0.6304.

Now we want the area between these two z-scores, which represents the probability that the test completion time is between 31 and 40 minutes. To find this, we can subtract the area for \( z_{31} \) from the area for \( z_{40} \):

Probability \( P(31 < X < 40) = P(z_{40}) - P(z_{31}) \)

Substituting the values,

\[ P(31 < X < 40) = 0.6304 - 0.3944 = 0.2360 \]

So the probability that it will take between 31 and 40 minutes to complete the test is approximately 0.2360, or 23.60%.

(a) Recursive formula for the number of cars sold, [tex]P_n[/tex], in the (n + 1)th week: [tex]P_n = P_{n-1} + 2[/tex]
(b) Explicit formula for the number of cars sold, [tex]P_n[/tex], in the (n + 1)th week:

[tex]P_n = 6 + 2n[/tex]

(c) In the fourth week, the dealership will sell 12 cars.

To find the recursive formula for the number of cars sold, [tex]P_n[/tex], in the (n + 1)th week, we can observe the pattern from the given information.

Given data:

[tex]P_0 = 6[/tex] (number of cars sold in the first week)

[tex]P_1 = 8[/tex] (number of cars sold in the second week)

We can see that each week, the number of cars sold increases by 2. So, the recursive formula can be expressed as:

[tex]P_n = P_{n-1} + 2[/tex]

This formula states that the number of cars sold in the nth week [tex](P_n)[/tex] is equal to the number of cars sold in the previous week [tex](P_{n-1})[/tex] plus 2.

Next, let's find the explicit formula for the number of cars sold, [tex]P_n[/tex], in the (n + 1)th week.

To do this, we need to identify the initial value ([tex]P_0[/tex]) and the common difference (d) in the arithmetic sequence. In this case, the initial value is 6 (P₀ = 6), and the common difference is 2 (the number of cars sold increases by 2 each week).

The explicit formula for an arithmetic sequence is given by:

[tex]P_n = P_0 + n * d[/tex]

Substitute the given values:

[tex]P_n = 6 + n * 2[/tex]

Therefore, the explicit formula for the number of cars sold, P_n, in the (n + 1)th week is [tex]P_n = 6 + 2n[/tex].

Now, let's find how many cars will be sold in the fourth week (n = 3):

[tex]P_3 = 6 + 2 * 3[/tex]

[tex]P_3 = 12[/tex]

In the fourth week, the dealership will sell 12 cars.

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The recursive formula for the number of cars sold weekly is Pn = Pn - 1 + 2, where Pn represents the number of cars in the nth week. The explicit formula is Pn = 6 + 2n. Based on this model, the dealership would sell 14 cars in the fourth week.

In your problem, the number of cars sold each week is increasing by a constant amount, following a linear growth model. The first week, 6 cars were sold and the second week, 8 cars were sold. This indicates an increase of 2 cars from week 1 to week 2.

To write the recursive formula for the number of cars sold, Pn, in the (n + 1)th week, we determine the growth by subtraction: P1 - P0 = 8 - 6 = 2. So, every week, the number of cars sold increases by 2. The recursive formula would therefore be Pn = Pn - 1 + 2.

For the explicit formula which gives us the number of cars sold in the n-th week directly, we observe that it started with 6 cars (base) and increments by 2 each week. Therefore, the explicit formula would be Pn = 6 + 2n.

By the fourth week, the number of cars sold would be P4 = 6 + 2 * 4 = 6 + 8 = 14 cars sold.

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Answer:

0.0082 or 0.82%

Step-by-step explanation:

Given:

Mean of jump time (μ) = 45 s

Standard deviation (σ) = 5 s

Time for jump required for accomplishment (x) = 33 s

The distribution is normal distribution.

So, first, we will find the z-score of the distribution using the formula:

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

Plug in the values and solve for 'z'. This gives,

[tex]z=\frac{33-45}{5}=-2.4[/tex]

So, the z-score of the distribution is -2.4.

Now, we need the probability [tex]P(x\leq 33)=P(z\leq -2.4)[/tex].

From the normal distribution table for z-score equal to -2.4, the value of the probability is 0.0082 or 0.82%.

Therefore, the probability of making a jump in 33 seconds or less is 0.0082 or 0.82%.

Using the normal distribution, it is found that there is a 0.0082 = 0.82% probability of such an accomplishment.

In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

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

In this problem:

The probability is the p-value of Z when X = 33, hence:

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

[tex]Z = \frac{33 - 45}{5}[/tex]

[tex]Z = -2.4[/tex]

[tex]Z = -2.4[/tex] has a p-value of 0.0082.

0.0082 = 0.82% probability of such an accomplishment.

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Answer:

B. x = -4/3 and x = 5

Step-by-step explanation:

2 / (x − 2) + 7 / (x² − 4) = 5 / x

2 / (x − 2) + 7 / ((x + 2) (x − 2)) = 5 / x

Multiply both sides by x − 2:

2 + 7 / (x + 2) = 5 (x − 2) / x

Multiply both sides by x + 2:

2 (x + 2) + 7 = 5 (x − 2) (x + 2) / x

Multiply both sides by x:

2x (x + 2) + 7x = 5 (x − 2) (x + 2)

Simplify:

2x² + 4x + 7x = 5 (x² − 4)

2x² + 4x + 7x = 5x² − 20

0 = 3x² − 11x − 20

Factor:

0 = (3x + 4) (x − 5)

x = -4/3 or x = 5

Answer: 9.16666666667

Step-by-step explanation:

To set up the iterated integrals for a given region, we need to find the limits of integration for both x and y. In this case, the region D is bounded by y = x - 42 and x = y^2. By visualizing the region, we can determine the limits of integration for both orders of integration. We can then choose the easier order to evaluate the double integral.

To set up the iterated integrals, we need to find the limits of integration for both x and y. First, let's visualize the region D bounded by y = x - 42 and x = y^2. The curve y = x - 42 intersects the parabola x = y^2 at two points: (7, -35) and (-7, -49).

To set up the iterated integral with dx dy order, the outer integral will have limits of integration for y that go from -49 to -35. The inner integral will have limits of integration for x that go from the parabola x = y^2 to the line x = y + 42.

The iterated integral with dy dx order can be set up by reversing the order of integration. The outer integral will have limits of integration for x that go from -7 to 7, and the inner integral will have limits of integration for y that go from the line y = x - 42 to the parabola y = sqrt(x).

To evaluate the double integral using the easier order, we can choose either order and find the corresponding iterated integral.

Let's choose the dy dx order. The limits of integration for the outer integral are x = -7 to x = 7, and for each x, the limits of integration for the inner integral are y = x - 42 to y = sqrt(x).

Now, we can evaluate the double integral.

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Answer:

A sampling method is  independent qualitative when an individual selected for one sample does not dictate which individual is to be in the second sample.

Step-by-step explanation:

individuals selected for one sample do not dictate which individuals are to be in the second sample, while dependent quantitative involves individuals selected to be in one sample are used to determine the individuals in the second sample.

Final answer:

An independent sampling method allows for selection of individuals for one sample without influencing the selection for another, maintaining the integrity of quantitative research.

Explanation:

A sampling method is considered independent when an individual selected for one sample does not dictate which individual is to be in the second sample. This type of sampling strategy ensures that the selection of participants for one sample does not influence the selection for another, which is crucial for maintaining the validity and reliability of research findings, particularly in quantitative research.

In research, particularly in studies involving statistical analysis, it is essential to use appropriate sampling methods to ensure that the results are representative of the wider population. Probability sampling methods are often employed in quantitative research as they provide a means to make generalizations from the sample to the population. Types of probability samples include simple random samples, stratified samples, and cluster samples.

The LCLR is approximately 14.5148.

We have,

To calculate the lower control limit (LCLR) for the range in a control chart, we use the formula:

LCLR = D3 * Average Range

Given the sample ranges of the last 5 samples:

9, 14, 8, 7, and 5, we can calculate the average range:

Average Range = (9 + 14 + 8 + 7 + 5) / 5 = 8.6

The D3 value for a sample size of 8 is typically 1.693.

The lower control limit (LCLR) for the range.

LCLR = D3 * Average Range

= 1.693 * 8.6 = 14.5148

(rounded to four decimal places)

Thus,

The LCLR is approximately 14.5148.

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To find LCLR, calculate the average range of the samples and subtract a constant (d2) from it. In this case, LCLR = 53.977 grams.

In this question, we are given the results from the last 5 samples of the weight of packets from the box-filling line. The process average weight per packet is 127 grams. We are asked to find the Lower Control Limit for Range (LCLR) to check if the process is in control. To find LCLR, we need to calculate the average range of the samples and subtract a constant (d2) from it.

First, we calculate the average range of the samples by summing the ranges and dividing by the number of samples. In this case, the sum of the ranges is 92 + 83 + 14 + 85 + 7 = 281. So, the average range is 281 / 5 = 56.2.

Next, we need to find the value of d2, which depends on the sample size. In this case, the sample size is 8. Looking up the value of d2 for a sample sisizeze of 8 in the control chart constants table, we find that it is 2.223. Finally, we can calculate LCLR by subtracting 2.223 from the average range: LCLR = 56.2 - 2.223 = 53.977 grams.

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The z-scores for the given data values are approximately:

560: 0.6

650: 1.5

500: 0

450: -0.5

300: -2

Given the mean (μ) of 500 and the standard deviation (σ) of 100, we can calculate the z-scores for the provided data values:

The z-score (also known as the standard score) measures how many standard deviations a data point is away from the mean. It is calculated using the formula:

z = (x - μ) / σ

Where:

x is the data value

μ is the mean of the sample

σ is the standard deviation of the sample

For x = 560:

z = (560 - 500) / 100 = 0.6

For x = 650:

z = (650 - 500) / 100 = 1.5

For x = 500:

z = (500 - 500) / 100 = 0

For x = 450:

z = (450 - 500) / 100 = -0.5

For x = 300:

z = (300 - 500) / 100 = -2

So, the z-scores for the given data values are approximately:

560: 0.6

650: 1.5

500: 0

450: -0.5

300: -2

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The z-scores for the data values 560, 650, 500, 450, and 300 in a sample with a mean of 500 and a standard deviation of 100 are 0.6, 1.5, 0, -0.5, and -2 respectively. They are computed using the formula z = (X - μ) / σ.

This question refers to the concept of z-scores in statistics, which is a part of Mathematics. A z-score indicates how many standard deviations a given data point is from the mean. The formula to calculate the z-score is: z = (X - μ) / σ, where X is the data point, μ is the mean, and σ is the standard deviation.

The given sample has a mean (μ) of 500 and a standard deviation (σ) of 100. Let's calculate the z-scores:

So the z-scores for the data values 560, 650, 500, 450, and 300 are 0.6, 1.5, 0, -0.5, and -2 respectively.

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Answer:

the minimum is located in x = -5/3 , y= -5/3

Step-by-step explanation:

for the function

f(x,y)=2x + 2y

we define the function g(x)=9x² - 9xy + 9y² - 25  ( for g(x)=0 we get the constrain)

then using Lagrange multipliers f(x) is maximum when

fx-λgx(x)=0 → 2 - λ (9*2x - 9*y)=0 →

fy-λgy(x)=0 → 2 - λ (9*2y - 9*x)=0

g(x) =0 → 9x² - 9xy + 9y² - 25 = 0

subtracting the second equation to the first we get:

2 - λ (9*2y - 9*x) - (2 - λ (9*2x - 9*y))=0

- 18*y + 9*x + 18*x - 9*y = 0

27*y = 27 x  → x=y

thus

9x² - 9xy + 9y² - 25 = 0

9x² - 9x² + 9x² - 25 = 0

9x² = 25

x = ±5/3

thus

y = ±5/3

for x=5/3 and y=5/3 →  f(x)= 20/3 (maximum) , while for x = -5/3 , y= -5/3 →  f(x)= -20/3  (minimum)

finally evaluating the function in the boundary , we know because of the symmetry of f and g with respect to x and y that the maximum and minimum are located in x=y

thus the minimum is located in x = -5/3 , y= -5/3

Answer:

(a) Probability that a single randomly selected element X of the population is between 57,000 and 58,000 = 0.46411

(b) Probability that the mean of a sample of size 100 drawn from this population is between 57,000 and 58,000 = 0.99621

Step-by-step explanation:

We are given that a normally distributed population has mean 57,800 and standard deviation 75, i.e.; [tex]\mu[/tex] = 57,800  and  [tex]\sigma[/tex] = 750.

Let X = randomly selected element of the population

The z probability is given by;

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

(a) So, P(57,000 <= X <= 58,000) = P(X <= 58,000) - P(X < 57,000)

P(X <= 58,000) = P( [tex]\frac{X-\mu}{\sigma}[/tex] <= [tex]\frac{58000-57800}{750}[/tex] ) = P(Z <= 0.27) = 0.60642

P(X < 57000) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{57000-57800}{750}[/tex] ) = P(Z < -1.07) = 1 - P(Z <= 1.07)

                                                          = 1 - 0.85769 = 0.14231

Therefore, P(31 < X < 40) = 0.60642 - 0.14231 = 0.46411 .

(b) Now, we are given sample of size, n = 100

So, Mean of X, X bar = 57,800 same as before

But standard deviation of X, s = [tex]\frac{\sigma}{\sqrt{n} }[/tex] = [tex]\frac{750}{\sqrt{100} }[/tex] = 75

The z probability is given by;

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

Now, probability that the mean of a sample of size 100 drawn from this population is between 57,000 and 58,000 = P(57,000 < X bar < 58,000)

P(57,000 <= X bar <= 58,000) = P(X bar <= 58,000) - P(X bar < 57,000)

P(X bar <= 58,000) = P( [tex]\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] <= [tex]\frac{58000-57800}{\frac{750}{\sqrt{100} } }[/tex] ) = P(Z <= 2.67) = 0.99621

P(X < 57000) = P( [tex]\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] < [tex]\frac{57000-57800}{\frac{750}{\sqrt{100} } }[/tex] ) = P(Z < -10.67) = P(Z > 10.67)

This probability is that much small that it is very close to 0

Therefore, P(57,000 < X bar < 58,000) = 0.99621 - 0 = 0.99621 .

Using the normal distribution and the central limit theorem, it is found that:

In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

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

In this problem:

The probability that a single randomly selected element X of the population is between 57,000 and 58,000 is the p-value of Z when X = 58000 subtracted by the p-value of Z when X = 57000, thus:

X = 58000:

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

[tex]Z = \frac{58000 - 57800}{750}[/tex]

[tex]Z = 0.27[/tex]

[tex]Z = 0.27[/tex] has a p-value of 0.6064.

X = 57000:

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

[tex]Z = \frac{57000 - 57800}{750}[/tex]

[tex]Z = -1.07[/tex]

[tex]Z = -1.07[/tex] has a p-value of 0.1423.

0.6064 - 0.1423 = 0.4641.

0.4641 = 46.41% probability that a single randomly selected element X of the population is between 57,000 and 58,000.

For samples of size 100, [tex]n = 100[/tex], and then:

[tex]s = \frac{750}{\sqrt{100}} = 75[/tex]

For samples of size 100, the mean is of 57800 and the standard deviation is 75.

Then, the probability is:

X = 58000:

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

By the Central Limit Theorem

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

[tex]Z = \frac{58000 - 57800}{75}[/tex]

[tex]Z = 2.7[/tex]

[tex]Z = 2.7[/tex] has a p-value of 0.9965.

X = 57000:

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

[tex]Z = \frac{57000 - 57800}{75}[/tex]

[tex]Z = -10.7[/tex]

[tex]Z = -10.7[/tex] has a p-value of 0.

0.9965 - 0 = 0.9965.

0.9965 = 99.65% probability that the mean of a sample of size 100 drawn from this population is between 57,000 and 58,000.

A similar problem is given at https://brainly.com/question/24663213

Answer:

[tex]126.07-1.64\frac{15}{\sqrt{72}}=123.17[/tex]    

[tex]126.07+1.64\frac{15}{\sqrt{72}}=128.97[/tex]    

So on this case the 90% confidence interval would be given by (123.17;128.97)    

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[/tex] represent the sample mean for the sample  

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

s represent the sample standard deviation

n represent the sample size  

Solution to the problem

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

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

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: "=-NORM.INV(0.05,0,1)".And we see that [tex]z_{\alpha/2}=1.64[/tex]

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

[tex]126.07-1.64\frac{15}{\sqrt{72}}=123.17[/tex]    

[tex]126.07+1.64\frac{15}{\sqrt{72}}=128.97[/tex]    

So on this case the 90% confidence interval would be given by (123.17;128.97)    

The probability that the chosen will be a multiple of 5 between 1 and 15 is [tex]\frac{2}{13}[/tex].

Solution:

Given data:

Number between 1 and 15 is 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13 and 14.

Total numbers between 1 and 15 = 13

N(S) = 13

Multiple of 5 between 1 and 15 = 5, 10

Number of multiples of 5 between 1 and 15 = 2

N(A) = 2

Probability of multiple of 5 between 1 and 15:

[tex]$P(A)=\frac{N(A)}{N(S)}[/tex]

[tex]$P(A)=\frac{2}{13}[/tex]

The probability that the chosen will be a multiple of 5 between 1 and 15 is [tex]\frac{2}{13}[/tex].

Final answer:

To determine the probability of selecting a multiple of 5 between 1 and 15, count the multiples (5 and 10) and divide by the total number (15), resulting in a probability of 2/15.

Explanation:

The question is related to the concept of probability in mathematics. To find the probability that a randomly chosen number between 1 and 15 is a multiple of 5, you simply count how many multiples of 5 are in that range and divide by the total number of possibilities (15). The multiples of 5 between 1 and 15 are 5 and 10, so there are 2 favorable outcomes. Hence, the probability is 2 divided by 15, which equals 2/15.

Answer:

a) P(B'|A') = P(B') = 1 - 0.7 = 0.3.

The reasoning is that: since, events A and B are independent, then, events A' and B' are also independent and for independent events, P(A|B) = P(A).

b) P(A u B) = 0.82

c) P[(A n B')|(A u B)] = 0.146

Step-by-step explanation:

P(A) = 0.4

P(B) = 0.7

A and B are independent events.

P(A') = 1 - 0.4 = 0.6

P(B') = 1 - 0.7 = 0.3

a) If the Asian project is not successful, what is the proba- bility that the European project is also not successful?

P(B'|A') = P(B') = 1 - 0.7 = 0.3

The reasoning is that: since, events A and B are independent, then, events A' and B' are also independent and for independent events, P(A|B) = P(A).

b) The probability that at least one of the two projects will be successful = P(A u B)

P(A u B) = P(A) + P(B) - P(A n B) = 0.4 + 0.7 - (0.4)(0.7) = 0.82

OR

P(A u B) = P(A n B') + P(A' n B) + P(A n B) = (0.4)(0.3) + (0.6)(0.7) + (0.4)(0.7) = 0.82

c) Given that at least one of the two projects is successful, what is the probability that only the Asian project is successful?

P[(A n B')|(A u B)] = }P(A n B') n P(A u B)]/P(A u B) = P(A n B')/P(A u B) = (0.4)(0.3)/(0.82)

P[(A n B')|(A u B)] = 0.146

Hope this Helps!!!

Answer:

20 liters of the 40% solution

60-20=40 liters of the 70% solution

Step-by-step explanation:

40%x+70%(60-x)=60%(60)

0.4x+0.7(60-x)=0.6(60)

0.4x+42-0.7x=36

-0.3x+42=36

-0.3x=-42+36

-0.3x=-6

0.3x=6

3x=60

x=60/3

x=20 liters of the 40% solution

60-20=40 liters of the 70% solution

check: 0.4*20=8

0.7*40=28

8+28=36 and 0.6*60=36 also

Answer:

[tex]3\sqrt{t} +\frac{2}{\sqrt{t}}[/tex] square inches per seconds.

Step-by-step explanation:

given that a rectangle is growing such that the length of a rectangle is 5t+4 and its height is √t, where t is time in seconds and the dimensions are in inches

Area of rectangle = length * width

A = [tex]\sqrt{t} (5t+4)\\= 5t\sqrt{t} +4\sqrt{t}[/tex]

To find rate of change of A with respect to t, we can find derivative of A with respect to t.

WE get

[tex]\frac{dA}{dt} =5(\frac{3}{2} )\sqrt{t} +\frac{4}{2\sqrt{t} } \\= 3\sqrt{t} +\frac{2}{\sqrt{t}}[/tex]

Hence rate of change of area with respect to time is

[tex]3\sqrt{t} +\frac{2}{\sqrt{t}}[/tex] square inches per seconds.

Answer:

Total mean= 27cm

Total standard deviation = 0.1849mm

Step-by-step explanation:

The mechanical assembly has three parts consisting of the length of the rod (the this be called x), the length of the bearing (be called y). You might be wondering where the third part is but the assembly usually has two bearings. Then the second length of the bearing (be called z).

From the problem,

μx = 23, μy =2, μz = 2

σx = 0.18, σy = 0.03, σz = 0.03

We can find the total Mean by adding all the means,

μxyz =  23 + 2 + 2 = 27cm

Since the length of the assembly are independent,

To find the total standard deviation, we must first find the total variance(square of standard deviation)

σ² = (0.18)² + (0.03)² + (0.03)² =  0.0342

Now, we find the standard deviation (square root of the variance)

σ = √0.0342 = 0.18493242

σ ≈ 0.1849mm (you can change to cm by multiplying by 10)

The mean total length of the mechanical assembly is 27 cm and the standard deviation is approximately 0.0183 cm when rounded to four decimal places.

The question is asking for the mean and standard deviation of the total length of a mechanical assembly with a rod and two bearings. To find the mean total length, we simply add the mean lengths of the rod and two bearings.

The mean length of the rod is 23 cm and of each bearing is 2 cm, which sums up to a mean total length of 27 cm for the assembly since there are two bearings (23 + 2 + 2).

For standard deviation, since the parts are manufactured independently, we apply the rule of variances: The variance of the sum of independent variables is the sum of their variances. The standard deviation of the rod is 0.18 mm (or 0.018 cm), and of each bearing is 0.03 mm (or 0.003 cm).

Thus, the total variance for the assembly is the sum of the variances: (0.0182 + 0.0032 + 0.0032). After calculating, we take the square root to find the total standard deviation, which is approximately 0.0183 cm, rounded to four decimal places.

Answer:

a) 4, 5, 7, 12

b) 7

c) 4.5, 6.5, 8, 6, 8.5, 9.5

d) 7.167              

Step-by-step explanation:

We are given the following in the question:

4, 5, 7, 12

a) unique values for x

4, 5, 7, 12

b) mean of the population

[tex]Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}[/tex]

[tex]\mu =\displaystyle\frac{28}{4} = 7[/tex]

c) sampling distribution for samples of size 2

Sample size, n = 2

Possible samples of size 2 are (4,5),(4,7),(4,12),(5,7),(5,12),(7,12)

Sample means are:

[tex]\bar{x_1} = \dfrac{4+5}{2} = 4.5\\\\\bar{x_2} = \dfrac{4+7}{2} = 6.5\\\\\bar{x_3} = \dfrac{4+12}{2} = 8\\\\\bar{x_4} = \dfrac{5+7}{2} = 6\\\\\bar{x_5} = \dfrac{5+12}{2} = 8.5\\\\\bar{x_6} = \dfrac{7+12}{2} = 9.5[/tex]

Thus, the list is 4.5, 6.5, 8, 6, 8.5, 9.5

d) mean of the sampling distribution

[tex]\bar{x} = \dfrac{4.5 + 6.5 + 8+ 6 + 8.5+ 9.5}{6} = \dfrac{43}{6} = 7.167[/tex]

The unique values for x are 4, 5, 7, and 12 pounds. The population mean is 7 pounds, and the sampling distribution for the average weight from a sample of two juvenile condors has possible outcomes of 4.5, 5.5, 6, 8, 8.5, and 9.5 pounds, with a mean of 7 pounds.

To answer the student's questions regarding juvenile condor weights:

Answer:

Yp = t[Asin(2t) + Acos(2t)]

Yp = t²[At² + Bt + C]

Step-by-step explanation:

The term "multiplicity" means when a given equation has a root at a given point is the multiplicity of that root.

(a) r1=-2i; r2=2i g(t)=2sin(2t) + 3cos(2t)

As you can notice the multiplicity of this equation is 1 since the roots r1 = 2i and r2 = 2i appear for only once.

The form of a particular solution will be

Yp = t[Asin(2t) + Acos(2t)]

where t is for multiplicity 1

(b) r1=r2=0; r3=1 g(t)= t² +2t + 3

As you can notice the multiplicity of this equation is 2 since the roots r1 = r2 = 0 appears 2 times.

The form of a particular solution will be

Yp = t²[At² + Bt + C]

where t² is for multiplicity 2

Answer:

the demand curve will shift upward by $20 and the price paid by buyers will decrease by $20.

Step-by-step explanation:

the reduction in the fixed amount of tax from $50 t0 $30 will bring about reduction  of $20 in the price of ticket. the reduction in the price of the ticket, other factors held constant, will brings about change in the demand curve.

The correct answer is D, as supply curve will shift downward by $20, and the effective price received by sellers will increase by $20.

This is so because by reducing the tax, the value that each airline will effectively increase. In this way, as they obtain greater income for each ticket, the airlines will in turn increase the offer of tickets.

Learn more about maths in https://brainly.com/question/25838173

COMPLETE QUESTION:

What is the expected number of schools of fish found in one day of searching?

Answer: 0.1

Step-by-step explanation:

One school of fish per 100,000 square miles of ocean

Plane can search 10,000 square miles

Therefore,

Expected number of schools of fish found in one day of searching

= 10,000 / 100,000

= 0.1

Answer:

Step-by-step explanation:

Hello!

The definition of the Central Limi Theorem states that:

Be a population with probability function f(X;μ,δ²) from which a random sample of size n is selected. Then the distribution of the sample mean tends to the normal distribution with mean μ and variance δ²/n when the sample size tends to infinity.

As a rule, a sample of size greater than or equal to 30 is considered sufficient to apply the theorem and use the approximation.

X[bar]≈N(μ;σ²/n)

If the variable of interest is X: the number of accidents per week at a hazardous intersection.

There is no information about the distribution of this variable, but a sample of n= 52 weeks was taken, and since the sample is large enough you can approximate the distribution of the sample mean to normal. With population mean μ= 2.2 and standard deviation σ/√n= 1.1/√52= 0.15

I hope it helps!

Answer:

A Not a probability B.Probability C.not a probability d.Probability E propability F.not a propability

Step-by-step explanation:

A propability of any event is a number between zero and one

A)175%  converted 1.75 so lies outside zero and one not a propability

B)5.91 lies outside the zero and one rage not a propability

C)0.66 lies in the zero and one rage therefore a propability

D)0.002 Lies within the zero and one range therefore a propability

E) -110% converted equals -1.1 so does not lies in this range not a propability

The following numbers could not be probabilities:

A. The number 175% could not be a probability because it is larger than 100%.

B. The number 5.91 could not be a probability because it is larger than 1.

E. The number negative 110% could not be a probability because it is negative.

Probabilities must be between 0 and 1, inclusive. This is because a probability represents the likelihood of an event occurring, and there is no event that can be more likely than certain (1) or less likely than impossible (0).

The numbers 0.66 and 0.002 are valid probabilities.

For such more question on probability

https://brainly.com/question/25839839

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Answer:

The answer to the given problem is given below.

Step-by-step explanation:

What values do p, q,n,E and p represents?

The value of p is the sample proportion.

The value of q is found from evaluating 1− p.

The value of n is the sample size.

The value of E is the margin of error.

The value of p is the population proportion.

If the confidence level is 90%, what is the value of α?

α = 1- 0.90

α = 0.10