Answer:
0.0062 = 0.62% probability that a randomly selected truck from the fleet will have to be inspected
Step-by-step explanation:
Z-score
Problems of normally distributed samples can be 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.
What is the probability that a randomly selected truck from the fleet will have to be inspected?
Values of Z above 2.5. So this probability is 1 subtracted by the pvalue of Z = 2.5.
Z = 2.5 has a pvalue of 0.9938
1 - 0.9938 = 0.0062
0.0062 = 0.62% probability that a randomly selected truck from the fleet will have to be inspected
Answer:
For this case we want this probability:
[tex] P(X > \mu +2.5 \sigma) [/tex]
And we can use the z score formula given by:
[tex] z = \frac{ X -\mu}{\sigma}[/tex]
And replacing we got:
[tex] z = \frac{\mu +2.5 \sigma -\mu}{\sigma}= 2.5[/tex]
We want this probability:
[tex] P(Z>2.5) = 1-P(Z<2.5)[/tex]
And using the normal standard table or excel we got:
[tex] P(Z>2.5) = 1-P(Z<2.5)= 1-0.9938= 0.0062[/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 mileage of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(\mu,\sigma)[/tex]
For this case we want this probability:
[tex] P(X > \mu +2.5 \sigma) [/tex]
And we can use the z score formula given by:
[tex] z = \frac{ X -\mu}{\sigma}[/tex]
And replacing we got:
[tex] z = \frac{\mu +2.5 \sigma -\mu}{\sigma}= 2.5[/tex]
We want this probability:
[tex] P(Z>2.5) = 1-P(Z<2.5)[/tex]
And using the normal standard table or excel we got:
[tex] P(Z>2.5) = 1-P(Z<2.5)= 1-0.9938= 0.0062[/tex]
Two married couples have purchased theater tickets and are seated in a row consisting of just 4 seats. If they take their seats in a completely random order, what is the probability that at least one of the wives ends up sitting next to her husband? Hint: Use probability of a complement event. That is, P(A) = 1 − P(A′), where A′ is the complement of an event A.
Answer:
The answer is 0.75.
Step-by-step explanation:
To use probability of a complement event, we need to find the situation where none of the couples are sitting next to each other which can happen either two husbands are sitting in the seats 2 and 3 or two wives are sitting in the seats 2 and 3. Also in the first scenario, the wives would need to be sitting at the opposite ends to their husbands and vice versa for the second scenario.
So there are in total 4 scenarios where no couple is sitting next to each other.
There are a total of 16 different ways that they can be seated so the probability of compelement is 16 - 4 = 12, which is 3/4 of the total, meaning that the probability of at least one of the wives sittting next to her husband is 0.75 or 75%.
I hope this answer helps.
An "x-bar" control chart is developed for recording the mean value of a quality characteristic by use of a sample size of three. The control chart has control limits (LCL and UCL) of 1.000 and 1.020 pounds, respectively. If a new sample of three items has weights of 1.023, 0.999, and 1.025 pounds, what can we say about the lot (batch) it came from?
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.
Explanation: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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- (3n - 2)(4n+1)=0
Solve by factoring
Answer:
n=2/3, -1/4
Step-by-step explanation:
From the Balance Sheet and Income Statement Information below, calculate the following ratios:
[a.] Return on Sales
[b.] Current Ratio
[c.] Inventory Turnover – If there are no beginning inventory or ending inventory figures, then use the Merchandise Inventory figure.
ABC INC. Income Statement Year Ended December 31, 2018
Net Sales Revenue $20,941
Cost of Goods Sold 7,055
Gross Profit 13,886
Operating Expenses 7,065
Operating Income 6,821
Interest Expense 210
Income Before Taxes 6,611
Income Tax Expense 2,563
Net Income $4,048
ABC INC. Balance Sheet December 31, 2018 Assets
Current Assets
Cash $2,450
Accounts Receivable 1,813
Merchandise Inventory 1,324
Prepaid Expenses 1,709
Total Current Assets 7,296
Long-Term Assets 18,500
Total Assets $25,796
Liabilities
Current Liabilities $7,320
Long-Term Liabilities 4,798
Total Liabilities 12,028
Stockholders’ Equity
Common Stock 6,568
Retained Earnings 7,200
Total Stockholders’ Equity 13,768
Total Liabilities & Stockholders’ Equity $25,796
1. NOTES: 1- Round up
2- Your responses should be in the following formats
a. XX% b. x.xx c. x.xx
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.
Explanation:To calculate the ratios requested, we'll be using the financial data from ABC Inc.'s income statement and balance sheet.
[a.] Return on SalesThe Return on Sales (ROS) is calculated by dividing the Net Income by the Net Sales Revenue. Here it's $4,048 / $20,941 = 0.193 or 19.3%.[b.] Current RatioThe Current Ratio represents a company's ability to repay its short-term liabilities with its short-term assets. It's calculated by dividing Total Current Assets by Total Current Liabilities: $7,296 / $7,320 = 0.997.[c.] Inventory TurnoverSince we don't have beginning inventory or ending inventory figures, we use the Merchandise Inventory of $1,324 in our calculation. Inventory Turnover is Cost of Goods Sold divided by Merchandise Inventory: $7,055 / $1,324 = 5.33.Learn more about Financial Ratios here:https://brainly.com/question/31531442
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how many cups are equal to 34 pint
By using multiplication, there are 68 cups is equal to 34 pints.
Given that,
Change the number 34 pint into cups.
Used the concept of measurement unit,
A measurement unit is a standard quality used to express a physical quantity. Also, it refers to the comparison between the unknown quantity with the known quantity.
Since we know that;
1 pint = 2 cups
Hence, we get;
34 pint = 34 x 2 cups
= 68 cups
Therefore, the number of cups in 34 pints is 68.
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Do flashcards help students retain the Chinese Language? Six randomly selected students are given a pretest, asked to study a set of flashcards once a day for a week and then given post test for their Chinese Language class. Is there evidence that flashcards can improve scores? Their data is below. What is the test statistic?(pretest – posttest)
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.
Csc^2 u -cos u sec u=cot^2 u
Step-by-step explanation:
csc²u − cos u sec u
csc²u − 1
cot²u
Write parametric equations of the line 9x + y = -1
a. X= 1, y = -9z+ 9
b. X= 1, y =--9
C. X= t_y= 9- 1
d. X= 1, y=-97- 1
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
the length of a classroom is 29 feet what is the measurements in yards and feet
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
14) World Wide Insurance Company found that 53% of the residents of a city had homeowner’s insurance (H) with the company. Of these clients, 27% also had automobile insurance (A) with the company. If a resident is selected at random, find the probabil- ity that the resident has both homeowner’s and automobile insurance with World Wide Insurance Company.
Answer:
The probability that the resident has both homeowner’s and automobile insurance with World Wide Insurance Company is 14% (P=0.1431).
Step-by-step explanation:
The probability of having homeowner’s insurance (H) with the company is P(H)=0.53.
The probability that, given that the resident has homeowner insurance with the company, also had a automobile insurance is P(A|H)=0.27.
We have to calculate P(A&H): probability of having automovile and homeowner:
[tex]P(A\&H)=P(H)*P(A|H)=0.53*0.27=0.1431[/tex]
Answer:
The probability that the resident has both homeowner’s and automobile insurance with World Wide Insurance Company is 0.143 OR 14.3%.
Step-by-step explanation:
We are given that:
53% of the residents had homeowner's insurance i.e. P(H) = 0.53
Of these people, 27% also had automobile insurance i.e. they had automobile insurance given that they had homeowner's insurance. So, P(A|H) = 0.27.
We need to find out the probability that the resident has both homeowner's and automobile insurance i.e. P(A and H) or P(A∩H). To compute P(A∩H) we will use the conditional probability formula:a
P(A|B) = P(A∩B)/P(B)
Here we have:
P(A|H) = P(A∩H)/P(H)
P(A∩H) = P(A|H) * P(H)
= 0.27 * 0.53
P(A∩H) = 0.143
The probability that the resident has both homeowner’s and automobile insurance with World Wide Insurance Company is 0.143 OR 14.3%.
What is the simplification if (x+2)(x+7)
Jim bought a new bicycle for $129. If the sales tax is 8%, how much did Jim pay in sales tax? What was the total cost?
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.
6.6.1. Rao (page 368, 1973) considers a problem in the estimation of linkages in genetics. McLachlan and Krishnan (1997) also discuss this problem and we present their model. For our purposes, it can be described as a multinomial model with the four categories C1, C2, C3 , and C4 . For a sample of size n, let X = (X1, X2, X3 , X4 )′ denote the observed frequencies of the four categories. Hence, n = +4 i=1 Xi. The probability model is
Answer:
Step-by-step explanation:
find attached the solution
Find the volume of a cone with a diameter of 40cm and a height of 40 cm.
Answer:
[tex]v = \pi {r}^{2} \frac{h}{3} \\ \: \: \: \: = \pi \times 20 \frac{40}{3} \\ \: \: \: \: = 16755.16 {cm}^{3} [/tex]
hope this helps you
solve the system equations -x-y=0 and 6x-5y=66 by combining the equations
Answer:
x,y(6.-6)
Step-by-step explanation:
{x=-y
6(-y)-5y=66
-6-5y=66
11y=66
y=6
-x-y=0
x=-y;
x=-6
If f(-3) = -8, then f(x)= 3x +
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!
A corporation is considering a new issue of convertible bonds, mandagemnt belives that the offer terns will be founf attractiv by 20% of all its current stockholders, suppoer that the belif is correct, a random sample of 130 stockholderss is taken.
a. What is the standard error of the sample proportion who find this offer attractive?
b. What is the probability that the sample proportion is more than 0.15?
c. What is the probability that the sample proportion is between 0.18 and 0.22?
d. Suppose that a sample of 500 current stockholders had been taken. Without doing the calculations, state whether the probabilities in parts (b) and (c) would have been higher, lower, or the same as those found.
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
Suppose that 73.2% of all adults with type 2 diabetes also suffer from hypertension. After developing a new drug to treat type 2 diabetes, a team of researchers at a pharmaceutical company wanted to know if their drug had any impact on the incidence of hypertension for diabetics who took their drug. The researchers selected a random sample of 1000 participants who had been taking their drug as part of a recent large-scale clinical trial and found that 718 suffered from h The researchers want to use a one-sample z-test for a population have hypertension while taking their new drug, p, is different from the proportion of all type 2 diabetics who have hypertension. They decide to use a significance level of a 0.01. to see if the proportion of type 2 diabetics who Determine if the requirements for a one-sample z-test for a proportion been met. If the requirements have not been met leave the rest of the questions blank. O The requirements have not been met because there are two samples: the type 2 diabetics with hypertension who are not taking the drug and the ones who are taking the drug O The requirements have been met because the sample was appropriately selected, the variable of interest is categorical with two possible outcomes, and the sampling distribution is approximately normal. O The requirements have not been met because the population standard deviation is unknown. O The requirements have been met because the sample was appropriately selected, the variable of interest is categorical with two possible outcomes, and the sample size is at least 30.
Using the normal approximation to the binomial, it is found that the correct option is:
The requirements have been met because the sample was appropriately selected, the variable of interest is categorical with two possible outcomes, and the sampling distribution is approximately normal.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:
Random sample of 1000, hence, appropriately selected.For each person, there are two possible outcomes, either they suffered from hypertension, or they did not, hence, the variable is categorical.718 have hypertension, 282 have not, at least 10 of each, hence, the sampling distribution is approximately normal.Thus, the correct option is:
The requirements have been met because the sample was appropriately selected, the variable of interest is categorical with two possible outcomes, and the sampling distribution is approximately normal.A similar problem is given at https://brainly.com/question/24261244
A circle has a radius of square root 98 units and is centered at (5.9, 6.7). Write the equation of this circle. Please help!
Given:
Given that the radius of the circle is √98 units.
The circle is centered at the point (5.9, 6.7)
We need to determine the equation of the circle.
Equation of the circle:
The general form of the equation of the circle is given by
[tex](x-h)^2+(y-k)^2=r^2[/tex]
where (h,k) is the center and r is the radius of the circle.
Substituting the center point (h,k) = (5.9, 6.7) and r = √98, we get;
[tex](x-5.9)^2+(y-6.7)^2=(\sqrt{98})^2[/tex]
Simplifying, we get;
[tex](x-5.9)^2+(y-6.7)^2=98[/tex]
Thus, the equation of the circle is [tex](x-5.9)^2+(y-6.7)^2=98[/tex]
the equation of the circle with radius square root of 98 centered at (5.9, 6.7) is: [tex](x - 5.9)^2 + (y - 6.7)^2 = 98[/tex]
We know that the equation of circle is given by the formula:
[tex](x - h)^2 + (y - k)^2 = r^2[/tex]
here h and k are the x and y coordinates of the center of circle and r is the radius. It is given to us that:
Radius = √98 units
Center at (5.9, 6.7)
Putting values into the equation we get:
[tex](x - 5.9)^2 + (y - 6.7)^2 = (\sqrt{98})^2\\\\ (x - 5.9)^2 + (y - 6.7)^2 = 98[/tex]
Therefore, the equation of the circle with radius square root of 98 centered at (5.9, 6.7) is: [tex](x - 5.9)^2 + (y - 6.7)^2 = 98[/tex]
A diet guide claims that you will consume 120 calories from a serving of vanilla yogurt. A random
sample of 22 brands of vanilla yogurt produced a mean of 118.3 calories with standard deviation 3.7
calories.
Suppose that mu is the true mean number of calories for a serving of vanilla yogurt. We want to test if
the diet guide’s claim is accurate. Use significance level 0.05.
1. What are the null and alternative hypotheses?
2. What is the value of the test statistic?
3. What is the rejection region?
4. Do we reject the null hypothesis?
5. Based on this hypothesis test, do we conclude the diet guide is accurate or inaccurate?
6. Should the p-value for this test be greater than or less than the significance level?
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]
Convert 30°to radians. Leave answer as a reduced fraction in terms of π.
a: pi/3
b: pi/6
c: pi/4
d:pi/2
Answer: B
Step-by-step explanation:
[tex]30(\frac{\pi}{180})=\frac{\pi}{6}[/tex]
A rectangular prism has a volume of 120 square feet a height of 15 feet and a width of 4 feet find the length
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!! :)
You wish to test the following claim ( H a ) at a significance level of α = 0.01 . H o : μ = 60.8 H a : μ ≠ 60.8 You believe the population is normally distributed, but you do not know the standard deviation. You obtain a sample of size n = 8 with mean M = 66.9 and a standard deviation of S D = 10.7 . What is the p-value for this sample? (Report answer accurate to four decimal places.) p-value = -.1228 Incorrect The p-value is... less than (or equal to) α greater than α Correct This p-value leads to a decision to... reject the null accept the null fail to reject the null Incorrect
Answer:
Step-by-step explanation:
This is a test of a single population mean since we are dealing with mean
For the null hypothesis,
µ = 60.8
For the alternative hypothesis,
µ ≠ 60.8
Since the number of samples is small and no population standard deviation is given, the distribution is a student's t.
Since n = 8,
Degrees of freedom, df = n - 1 = 8 - 1 = 7
t = (x - µ)/(s/√n)
Where
x = sample mean = 66.9
µ = population mean = 60.8
s = samples standard deviation = 10.7
t = (66.9 - 60.8)/(10.7/√8) = 1.61
We would determine the p value using the t test calculator. It becomes
p = 0.076
Since alpha, 0.01 < than the p value, 0.076, then we would fail to reject the null hypothesis. Therefore, At a 1 % level of significance, the sample data showed that there is no significant evidence that μ ≠ 60.8
Therefore, this p-value leads to a decision to accept the null hypothesis
In the 1992 presidential election Alaska's 40 election districts averaged 1955 votes per district for President Clinton. The standard deviation was 556. (There are only 40 election districts in Alaska.) The distribution of the votes per district for President Clinton was bell-shaped. Let X = number of votes for President Clinton for an election district. (Source: The World Almanac and Book of Facts) Round all answers except part e. to 4 decimal places.
a. What is the distribution of X? X-NO
b. Is 1955 a population mean or a sample mean? Select an answer
C. Find the probability that a randomly selected district had fewer than 1911 votes for President Clinton.
d. Find the probability that a randomly selected district had between 1868 and 2105 votes for President Clinton
e. Find the 95th percentile for votes for President Clinton, Round your answer to the nearest whole number.
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.
Explanation:a. The approximate distribution of X is a normal distribution.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.
Learn more about Normal distribution here:https://brainly.com/question/34741155
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Solve for x: 3x + 3 = -2x + 13
Please add step by step! I am here to understand, not just for answers.
Please and thank you!!
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
Riddle me this. A New Hotel containing 100 rooms. Tom was Hired to paint the numbers from 1-100 on the doors. How many times will Tom have to paint the number 8?
I beat Desi Velasquez
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
Paul plays a video game that is scored by round . In about 500 rounds his career average is 20 points per round with a standard deviation of 5 points per round . Suppose we take random samples of past 3 rounds and calculate the mean number of points scored in each sample . What is the mean and sd of sampling distribibution
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:
The mean of the sampling distribution (μX) is equal to the population mean (μ), which is 20 points per round.The standard deviation of the sampling distribution (σX), also known as the standard error, is equal to the population standard deviation (σ) divided by the square root of the sample size (n), so σX = σ/√n. In this case, it would be 5/√3, approximately 2.89 points per round.Therefore, the mean of the sampling distribution is 20, and the standard deviation is approximately 2.89.
About 8 million tons of plastic waste enters the world’s oceans every year. Write a y = mx equation to show this
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]
Find the values of x, y, and λ that satisfy the system of equations. Such systems arise in certain problems of calculus, and λ is called the Lagrange multiplier. (Enter your answers as a comma-separated list. If there is no solution, enter NO SOLUTION. If the system has an infinite number of solutions, set λ = t and solve for x and y in terms of t.) 3x + λ = 0 3y + λ = 0 x + y − 4 = 0 (x, y, λ) =
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
The supervisor of a production line wants to check if the average time to assemble an electronic component is different from 14 minutes. Assume that the population of assembly time is normally distributed with a standard deviation of 3.4 minutes. The supervisor times the assembly of 14 components, and finds that the average time for completion is 11.6 minutes. How would you calculate the p-value for the hypothesis test
Answer:
The p-value for the hypothesis test is 0.0042.
Step-by-step explanation:
We are given that the supervisor of a production line wants to check if the average time to assemble an electronic component is different from 14 minutes.
Assume that the population of assembly time is normally distributed with a standard deviation of 3.4 minutes. The supervisor times the assembly of 14 components, and finds that the average time for completion is 11.6 minutes.
Let [tex]\mu[/tex] = average time to assemble an electronic component.
SO, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu[/tex] = 14 minutes {means that the average time to assemble an electronic component is equal to 14 minutes}
Alternate Hypothesis, [tex]H_A[/tex] : [tex]\mu[/tex] [tex]\neq[/tex] 14 minutes {means that the average time to assemble an electronic component is different from 14 minutes}
The test statistics that will be used here is One-sample z test statistics as we know about the population standard deviation;
T.S. = [tex]\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] ~ N(0,1)
where, [tex]\bar X[/tex] = sample average time for completion = 11.6 minutes
[tex]\sigma[/tex] = population standard deviation = 3.4 minutes
n = sample of components = 14
So, test statistics = [tex]\frac{11.6-14}{\frac{3.4}{\sqrt{14} } }[/tex]
= -2.64
Now, P-value of the hypothesis test is given by the following formula;
P-value = P(Z < -2.64) = 1 - P(Z [tex]\leq[/tex] 2.64)
= 1 - 0.99585 = 0.0042
Hence, the p-value for the hypothesis test is 0.0042.
The p-value for the hypothesis test is approximately 0.0082
To calculate the p-value for the hypothesis test, follow these steps:
1. State the null and alternative hypotheses:
- Null hypothesis (\(H_0\)): The average assembly time is 14 minutes. \( \mu = 14 \)
- Alternative hypothesis (\(H_1\)): The average assembly time is different from 14 minutes. \( \mu \ne 14 \)
2. Determine the test statistic:
Since the population standard deviation (\(\sigma\)) is known, use the \( z \)-test for the mean. The test statistic is calculated using the formula:
[tex]\[ z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}} \][/tex]
where:
- \(\bar{x}\) is the sample mean,
- \(\mu\) is the population mean under the null hypothesis,
- \(\sigma\) is the population standard deviation,
- \(n\) is the sample size.
Given:
[tex]\(\bar{x} = 11.6\) minutes,\\ \(\mu = 14\) minutes,\\ \(\sigma = 3.4\) minutes,\\ \(n = 14\).[/tex]
Plug these values into the formula:
[tex]\[ z = \frac{11.6 - 14}{3.4 / \sqrt{14}} \][/tex]
3. Calculate the value of the test statistic:
[tex]\[ z = \frac{11.6 - 14}{3.4 / \sqrt{14}} = \frac{-2.4}{3.4 / \sqrt{14}} \approx \frac{-2.4}{0.9087} \approx -2.64 \][/tex]
4. Find the p-value:
The p-value for a two-tailed test is calculated by finding the probability that the test statistic is at least as extreme as the observed value, under the null hypothesis.
[tex]\[ p\text{-value} = 2 \times P(Z \leq z) \][/tex]
Since we have a \( z \)-score of -2.64, we look up the cumulative probability for \( Z = -2.64 \) in the standard normal distribution table or use a calculator to find:
[tex]\[ P(Z \leq -2.64) \approx 0.0041 \][/tex]
Therefore:
[tex]\[ p\text{-value} = 2 \times 0.0041 = 0.0082 \][/tex]
5. Interpret the results:
Compare the p-value to the significance level (\(\alpha\)) to make a decision. Typically, \(\alpha\) is set at 0.05. If the p-value is less than \(\alpha\), we reject the null hypothesis.
In this case:
[tex]\[ p\text{-value} = 0.0082 \][/tex]
Since \(0.0082 < 0.05\), we reject the null hypothesis. There is significant evidence to suggest that the average assembly time is different from 14 minutes.
So, the p-value for the hypothesis test is approximately 0.0082.