A 95% confidence interval was computed using a sample of 16 lithium batteries, which had a sample mean life of 645 hours. The confidence interval was (628.5, 661.5) hours. Which of the following would produce a confidence interval that is wider than the one originally computed (assuming everything else remained the same)? Select ALL that are correct. Having a sample with a larger standard deviation. Using a 99% confidence level instead of 95%. Removing an outlier from the data. Using a 90% confidence level instead of 95%. Testing 10 batteries instead of 16. Testing 24 batteries instead of 16.

Volume = length x width x height

120 = 15 x 4 x length

Simplify:

120 = 60 x length

Divide both sides by 60:

Length = 2 feet

Formula for the volume of a rectangular prism: V = length x width x height

120 = l x 4 x 15

120 = l x 60

l = 2

The length of the rectangular prism is 2 feet.

Hope this helps!! :)

Answer:

You can get it by doing this

Step-by-step explanation:

The Probability

Answer:

0.08 percent chance or 1/12

Answer:

x=2

Step-by-step explanation:

To solve, we need to isolate the variable (x)

To do this, we must get all the variables to one side of the equation, and all the numbers to the other

3x+3= -2x+13

Add 2x to both sides to "reverse" the subtraction

5x+3=13

Subtract 3 from both sides to "reverse" the addition

5x=10

Divide both sides by 5 to "reverse" the multiplication

x=2

Answer:

36.88% probability that her pulse rate is between 69 beats per minute and 81 beats per minute.

Step-by-step explanation:

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.

In this problem, we have that:

[tex]\mu = 75, \sigma = 12.5[/tex]

Find the probability that her pulse rate is between 69 beats per minute and 81 beats per minute.

This is the pvalue of Z when X = 81 subtracted by the pvalue of Z when X = 69.

X = 81

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

[tex]Z = \frac{81 - 75}{12.5}[/tex]

[tex]Z = 0.48[/tex]

[tex]Z = 0.48[/tex] has a pvalue of 0.6844

X = 69

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

[tex]Z = \frac{69 - 75}{12.5}[/tex]

[tex]Z = -0.48[/tex]

[tex]Z = -0.48[/tex] has a pvalue of 0.3156

0.6844 - 0.3156 = 0.3688

36.88% probability that her pulse rate is between 69 beats per minute and 81 beats per minute.

Answer:

[tex]P(69<X<81)=P(\frac{69-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{81-\mu}{\sigma})=P(\frac{69-75}{12.5}<Z<\frac{81-75}{12.5})=P(-0.48<z<0.48)[/tex]

And we can find this probability with this difference:

[tex]P(-0.48<z<0.48)=P(z<0.48)-P(z<-0.48)=0.684-0.316= 0.368 [/tex]

Step-by-step explanation:

Previous concepts

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

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

Solution to the problem

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

[tex]X \sim N(75,12.5)[/tex]  

Where [tex]\mu=75[/tex] and [tex]\sigma=12.5[/tex]

We are interested on this probability

[tex]P(69<X<81)[/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(69<X<81)=P(\frac{69-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{81-\mu}{\sigma})=P(\frac{69-75}{12.5}<Z<\frac{81-75}{12.5})=P(-0.48<z<0.48)[/tex]

And we can find this probability with this difference:

[tex]P(-0.48<z<0.48)=P(z<0.48)-P(z<-0.48)=0.684-0.316= 0.368 [/tex]

Answer:

[tex]v = \pi {r}^{2} \frac{h}{3} \\ \: \: \: \: = \pi \times 20 \frac{40}{3} \\ \: \: \: \: = 16755.16 {cm}^{3} [/tex]

hope this helps you

Answer: [tex]y=8x[/tex]

Step-by-step explanation:

For this exercise it is important to remember that the equation of a line that passes through the origin (this is located at the point [tex](0,0)[/tex]), has the following form:

[tex]y=mx[/tex]

Where "m" is the slope of the line.

In this case, according to the information given in the exercise, you know that 8 million tons of plastic waste enters the world’s oceans every year, apprdoximately.

Then, let the independent variable "x" represents the number of years.

 Analizing the data given, you can identify that the slope of that line is the following:

[tex]m=8[/tex]

So, knowing the value of "m", you can subsitute them into the equation  [tex]y=mx[/tex], in order to get the equation that shows that situation.

This is:

[tex]y=8x[/tex]

Answer:

f(x)= 3x + 1

Step-by-step explanation:

Let's set the unknown variable as b, since it's a linear equation.

Using the point given (-3, -8), we can find b by plugging in the known numbers and solving.

-8 = -3 (3) + b

-8 = -9 + b

b = 1

If you want, you can check it by forming the complete equation and plugging in -3.

f(x)= 3x + 1

f(-3) = -9 + 1

f(-3) = -8 = -8

I hope this helped!

Answer:

Solution (x, y, λ) = (2, 2, - 6)

Step-by-step explanation:

From the above information given:

3x + λ = 0............ eqn (1)

3y + λ = 0 ........... eqn (2)

x + y − 4 = 0 ......... eqn (3)

Now, from equation 1 and 2:

x= - λ / 3

y= - λ / 3

Substitute the values of X and y into equation 3

-λ/ 3 - λ/ 3 - 4 =0

-λ - λ - 12= 0

-2λ - 12= 0

-2λ = 12

λ = - 6

x= - λ/ 3

x = - (- 6) / 3

x= 2

y= - λ/3

y= - (-6)/3

y= 2

Solution (x, y, λ) = (2, 2, - 6)

Therefore,

x = 2

y= 2

λ = - 6

80% of the times we are confident that the mean number of oocytes retrieved will fall in interval  ( 8.8 , 10.6 ).

A confidence interval is the probability that an estimate will fall in a particular range for a certain number of times.

From the given scenario, women are selected from the age group 36-40.

The mean number of oocytes retrieved from the 98 participants was 9.7 with an 80 % confidence level z ‑interval of ( 8.8 , 10.6 )

From the given condition we can conclude that if we repeat the experiment 80% of the times we are confident that the mean number of oocytes retrieved will fall in interval  ( 8.8 , 10.6 ).

Thus, the mean number of oocytes retrieved from the 98 participants in Group A was 9.7 with an 80 % confidence level has z ‑interval of ( 8.8 , 10.6 )

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The question discusses a study in reproductive endocrinology involving hormone-assisted retrieval of oocytes for assisted reproduction, like IVF. Two hormones, likely follicle-stimulating hormone and luteinizing hormone, are used from day one to boost ovulation. The reported average number of oocytes and the confidence interval indicate the effectiveness of this treatment strategy.

The study described touches upon a field within biology known as reproductive endocrinology with a focus on hormone-assisted techniques in assisted reproduction like IVF (In Vitro Fertilization). The two hormones administered from day one are likely gonadotropins, consisting of follicle-stimulating hormone (FSH) and luteinizing hormone (LH), as per the common methods in IVF procedures. FSH supports the development of multiple follicles while LH triggers ovulation. Their administration significantly boosts the usual number of oocytes (eggs) produced in a woman's ovulation cycle.

Moreover, the mean number of oocytes (9.7) retrieved from the participants in Group A gives an indication of the effectiveness of the treatment strategy. The 80% confidence level z-interval of (8.8, 10.6) suggests that the actual mean number of oocytes retrieved for the overall population of women, aged 36-40 years, following this treatment strategy, is likely to lie within this range with an 80% probability.

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Answer:

Given:

Mean, x' = 94.32

s.d = 1.20

n = 16

a) For null hypothesis:

H0 : u= 95

For alternative hypothesis:

Ha : u≠95

Level of significance, a = 0.01

(Two tailed test)

We reject null hypothesis, h0, if P<0.01 level of significance.

Calculating the test statistic, we have:

[tex] Z = \frac{x' - u_0}{s.d / \sqrt{n}} = \frac{94.32-95}{1.20/ \sqrt{16}}[/tex]

= -2.266

= -2.27

Calculating the P value:

P value = 2P(Z < -2.27)

Using the standard normal table,

NORMSDIST(-2.27)

= 0.0116

P value= 2(0.0116)

= 0.0232

Since P value(0.0232) is greater than level of significance (0.01), we fail to reject the null hypothesis H0.

We can say that there is enough evidence to conclude the data does not support that average melting point differs from the level of 95 at the level of 0.01 significance level.

b) B(u = 94)

= P(when u=94, do not reject H0)

Using the standard nkrmal table, the z-score corresponding to

Z(0.01/2)= 0.005 will be

[tex]Z_a_/_2 = 2.58[/tex]

[tex]B(94) = ∅[Z_a_/_2+ \frac{u_0-u}{s.d- \sqrt{n}}] - ∅[-Z_a_/_2+ \frac{u_0-u}{s.d/ \frac{n}}] [/tex]

[tex]B(94) = ∅[2.58+ \frac{95-94}{1.20- \sqrt{16}}] - ∅[-2.58+ \frac{95-94}{1.2/ \frac{16}}] [/tex]

= ∅(5.91)-∅(0.75)

P(Z≤5.91)-P(Z≤0.75)

From standard normal table, we have:

P(Z≤0.75)=0.7734, P(Z≤5.91)=1

= 1 - 0.7734

= 0.2266

Probability of making type II error when u=94 is 0.2266

c) Probability of committing type II error when u= 94 at a significance level of 0.01 will be =0.10.

B(94) = 0.10

Finding sample size, n for a two tailed test:

[tex] n = [\frac{s.d(Z_a_/_2+Z_B)}{u_0-u}]^2 [/tex]

Using standard normal table, Z score corresponding to a/2 = 0.005 cummulative area(1-0.005 = 0.995) is Z= 2.58

Z score corresponding to 0.10 cummulative area(1-0.10 = 0.90) is Z = 1.28

Our sample size n, wil be=

[tex] n = [\frac{1.2(2.58+1.28)}{95-94}]^2 [/tex]

[tex] = [\frac{4.632}{1}]^2 [/tex]

= 21.46

= 22

Sample size = 22

The z-score for the hypothesis test can be calculated and compared with the critical values for a two-tailed level of .01. The probability of a Type II error can be calculated using the standard normal table, and the required sample size to ensure β(94)=0.1 when α=0.01 can be calculated using the appropriate formula. Remember, in the calculations, to use the values provided in the question.

The null hypothesis will be H0: μ = 95 and the alternative hypothesis Ha: μ ≠ 95. Since we are given σ = 1.20 and the sample mean ¯x = 94.32, we can perform a z-test.

The z-score is calculated as: z = (¯x - μ) / (σ / √n), here n is the number of samples i.e., 16. Plug in the given values to calculate the z-score.

If the obtained z-score lies within the critical values of a two-tailed level of .01 (which for a normal distribution is approximately -2.576 and 2.576), we can't reject the null hypothesis.

Beta error, β(94), is the probability of a Type II error when actual mean (µ) is 94. To calculate this, you first need to find the z value for the α level .01, and then use the standard normal table to find the corresponding values for β(94).

To ensure that β(94)=0.1 when α=0.01, use the formula for sample size in hypothesis testing. You would need to find the z-scores associated with α and β, and then use these in the formula: n = [(Zα √2π + Zβ)^2 σ^2]/ (μ - μ0)^2.

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Answer:

$10.32 is the sales tax.

$139.32 is the total price.

Step-by-step explanation:

Sales tax is 8% of the original price.

First, change the percentage into a decimal, by moving the decimal point to the left two place values: 8% = 0.08

Next, multiply 129 with 0.08:

129 x 0.08 = 10.32

$10.32 is the sales tax.

Then, add 10.32 to the original price:

10.32 + 129 = 139.32

$139.32 is the total price.

Answer: B

Step-by-step explanation:

[tex]30(\frac{\pi}{180})=\frac{\pi}{6}[/tex]

Answer:

a) X = N(1955,556)

b) 1955 is the population mean

c) P(X <1911) = 0.4685

d) P(1868 < X <2105) = 0.1685

e) 95th percentile = 2870 (nearest whole number)

Step-by-step explanation:

a) Average number of votes per district, μ = 1955

The standard deviation, [tex]\sigma = 556[/tex]

The distribution of X will take the form [tex]N(\mu, \sigma)[/tex]

Therefore the distribution of X is N(1955,556)

b)1955 is the average of all the votes president Clinton had per district(i.e. the mean of all the values of the population) and not the mean of collected samples.

1955 is the population mean

c) Probability that a randomly selected district had fewer than 1911 votes for president Clinton

[tex]P(X < 1911) = P ( z < \frac{x - \mu}{\sigma} )[/tex]

z=(1911-1955)/556

z=-0.079

From the probability distribution table

P(z<-0.079)=0.4685

Therefore, P(X <1911) = 0.4685

d)probability that a randomly selected district had between 1868 and 2105 votes for President Clinton

[tex]P(x_{1} < X <x_{2}) = P(z_{2} < \frac{x_{2} - \mu }{\sigma} )- P(z_{1} < \frac{x_{1} - \mu }{\sigma} )[/tex]

[tex]P(1868 < X <2105) = P(z_{2} < \frac{2105 - 1955 }{556} )- P(z_{1} < \frac{1868 - 1955 }{556} )\\P(1868 < X <2105) = P(z_{2} < 0.27)- P(z_{1} < -0.16 )[/tex]

P(1868 < X <2105) = 0.6063 - 0.4378

P(1868 < X <2105) = 0.1685

e)  Find the 95th percentile for votes for President Clinton

[tex]P_{95} = \mu + 1.645 \sigma\\P_{95} = 1955 + 1.645(556)\\P_{95} = 2869.62\\P_{95} = 2970[/tex]

P95=1955+1.645*556=2870

The distribution of X is a normal distribution with a sample mean of 1,956.8. The probability of a randomly selected district having fewer than 1,600 votes is approximately 0.267. The probability of a randomly selected district having between 1,800 and 2,000 votes is approximately 0.161. The 95th percentile for votes is approximately 2,885.

b. 1,956.8 is a sample mean because it is calculated from the data of the 40 election districts in Alaska.

c. To find the probability that a randomly selected district had fewer than 1,600 votes for the candidate, you need to standardize the value and use the standard normal distribution table. First, calculate the z-score using the formula z = (x - mean) / standard deviation. Plugging in the values, we get z = (1600 - 1956.8) / 572.3 = -0.621. Then, we can use the standard normal distribution table to find the probability associated with a z-score of -0.621, which is approximately 0.267.

d. To find the probability that a randomly selected district had between 1,800 and 2,000 votes for the candidate, you need to calculate the z-scores for both values and find the area between them on the standard normal distribution. Use the same formula as before to get z1 = (1800 - 1956.8) / 572.3 = -0.275 and z2 = (2000 - 1956.8) / 572.3 = 0.075. Then, use the standard normal distribution table to find the area between z1 and z2, which is approximately 0.161.

e. The 95th percentile represents the value below which 95% of the data falls. To find the 95th percentile for votes for the candidate, you can use the z-score associated with a cumulative probability of 0.95 on the standard normal distribution table. The z-score is approximately 1.645. Then, use the formula x = z * standard deviation + mean to calculate the value. Plugging in the values, we get x = 1.645 * 572.3 + 1956.8 ≈ 2,885 votes.

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Answer:

1) Null hypothesis:[tex]\mu = 120[/tex]  

Alternative hypothesis:[tex]\mu \neq 120[/tex]  

2) [tex]t=\frac{118.3-120}{\frac{3.7}{\sqrt{22}}}=-2.155[/tex]  

3) [tex] t_{cric}=\pm 2.08[/tex]

And the rejection zone of the null hypothesis would be [tex] |t_{calc}|>2.08[/tex]

4) Since our statistic calculated is higher than the critical value we have enough evidence to reject the null hypothesis at 5% of significance

5) Since we reject the null hypothesis of accuracy we have enough evidence to conclude at 5% of significance that the procedure is not accurate

6) Since we reject the null hypothesi we need to expect that the p value would be lower than the significance level provided and that means:

[tex] p_v <\alpha[/tex]

Step-by-step explanation:

Data given and notation  

[tex]\bar X=118.3[/tex] represent the sample mean

[tex]s=3.7[/tex] represent the sample standard deviation

[tex]n=22[/tex] sample size  

[tex]\mu_o =120[/tex] represent the value that we want to test

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

t would represent the statistic (variable of interest)  

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

Part 1: State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true mean is equal to 120 because that means accurate, the system of hypothesis would be:  

Null hypothesis:[tex]\mu = 120[/tex]  

Alternative hypothesis:[tex]\mu \neq 120[/tex]  

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

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

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

Part 2: Calculate the statistic

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

[tex]t=\frac{118.3-120}{\frac{3.7}{\sqrt{22}}}=-2.155[/tex]    

Part 3: Rejection region

We calculate the degrees of freedom given by:

[tex] df = n-1= 22-1 =21[/tex]

We need to find a critical value who accumulates [tex]\alpha/2 = 0.025[/tex] of the area in the tails of the t distribution with 21 degrees of freedom and we got:

[tex] t_{cric}=\pm 2.08[/tex]

And the rejection zone of the null hypothesis would be [tex] |t_{calc}|>2.08[/tex]

Part 4

Since our statistic calculated is higher than the critical value we have enough evidence to reject the null hypothesis at 5% of significance

Part 5

Since we reject the null hypothesis of accuracy we have enough evidence to conclude at 5% of significance that the procedure is not accurate

Part 6

Since we reject the null hypothesi we need to expect that the p value would be lower than the significance level provided and that means:

[tex] p_v <\alpha[/tex]

To cover a rectangle gift box with dimensions of 8 inches by 10 inches by 7 inches, you need to calculate the surface area by adding the area of all six faces. After calculation, you would need 412 square inches of wrapping paper.

To calculate how much wrapping paper is needed to cover a rectangle gift box exactly, we need to find the surface area of the box. The box dimensions are 8 inches wide, 10 inches long, and 7 inches tall.

Surface area of a rectangle box is calculated by adding the areas of all six faces. The areas of the faces are:

Calculating each area:

Adding all of these areas together gives us:

160 in² + 112 in² + 140 in² = 412 in².

Therefore, you would need 412 square inches of wrapping paper to cover the box exactly.

Answer:

Benny's mom's age is 400% of Benny's age.

Step-by-step explanation:

Let p represent the percent of his mom's age to Benny's age.

48=P(12)

p=4=400%

Answer:

n=2/3, -1/4

Step-by-step explanation:

Answer:

20 times

Step-by-step explanation:

8, 18, 28, 38, 48, 58, 68, 78, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 98

1     2    3   4   5   6      7   8     9   10  11    12   13   14   15   16  17 18 19 20

Answer:

-1.64

Explanation:

Multiple Choice Answers that were not provided in the question:

a. 0.103 .

b. None of the above.

c. -4.16

d. -1.45

e. 2.86

f .-1.64

The provided data is as below:

             Student 1  Student 2  Student 3  Student 4  Student 5  Student 6

Pre- test    80              76              76              50             48              67

Post- test   75              86              87              60            48               68

Further Explanation:

                    Student 1  Student 2  Student 3  Student 4  Student 5  Student 6

Pre- test             80              76              76              50             48              67

Post- test            75              86              87              60            48               68

Pretest-Postest   5               -10              -11              -10            0                 -1

Calculating the mean of the difference (Pretest-postest) we get d bar = -4.5

Calculating the standard deviation  of the difference  (Pretest-postest) we get sd= 6.716.

Since sample score is under 30 and population standard deviation is unknown, we will use a tscore to find the standardized test statistic

t = dbar/(sd/√n) = -4.5/(6.716√6) = -1.642 ≈ -1.64

The test statistic is -1.64.

Final answer:

To find the number of kilometers in a 95-mile drive, multiply 95 by the conversion factor of 1.61 kilometers per mile, giving you 152.95 kilometers, which rounds to 153.0 kilometers when rounded to the nearest tenth.

Explanation:

To calculate how many kilometers are in a 95-mile drive when there are approximately 0.62 miles in a kilometer, we need to use the conversion factor between miles and kilometers. The conversion factor is 1 mile equals to 1.61 kilometers. Therefore, to convert 95 miles to kilometers, you multiply 95 miles by 1.61 kilometers per mile.

The calculation will be:

95 miles × 1.61 km/mile = 152.95 kilometers

After calculating, we round the answer to the nearest tenth, resulting in:

153.0 kilometers.

Answer:

Six more than half of a number is 20

Step-by-step explanation:

The given expression in word problem can be translated as:

Six more than half of a number is 20

Answer:

BURRITOS

Step-by-step explanation:

Answer:

9.66666666 yards  repeating

Step-by-step explanation:

The formula is to divide feet by 3,so 29/3=9.666666 repeating.

Mark me brainliest if I helped:D

Using the normal approximation to the binomial, it is found that the correct option is:

The binomial distribution, which is the probability of exactly x successes on n repeated trials, with p probability of success on each trial(each trial either is a success or it is a failure), can be approximated to the normal if in a sample, there are at least 10 successes and at least 10 failures.

In this problem:

Thus, the correct option is:

A similar problem is given at https://brainly.com/question/24261244

The question is faulty written. I'll solve it assuming values for one variable and provide the answer to guide the student in their own question

Answer:

x=t

y=-1-9t

Step-by-step explanation:

Parametric Equations

Given an explicit relation between variables x and y, we can find expression for both of them in term of a third parameter t, such as

x=f(t)

y=g(t)

provided the main relation holds.

we have the equation

9x+y=-1

There are infinitely many forms to find parametric expressions for the variables, let's assume

x=t

Solving the equation for y

y=-1-9x=-1-9t

Thus the parametric equations are

x=t

y=-1-9t

Answer:

a)   32.57%

b)   0.9967

c)   5.3285 times.

Step-by-step explanation:

a. Return on sales is a simple ratio that is calculated by dividing the operating profit/income  by the net sales revenue.

-Given net sales revenue is $20,941 and the operating income is $6,821:

[tex]Return \ on \ sales(ROS)=\frac{Operating \ Income}{Net \ Sales \ Revenue}\\\\=\frac{6821}{20941}\\\\=0.3257\\\\=32.57\%[/tex]

Hence, the return on sales is 32.57%

b. Current ratio is a simple ratio that compares the current assets to the current liabilities.

-Given that current assets=$7,296 and current liabilities=$7,320, the current ratio is calculated as below:

[tex]Current \ Ratio=\frac{Current \ Assets}{Current \ Liabilities}\\\\=\frac{7296}{7320}\\\\=0.9967[/tex]

Hence, the current ratio is 0.9967

c. Inventory turnover is a measure of the frequency with which a company's goods is used or sold and subsequently restocked in given period.

-It's calculated by dividing the cost of goods sold by the average invenory as below:

[tex]Inventory \ Turnover=\frac{Cost \ of \ goods \ sold}{mean \ Invenory}\\\\\\=\frac{7055}{1324}\\\\=5.3285\ times[/tex]

Hence, the inventory turnover is 5.3285 times.

The calculated ratios for ABC Inc. are: Return on Sales of 19.3%, a Current Ratio of 0.997, and an Inventory Turnover of 5.33.

To calculate the ratios requested, we'll be using the financial data from ABC Inc.'s income statement and balance sheet.

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Answer:

a. What is the standard error of the sample proportion who find this offer attractive? =  0.0351

b. What is the probability that the sample proportion is more than 0.15? = 0.9222

c. What is the probability that the sample proportion is between 0.18 and 0.22? =  0.4314

Step-by-step explanation:

see attachment for explanation

Answer:

Step-by-step explanation:

Given that,

p = 0.20

1 - p = 1 - 0.20 = 0.80

n = 130

\mu\hat p = p = 0.20

A) \sigma \hat p =  \sqrt[p( 1 - p ) / n] = \sqrt [(0.20 * 0.80 ) / 130] = 0.0351

B) P( \hat p > 0.15) = 1 - P( \hat p < 0.15)

= 1 - P(( \hat p - \mu \hat p ) / \sigma \hat p < (0.15 - 0.20) / 0.0351 )

= 1 - P(z < -1.42)

Using z table

= 1 - 0.0778

= 0.9222

C) P(0.18 < \hat p < 0.22)

= P[(0.18 - 0.20) / 0.0351 < ( \hat p - \mu \hat p ) / \sigma \hat p < (0.22 - 0.20) / 0.0351]

= P(-0.57 < z < 0.57)

= P(z < 0.57) - P(z < -0.57)

Using z table,    

= 0.7157 - 0.2843

= 0.4314

Answer:

Step-by-step explanation:

find attached the solution

Answer:

  x = 10

Step-by-step explanation:

After your second step, simplify the result:

  [tex]x=\dfrac{6}{3}\cdot\dfrac{5}{1}=2\cdot 5\\\\x=10[/tex]

Answer:

10

Step-by-step explanation:

Answer:

0.0879 or 8.79 % is the process capability of the batch and the batch doesn't meet the control limits.

Step-by-step explanation:

mean of three readings=(1.023+0.999+1.025)/3= 1.0157

standard deviation=√((1.023-1.0157)² + (0.999-1.0157)²+ (1.025-1.0157)²)/3)

                             = 0.0163

Process Capability= min((USL-mean)/3SD, (Mean-LSL)/3SD)

(USL-mean)/3SD= (1.02-1.0157)/(3×0.0163)= 0.0879

(Mean-LSL)/3SD)= (1.0157-1.00)/(3×0.0163)= 0.321

Process capability=(0.0879,0.321)

0.0879 or 8.79 % is the process capability

The x-bar control chart and other statistical tools are used to assess whether product weights are within acceptable limits, thereby determining if quality standards are met and if equipment recalibration might be necessary.

The question involves evaluating if a batch of products (in this case, potentially cereal boxes, soda servings, lifting weights, apples, or lettuce) meets certain quality control standards, specifically relevant to the weights and measurements being within predefined control limits or tolerances. Using an x-bar control chart and statistical methods like hypothesis testing or calculating percent uncertainty, quality control specialists can determine whether a product's quality characteristic, such as weight, is within acceptable ranges, thereby assessing if equipment recalibration is needed or if a batch complies with quality standards.

For instance, if a new sample of items has weights that fall outside of the established control limits on the x-bar control chart, it would suggest that there may be an issue with the process control, indicating potential problems with the lot from which the sample was taken.

Concerning the exercises provided: In the case of the cereal boxes with a standard deviation higher than the acceptable limit, it suggests that recalibration might be needed. For the Class A apples, a hypothesis test at the specified significance levels would help determine compliance with weight tolerance. Calculating the percent uncertainty of a bag's weight informs us about the relative size of the uncertainty in measurements.

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Answer:

The mean of the sampling distribution is 20 and the standard deviation is 2.89.

Step-by-step explanation:

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

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Population:

Mean 20, standard deviation of 5.

Sampling distribution:

3 rounds

Mean = 20

[tex]s = \frac{5}{\sqrt{3}} = 2.89[/tex]

The mean of the sampling distribution is 20 and the standard deviation is 2.89.

Final answer:

The mean of Paul's sampling distribution is 20 points per round, and the standard deviation is approximately 2.89 points per round, calculated using the formula for the standard error of the mean.

Explanation:

The scenario involves a sampling distribution of the mean scores over random samples of 3 rounds from Paul's video game scores. To calculate the mean and standard deviation (sd) of the sampling distribution, we use the following formulas:

Therefore, the mean of the sampling distribution is 20, and the standard deviation is approximately 2.89.