You collect 100 samples from a large butterfly population. Fifty specimens are dark brown, 20 are speckled, and 30 are white. Coloration in this species of butterfly is controlled by one gene locus: BB individuals are brown, Bb are speckled, and bb are white. What are the allele frequencies for the coloration gene in this population

Answer:

The standard error of the mean is 1.3.

87.64% probability that the sample mean age of the employees will be within 2 years of the population mean age

Step-by-step explanation:

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

Normal probability distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], a large sample size can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation, which is also called standard error [tex]s = \frac{\sigma}{\sqrt{n}}[/tex]

In this problem, we have that:

[tex]\sigma = 8.2, n = 40[/tex]

Computer the standard error of the mean

[tex]s = \frac{8.2}{\sqrt{40}} = 1.3[/tex]

The standard error of the mean is 1.3.

What is the probability that the sample mean age of the employees will be within 2 years of the population mean age

This is the pvalue of Z when [tex]X = \mu + 2[/tex] subtracted by the pvalue of Z when [tex]X = \mu - 2[/tex]. So

[tex]X = \mu + 2[/tex]

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

By the Central Limit Theorem

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

[tex]Z = \frac{\mu + 2 - \mu}{1.3}[/tex]

[tex]Z = 1.54[/tex]

[tex]Z = 1.54[/tex] has a pvalue of 0.9382

-----

[tex]X = \mu - 2[/tex]

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

[tex]Z = \frac{\mu - 2 - \mu}{1.3}[/tex]

[tex]Z = -1.54[/tex]

[tex]Z = -1.54[/tex] has a pvalue of 0.0618

0.9382 - 0.0618 = 0.8764

87.64% probability that the sample mean age of the employees will be within 2 years of the population mean age

Answer:

19.625 feet²

Step-by-step explanation:

Max diameter = 5 feet

Radius = 2.5 feet

Area = 3.14×2.5² = 19.625 feet²

Answer:

The probability that the service desk will have at least 100 customers with returns or exchanges on a randomly selected day is P=0.78.

Step-by-step explanation:

With the weekly average we can estimate the daily average for customers, assuming 7 days a week:

[tex]M=756/7=108[/tex]

We can model this situation with a Poisson distribution, with parameter λ=108. But because the number of events is large, we use the normal aproximation:

[tex]P(\lambda)\approx N(\lambda,\lambda)[/tex]

Then we can calculate the z value for x=100:

[tex]z=\frac{x-\mu}{\sigma}=\frac{100-108}{\sqrt{108}}=\frac{-8}{10.4} =-0.77[/tex]

Now we calculate the probability of x>100 as:

[tex]P(x>100)=P(z>-0.77)=0.78[/tex]

The probability that the service desk will have at least 100 customers with returns or exchanges on a randomly selected day is P=0.78.

Answer:

[tex]A^{-1}=\left[\begin{array}{cc}-3&-4\\1&1\end{array}\right][/tex] message SHOW_ME_THE_MONEY_  

Step-by-step explanation:

The matrix

[tex]A=\left[\begin{array}{cc}1&4\\-1&-3\end{array}\right]\rightarrow |A|=(1 \times -3)-(-1\times 4)=1\\\rightarrow A^{-1}=\left[\begin{array}{cc}-3&-4\\1&1\end{array}\right] \\[/tex]

We can check that in fact A*A^⁻1=I_2 the identity matrix of size 2 x 2.

Now the message was divided in 1 x 2 matrices, then we have that the sequence given is the result of multiplying m by A, so to get m again we multiply now by A^⁻1. and we get the next table

Encoded message    Decoded message    message in letters by association

11 52        19 8    S H

-8 -9         15 23  O W

-13 -39        0 13   _ M

5 20         5 0   E _

12 56        20 8    T H

5 20         5 0    E _

-2 7           13 15  M O

9 41         14 5   N E

25 100       25 0    Y _

Then the message decoded is SHOW_ME_THE_MONEY_                                

Final answer:

To decode the message, first find the inverse of the encoding matrix A, then multiply the encoded message matrices by this inverse. Match the resulting numbers to letters using the given table to reveal the original message.

Explanation:

The question involves the process of decoding a coded message that was encrypted using matrix multiplication with a given encoding matrix. Firstly, we need to decode the message by multiplying the encoded row matrices by the inverse of the encoding matrix A, which is provided as A = [[1 47] [-1 -3]]. Then, we will use the inverse of matrix A to transform the coded row matrices back to their original message form.

Once we have the row matrices that represent our numbers, we match these numbers to the corresponding letters using the given table, ultimately revealing the decoded message.

For example, if we have a decoded row matrix of [3 1], it would correspond to the letters "C" and "A" according to the provided letter to number mapping.

Answer:

Step-by-step explanation:

The detailed steps and appropriate formular is as shown in the attached file.

The answers are :

(a) [tex]\text{Sample Mean is}[/tex] [tex]49[/tex] [tex]\text{and Sample Standard Deviation is}[/tex] [tex]11.7[/tex].

(b) [tex]95\%[/tex][tex]\text{Confidence Interval for True Mean} (36.72, 61.28).[/tex]

(c) [tex]95\% \text{Prediction Interval for a New Sample is} (16.54, 81.46)[/tex]

(a) Calculate the Sample Mean and Sample Standard Deviation

First, let's calculate the sample mean (\(\bar{x}\)) and the sample standard deviation (s).

Given data: [tex]34, 40, 46, 49, 61, 64[/tex]

[tex]\text{Sample Mean} (\(\bar{x}\))[/tex]

[tex]\[\bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i = \frac{1}{6} (34 + 40 + 46 + 49 + 61 + 64) = \frac{294}{6} = 49\][/tex]

Sample Standard Deviation (s)

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

[tex]\[\begin{aligned}s & = \sqrt{\frac{1}{5} ((34-49)^2 + (40-49)^2 + (46-49)^2 + (49-49)^2 + (61-49)^2 + (64-49)^2)} \\& = \sqrt{\frac{1}{5} (225 + 81 + 9 + 0 + 144 + 225)} \\& = \sqrt{\frac{1}{5} \times 684} \\& = \sqrt{136.8} \\& = 1.7\end{aligned}\][/tex]

(b) Determine the Range of the True Mean at [tex]95\%[/tex] Confidence Level

To find the 95% confidence interval for the mean, we use the formula:

[tex]\[\bar{x} \pm t_{\alpha/2, n-1} \frac{s}{\sqrt{n}}\][/tex]

For [tex]\(n = 6\)[/tex], degrees of freedom [tex]= \(n-1 = 5\)[/tex]. Using a t-table, the critical value for [tex]95\%[/tex] confidence and [tex]5[/tex] degrees of freedom is approximately [tex]2.571[/tex]

[tex]\[\begin{aligned}\text{Margin of Error} & = t_{\alpha/2, n-1} \frac{s}{\sqrt{n}} \\& = 2.571 \times \frac{11.7}{\sqrt{6}} \\& = 2.571 \times 4.776 \\& = 12.28\end{aligned}\][/tex]

So, the [tex]95\%[/tex] confidence interval is:

[tex]\[49 \pm 12.28 \Rightarrow (36.72, 61.28)\][/tex]

(c) Prediction Interval ([tex]95\%[/tex] Confidence Level) for a Seventh Sample

The prediction interval for a new observation is calculated using:

[tex]\[\bar{x} \pm t_{\alpha/2, n-1} s \sqrt{1 + \frac{1}{n}}\][/tex]

[tex]\[\begin{aligned}\text{Prediction Interval} & = 49 \pm 2.571 \times 11.7 \times \sqrt{1 + \frac{1}{6}} \\& = 49 \pm 2.571 \times 11.7 \times \sqrt{1.1667} \\& = 49 \pm 2.571 \times 11.7 \times 1.080 \\& = 49 \pm 32.46\end{aligned}\][/tex]

So, the prediction interval is:

[tex]\[(16.54, 81.46)\][/tex]

Answer:

18

Step-by-step explanation:

if neither wants to go, from 20 it will be 18

we conclude that there is sufficient evidence to suggest that there is a change in biomass of rainforest areas following clear-cutting.

To test whether there is a change in biomass of rainforest areas following clear-cutting, we can perform a one-sample t-test.

The null hypothesis (H0) would be that there is no change in biomass, meaning the mean change in biomass[tex](\(\mu\))[/tex] is equal to zero. The alternative hypothesis (H1) would be that there is a change in biomass, meaning the mean change in biomass [tex](\(\mu\))[/tex] is not equal to zero.

Given the data provided, let's perform the one-sample t-test:

1. Calculate the mean [tex](\(\bar{x}\))[/tex] and standard deviation (s) of the sample.

2. Calculate the t-statistic using the formula:

[tex]\[ t = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}} \][/tex]

where [tex]\(\mu_0\)[/tex] is the hypothesized population mean (in this case, 0), s is the sample standard deviation, and n is the sample size.

3. Determine the degrees of freedom (df = n - 1).

4. Determine the critical t-value for the desired significance level (e.g., [tex]\(α = 0.05\))[/tex] and degrees of freedom.

5. Compare the calculated t-statistic to the critical t-value.

6. Make a decision: if the calculated t-statistic is greater than the critical t-value, reject the null hypothesis; otherwise, fail to reject the null hypothesis.

Let's start by performing these calculations:

Let's denote the given data points as follows:

Data = -10.8, -4.9, -2.6, -1.6, -3, -6.2, -6.5, -9.2, -3.6, -1.8, -1, 0.2, 0.2, 0.1, -0.3, -1.4, -1.5, -0.8, 0.3, 0.6, 1, 1.2, 2.9, 3.5, 4.3, 4.7, 2.9, 2.8, 2.5, 1.7, 2.7, 1.2, 0.1, 1.3, 2.3, 0.5

Now, let's calculate the mean [tex](\(\bar{x}\))[/tex] and standard deviation (s) of the sample.

First, let's calculate the mean [tex](\(\bar{x}\))[/tex] of the sample:

[tex]\[ \bar{x} = \frac{\sum_{i=1}^{n} x_i}{n} \][/tex]

where xi represents each individual data point and n is the sample size.

Using the given data:

[tex]\[ \text{Total number of data points (} n \text{)} = 36 \]\[ \bar{x} = \frac{-10.8 + (-4.9) + (-2.6) + \ldots + 2.3 + 0.5}{36} \]\[ \bar{x} = \frac{-67.1}{36} \]\[ \bar{x} \approx -1.8639 \][/tex]

Next, let's calculate the sample standard deviation (s):

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

Using the given data and the calculated mean:

[tex]\[ s = \sqrt{\frac{(-10.8 - (-1.8639))^2 + (-4.9 - (-1.8639))^2 + \ldots + (0.5 - (-1.8639))^2}{36 - 1}} \]\[ s = \sqrt{\frac{ \sum_{i=1}^{36} (x_i - (-1.8639))^2}{35}} \]\[ s \approx 3.1938 \][/tex]

Now, let's use these values to calculate the t-statistic.

To perform the one-sample t-test, we need to calculate the t-statistic using the formula:

[tex]\[ t = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}} \][/tex]

where:

- [tex]\( \bar{x} \)[/tex] is the sample mean,

- [tex]\( \mu_0 \)[/tex] is the hypothesized population mean (in this case, 0),

- s is the sample standard deviation,

- n is the sample size.

Let's plug in the values:

[tex]\[ t = \frac{-1.8639 - 0}{\frac{3.1938}{\sqrt{36}}} \]\[ t = \frac{-1.8639}{\frac{3.1938}{6}} \]\[ t = \frac{-1.8639}{0.5323} \]\[ t \approx -3.5008 \][/tex]

Now, we need to determine the degrees of freedom (df). Since we have a sample size of 36, the degrees of freedom is df = n - 1 = 36 - 1 = 35.

Next, we need to determine the critical t-value for the desired significance level (e.g., [tex]\( \alpha = 0.05 \)[/tex]) and degrees of freedom (35). We can look this up in a t-table or use statistical software.

Finally, we'll compare the calculated t-statistic to the critical t-value to make a decision about the null hypothesis. Let's proceed with these steps.

For a two-tailed test at a significance level of [tex]\( \alpha = 0.05 \)[/tex] and df = 35, the critical t-value is approximately [tex]\( \pm 2.0301 \).[/tex]

Since |t| = 3.5008 > 2.0301, we reject the null hypothesis (H0).

Therefore, we conclude that there is sufficient evidence to suggest that there is a change in biomass of rainforest areas following clear-cutting.

we conclude that there is sufficient evidence to suggest that there is a change in biomass of rainforest areas following clear-cutting

To test whether there is a change in biomass of rainforest areas following clear-cutting, we can perform a one-sample t-test.

The null hypothesis (H0) would be that there is no change in biomass, meaning the mean change in biomass[tex](\(\mu\))[/tex] is equal to zero.

The alternative hypothesis (H1) would be that there is a change in biomass, meaning the mean change in biomass [tex](\(\mu\))[/tex] is not equal to zero.

Given the data provided, let's perform the one-sample t-test:

1. Calculate the mean ([tex]\\(\bar{x}\))[/tex]and standard deviation (s) of the sample.

2. Calculate the t-statistic using the formula:

[tex]\[ t = \frac{\bar{x} - \mu_0}{(s)/(√(n))} \][/tex]

where[tex]\(\mu_0\)[/tex] is the hypothesized population mean (in this case, 0), s is the sample standard deviation, and n is the sample size.

3. Determine the degrees of freedom (df = n - 1).

4. Determine the critical t-value for the desired significance level (e.g., \(α = 0.05\)) and degrees of freedom.

5. Compare the calculated t-statistic to the critical t-value.

6. Make a decision: if the calculated t-statistic is greater than the critical t-value, reject the null hypothesis; otherwise, fail to reject the null hypothesis.

Let's start by performing these calculations:

Let's denote the given data points as follows:

Data = -10.8, -4.9, -2.6, -1.6, -3, -6.2, -6.5, -9.2, -3.6, -1.8, -1, 0.2, 0.2, 0.1, -0.3, -1.4, -1.5, -0.8, 0.3, 0.6, 1, 1.2, 2.9, 3.5, 4.3, 4.7, 2.9, 2.8, 2.5, 1.7, 2.7, 1.2, 0.1, 1.3, 2.3, 0.5

Now, let's calculate the mean[tex](\(\bar{x}\))[/tex] and standard deviation (s) of the sample.

First, let's calculate the mean ([tex]\\(\bar{x}\))[/tex] of the sample:

[tex]\[ \bar{x} = (\sum_(i=1)^(n) x_i)/(n) \][/tex]

where xi represents each individual data point and n is the sample size.

Using the given data:

[tex]\[ \text{Total number of data points (} n \text{)} = 36 \]\[ \bar{x} = (-10.8 + (-4.9) + (-2.6) + \ldots + 2.3 + 0.5)/(36) \]\[ \bar{x} = (-67.1)/(36) \]\[ \bar{x} \approx -1.8639 \][/tex]

Next, let's calculate the sample standard deviation (s):

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

Using the given data and the calculated mean:

[tex]\[ s = \sqrt{((-10.8 - (-1.8639))^2 + (-4.9 - (-1.8639))^2 + \ldots + (0.5 - (-1.8639))^2)/(36 - 1)} \]\[ s = \sqrt{( \sum_(i=1)^(36) (x_i - (-1.8639))^2)/(35)} \]\[ s \approx 3.1938 \][/tex]

Now, let's use these values to calculate the t-statistic.

To perform the one-sample t-test, we need to calculate the t-statistic using the formula:

[tex]\[ t = \frac{\bar{x} - \mu_0}{(s)/(√(n))} \][/tex]

where:

- [tex]\( \bar{x} \)[/tex] is the sample mean,

-[tex]\( \mu_0 \)[/tex] is the hypothesized population mean (in this case, 0),

- s is the sample standard deviation,

- n is the sample size.

Let's plug in the values:

[tex]\[t = (-1.8639 - 0)/((3.1938)/(√(36))) \]\\[/tex]

[tex]\[ t = (-1.8639)/((3.1938)/(6)) \][/tex]

[tex]\[ t = (-1.8639)/(0.5323) \]\\\[/tex]

Now, we need to determine the degrees of freedom (df). Since we have a sample size of 36, the degrees of freedom is df = n - 1 = 36 - 1 = 35.

Next, we need to determine the critical t-value for the desired significance level (e.g., [tex]\( \alpha = 0.05[/tex]) and degrees of freedom (35). We can look this up in a t-table or use statistical software.

Finally, we'll compare the calculated t-statistic to the critical t-value to make a decision about the null hypothesis. Let's proceed with these steps.

For a two-tailed test at a significance level of [tex]\( \alpha = 0.05 \)[/tex] and df = 35, the critical t-value is approximately [tex]\( \pm 2.0301 \).[/tex]

Since |t| = 3.5008 > 2.0301, we reject the null hypothesis (H0).

Therefore, we conclude that there is sufficient evidence to suggest that there is a change in biomass of rainforest areas following clear-cutting.

Answer:

Step-by-step explanation:

Hello!

A double-blind experiment is a type of experiment where both the experimental units and the researchers that analyze the data don't know what kind of treatment was applied to each subject.

This means, that if it is a placebo-controlled experiment. The subjects will be randomly assigned either the medicament to test ("treatment" group) or the placebo ( "control" group) but they will not know which one they are taking. This way the placebo effect is eliminated.

On the other hand, in other to eliminate the observer bias, the researchers will also not know which patient belongs to the "control" group and wich patient belongs to the "treatment" group.

I hope it helps!

Answer:

Step-by-step explanation:

Given that the owner of a motel has 2900 m of fencing and wants to enclose a rectangular plot of land that borders a straight highway.

Fencing is used for 2times length and 1 width if highway side is taken as width

So we have 2l+w = 2900

Or w = 2900-2l

Area of the rectangular region = lw

[tex]A(l) = l(2900-2l) = 2900l-2l^2\\[/tex]

Use derivative test to find the maximum

[tex]A'(l) = 2900-4l\\A"(l) = -4<0[/tex]

So maximum when I derivative =0

i.e when [tex]l =\frac{2900}{4} =725[/tex]

Largest area = A(725)

= [tex]725(2900-2*725)\\= 1051250[/tex]

1051250 sqm is area maximum

Final answer:

To maximize the area enclosed with 2900 m of fencing along a highway, the motel owner should use a width of 725 m, resulting in a rectangular area of 1,051,250 m².

Explanation:

The motel owner wants to enclose the largest area possible with 2900 m of fencing, without fencing the side along the highway. We can determine the maximum area by recognizing this is an optimization problem that can be solved using calculus or by understanding the properties of geometrical shapes. The most efficient use of the fence, to enclose the maximum area, is to create a shape where two sides are of equal length, essentially a rectangle with one side being the highway. Let's denote the two sides perpendicular to the highway as width (W), and the side opposite the highway as length (L). So, we have 2W + L = 2900. To find the largest enclosed area (A), we use the formula A = W * L. Now, we can express L in terms of W from the fencing constraint as L = 2900 - 2W, and thus express A in terms of W only: A = W * (2900 - 2W).

To maximize the area, we take the derivative of A with respect to W, set it to zero, and solve for W, which will give us the width that maximizes the area. Doing this yields W = 725 m. Therefore, the length (L) will also be 1450 m. Hence, the largest area that can be enclosed is A = 725 m * 1450 m = 1,051,250 m2.

Answer:

[tex] ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex] (1)  

[tex]n=(\frac{z_{\alpha/2} s}{ME})^2[/tex] (2)  

The critical value for 95% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.025,0,1)", and we got [tex]z_{\alpha/2}=1.96[/tex], replacing into formula (2) we got:  

[tex]n=(\frac{1.96(0.05)}{0.02})^2 =24.01 [/tex]  

So the answer for this case would be n=25 rounded up to the nearest integer  

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)  

[tex]\sigma=0.05[/tex] represent the population standard deviation  

n represent the sample size (variable of interest)  

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

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

The margin of error is given by this formula:  

[tex] ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex] (1)  

And on this case we have that ME =0.02 and we are interested in order to find the value of n, if we solve n from equation (1) we got:  

[tex]n=(\frac{z_{\alpha/2} s}{ME})^2[/tex] (2)  

The critical value for 95% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.025,0,1)", and we got [tex]z_{\alpha/2}=1.96[/tex], replacing into formula (2) we got:  

[tex]n=(\frac{1.96(0.05)}{0.02})^2 =24.01 [/tex]  

So the answer for this case would be n=25 rounded up to the nearest integer  

Answer:

a) Figure attached

b) [tex]P(234<X<298)[/tex]

And the best way to solve this problem is using the normal standard distribution and the z score given by:

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

If we apply this formula to our probability we got this:

[tex]P(234<X<298)=P(\frac{234-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{298-\mu}{\sigma})=P(\frac{234-266}{16}<Z<\frac{298-266}{16})=P(-2<z<2)[/tex]

And we can find this probability with this difference:

[tex]P(-2<z<2)=P(z<2)-P(z<-2)[/tex]

An we know using the graph in part a that this area correspond to 0.95 or 95%

c) [tex]P(250<X<314)=P(\frac{250-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{314-\mu}{\sigma})=P(\frac{250-266}{16}<Z<\frac{314-266}{16})=P(-1<z<3)[/tex]

And we can find this probability with this difference:

[tex]P(-1<z<3)=P(z<3)-P(z<-1)[/tex]

An we know using the graph in part a that this area correspond to 0.95 or 34+34+13.5+2.35%=83.85%

d) [tex] P(X<218)[/tex]

And using the z score we got:

[tex]P(X<218) = P(Z< \frac{218-266}{16}) = P(Z<-3) [/tex]

And that correspond to approximately 0.15%

Step-by-step explanation:

Part a

For this case we can see the figure attached.

Previous concepts

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

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

Part b

Let X the random variable that represent the length of human pregnancy of a population, and for this case we know that:

Where [tex]\mu=266[/tex] and [tex]\sigma=16[/tex]

We are interested on this probability

[tex]P(234<X<298)[/tex]

And the best way to solve this problem is using the normal standard distribution and the z score given by:

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

If we apply this formula to our probability we got this:

[tex]P(234<X<298)=P(\frac{234-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{298-\mu}{\sigma})=P(\frac{234-266}{16}<Z<\frac{298-266}{16})=P(-2<z<2)[/tex]

And we can find this probability with this difference:

[tex]P(-2<z<2)=P(z<2)-P(z<-2)[/tex]

An we know using the graph in part a that this area correspond to 0.95 or 95%

Part c

[tex]P(250<X<314)=P(\frac{250-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{314-\mu}{\sigma})=P(\frac{250-266}{16}<Z<\frac{314-266}{16})=P(-1<z<3)[/tex]

And we can find this probability with this difference:

[tex]P(-1<z<3)=P(z<3)-P(z<-1)[/tex]

An we know using the graph in part a that this area correspond to 0.95 or 34+34+13.5+2.35%=83.85%

Part d

We want this probability:

[tex] P(X<218)[/tex]

And using the z score we got:

[tex]P(X<218) = P(Z< \frac{218-266}{16}) = P(Z<-3) [/tex]

And that correspond to approximately 0.15%

Answer:

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

[tex]z=\frac{0.36 -0.41}{\sqrt{\frac{0.41(1-0.41)}{100}}}=-1.017[/tex]  

Step-by-step explanation:

Data given and notation

n=100 represent the random sample taken

[tex]\hat p=0.36[/tex] estimated proportion with the survey

[tex]p_o=0.41[/tex] is the value that we want to test

[tex]\alpha[/tex] represent the significance level

z would represent the statistic (variable of interest)

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

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion is lower than 0.41.:  

Null hypothesis:[tex]p\geq 0.41[/tex]  

Alternative hypothesis:[tex]p < 0.41[/tex]  

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

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

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

Calculate the statistic  

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

[tex]z=\frac{0.36 -0.41}{\sqrt{\frac{0.41(1-0.41)}{100}}}=-1.017[/tex]  

Answer:

z test statistic is -1.042 .

Step-by-step explanation:

We are given that based on the Nielsen ratings, the local CBS affiliate claims its 11 p.m. newscast reaches 41% of the viewing audience in the area. In a survey of 100 viewers, 36% indicated that they watch the late evening news on this local CBS station.

Let Null Hypothesis, [tex]H_0[/tex] : p = 0.41 {means that % of the viewing audience in the area is 41%}

Alternate Hypothesis, [tex]H_1[/tex] : p [tex]\neq[/tex] 0.41 {means that % of the viewing audience in the area is different from 41%}

The z-test statistics we will use here is One sample proportion test ;

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

where, p = % of the viewing audience based on the Nielsen ratings = 41%

       [tex]\hat p[/tex]  = % of the viewing audience based on a survey of 100 viewers = 36%

       n = sample of viewers = 100

So, test statistics = [tex]\frac{0.36 - 0.41}{\sqrt{\frac{0.36(1-0.36)}{100} } }[/tex]

                             = -1.042

Therefore, the z test statistic is -1.042 .

Answer:

Step-by-step explanation:

Hello!

a.

The objective is to study the relationship between the shape of an ibuprofen tablet and its dissolution time.

For these two independent samples of tablets from different shapes where taken and their dissolution times measured:

Sample 1: Disk.shaped tablets

n₁= 6

X[bar]₁= 255.8

S₁= 8.22

Sample 2: Oval-shaped tablets

n₂=8

X[bar]₂= 270.74

S₂= 11.90

Assuming that the population variances are equal and both samples come from normal distributions you need to test if the average dissolution time of the disk-shaped tablets is less than the average dissolution time of the oval-shaped tablets, symbolically:

H₀: μ₁ ≥ μ₂

H₁: μ₁ < μ₂

α: 0.05

Considering the given information about both populations, the statistic to use for this test is a Student t for independent samples with pooled sample variance:

[tex]t= \frac{(X[bar]_1-X[bar]_2)-(Mu_1-Mu_2)}{Sa\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } } ~~t_{n_1+n_2-2}[/tex]

[tex]Sa^2= \frac{(n_1-1)S^2_1+(n_2-1)S^2_2}{n_1+n_2-2}= \frac{5*(8.22)^2+7*(11.9)^2}{6*8-2}[/tex]

Sa²= 110.76

Sa= 10.52

[tex]t_{H_0}= \frac{(255.8-270.74)-0}{10.52\sqrt{\frac{1}{6} +\frac{1}{8} } } = -2.629= -2.63[/tex]

This test is one-tailed to the left, meaning that you will reject the null hypothesis to small values of t, the p-value has the same direction and you can calculate it as:

P(t₁₂≤-2.63)= 0.0110

Since the p-value= 0.0110 is less than the significance level α: 0.05, the decision is to reject the null hypothesis.

At a 5% significance level you can conclude that the average dissolution time of the disk-shaped ibuprofen tablets is less than the average dissolution time of the oval-shaped ibuprofen tablets.

b.

(X[bar]₁-X[bar]₂)+Sa[tex]\sqrt{\frac{1}{n_1}+\frac{1}{n_2} }* t_{n_1+n_2-2; 0.95}[/tex]

(255.8-270.74)+ 10.52*[tex]\sqrt{\frac{1}{6} +\frac{1}{8} } * 1.782[/tex]

(-∞;-4.815)

I hope it helps!

The test of comparison between the mean dissolve time of each tablet can

be made using a t-test given that the sample size is small.

The correct responses are;

Reasons:

The given data is presented as follows;

[tex]\begin{array}{|c|cccccc}&&Time & of & disolution\\Disk&269.0&249.3 &255.2&252.7&247.0&261.6\end{array}\right][/tex]

[tex]\begin{array}{|c|cccccccc}&&Time & of & disolution\\Oval&268.8&260.0&273.5&253.9&278.5&289.4&261.6&280.2\end{array}\right][/tex]

The mean for Disks, [tex]\overline x_1[/tex] = 255.8

The standard deviation for Discs, s₁ ≈ 8.22

Sample size of the Disks, n₁ = 6

Mean for Oval, [tex]\overline x_2[/tex] ≈ 270.74

Standard deviation for Oval, s₂ ≈ 11.9

Sample size of the Oval, n₂ = 8

a. Null hypothesis is H₀; [tex]\overline x_1[/tex] = [tex]\overline x_2[/tex] (there is no difference between the mean of the samples)

Alternative hypothesis is Hₐ; [tex]\overline x_1[/tex] < [tex]\overline x_2[/tex]

The standard deviation of the two populations are equal; [tex]\sigma_{disk} = \mathbf{\sigma_{oval}}[/tex]

The pooled standard deviation, [tex]s_p[/tex], is given as follows;

[tex]s_p = \mathbf{\sqrt{\dfrac{\left ( n_{1}-1 \right )\cdot s_{1}^{2} +\left ( n_{2}-1 \right )\cdot s_{2}^{2}}{n_{1}+n_{2}-2}}}[/tex]

[tex]s_p =\sqrt{\dfrac{\left ( 6-1 \right )\times 8.22^{2} +\left ( 8-1 \right )\times 11.9^{2}}{6+8-2}} \approx 10.524[/tex]

The test statistic is found using the following formula;

[tex]\displaystyle t = \mathbf{ \frac{\overline x_1 - \overline x_2}{s_p \cdot \sqrt{\dfrac{1}{n_1} +\dfrac{1}{n_2} } }}[/tex]

Which gives;

[tex]\displaystyle t = \frac{255.8 - 270.74}{10.524 \times \sqrt{\dfrac{1}{6} +\dfrac{1}{8} } } \approx -2.629[/tex]

The degrees of freedom, df = n₁ + n₂ - 2

Therefore;

df = 8 + 6 - 2  = 12

From the t-test table, we have; 0.005 < p-value < 0.01

Given that the p-value is less than the alpha level of α = 5% = 0.05, we reject the null hypothesis.

Therefore;

b. The 95% one sided confidence interval is presented as follows;

[tex]\displaystyle \left (\bar{x}_{1}- \bar{x}_{2} \right )\pm \mathbf{t_{(\alpha /2, \, df)} \cdot s_p \cdot \sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}}}}[/tex]

[tex]\displaystyle t_{(\alpha /2, \, df)}[/tex] = [tex]t_{(0.025, \, 12)}[/tex] = 2.18

Which gives;

[tex]\mathbf{\overline x_1 - \overline x_2} <\displaystyle \left (255.8- 270.74 \right )+2.18 \times 10.524 \cdot \sqrt{\frac{1}{6}+\frac{1}8}}[/tex]

The one sided 95% confidence interval is therefore;

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Answer:

Maya got more than Sam with $630

Step-by-step explanation:

Sam, Aria and Maya in a ratio of 4:9:11

4+9+11 = 24

Sam will get 4/24 x 2160 = 360

Maya will get 11/24 x 2160 =990

Maya got more than Sam with 990-360 = $630

Answer: Maya made $630 more than Sam.

Step-by-step explanation:

The total amount that Sam, Aria and Maya got paid for painting the house is $2160.

Since they worked for different number of hours on the job, the money was split in the ratio of

4 : 9 : 11

The total ratio is the sum of the proportions. It becomes

4 + 9 + 11 = 24

Therefore, the amount that Sam made was

4/24 × 2160 = $360

The amount that Aria made was

9/24 × 2160 = $810

The amount that Maya made was

11/24 × 2160 = $990

The difference in the amounts made by Maya and Sam is

990 - 360 = $630

Answer:

[tex] z= \frac{0.6-0.637}{0.0439} =-0.843[/tex]

So then we can find the probability like this:

[tex]P(p<0.6) = P(Z<-0.843)[/tex]

And using the normal standard table or excel we got:

[tex]P(p<0.6) = P(Z<-0.843)=0.1996[/tex]

Step-by-step explanation:

For this case we can check if we can use the normal approximation for the proportion and we have this:

[tex] np = 120*0.637 =76.44 >10[/tex]

[tex] n(1-p) = 120*(1-0.637) = 43.56>10[/tex]

Then we can conclude that we can use the normal approximation. And we have this:

[tex] p\sim N (p, \sqrt{\frac{p(1-p)}{n}})[/tex]

So the mean is given by:

[tex]\mu_p = 0.637[/tex]

And the deviation is given by:

[tex]\sigma_p = \sqrt{\frac{0.637*(1-0.637)}{120}}= 0.0439[/tex]

And for this case we want to find this probability:

[tex] P( p<0.6)[/tex]

And we can use the z score given by:

[tex] z = \frac{p -\mu}{\sigma_p}[/tex]

And for this case the z score is:

[tex] z= \frac{0.6-0.637}{0.0439} =-0.843[/tex]

So then we can find the probability like this:

[tex]P(p<0.6) = P(Z<-0.843)[/tex]

And using the normal standard table or excel we got:

[tex]P(p<0.6) = P(Z<-0.843)=0.1996[/tex]

Answer:

XYZ is NOT a good buy.

Step-by-step explanation:

Calculate the market price of stock:

[tex]\frac{Next year's Dividend}{Reqd.return - Growth rate}[/tex]

[tex]= \frac{(2.4)(1.065)}{0.115-0.065}[/tex]

[tex]= 51.12[/tex]

The Market price of the stock is $51. Therefore, buying the stock at $60 is overpriced and is NOT a good buy.

Answer:

0.4

Step-by-step explanation:

Probability of rain = P(R)

Probability of late plane = P(L)

So, the probability of no rain = P(R')

Breaking it down

If it rains, 40% chance, P(R) = 0.4

That the plane would be late if it rains = 70% × 40%, that is, P(R n L) = 0.7 × 0.4 = 0.28, 28% of the total chance.

That the plane would be on time if it rains = 30% × 40%, that is, P(R n L') = 0.3 × 0.4 = 0.12, 12% of the total chance.

If it doesn't rain, 60% chance, P(R') = 1 - P(R) = 1 - 0.4 = 0.6

That the plane would be late if it doesn't rain = 20% × 60%, that is, P(R n L') = 0.2 × 0.6 = 0.12, 12% of the total chance.

That the plane would be on time if it doesn't rain = 80% × 60%, that is, P(R' n L') = 0.8 × 0.6 = 0.48, 48% of the total chance.

So, probability that the plane would be late = P(L) = P(R n L) + P(R' n L) = 0.28 + 0.12 = 0.4 = 40%

Answer:

1/120

Step-by-step explanation:

Since the players have different names, there is only one possible arrangement in which they are in alphabetical order. The total number of ways to order 5 basketball players (n) is:

[tex]n = 5!=5*4*3*2*1\\n=120[/tex]

Therefore, there is a 1/120 probability that they shoot free throws in alphabetical​ order.

Answer:

Expected Number=71

Variance =4959

Standard Deviation= 70.42

Step-by-step explanation:

Expected Number = E (x) = np

Here p = 0.71 and q= 1- 0.71= 0.29 n= 100

Expected Number = E (x) = np = 0.71*100= 71

b) The Variance and Standard Deviation for the number of these tweets with no reaction

Suppose the number of tweets are hundred then the number of tweets with no reaction would be 71 .

Variance = E(x²) - [E(x)]² =  (100)²- (71)²= 10000- 5041= 4959

Standard Deviation= √4959= 70.42

Answer:

the prduct of 1/2*6/5 is bigger

Step-by-step explanation:

1/2 x 6/5 = 6/10 = 3/5

1/2 x 5/6 = 5/12

Answer:

[tex]163.83-4.03\frac{20.094}{\sqrt{6}}=130.77[/tex]    

[tex]163.83+4.03\frac{20.094}{\sqrt{6}}=196.89[/tex]    

So on this case the 99% confidence interval would be given by (130.77;196.89)    

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".

Data: 175 177 175 180 138 138

We can calculate the mean and the deviation from these data with the following formulas:

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

[tex]s=\sqrt{\frac{\sum_{i=1}^n (x_i -\bar X)^2}{n-1}}[/tex]

[tex]\bar X=163.83[/tex] represent the sample mean for the sample  

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

s=20.093 represent the sample standard deviation

n=6 represent the sample size  

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

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

In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:

[tex]df=n-1=6-1=5[/tex]

Since the Confidence is 0.99 or 99%, the value of [tex]\alpha=0.01[/tex] and [tex]\alpha/2 =0.005[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,5)".And we see that [tex]t_{\alpha/2}=4.03[/tex]

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

[tex]163.83-4.03\frac{20.094}{\sqrt{6}}=130.77[/tex]    

[tex]163.83+4.03\frac{20.094}{\sqrt{6}}=196.89[/tex]    

So on this case the 99% confidence interval would be given by (130.77;196.89)    

Answer:

a) k=2.07944 (1/hour)

b) [tex]P(t)=40e^{2.0794t}[/tex]

c) P(7)=83,886,080

Step-by-step explanation:

We know that the cells duplicates after 20 minutes (t=1/3 hours).

We can write a model of that as:

[tex]\frac{dP}{dt}=kP\\\\\frac{dP}{P}=kdt\\\\\int \frac{dP}{P}=k\int dt\\\\ln(P)+C_1=kt\\\\P=Ce^{kt}\\\\\\P(0)=40=Ce^0=C\\\\C=40\\\\\\P(1/3)=80=40e^{k*(1/3)}\\\\e^{k*(1/3)}=80/40=2\\\\k/3=ln(2)\\\\k=3*ln(2)=2.07944[/tex]

a) k=2.0794 h^(-1)

b) [tex]P(t)=40e^{2.0794t}[/tex]

c) [tex]P(7)=40e^{2.0794*7}=40*e^{14.556}=40*2,097,152=83,886,080[/tex]

The expected value is

[tex]E[X]=\displaystyle\sum_xx\,P(X=x)=\frac1{11}\sum_{x=0}^{10}x=\dfrac{0+1+\cdots+9+10}{11}=\dfrac{55}{11}=\boxed{5}[/tex]

The standard deviation is the square root of the variance, which is

[tex]V[X]=E[(X-E[X])^2]=E[X^2]-E[X]^2[/tex]

where

[tex]E[X^2]=\displaystyle\sum_xx^2\,P(X=x)=\frac1{11}\sum_{x=0}^{10}x^2=\dfrac{0^2+1^2+\cdots+9^2+10^2}{11}=\dfrac{385}{11}=35[/tex]

so that

[tex]V[X]=35-5^2=10[/tex]

making the standard deviation

[tex]\sqrt{V[X]}=\sqrt{10}\approx\boxed{3.16}[/tex]

The expected value of the random variable is 5 and the standard deviation is approximately 1.674.

To find the expected value of this probability distribution, we multiply each possible outcome by its respective probability and sum the results. The expected value is given by the formula E(X) = ∑(x * P(X=x)). In this case, the expected value is (0 * 1/11) + (1 * 1/11) + (2 * 1/11) + ... + (10 * 1/11) = 5.

To find the standard deviation, we first calculate the variance. The variance is given by the formula Var(X) = ∑((x - E(X))2 * P(X=x)). After calculating the variance, the standard deviation is the square root of the variance. In this case, the variance is ((0 - 5)2 * 1/11) + ((1 - 5)2 * 1/11) + ... + ((10 - 5)2 * 1/11) = 20/11. Taking the square root of 20/11 gives us a standard deviation of approximately 1.674.

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Answer:

The angle of depression of the lens must be 26.3°

Step-by-step explanation:

Here we have a right triangle with the opposite side to the angle equal to 5.93 feet and the adjacent side to the angle equal to 12.02 feet. Therefore we just need to use the tangent definition to find the angle.

[tex]tan(\alpha)=\frac{5.93 ft}{12.02 ft}=0.49[/tex]

[tex]\alpha=tan^{-1}(0.49)=26.3^{\circ}[/tex]

The angle of depression of the lens must be 26.3°

I hope it helps you!  

The angle of depression is approximately 26.23°.

To determine the angle of depression from the camera to the teller, we can use trigonometry. The camera is mounted 5.93 feet off the ground, and the teller is 12.02 feet away horizontally from the point directly below the camera.

We will use the tangent function, which relates the opposite side (the height difference) to the adjacent side (the horizontal distance).

The formula for the tangent of an angle is:

tan(θ) = opposite / adjacent

So, tan(θ) = 5.93 / 12.02

Calculating this gives:

tan(θ) ≈ 0.4935

To find the angle θ, we take the arctangent (inverse tangent) of 0.4935:

θ = arctan(0.4935)

Using a calculator, we find:

θ ≈ 26.23°

Thus, the angle of depression that the camera lens should make is approximately 26.23°  .

Answer:

Permutation count the number of different arrangements pf r out of n items, while combination count  the number of group of r out of n items.

Step-by-step explanation:

Permutation is the different possible arrangements or different possible order taking by the given things, objects ,words and numbers. it is also know rearranging.

Result are vary with different conditions Like Repetition is allowed or Repetition is not allowed

In mathematics we denote permutation by   [tex]{\textup{n}p_{r}}[/tex] no of permutation of n taken r at a  time.

Combination is a selection of some specific item or all items at a time from a collection is known as combination. It is denote by [tex]{\textup{n}c_{r}}[/tex] number of combination of n  different things taken r at a time

Eg. We have to choose 2 boys in group of 5 so, we can choose by many ways

Combination is widely used in lottery system.

So

Permutation count the number of different arrangements pf r out of n items, while combination count  the number of group of r out of n items.

Permutations count arrangements with order, combinations count groups without order.

The main difference between a situation in which the use of the permutations rule is appropriate and one in which the use of the combinations rule is appropriate is:

In summary, permutations focus on arrangements where order matters, while combinations focus on groups where order doesn't matter.

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Final answer:

Hosea can fulfill his daily dietary requirements in C(19, 5) * C(7, 5) * C(9, 4) ways by choosing from 19 vegetables, 7 fruits, and 9 whole grains without repetition.

Explanation:

The question asks how many ways Hosea can fulfill his daily dietary requirements recommended by his doctor, choosing from a variety of vegetables, fruits, and whole grains without eating two servings of the same thing in one day. To find the solution, we can use combinations since the order of choosing the items doesn't matter. Hosea has a choice of 19 vegetables, 7 fruits, and 9 whole grains and needs to select 5 vegetables, 5 fruits, and 4 whole grains.

Therefore, Hosea has C(19, 5) * C(7, 5) * C(9, 4) ways to choose his daily servings of vegetables, fruits, and whole grains without repetitions.

Answer:

The probability mass function is expressed as:

P(x) = [(x+r-1)C(r-1)]*[p^r]*[(1-p)^x]

Step-by-step explanation:

This is not a binomial distribution. It is actually a negative binomial distribution. The probability mass function is expressed below:

P(x) = [(x+r-1)C(r-1)]*[p^r]*[(1-p)^x]

where:

x = number of failures

r-1 = number of successes (10 in this scenario)

p = probability of a success

nCr = n!/[r!(n-r)!]

The main formula difference in the positive binomial versus negative binomial is this: With respect to the negative binomial, it is obviously  known that the last event will be: when we reach our 10th "head", we stop .

Thus, the last flip will ALWAYS be a "head".

Answer:

(a) The mean is 4.31 pounds. The variance is 51.42 pounds.

(b) The average shipping cost of a package is $10.78.

(c) The probability that the weight of a package exceeds 59 pounds is 0.0027.

Step-by-step explanation:

The probability density function of the weight of packages is:

[tex]f(x) = \frac{70}{69x^{2}};\ 1 < x < 70[/tex]

(a)

The formula for expected value (or mean) of X is:

[tex]E(X)=\int\limits^a_b {x\times f(x)} \, dx[/tex]

Compute the expected value of X as follows:

[tex]E(X)=\int\limits^{70}_{1} {x\times \frac{70}{69x^{2}}} \, dx=\frac{70}{69} \int\limits^{70}_{1} {x \times x^{-2}} \, dx\\=\frac{70}{69} \int\limits^{70}_{1} {x^{-1}} \, dx=\frac{70}{69} |\ln x|^{70}_{1}\\=\frac{70}{69}\times\ln 70\\=4.31[/tex]

Thus, the mean is 4.31 pounds.

The formula to compute the variance is:

[tex]V(X)=E(X^{2})-[E(X)]^{2}[/tex]

Compute the E (X²) as follows:

[tex]E(X^{2})=\int\limits^{70}_{1} {x^{2}\times \frac{70}{69x^{2}}} \, dx=\frac{70}{69} \int\limits^{70}_{1} {x^{2} \times x^{-2}} \, dx\\=\frac{70}{69} \int\limits^{70}_{1} {1} \, dx=\frac{70}{69} | x|^{70}_{1}\\=\frac{70}{69}\times69\\=70[/tex]

The variance is:

[tex]V(X)=E(X^{2})-[E(X)]^{2}\\=70-(4.31)^{2}\\=51.4239\\\approx51.42[/tex]

Thus, the variance is 51.42 pounds.

(b)

It is provided that the shipping cost for per pound is, C = $2.50.

Compute the average shipping cost of a package as follows:

[tex]Average\ cost=Cost\ per\ pound\times E(X)\\=2.50\times4.31\\=10.775\\\approx10.78[/tex]

Thus, the average shipping cost of a package is $10.78.

(c)

Compute the probability that the weight of a package exceeds 59 pounds as follows:

[tex]P(59<X<70)=\int\limits^{70}_{59} {\frac{70}{69x^{2}}} \, dx=\frac{70}{69} \int\limits^{70}_{59} {x^{-2}} \, dx\\=\frac{70}{69} |-\frac{1}{x}|^{70}_{59}=\frac{70}{69} [-\frac{1}{70}+\frac{1}{59}]\\=\frac{70}{69}\times0.0027\\=0.0027[/tex]

Thus, the probability that the weight of a package exceeds 59 pounds is 0.0027.

The weights of the package follows a probability density function

The probability density function is given as:

[tex]\mathbf{f(x) = \frac{70}{69x^2},\ 1 < x < 70}[/tex]

(a) The mean and the variance

The mean is calculated as:

[tex]\mathbf{E(x) = \int\limits^a_b {x \cdot f(x)} \, dx }[/tex]

So, we have:

[tex]\mathbf{E(x) = \int\limits^{70}_1 {x \cdot \frac{70}{69x^2} } \, dx }[/tex]

[tex]\mathbf{E(x) = \int\limits^{70}_1 {\frac{70}{69x} } \, dx }[/tex]

Rewrite as:

[tex]\mathbf{E(x) = \frac{70}{69}\int\limits^{70}_1 {x^{-1} } \, dx }[/tex]

Integrate

[tex]\mathbf{E(x) = \frac{70}{69} {ln(x)}|\limits^{70}_1 } }[/tex]

Expand

[tex]\mathbf{E(x) = \frac{70}{69} \cdot {(ln(70) - ln(1)) }}[/tex]

[tex]\mathbf{E(x) = 4.31 }}[/tex]

The variance is calculated as:

[tex]\mathbf{Var(x) = E(x^2) - (E(x))^2}[/tex]

Where:

[tex]\mathbf{E(x^2) = \int\limits^a_b {x^2 \cdot f(x)} \, dx }[/tex]

So, we have:

[tex]\mathbf{E(x^2) = \int\limits^{70}_1 {x^2 \cdot \frac{70}{69x^2} } \, dx }[/tex]

[tex]\mathbf{E(x^2) = \int\limits^{70}_1 {\frac{70}{69} } \, dx }[/tex]

Rewrite as:

[tex]\mathbf{E(x^2) = \frac{70}{69}\int\limits^{70}_1 {1 } \, dx }[/tex]

Integrate

[tex]\mathbf{E(x^2) = \frac{70}{69} x|\limits^{70}_1 } }[/tex]

Expand

[tex]\mathbf{E(x^2) = \frac{70}{69} \cdot {(70 - 1) }}[/tex]

[tex]\mathbf{E(x^2) = 70 }}[/tex]

So, we have:

[tex]\mathbf{Var(x) = E(x^2) - (E(x))^2}[/tex]

[tex]\mathbf{Var(x) = 70 - 4.31^2}[/tex]

[tex]\mathbf{Var(x) = 51.42}[/tex]

Hence, the mean is 4.31 and the variance is 51.42, respectively.

(b) The average cost of shipping a package

In (a), we have:

[tex]\mathbf{E(x) = 4.31 }}[/tex] ---- Mean

So, the average cost of shipping a package is:

[tex]\mathbf{Average =4.31 \times 2.50}[/tex]

[tex]\mathbf{Average =10.78}[/tex]

Hence, the average cost of shipping a package is $10.78

(c) The probability a package weighs over 59 pounds

This is represented as: P(x > 59)

So, we have:

[tex]\mathbf{P(x > 59) = P(59 < x < 70)}[/tex]

So, we have:

[tex]\mathbf{P(x > 59) = \int\limits^{70}_{59} { \frac{70}{69x^2}} \, dx }[/tex]

Rewrite as:

[tex]\mathbf{P(x > 59) = \frac{70}{69}\int\limits^{70}_{59} { x^{-2}} \, dx }[/tex]

Integrate

[tex]\mathbf{P(x > 59) = \frac{70}{69} \cdot { -\frac 1x}|\limits^{70}_{59}}[/tex]

Expand

[tex]\mathbf{P(x > 59) = \frac{70}{69} \cdot (-\frac{1}{70} + \frac{1}{59})}[/tex]

[tex]\mathbf{P(x > 59) = 0.0027}[/tex]

Hence, the probability a package weighs over 59 pounds is 0.0027

Read more about probability density functions at:

https://brainly.com/question/14749588

Answer:

A) Yes because two pairs in the corresponding angles are congruent.

Step-by-step explanation:

Let's analyse similar angles to have some Understanding.

Let's go now!

Similar angles are angles whose:

i. Corresponding angles in the both triangles are equal

ii. Ratio of corresponding sides are constant

iii. Two pairs of corresponding sides are in thesame ratio and the angles between them are equal.

Pls see the attached file.

Enjoy math!

Answer:

It would be significant

Step-by-step explanation:

Population proportion of consumers who recognize the company name = 68% = 0.68

If 634 consumers out of the 800 randomly selected consumers recognize the company name, sample proportion = 634/800 = 0.7925.

It is significant to get 634 because the sample proportion of consumers who recognize the company name is greater than the population proportion.

Given Information:

Probability = p = 68% = 0.68

Population = n = 800

Answer:

it would not be significant to get 634 consumers who might recognize the name of Dull Computer Company.

Step-by-step explanation:

We can check whether it would be significant to get 635 consumers who recognize the Dull Computer Company by finding out the mean and standard deviation.

mean = μ = np

μ = 800*0.68

μ = 544

standard deviation = σ = √np(1-p)

σ = √800*0.68(1-0.68)

σ = 13.2 ≈ 13

we know that 99% of data fall within 3 standard deviations from the mean

μ ± 3σ = 544+3*13, 544-2*13

μ ± 3σ = 544+39, 544-39

μ ± 3σ = 583, 505

So we can say with 99% confidence that the number of consumers who can  recognize the name of Dull Computer Company will be from 505 to 583 and since 583 < 634 we can conclude that it would not be significant to get 634 consumers who might recognize the name of Dull Computer Company.