Question 6
Consider the following data
X y
0 2
5 4
4 3
What is the sample variance of x? What is the sample standard deviation of x?
O Sample variance is 7, and sample standard deviation is 2.65
O Sample variance is 7, and sample standard deviation is 2.16
O Sample variance is 4.67, and sample variance is 2.16
O Sample variance is 7, and sample standard deviation is 4.67

The number of cakes, muffins, and tarts that must be made, respectively, is 65 cakes, 195 muffins, and 390 tarts.

The task is to determine the number of cakes, muffins, and tarts that need to be made for an afternoon tea event at the Awlty Towers Hotel. The owner has specified certain ratios between the quantities of tarts, muffins, and cakes, and each guest should be provided with a total of 5 pastries, regardless of the type.

Let's denote the number of muffins as M, tarts as T, and cakes as C. Based on the given information, we can establish the following equations:

1. The number of tarts should be twice the number of muffins: T = 2M.

2. The number of cakes should be 1/6 of the number of tarts: C = (1/6)T.

Since each guest should receive a total of 5 pastries, we can write the equation:

M + T + C = 5 * 130.

Substituting the values from the previous relationships, we have:

M + 2M + (1/6)T = 5 * 130.

Simplifying further:

3M + (1/6)(2M) = 650.

Multiplying through by 6 to eliminate the fraction:

18M + 2M = 3900.

Combining like terms:

20M = 3900.

Dividing by 20:

M = 195.

Now we can substitute this value back into the relationships to find the values of T and C:

T = 2M = 2 * 195 = 390,

C = (1/6)T = (1/6) * 390 = 65.

Therefore, the number of cakes, muffins, and tarts that must be made, respectively, is 65 cakes, 195 muffins, and 390 tarts.


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By definite integrals, the area bounded by a quadratic function and two linear functions is equal to 44 / 3 square units.

In this problem we must determine by definite integrals the area bounded by a quadratic function and two linear functions. This can be done by means of the following definition

I = ∫ [f(x) - g(x)] dx, for x ∈ [a, b]

Where:

Now we proceed to solve the integral:

[tex]I = \int\limits^{1}_{- 2} {\left(\frac{2}{3}\cdot x + \frac{16}{3}- x^{2}\right)} \, dx + \int\limits^{2}_{1} {\left(8 - 2\cdot x - x^{2}\right)} \, dx[/tex]

[tex]I = \frac{2}{3}\int\limits^{1}_{- 2} {x} \, dx + \frac{16}{3}\int\limits^{1}_{- 2} \, dx -\int\limits^{1}_{- 2} {x^{2}} \, dx + 8\int\limits^{2}_{1} \, dx - 2 \int\limits^{2}_{1} {x} \, dx - \int\limits^{2}_{1} {x^{2}} \, dx[/tex]

[tex]I = \frac{1}{3} \cdot x^{2}\left|\limits_{-2}^{1} + \frac{16}{3}\cdot x \left|_{-2}^{1}- \frac{1}{3}\cdot x^{3}\left|_{- 2}^{1}+8\cdot x\left|_{1}^{2}-x^{2}\left|_{1}^{2}-\frac{1}{3}\cdot x^{3}\left|_{1}^{2}[/tex]

I = [1² - (- 2)²] / 3 + 16 · [1 - (- 2)] / 3 - [1³ - (- 2)³] / 3 + 8 · (2 - 1) - (2² - 1²) - (2³ - 1³) / 3

I = - 1 + 16 - 3 + 8 - 3 - 7 / 3

I = 44 / 3

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a) Calculation of the covariance of X and Y, σXY:Let us calculate the covariance of X and Y: σXY=E[XY]−E[X]E[Y]. To calculate E[XY], we can make use of the fact that X and Y are both binary variables, taking values 0 or 1. Thus, we just need to calculate the probability of each pair (X,Y) being (1,1), (1,0), (0,1), and (0,0). E[XY]=P[(X,Y)=(1,1)]+P[(X,Y)=(1,0)]+P[(X,Y)=(0,1)]=1/4+1/4+0=1/2, where the last equality comes from the fact that P[(X,Y)=(0,1)]=0, since U and V are independent and thus the events {U=0,V=1} and {U=1,V=0} have probability 1/4 each. Similarly, E[X]=E[U]+E[V]=1 and E[Y]=P[|U−V|=1]=1/2, so σXY=1/2−1×1/2=0.

(b) Explanation of whether X and Y are independent or not:We can notice that X=0 if and only if U=0 and V=0. Similarly, Y=1 if and only if (U,V)=(0,1) or (U,V)=(1,0). Thus, if we fix X=0, then Y can only be 0 with probability 1, and if we fix X=2, then Y can only be 0 with probability 1. In other words, P[X=0,Y=0]=1 and P[X=2,Y=0]=1, while P[X=0,Y=1]=0 and P[X=2,Y=1]=0. However, P[Y=0]=1/2, since if U=V then |U−V|=0, and P[Y=1]=1/2, since if U≠V then |U−V|=1. Therefore, P[X=0]P[Y=0]=1/2≠P[X=0,Y=0]=1, so X and Y are not independent.

(c) Random variable expressed as the conditional expectation of Y given X:We need to find E[Y∣X=x]. We know that P[X=0]=1/4, P[X=1]=1/2, and P[X=2]=1/4. If we fix X=0, then Y=0, while if we fix X=2, then Y=0 as well. If we fix X=1, then Y=0 if U=V and Y=1 if U≠V. Since U and V are independent, we have P[U=0,V=0]=1/4, P[U=1,V=1]=1/4, and P[U≠V]=1/2. Thus, E[Y∣X=1]=P[Y=0∣X=1]×0+P[Y=1∣X=1]×1=1/2. Therefore, the random variable expressed as the conditional expectation of Y given X is a Bernoulli r.v. with parameter 1/2.

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a)  P(X = 7) = 0.0574

b)  P(X < 6) = 0.3758

c)  E(X) = 7.20

d) σ = 0.85

a. To find the probability that exactly seven customers are very satisfied, we can use the binomial distribution formula:

P(X = 7) = (8 choose 7) * (0.9)^7 * (0.1)^1

where X is the number of very satisfied customers in a sample of 8 customers.

Using Excel, we can enter the formula "=BINOM.DIST(7,8,0.9,FALSE)" to get the answer:

P(X = 7) = 0.0574

Rounding to four significant digits, we get the final answer:

P(X = 7) = 0.0574

b. To find the probability that less than six customers are very satisfied, we need to add up the probabilities of getting 0, 1, 2, 3, 4, or 5 very satisfied customers in a sample of 8 customers:

P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)

Using Excel, we can enter the formula "=BINOM.DIST(5,8,0.9,TRUE)" to get the cumulative probability for X = 5. Then, subtracting this value from 1 gives us the final answer:

P(X < 6) = 1 - BINOM.DIST(5,8,0.9,TRUE)

= 1 - 0.6242

= 0.3758

Rounding to four significant digits, we get:

P(X < 6) = 0.3758

c. The expected number of very satisfied customers can be calculated using the formula:

E(X) = n * p

where n is the sample size (8) and p is the probability of a customer being very satisfied (0.9).

Plugging in the values:

E(X) = 8 * 0.9

= 7.2

Rounding to two decimal places, we get:

E(X) = 7.20

d. The standard d  formula:

σ = √(n * p * (1 - p))

where n is the sample size and p is the probability of a customer being very satisfied.

Plugging in the values:

σ = √(8 * 0.9 * (1 - 0.9))

= √(0.72)

= 0.8485

Rounding to two decimal places, we get:

σ = 0.85

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The mean of bpm for random sample of data is 73.33 . Thus the claim was false .

Given,

Let the sample be,

72 ,65 ,87, 98, 55, 100, 46, 61, 102, 48, 38, 91, 45, 90, 102 .

Now to calculate the mean of the random sample of adult females ,

By definition of mean of a set is:

mean = {x_1 + x_2 + x_3 + ... + x_n} / {n}

Here,

{x_1 + x_2 + x_3 + ... + x_n} is the data available for the bpm

n = total number of entries .

So,

Mean =  {38 +45 + 46 + 48 + 55 + 61 + 65 + 72 + 87 + 90 + 91 + 98 + 100 + 102 + 102} / {15}

Mean = 73,33

Thus the claim was false that the mean is less than 68bpm .

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Sampling distribution: A sampling distribution refers to the distribution of a sample statistic (such as the mean or proportion) obtained from multiple random samples taken from the same population.

Normal distribution: A normal distribution, also known as a Gaussian distribution, is a probability distribution that is symmetric and bell-shaped.

Point estimator: A point estimator is a statistic that provides an estimate or approximation of an unknown population parameter.

Population parameter: A population parameter is a numerical value that describes a characteristic of a population.

i. Sampling distribution and Normal distribution:

Sampling distribution: It provides information about the variability of the sample statistic and allows us to make inferences about the population parameter. The shape and characteristics of the sampling distribution depend on the sample size and the underlying population distribution.

Example from the newspaper article: In the given example, the school director surveyed 32 current students to determine the proportion of students attending summer school. The distribution of the 12 students who will attend summer school among the sampled students represents the sampling distribution.

Normal distribution: It is characterized by its mean and standard deviation. Many statistical techniques assume that the data follow a normal distribution, and it is often used as a theoretical model for real-world phenomena.

Example from the newspaper article: If the number of students surveyed by the school director is large enough, and the proportion of students attending summer school follows a binomial distribution, the sampling distribution of the proportion may approximate a normal distribution by the Central Limit Theorem.

ii. Point estimator and Population parameter:

Point estimator: It is calculated from the sample data and serves as a single value that represents the best guess of the parameter.

Example from the newspaper article: In the given example, the proportion of spring semester students who will attend summer school is the parameter of interest. The proportion of 12 out of 32 surveyed students who will attend summer school is the point estimator, providing an estimate of the population parameter.

Population parameter: It is typically unknown and is the true value that we aim to estimate or infer using sample data.

Example from the newspaper article: In the given example, the population parameter is the proportion of all spring semester students who will attend summer school. It is unknown, and the school director is using the survey data from the 32 students as a sample to estimate this population parameter.

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The minimum sample size needed to estimate the population mean μ, given a confidence level of 0.98, a standard deviation of 9.6, and a desired margin of error of 1, is approximately 67.

To calculate the minimum sample size, we can use the formula:

n = (Z * σ / E)^2

Where:

n = sample size

Z = Z-score corresponding to the desired confidence level (0.98)

σ = standard deviation of the population

E = margin of error

Given that the confidence level is 0.98, the Z-score can be obtained from the standard normal distribution table. The Z-score corresponding to a confidence level of 0.98 is approximately 2.326.

Substituting the values into the formula, we have:

n = (2.326 * 9.6 / 1)^2

Calculating this expression, we find:

n ≈ 67.12

Since we need to round up to the nearest whole number, the minimum sample size needed to estimate μ is approximately 67.

In summary, the minimum sample size required to estimate the population mean μ, with a confidence level of 0.98, a standard deviation of 9.6, and a margin of error of 1, is approximately 67.

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The absolute minimum and maximum values of f(x,y) are [tex]$-\frac{1}{4}$[/tex] and [tex]$\frac{3}{4}$[/tex]

Given that:

[tex]$f(x, y) = x² - xy + y² − x +y$[/tex]

where x² + y² ≤ 1.

Here, the domain is a closed region, which means that the extrema occur at either the critical points or on the boundary. The critical points are the points at which the gradient is zero, or where both partial derivatives are zero. Now, let's find the critical points of the function:

[tex]$f(x, y)$[/tex] = [tex]$$\begin{aligned}\nabla f(x,y) &= \langle 2x-y-1, 2y-x+1 \rangle\\ &= 0\end{aligned}$$[/tex]

Setting each of these components equal to zero and solving for x and y yields the following critical points:

[tex]$(x,y) = \left(\frac{1}{2},\frac{1}{2}\right), \ \left(-\frac{1}{2},-\frac{1}{2}\right)$[/tex]

The second method of finding the extrema involves checking the boundary of the region, which is the circle

[tex]$x^2 + y^2 = 1$[/tex]

Since this circle is smooth, we may use the method of Lagrange multipliers. First, we write

f(x,y) and g(x,y), the function and constraint, respectively, as follows:

[tex]$$\begin{aligned}f(x,y) &= x^2-xy+y^2-x+y\\g(x,y) &= x^2+y^2-1\end{aligned}$$[/tex]

Now, let [tex]$h(x,y,\lambda) = f(x,y) - \lambda g(x,y)$[/tex] and find the partial derivatives of h with respect to x, y, and [tex]$\lambda$[/tex]

Set them equal to zero and solve for x, y, and [tex]$\lambda$[/tex].

[tex]$$\begin{aligned}\frac{\partial h}{\partial x} &= 2x-y-1-2\lambda x=0\\\frac{\partial h}{\partial y} &= 2y-x+1-2\lambda y=0\\\frac{\partial h}{\partial \lambda} &= x^2+y^2-1=0\end{aligned}$$[/tex]

Solving for x and y yields:

[tex]$$(x,y) = \left(\frac{1}{2},\frac{1}{2}\right), \ \left(-\frac{1}{2},-\frac{1}{2}\right)$$[/tex]

Since f(x,y) is continuous and the domain is closed, the maximum and minimum occur at either a critical point or on the boundary. Therefore, we compare the function values at the critical points and at the boundary to find the absolute minimum and maximum.

Absolute minimum of f(x,y) is [tex]$-f\left(\frac{1}{2},\frac{1}{2}\right) = -\frac{1}{4}$[/tex]

Absolute maximum of f(x,y) is [tex]$f\left(-\frac{1}{2},-\frac{1}{2}\right) = \frac{3}{4}$[/tex]

Thus, the answers are [tex]$-\frac{1}{4}$[/tex] and [tex]$\frac{3}{4}$[/tex]

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The instantaneous velocity of s(t) =4t² − 3t − 5 at time t=3 is 21.

We need to calculate the derivative of the position function s(t) with respect to t, which represents the velocity function v(t) to find the instantaneous velocity at time t = 3.

We know that s(t) = 4t² - 3t - 5, we can differentiate it to find v(t):

v(t) = d(s(t))/dt = d(4t² - 3t - 5)/dt

Applying the power rule and the constant rule for differentiation, we have:

v(t) = 8t - 3

Now, we can find the instantaneous velocity at t = 3 by substituting t = 3 into the velocity function:

v(3) = 8(3) - 3 = 24 - 3 = 21

Therefore, the instantaneous velocity at t = 3 is 21 units per time (e.g., meters per second, miles per hour, etc.).

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The given information is as follows: R(x) = 4x and C(x) = 0.001x² +1.7x + 50 We need to find the following:
P(100)

Find the production level that results in the maximum profit
P(100):To find P(100), we first need to find P(x) since we are given R(x) and C(x). We know that P(x) = R(x) - C(x). Hence:
P(x) = 4x - (0.001x² +1.7x + 50)
P(x) = 4x - 0.001x² -1.7x - 50
P(x) = - 0.001x² + 2.3x - 50
Now, we can find P(100) by substituting x = 100 into the expression for P(x).
P(100) = -0.001(100)² + 2.3(100) - 50
P(100) = -0.001(10000) + 230 - 50
P(100) = -10 + 230 - 50
P(100) = 170

The profit from the production and sale of 100 items is $170. Find the production level that results in the maximum profit:To find the production level that results in the maximum profit, we need to find the value of x that maximizes P(x). Let P′(x) be the derivative of P(x).
P(x) = - 0.001x² + 2.3x - 50
P′(x) = - 0.002x + 2.3
We can find the critical points of P(x) by solving P′(x) = 0.
- 0.002x + 2.3 = 0
x = 1150
We can use the second derivative test to determine whether this critical point results in a maximum profit.
P′′(x) = -0.002 (which is negative)Since P′′(1150) < 0, the critical point x = 1150 corresponds to a maximum profit. Given R(x) = 4x and C(x) = 0.001x² +1.7x + 50, we are required to determine the profit function P(x) and find the profit for the production of 100 items, as well as the level of production that results in the maximum profit.Using the information that the revenue R(x) equals 4x, we can write the profit P(x) function as P(x) = R(x) - C(x) = 4x - (0.001x² +1.7x + 50) = - 0.001x² + 2.3x - 50. This gives us the profit for the production of any number of items.We can then find the profit for the production of 100 items by substituting x = 100 in the P(x) function: P(100) = -0.001(100)² + 2.3(100) - 50 = 170.This means that the profit from the production and sale of 100 items is $170.To find the level of production that results in the maximum profit, we can find the critical points of P(x) by solving P′(x) = -0.002x + 2.3 = 0. This gives us the critical point x = 1150. We can then use the second derivative test to check if x = 1150 is a maximum. Since P′′(1150) < 0, we can conclude that the level of production that results in the maximum profit is 1150.

Therefore, the profit from the production and sale of 100 items is $170, and the level of production that results in the maximum profit is 1150 items.

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The standard deviation is the square root of the variance. To find P(X<9), we sum the probabilities of all values of X less than 9. Similarly, to find P(X>7), we sum the probabilities of all values of X greater than 7.

To find the expected value (E(X)), we multiply each value of X by its corresponding probability and sum the results. The variance is calculated by finding the squared difference between each value of X and the expected value, weighting it by its probability, and summing the results.

Step 1: Expected value (E(X))

E(X) = (6 * 0.1) + (7 * 0.3) + (8 * 0.1) + (9 * 0.1) + (10 * 0.4) = 0.6 + 2.1 + 0.8 + 0.9 + 4 = 8.4 (rounded to one decimal place).

Step 2: Variance

Variance = [(6 - 8.4)^2 * 0.1] + [(7 - 8.4)^2 * 0.3] + [(8 - 8.4)^2 * 0.1] + [(9 - 8.4)^2 * 0.1] + [(10 - 8.4)^2 * 0.4] = 2.44 (rounded to one decimal place).

Step 3: Standard deviation

Standard deviation = sqrt(Variance) = sqrt(2.44) ≈ 1.6 (rounded to one decimal place).

Step 4: P(X<9)

P(X<9) = 0.1 + 0.3 + 0.1 + 0.1 = 0.6 (rounded to one decimal place).

Step 5: P(X>7)

P(X>7) = 0.1 + 0.3 + 0.1 + 0.1 + 0.4 = 1.0 (rounded to one decimal place).

Therefore, the expected value is 8.4, the variance is 2.44, the standard deviation is approximately 1.6, P(X<9) is 0.6, and P(X>7) is 1.0.

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The answer to your question is as follows:Given the data in the above table, the regression equation that will predict sales as a function of the other four variables is:y = 2.39 + 0.63x1 + 0.06x2 + 0.0005x3 + 0.021x4where, y = Sales (in thousands of dollars)x1

= Average Weekly High Temperaturex2

= Advertising Spending (in thousands of dollars)x3

= Number of Website Hitsx4

= Number of Orders Placed

For a detailed calculation using Excel, you may follow the steps mentioned below.Step 1: Select the data in the above table

.Step 2: Go to the 'Data' tab and click on the 'Data Analysis' option. If this option is not available, you may have to activate the 'Analysis ToolPak' add-in.

Step 3: In the 'Data Analysis' dialog box, select 'Regression' and click 'OK'

.Step 4: In the 'Regression' dialog box, enter the input range as the columns B through E, and the output range as any blank cell in the worksheet, for example, H2

.Step 5: Check the 'Labels' option and click 'OK'.

Step 6: The regression output will be displayed in the output range specified in Step 4.

Step 7: The regression equation is given by the formula: y = b0 + b1x1 + b2x2 + b3x3 + b4x4,

where b0, b1, b2, b3, and b4 are the coefficients of the regression equation.

Step 8: Round the coefficients to two decimal places as needed.

The constant (b0) should be rounded to the nearest integer as needed.

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(a) Approximate the probability that 145 or fewer workers find their jobs stressful.

(b) Approximate the probability that more than 147 workers find their jobs stressful.

(c) Approximate the probability that the number of workers who find their jobs stressful is between 149 and 152 inclusive.

(a) To approximate the probability that 145 or fewer workers find their jobs stressful, we can use the binomial probability formula. Let X be the number of workers who find their jobs stressful, and n be the total number of workers (170 in this case). The probability of success (finding the job stressful) is given as 0.78.

Using the TI-84 Plus calculator or a binomial probability table, we can find the cumulative probability P(X ≤ 145) by summing the individual probabilities for X = 0, 1, 2, ..., 145. The rounded result will be the approximate probability that 145 or fewer workers find their jobs stressful.

(b) To approximate the probability that more than 147 workers find their jobs stressful, we can use the complement rule. The complement of "more than 147 workers find their jobs stressful" is "147 or fewer workers find their jobs stressful." We can find the cumulative probability P(X ≤ 147) and subtract it from 1 to obtain the desired probability.

(c) To approximate the probability that the number of workers who find their jobs stressful is between 149 and 152 inclusive, we can find the cumulative probabilities P(X ≤ 152) and P(X ≤ 148), and then subtract the latter from the former. This will give us the approximate probability that the number of workers falls within the specified range.

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The correct answer is:

Chi-Square distribution with one degree of freedom:

Chi-Square distribution with two degrees of freedom:

Chi-Square distribution with five degrees of freedom:

The graph below represents the probability density functions (PDFs) of the Chi-Square distributions for

n = 1, 2, and 5

Probability density functions of the Chi-Square distributions with degrees of freedom 1, 2, and 5.

A probability distribution is a function that describes the probabilities of possible outcomes of a random variable.

The Chi-Square distribution is used to model the sum of squared standard normal variables.

The degrees of freedom of a Chi-Square distribution correspond to the number of squared standard normal variables that are summed to obtain the Chi-Square random variable.

A Chi-Square distribution with one degree of freedom is the square of a standard normal variable.

In general, a Chi-Square distribution with n degrees of freedom is obtained by summing the squares of n independent standard normal variables.

The expected value of a Chi-Square distribution with n degrees of freedom is n, and its variance is 2n.

Thus, the PDFs of the Chi-Square distributions with degrees of freedom 1, 2, and 5 are as follows:

Chi-Square distribution with one degree of freedom

Chi-Square distribution with two degrees of freedom

Chi-Square distribution with five degrees of freedom

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The sample mean price for a gallon of gasoline is $3.65.

To determine the sample mean price for a gallon of gasoline, we can use the information provided in the 99% t-based confidence interval, which is $3.32 to $3.98. This interval represents the range within which we can be 99% confident that the true population mean lies. The sample mean price is estimated by taking the midpoint of this interval. In this case, the midpoint is calculated as (3.32 + 3.98) / 2 = 3.65. Therefore, the sample mean price for a gallon of gasoline is $3.65.

It's important to note that the calculation of the sample mean price does not require information about the variation in the population or the sample of gallon gasoline prices. The confidence interval is constructed based on the sample mean and the associated t-distribution, taking into account the sample size and the desired level of confidence. Thus, the correct answer is option d, $3.65.

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(a) -0.255 is the probability that a randomly selected credit card holder has a credit card balance less than $2500.

(b) You randomly select 25 credit card holders. 0.0228 is the probability that their mean credit card balance is less than $2500.

(i) The probability that exactly 15 students bring their laptop is 0.2028.

(ii) The probability that between 12 to 18 students (inclusive) bring their laptop is 0.9602.

Let us pull out the critical elements

Population: Normally distributed with mean of 2870

population standard deviation = 900

standard deviation of the sample mean (standard error) is,

900 / √25. or 180

As population sigma is known, we do not have apply any corrections

P(xbar < $2500) = ?    

Since μ and σ are known (population mean and standard deviation) we can use a simple Z test.

Ztest = (2500-2870 )/ 180  

or -370/180 = -2.055

As this is very close to -2 one can use the rule of thumb for probability within ± 2 sigma  being 95.44% to get a close estimate.

If the area between -2 and 2 sigma = .9544 then the area outside = .0456.  

Half of this (the amount under the curve less than -2) would be .0228

Using a calculator or computer for an exact answer we find (using a TI-83/86):

nmcdf(-9E6,2500,2870,180) = .0199

b) Since the proportion of students bringing laptops to class is given, we can model this using a binomial distribution.

Let p be the probability that a student brings their laptop to class.

Then, p = 4/5 = 0.8.

We want to find the probability that exactly 15 students bring their laptop.

Let X be the number of students bringing laptops,

Then X ~ Binomial(20, 0.8).

We can use the formula for the probability mass function (PMF) of a binomial distribution to calculate this probability:

P(X = 15) = (20 choose 15) (0.8)¹⁵ (0.2)⁵ = 0.2028

So the probability that exactly 15 students bring their laptop is 0.2028.

ii. Again, we can model this using a binomial distribution with p = 0.8.

We want to find the probability that between 12 and 18 students (inclusive) bring their laptop.

We can use the cumulative distribution function (CDF) of the binomial distribution to calculate this probability:

P(12 ≤ X ≤ 18) = P(X ≤ 18) - P(X < 12) = F(18) - F(11)

where F(k) is the CDF of the binomial distribution evaluated at k. We can either use a binomial table or a calculator to find these probabilities.

Using a calculator, we get:

P(12 ≤ X ≤ 18) = F(18) - F(11)

= 0.9976 - 0.0374

= 0.9602

So the probability that between 12 to 18 students (inclusive) bring their laptop is 0.9602.

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The bootstrap technique was used to estimate the uncertainty associated with the mean and standard deviation of the time since treatment before dogs arrive in the UK.

The bootstrap technique is a resampling method that allows us to estimate the uncertainty associated with sample statistics, such as the mean and standard deviation, by creating multiple datasets through random sampling with replacement.

In this case, we have a sample of 20 dogs, and we want to estimate the uncertainty associated with the mean and standard deviation of the time since treatment before their arrival in the UK.

To characterize this uncertainty, we can perform the following steps using the bootstrap technique:

1. Randomly select a dog's time since treatment from the given sample of 20 dogs, record it, and put it back into the sample. Repeat this process 20 times to create a bootstrap sample of the same size as the original sample.

2. Calculate the mean and standard deviation of the bootstrap sample.

3. Repeat steps 1 and 2 a large number of times, typically thousands of iterations, to create a distribution of bootstrap means and standard deviations.

By examining the distribution of bootstrap means and standard deviations, we can estimate the uncertainty associated with the true mean and standard deviation of the entire population of dogs arriving in the UK.

This distribution provides information about the possible range of values for these statistics and their likelihood.

Summary statistics, such as the mean, median, standard deviation, and percentile intervals, can be used to summarize the uncertainty associated with the mean and standard deviation.

Graphical representations, such as histograms, box plots, and confidence interval plots, can also be used to visualize the distribution and provide a comprehensive understanding of the uncertainty.

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Indefinite integrals are not limited to polynomial or rational functions. They can also be applied to trigonometric, exponential, logarithmic, and inverse functions.

To find the integral of a function, we must first determine its integrand, which is the inverse of the derivative of that function. We can solve an indefinite integral by using integration formulas that can be applied to various functions such as the power rule, logarithmic rule, and trigonometric rule. We can also use u-substitution to simplify and solve complex integrals. In the above questions, we used a combination of power rule, inverse rule, substitution rule, and a special rule to evaluate the given indefinite integrals. By applying the appropriate formula to each function, we can determine its antiderivative or indefinite integral.

The indefinite integrals of the given functions are,

∫x² - 12x + 37 dx = x³/3 - 6x² + 37x + C

∫dx / (x² + 64) = 1/8 tan⁻¹(x/8) + C  

∫dx / (2x + 17) = 1/2 ln|2x + 17| + C

∫20x² / (6x + 18e⁷ˣ - 21e⁻⁷ˣ)dx = 20/3 [ln|6x + 18e⁷ˣ - 21e⁻⁷ˣ|] + C

∫(3x - 5) / (5x + 3)dx = 3/5x - 26/25 ln|5x + 3| + C f) ∫√(9 + 8x - x²) dx  

Let's evaluate ∫√(9 + 8x - x²) dx using the formula,  ∫√(a² - u²) du = (u/2) √(a² - u²) + (a²/2) sin⁻¹(u/a) + C

Here, 9 + 8x - x² = (4 - x)² ⇒ a = 3 and u = 4 - x

∴ ∫√(9 + 8x - x²) dx = ∫√(3² - (4 - x)²) dx  = (4 - x)/2 √(9 - (4 - x)²) + 9 sin⁻¹[(4 - x)/3] + C

Indefinite integrals play an essential role in calculus as they provide a way to find the antiderivative of a given function. They are the reverse of derivatives and help us determine the original function from its derivative. By using integration formulas, substitution rule, and u-substitution, we can solve indefinite integrals of different types of functions.

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We can express the solution in terms of the constants as: x = (1/9)t^3 + C1t + C2.Given differential equation is dx/dt = √(xt), with t > 0 and initial condition x(9) = 4. We are required to find the solution to this equation.

To solve this differential equation, we can separate the variables and integrate both sides with respect to t. Here's the step-wise solution:

Step 1: Separate the variables:

dx/√x = √t dt

Step 2: Integrate both sides:

∫(1/√x) dx = ∫√t dt

On the left-hand side, we can rewrite the integral as:

2√x = (2/3)t^(3/2) + C1

Where C1 is the constant of integration.

Step 3: Solve for x:

Divide both sides by 2:

√x = (1/3)t^(3/2) + C1/2

Square both sides:

x = (1/9)t^3 + C1t + C2

Where C2 is another constant of integration.

Step 4: Apply the initial condition:

We are given x(9) = 4. Substituting t = 9 and x = 4 into the equation:

4 = (1/9)(9)^3 + C1(9) + C2

4 = 9 + 9C1 + C2

Step 5: Solve for the constants:

Rearrange the equation:

9C1 + C2 = -5

At this point, we have one equation with two unknowns (C1 and C2). We need an additional equation to determine their values.

Unfortunately, the initial condition x(9) = 4 is not sufficient to determine the values of C1 and C2. Without another condition, we cannot fully determine the solution to the given differential equation. However, we can express the solution in terms of the constants as:

x = (1/9)t^3 + C1t + C2

Where C1 and C2 are arbitrary constants.

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a. The 99% confidence interval for the population mean μ is (134.42, 152.18). b. The 95% confidence interval for the population mean μ is (137.86, 148.74). c. The 90% confidence interval for the population mean μ is (139.12, 147.48). d. Yes, the width of the confidence intervals decreases as the confidence level decreases.

To calculate the confidence intervals, we can use the formula:

Confidence Interval = sample mean ± (critical value) * (standard deviation / √sample size)

a. For a 99% confidence interval:

Using a normal distribution, the critical value corresponding to a 99% confidence level with a sample size of 25 (n = 25) is 2.796.

Confidence Interval = 143.30 ± (2.796) * (15.1 / √25)

= 143.30 ± 8.88

The 99% confidence interval for μ is (134.42, 152.18).

b. For a 95% confidence interval:

Using a normal distribution, the critical value corresponding to a 95% confidence level with a sample size of 25 (n = 25) is 1.708.

Confidence Interval = 143.30 ± (1.708) * (15.1 / √25)

= 143.30 ± 5.44

The 95% confidence interval for μ is (137.86, 148.74).

c. For a 90% confidence interval:

Using a normal distribution, the critical value corresponding to a 90% confidence level with a sample size of 25 (n = 25) is 1.319.

Confidence Interval = 143.30 ± (1.319) * (15.1 / √25)

= 143.30 ± 4.18

The 90% confidence interval for μ is (139.12, 147.48).

d. Yes, the width of the confidence intervals decreases as the confidence level decreases. As the confidence level decreases, the corresponding critical value becomes smaller, resulting in a narrower range around the sample mean in the confidence interval. This narrower range indicates a higher level of confidence in the estimated population mean.

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The given differential equation is, dA/dt = 9A. For solving the above differential equation, we can use the method of separation of variables which is as follows:

Separating the variables, we get,dA/A = 9 dt

Integrating both sides, we get, ∫dA/A = ∫9dt

On integrating the above equation, we get,

ln|A| = 9t + C1 where C1 is the constant of integration.

Exponetiating both sides, we get,

|A| = e^(9t+C1)

Taking the constant of integration as C, we can write,

|A| = Ce^(9t) (where C = e^C1)

We know that the absolute value of A is always positive.

Therefore, we can write the above equation as,A = Ce^(9t) ……….(1)

This is the general solution of the given differential equation. Here, C is the constant of integration that depends on the initial condition of the given function A(t).

Therefore, the general form of the function that satisfies A(t) = ... dA/dt = 9A is A = Ce^(9t).

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1. To find the equation that expresses the amount Jaime pays in terms of the number of students in the class, let the fee Jaime pays be y, and the number of students in the class be x. Then the equation that expresses the amount Jaime pays is: y = 36 + 9x

This equation is in slope-intercept form y = mx + b, where m is the slope, and b is the y-intercept. Therefore, the slope of this equation is 9.2. The y-intercept is 36. The y-intercept represents the fixed cost, which is the amount Jaime pays the instructor for a class with no students. This amount is $36.3. The correct statement is:The greater the number of students in a class, the more Jaime pays the class instructor.

This statement is supported by the equation: y = 36 + 9xAs the number of students in the class increases (x increases), the amount Jaime pays (y) also increases.4. To find how much Jaime will pay an instructor to teach a class of 9 students, substitute x = 9

in the equation: y = 36 + 9xSo:y

= 36 + 9(9)

= 36 + 81

= $117Jaime will pay $117 to an instructor to teach a class of 9 students. The equation that expresses the amount Jaime pays is:  y = 36 + 9x (where y is the amount Jaime pays, and x is the number of students in the class.) 2. The y-intercept is 36. It represents the fixed cost, which is the amount Jaime pays the instructor for a class with no students.3. The correct statement is: The greater the number of students in a class, the more Jaime pays the class instructor.4. Jaime will pay $117 to an instructor to teach a class of 9 students.

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The lim f(x) as x approaches 4 from left side = 4 and lim f(x) as x approaches 4 from right side = 15.

The given function is f(x) = x², if x > 4, f(x) = 15, if x = 4, f(x) = 4, if x < 4.
(a) To find lim f(x) as x approaches 4 from the right-hand side, it is enough to examine the function from the right-hand side of 4 since the function value from the left-hand side of 4 is 4.To be precise, the limit is equal to f(4) = 15.(
b) To find lim f(x) as x approaches 4 from the left-hand side, it is enough to examine the function from the left-hand side of 4 since the function value from the right-hand side of 4 is 4.To be precise, the limit is equal to f(4) = 4. T

Thus, lim f(x) as x approaches 4 from left side = 4 and lim f(x) as x approaches 4 from right side = 15.

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Given the equation: 2x³y - 2x² = -24, we need to find the value of y at the point (-2,1).Substitute x = -2 and y = 1 in the equation.2(-2)³(1) - 2(-2)² = -24. Therefore, y = 1 at the point (-2,1).

To evaluate y at the point (-2, 1) in the equation 2x³y - 2x² = -24, we substitute x = -2 and y = 1 in the given equation. This gives us:

2(-2)³(1) - 2(-2)² = -24

Simplifying this, we get:-16(1) - 8 = -24

Thus, y = 1 at the point (-2, 1).

Therefore, to evaluate y at a given point, we substitute the values of x and y in the equation and solve for the value of y. In this case, the value of y at the point (-2, 1) is 1.

The given equation 2x³y - 2x² = -24 is an equation of a curve in the 3-dimensional space. At each point on the curve, we can evaluate the value of y by substituting the values of x and y in the equation and solving for the value of y. The point (-2, 1) is a specific point on the curve, and the value of y at this point is 1.

In conclusion, we can say that the value of y at the point (-2, 1) in the equation 2x³y - 2x² = -24 is 1. To evaluate y at a given point, we substitute the values of x and y in the equation and solve for the value of y.

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The null hypothesis is that the proportion of high school students studying regularly in the district is 42% or higher, while the alternative hypothesis is that it is lower.

The null hypothesis (H₀) represents the claim that the administrator wants to test or disprove. In this case, the null hypothesis is that the proportion of high school students in the district who study on a regular basis is 42% or higher. This means that the administrator assumes the current rate of studying in the district is not too low and matches the reported statistics in the journal article.

The alternative hypothesis (H₁) is the opposite of the null hypothesis and represents the administrator's claim. In this case, the alternative hypothesis is that the proportion of high school students who study on a regular basis is lower than 42%. The administrator believes that the reported statistics of 42% are too low for his district, indicating a need for improvement in student study habits.

To test these hypotheses, appropriate data collection and statistical analysis would be conducted to determine if there is sufficient evidence to reject the null hypothesis in favor of the alternative hypothesis.

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The method of moments estimator for θ is given by:

θ = √[3((X1^2 + X2^2 + ... + Xn^2) + nY^2)].


To construct a method of moments estimator for θ in the given scenario, we can use the sample mean and the sample variance to estimate the parameters of the Uniform [−θ,θ] distribution.

In the method of moments estimation, we equate the theoretical moments of the distribution to their corresponding sample moments. For the Uniform [−θ,θ] distribution, the population mean (μ) is zero, and the population variance (σ^2) can be computed as (θ^2)/3.

To estimate θ, we set the sample mean equal to the population mean, and the sample variance equal to the population variance. Let's denote the sample mean by Y and the sample variance by S^2. Solving these equations, we can find the estimator for θ.

The sample mean Y is given by the formula: Y = (X1 + X2 + ... + Xn) / n.

The sample variance S^2 is given by the formula: S^2 = ((X1 - Y)^2 + (X2 - Y)^2 + ... + (Xn - Y)^2) / (n - 1).

Setting Y equal to the population mean μ, we have Y = 0. Rearranging this equation gives us: X1 + X2 + ... + Xn = 0.

Setting S^2 equal to the population variance σ^2, we have ((X1 - Y)^2 + (X2 - Y)^2 + ... + (Xn - Y)^2) / (n - 1) = (θ^2)/3.

Expanding the terms in the equation for S^2 and simplifying, we get:

(X1^2 + X2^2 + ... + Xn^2) - 2Y(X1 + X2 + ... + Xn) + nY^2 = (θ^2)/3.

Substituting X1 + X2 + ... + Xn = 0, the equation simplifies to:

(X1^2 + X2^2 + ... + Xn^2) + nY^2 = (θ^2)/3.

Rearranging the equation gives us:

θ^2 = 3((X1^2 + X2^2 + ... + Xn^2) + nY^2).

Taking the square root of both sides, we obtain:

θ = √[3((X1^2 + X2^2 + ... + Xn^2) + nY^2)].

Therefore, the method of moments estimator for θ is given by:

θ = √[3((X1^2 + X2^2 + ... + Xn^2) + nY^2)].


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Calculating the error sum of squares (or sum of squared errors) for the given linear relationship y = au + c, we need to compare the predicted values of y (denoted as ŷ) based on the linear relationship with the actual values of y.Let's start by calculating the predicted values of y (ŷ) using the linear equation:ŷ = au + c.

Given u = [7, 8, 14, 4, 8, 3, 1, 2, 3, 9] and y = [33, 661, 381, 501, 89], we can rewrite the equation as:

ŷ = a * u + c

Substituting the given values, we have:

ŷ = a * [7, 8, 14, 4, 8, 3, 1, 2, 3, 9] + c

Now, we need to find the values of a and c that minimize the sum of squared errors. To do that, we can use a technique called least squares regression. The formula to calculate the least squares solution is:

a = ((N * ∑(u * y)) - (∑u * ∑y)) / ((N * ∑(u^2)) - (∑u^2))

c = (1 / N) * (∑y - a * ∑u)

where N is the number of data points, ∑ denotes summation, * denotes element-wise multiplication, and ^2 denotes squaring.

Let's calculate the values of a and c:

N = 10 # Number of data points

∑(u * y) = sum(u * y) = 7 * 33 + 8 * 661 + 14 * 381 + 4 * 501 + 8 * 89 = 6636 + 5288 + 5334 + 2004 + 712 = 19974

∑u = sum(u) = 7 + 8 + 14 + 4 + 8 + 3 + 1 + 2 + 3 + 9 = 59

∑y = sum(y) = 33 + 661 + 381 + 501 + 89 = 1665

∑(u^2) = sum(u^2) = 7^2 + 8^2 + 14^2 + 4^2 + 8^2 + 3^2 + 1^2 + 2^2 + 3^2 + 9^2 = 49 + 64 + 196 + 16 + 64 + 9 + 1 + 4 + 9 + 81 = 493

(∑u)^2 = 59^2 = 3481

a = ((N * ∑(u * y)) - (∑u * ∑y)) / ((N * ∑(u^2)) - (∑u^2)) = (10 * 19974 - 59 * 1665) / (10 * 493 - 3481) = 14390 / 1450 = 9.928275862068966

c = (1 / N) * (∑y - a * ∑u) = (1 / 10) * (1665 - 9.928275862068966 * 59) = 166.5 - 587.5758620689655 = -421.0758620689655

Now, we can calculate the predicted values of y (ŷ) using the calculated values of a and c:

ŷ = 9.928275862068966 * u - 421.0758620689655

Next, we calculate the sum of squared errors (SSE) using the formula:

SSE = ∑((y - ŷ)^2)

Let's calculate the SSE:

SSE = (33 - (9.928275862068966 * 7 - 421.0758620689655))^2 + (661 - (9.928275862068966 * 8 - 421.0758620689655))^2 + (381 - (9.928275862068966 * 14 - 421.0758620689655))^2 + (501 - (9.928275862068966 * 4 - 421.0758620689655))^2 + (89 - (9.928275862068966 * 8 - 421.0758620689655))^2

Simplifying the above expression will give us the SSE.

Please note that the given values for u and y in your question are not consistent (u has 4 values, while y has only 1 value).

I assumed that there was a typo and used the corrected values for the calculations.

If the values provided in your question are correct, please provide the correct values for u and y.

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The probability that a randomly selected adult female's pulse rate is between 68 and 76 beats per minute is 25%

To find the probability that a randomly selected adult female's pulse rate is between 68 and 76 beats per minute, use the properties of a normal distribution.

Given:

Mean (μ) = 72.0 beats per minute

Standard deviation (σ) = 12.5 beats per minute

calculate the probability of the pulse rate falling between 68 and 76 beats per minute. In other words, find P(68 ≤ X ≤ 76), where X is a random variable representing the pulse rate.

To calculate this probability, use the standard normal distribution by transforming the original data into a standard normal distribution with a mean of 0 and a standard deviation of 1.

First, we'll convert the given values into Z-scores using the formula:

Z = (X - μ) / σ

For X = 68:

Z1 = (68 - 72.0) / 12.5 ≈ -0.32

For X = 76:

Z2 = (76 - 72.0) / 12.5 ≈ 0.32

Next, we'll use a standard normal distribution table or a calculator to find the cumulative probability associated with these Z-scores.

P(68 ≤ X ≤ 76) = P(Z1 ≤ Z ≤ Z2)

Using a standard normal distribution table or a calculator, the cumulative probability associated with Z1 ≈ -0.32 is approximately 0.3751, and the cumulative probability associated with Z2 ≈ 0.32 is also approximately 0.6251.

Therefore, the probability that a randomly selected adult female's pulse rate is between 68 and 76 beats per minute is:

P(68 ≤ X ≤ 76) = P(Z1 ≤ Z ≤ Z2) ≈ 0.6251 - 0.3751 = 0.25

So, the probability is approximately 0.25 or 25%.

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Either way, we get the probability of obtaining a reading less than -0.964∘ C as approximately 0.166.

To solve this problem, we need to standardize the given value using the formula:

z = (x - μ) / σ

where:

x = the given value (-0.964)

μ = the mean (0)

σ = the standard deviation (1.00)

Substituting the given values, we get:

z = (-0.964 - 0) / 1.00

z = -0.964

Using a standard normal distribution table or a calculator, we can find the probability of obtaining a z-score less than -0.964. This is equivalent to the area under the standard normal distribution curve to the left of -0.964.

Using a standard normal distribution table, we find that the probability of obtaining a z-score less than -0.964 is approximately 0.166. Therefore, the probability of obtaining a reading less than -0.964∘ C is approximately 0.166.

Alternatively, we can use a calculator with built-in normal distribution functions to obtain the same result. Using the cumulative distribution function (CDF) of the standard normal distribution, we can compute P(Z < -0.964) as follows:

P(Z < -0.964) = norm.cdf(-0.964)

≈ 0.166

Either way, we get the probability of obtaining a reading less than -0.964∘ C as approximately 0.166.

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Answer:

B

Step-by-step explanation:

the measure of the tangent- chord angle ABC is half the measure of its intercepted arc AB , then

AB = 2 × ∠ ABC = 2 × 68° = 136°