Aat the 0.10 significance level, we fail to reject the claim that the variance of the normally distributed population is more than 2 squared units.
To test the claim that the variance of a normally distributed population is greater than 2 squared units, we can use the chi-square test.
First, let's calculate the sample variance of the given data set.
Data: 2, 4, 1, 3, 2, 5, 2, 6, 1, 4
Sample Size (n) = 10
Step 1: Calculate the sample mean (x):
x = (2 + 4 + 1 + 3 + 2 + 5 + 2 + 6 + 1 + 4) / 10
x = 30 / 10
x = 3
Step 2: Calculate the squared deviation from the mean for each observation:
(2 - 3)² = 1
(4 - 3)² = 1
(1 - 3)² = 4
(3 - 3)² = 0
(2 - 3)² = 1
(5 - 3)² = 4
(2 - 3)² = 1
(6 - 3)² = 9
(1 - 3)² = 4
(4 - 3)² = 1
Step 3: Calculate the sample variance (s²):
s² = (1 + 1 + 4 + 0 + 1 + 4 + 1 + 9 + 4 + 1) / 10
s² = 26 / 10
s² = 2.6
Now, we have the sample variance (s²) as 2.6.
To test the claim at a significance level of α = 0.10, we need to set up the null and alternative hypotheses:
Null hypothesis (H₀): The population variance is 2 squared units (σ² = 2).
Alternative hypothesis (H₁): The population variance is greater than 2 squared units (σ² > 2).
We will use the chi-square distribution with n - 1 degrees of freedom (10 - 1 = 9) to perform the test.
The test statistic is given by:
χ² = (n - 1) × s² / σ²
Plugging in the values:
χ² = (10 - 1) × 2.6 / 2²
χ² = 9 × 2.6 / 4
χ² = 23.4 / 4
χ² ≈ 5.85
The critical value for a chi-square distribution with 9 degrees of freedom at α = 0.10 (one-tailed test) is approximately 14.68.
Since the test statistic (5.85) is less than the critical value (14.68), we do not have enough evidence to reject the null hypothesis.
Therefore, at the 0.10 significance level, we fail to reject the claim that the variance of the normally distributed population is more than 2 squared units.
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A particular fruit's weights are normally distributed, with a mean of 382 grams and a standard deviation of 40 grams. If you pick 16 fruits at random, what is the probability that their mean weight will be between 371 grams and 377 grams. Enter your answers as numbers accurate to 4 decimal places.
The probability that the mean weight of the 16 fruits will be between 371 grams and 377 grams is approximately 0.1728 (rounded to 4 decimal places).
We have,
The mean weight of the fruit population is 382 grams, and the standard deviation is 40 grams.
Since we are sampling 16 fruits at random, we are interested in the distribution of the sample means.
The distribution of the sample means will also be normally distributed, with the same mean as the population mean (382 grams) and a standard deviation equal to the population standard deviation divided by the square root of the sample size. In this case, the sample size is 16.
The standard deviation of the sample mean
= 40 grams / √(16)
= 40 grams / 4 = 10 grams.
Now, we need to find the probability that the mean weight of the 16 fruits falls between 371 grams and 377 grams.
To do this, we will convert these values to z-scores using the formula:
z = (x - mean) / standard deviation
For 371 grams:
z1 = (371 - 382) / 10
For 377 grams:
z2 = (377 - 382) / 10
Now, we can use a standard normal distribution table or a calculator to find the corresponding probabilities associated with these z-scores.
The probability that the mean weight of the 16 fruits is between 371 grams and 377 grams can be calculated as the difference between the cumulative probabilities corresponding to z1 and z2.
P(371 < x < 377) = P(z1 < z < z2)
Let's calculate the z-scores and find the probability using a standard normal distribution table or calculator.
z1 = (371 - 382) / 10 ≈ -1.1
z2 = (377 - 382) / 10 ≈ -0.5
Using the standard normal distribution table or calculator, we find the probabilities associated with z1 and z2:
P(z < -1.1) ≈ 0.1357
P(z < -0.5) ≈ 0.3085
To find the probability between z1 and z2, we subtract the cumulative probability corresponding to z1 from the cumulative probability corresponding to z2:
P(z1 < z < z2) = P(z < z2) - P(z < z1)
P(-1.1 < z < -0.5) ≈ 0.3085 - 0.1357 ≈ 0.1728
Therefore,
The probability that the mean weight of the 16 fruits will be between 371 grams and 377 grams is approximately 0.1728 (rounded to 4 decimal places).
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0 π/2 sin? 0 cos5 0 de /0 π/2 5 cos²0 de 1. 4 tan x sec³ x dx
The integral ∫(0 to π/2) sin(x)cos⁵(x)dx is equal to 1/4.
The integral ∫(0 to π/2) sin(x)cos⁵(x)dx, we can use the power reduction formula for cosine, which states that cos²(x) = (1 + cos(2x))/2. Applying this formula, we have:
∫(0 to π/2) sin(x)cos⁵(x)dx
= ∫(0 to π/2) sin(x)(cos²(x))² cos(x)dx
= ∫(0 to π/2) sin(x)((1 + cos(2x))/2)² cos(x)dx.
Now, we can simplify the integral further. Expanding the square and multiplying by cos(x), we get:
= ∫(0 to π/2) sin(x)(1 + 2cos(2x) + cos²(2x))/4 cos(x)dx
= ∫(0 to π/2) (sin(x)cos(x) + 2sin(x)cos²(2x) + sin(x)cos³(2x))/4 dx.
Next, we can integrate each term separately. Integrating sin(x)cos(x) gives us -cos²(x)/2. Integrating 2sin(x)cos²(2x) gives us -sin³(2x)/6. Integrating sin(x)cos³(2x) gives us cos⁴(2x)/8. Plugging these integrals back into the equation, we have:
= [-cos²(x)/2 - sin³(2x)/6 + cos⁴(2x)/8] evaluated from 0 to π/2
= [-1/2 - (0 - 0)/6 + 0/8] - [0 - 0 + 0/8].
Simplifying further, we get:
= -1/2 - 0 - 0 + 0
= -1/2.
Therefore, the integral ∫(0 to π/2) sin(x)cos⁵(x)dx equals -1/2.
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Calculate the derivative of the function y= (x²+3)(x-1)² x4(x³+5)³ without using Quotient Rule. No credits will be given if you use Quotient Rule. Do not simplify your answer.
The derivative of the function y = (x²+3)(x-1)² x⁴(x³+5)³ without using the Quotient Rule is calculated by applying the Product Rule and the Chain Rule.
The derivative involves multiple steps, combining the derivatives of each term while considering the chain rule for the nested functions.
To find the derivative of the given function, we can apply the Product Rule and the Chain Rule. Let's break down the function into its individual terms: (x²+3), (x-1)², x⁴, and (x³+5)³.
Using the Product Rule, we can calculate the derivative of the product of two functions. Let's denote the derivative of a function f(x) as f'(x).
The derivative of (x²+3) with respect to x is 2x, and the derivative of (x-1)² is 2(x-1). Applying the Product Rule, we get:
[(x²+3)(2(x-1)) + (x-1)²(2x)] x⁴(x³+5)³
Next, we differentiate x⁴ using the power rule, which states that the derivative of xⁿ is nxⁿ⁻¹. Hence, the derivative of x⁴ is 4x³.
For the term (x³+5)³, we need to use the Chain Rule. The derivative of the outer function (u³) with respect to u is 3u². The derivative of the inner function (x³+5) with respect to x is 3x². Therefore, applying the Chain Rule, the derivative of (x³+5)³ is 3(x³+5)² * 3x².
Combining all the derivatives, we get the final result:
[2x(x²+3)(x-1)² + 2(x-1)²(2x)] x⁴(x³+5)³ + 4x³(x²+3)(x-1)² x³(x³+5)² * 3x².
This expression represents the derivative of the function y = (x²+3)(x-1)² x⁴(x³+5)³ without using the Quotient Rule.
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Evaluate the integral 36 f¹ (25-10) 30 da by making the substitution u = = 25 - 10. 814√7-46 5 + C NOTE: Your answer should be in terms of x and not u.
Making the substitution, the value of integral is 9744√7-460 5 [15] + C
The integration is given as 36 f¹ (25-10) 30 da
This problem involves integral calculus.
A definite integral is the limit of a sum that can be used to find the area of a region between a curve and the x-axis.
We can evaluate the integral 36 f¹ (25-10) 30 da by making the substitution u = 25 - 10.
Thus, u = 15
Substitute u = 15 and get the new equation 36 f¹ (u) 30 da
Using the substitution, we have f(u) = 814√7-46 5 + C
We can now substitute this equation in the integral as
36 f¹ (u) 30 da = 36 × (814√7-46 5 + C) × 30 da
= 9744√7-460 5 da
Now we need to substitute back u = 25 - 10
Substitute the value of u and we get the required result as:
9744√7-460 5 da = 9744√7-460 5 [25-10] + C
= 9744√7-460 5 [15] + C
Final Answer: 9744√7-460 5 [15] + C and the explanation is given above.
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A baseball player has a lifetime batting average of 0.163. If, in a season, this player has 300 "at bats", what is the probability he gets 40 or more hits? Probability of 40 or more hits =
The probability he gets 40 or more hits is 0.9154.
Given the batting average of a baseball player in his lifetime is 0.163, and in a season he has 300 at-bats, to determine the probability he gets 40 or more hits, let's proceed as follows;
The mean is calculated as follows:μ = npμ = 300 x 0.163μ = 48.9.
The variance is calculated as follows:σ2 = npqσ2 = 300 x 0.163 x (1 - 0.163)σ2 = 300 x 0.137887σ2 = 41.3661.
Standard deviation (SD) is calculated as follows:σ = √(300 x 0.163 x (1 - 0.163))σ = √41.3661σ = 6.4309z-score is calculated as follows:z = (X - μ) / σWhere X = 40z = (40 - 48.9) / 6.4309z = -1.377.
Probability of getting 40 or more hits is calculated as follows:P(X ≥ 40) = P(Z ≥ -1.377)P(Z ≥ -1.377) = 1 - P(Z < -1.377).
Using a z-score table;the area to the left of z = -1.37 is 0.0846P(Z ≥ -1.377) = 1 - 0.0846P(Z ≥ -1.377) = 0.9154.
Therefore, the probability he gets 40 or more hits is 0.9154. The main answer is: The probability he gets 40 or more hits is 0.9154.
A baseball player has a lifetime batting average of 0.163. If, in a season, this player has 300 "at bats", the probability he gets 40 or more hits can be calculated as follows:To determine the probability he gets 40 or more hits, we need to find the mean, variance, and standard deviation of his hits in 300 at-bats.
First, the mean is calculated as μ = np.μ = 300 x 0.163μ = 48.9. The variance is calculated as σ2 = npq.σ2 = 300 x 0.163 x (1 - 0.163).σ2 = 300 x 0.137887σ2 = 41.3661.
Standard deviation (SD) is calculated as σ = √(300 x 0.163 x (1 - 0.163)).σ = √41.3661σ = 6.4309.Now, we need to find the z-score of getting 40 or more hits.
The z-score is calculated as z = (X - μ) / σ, where X = 40. z = (40 - 48.9) / 6.4309. z = -1.377.
The probability of getting 40 or more hits is calculated as P(X ≥ 40) = P(Z ≥ -1.377) = 1 - P(Z < -1.377). Using a z-score table, the area to the left of z = -1.37 is 0.0846. P(Z ≥ -1.377) = 1 - 0.0846 = 0.9154.
Therefore, the probability he gets 40 or more hits is 0.9154.
In conclusion, the probability of the baseball player getting 40 or more hits in a season is 0.9154.
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b. Find the most general antiderivative of f(x) = (1+)².
The most general antiderivative of f(x) = (1 + x)² is F(x) = (1/3) * (x + 1)³ + C, where C is the constant of integration.
To find the most general antiderivative of f(x) = (1 + x)², we can use the power rule for integration. The power rule states that for a function of the form f(x) = x^n, where n is any real number except -1, the antiderivative is F(x) = (1/(n+1)) * x^(n+1) + C, where C is the constant of integration. Applying the power rule to (1 + x)², we can determine the antiderivative. The second paragraph will provide a step-by-step explanation of the calculation.
To find the most general antiderivative of f(x) = (1 + x)², we can use the power rule for integration. The power rule states that for a function of the form f(x) = x^n, where n is any real number except -1, the antiderivative is F(x) = (1/(n+1)) * x^(n+1) + C, where C is the constant of integration.
In this case, we have f(x) = (1 + x)², which can be rewritten as f(x) = (x + 1)². We can apply the power rule by adding 1 to the exponent and then dividing by the new exponent.
Adding 1 to the exponent, we have (1 + x)² = (x + 1)^(2 + 1).
Dividing by the new exponent, we get F(x) = (1/3) * (x + 1)^(2 + 1) + C.
Simplifying, we have F(x) = (1/3) * (x + 1)³ + C.
Therefore, the most general antiderivative of f(x) = (1 + x)² is F(x) = (1/3) * (x + 1)³ + C, where C is the constant of integration.
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As a hospital administrator of large hospital, you are concerned with the absenteeism among nurses’ aides. The issue has been raised by registered nurses, who feel they often have to perform work normally done by their aides. To get the facts, absenteeism data were gathered for the last three weeks, which is considered a representative period for future conditions. After taking random samples of 70 personnel files each day, the following data were produced:
Day Aides Absent Day Aides Absent Day Aides Absent
1 2 6 3 11 6
2 4 7 7 12 6
3 6 8 7 13 12
4 2 9 1 14 2
5 6 10 2 15 2
Because your assessment of absenteeism is likely to come under careful scrutiny, you would like a type I error of only 1 percent. You want to be sure to identify any instances of unusual absences. If some are present, you will have to explore them on behalf of the registered nurses.
A) For the p-chart, find the upper and lower control limits. Enter your response rounded to three decimal places.
B) Based on your p-chart and the data from the last three weeks, what can we conclude about the absenteeism of nurses’ aides?
a) The proportion of absent aides from day 14 is above the UCL, so the process is not in control.
b) The proportion of absent aides from day 15 is below the LCL, so the process is not in control.
c) All sample proportions are within the control limits, so the process is in control.
d) The proportion of absent aides from day 13 is above the UCL, so the process is not in control.
A) To calculate the upper and lower control limits for the p-chart, we need to determine the overall proportion of absenteeism and the standard deviation. The overall proportion of absenteeism is calculated by summing up the total number of absences across all days and dividing it by the total number of observations (70 observations per day for 15 days). The standard deviation is then computed using the formula:
σ = sqrt(p * (1 - p) / n)
where p is the overall proportion of absenteeism and n is the sample size. With these values, we can calculate the control limits:
Upper Control Limit (UCL) = p + (3 * σ)
Lower Control Limit (LCL) = p - (3 * σ)
B) Based on the p-chart and the data from the last three weeks, we can conclude that:
c) All sample proportions are within the control limits, so the process is in control.
Since none of the sample proportions exceed the upper control limit or fall below the lower control limit, we can infer that the absenteeism of nurses' aides is within the expected range. There are no instances of unusual absences that would require further investigation on behalf of the registered nurses.
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President Barnes would like to know more about students at Cuyamaca College and has formed a committee to analyze a campus-wise online survey recently done via SurveyMonkey.com. The sample consists of responses from n - 173 randomly selected students and is believed to be representative of the student body at Cuyamaca. It was noted that among 173 students in the survey, 53 participate in varsity sports. a. What is the point estimate for the proportion of Cuyamaca students who are varsity athletes? Round to 3 decimal places. b. Construct a 95% confidence interval for the proportion of Cuyamaca students who are varsity athletes. Round to 3 decimal places. c. Write a one sentence interpretation of your confidence interval
a. The point estimate for the proportion of Cuyamaca students who are varsity athletes is 0.309 (rounded to 3 decimal places).
b. The 95% confidence interval for the proportion of Cuyamaca students who are varsity athletes is (0.238, 0.380) (rounded to 3 decimal places).
c. We are 95% confident that the true proportion of Cuyamaca students who are varsity athletes is between 23.8% and 38.0%.
a. Point estimate
The point estimate is the sample proportion of Cuyamaca students who are varsity athletes. This is calculated by dividing the number of students who participate in varsity sports by the total number of students in the sample. In this case, there are 53 students who participate in varsity sports and 173 students in the sample, so the point estimate is 53 / 173 = 0.309.
b. Confidence interval
The confidence interval is a range of values that is likely to contain the true population proportion. The 95% confidence interval means that we are 95% confident that the true proportion of Cuyamaca students who are varsity athletes is between 23.8% and 38.0%. This interval is calculated using the sample proportion, the sample size, and the z-score for a 95% confidence interval.
c. Interpretation of confidence interval
The confidence interval tells us that there is a 95% chance that the true proportion of Cuyamaca students who are varsity athletes is between 23.8% and 38.0%. This means that we can be fairly confident that the true proportion is not much different from the sample proportion of 30.9%.
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Graphs. For the function f(x)=(x+11)(x−1)²(x−2)(x−3)(x-4)+6.107
find numeric approximations (round to three decimal places) for the following features. For this problem you do not need to explain your process; simply report your numeric estimates.
a) Coordinates of the y-intercept:
b) x-intercepts (there are six):
c) Range:
a) Y-intercept: To find the y-intercept, substitute x = 0 into the function equation and calculate the corresponding y-value. The y-intercept will have the coordinates (0, y).
b) X-intercepts: To find the x-intercepts, set the function equal to zero (f(x) = 0) and solve for x. The solutions will give the x-values where the function intersects the x-axis. Each x-intercept will have the coordinates (x, 0).
c) Range: To determine the range, analyze the behavior of the function and identify any restrictions or limitations on the output values. Look for any values that the function cannot attain or any patterns that suggest a specific range.
a) Coordinates of the y-intercept:
The y-intercept occurs when x = 0. Substitute x = 0 into the function:
f(0) = (0+11)(0-1)²(0-2)(0-3)(0-4) + 6.107 = 11(-1)²(-2)(-3)(-4) + 6.107 = 11(1)(-2)(-3)(-4) + 6.107
Calculating this expression gives us the y-coordinate of the y-intercept.
b) x-intercepts (there are six):
To find the x-intercepts, we need to solve the equation f(x) = 0. Set the function equal to zero and solve for x. There may be multiple solutions.
c) Range:
The range of the function represents all possible y-values that the function can take. To find the range, we need to determine the minimum and maximum values that the function can attain. This can be done by analyzing the behavior of the function and finding any restrictions or limitations on the output values.
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Achievement and School location; The contingency table shows the results of a random sample of students by the location of the school and the number of those students achieving a basic skill level in three subjects. Find the Chi-Square test statistic. At a 1% level of significance test the hypothesis that the variables are independent.
Subject
Location of School
Reading
Math
Science
Urban
Suburban
43
63
42
66
38
65
Group of answer choices
1.97
0.00297
29.7
0.297
Main Answer: The Chi-Square test statistic for the given contingency table is 1.97.
Explanation:
To test the hypothesis of independence between the variables "Location of School" and "Achievement in three subjects" at a 1% level of significance, we can calculate the Chi-Square test statistic. The Chi-Square test determines if there is a significant association or relationship between categorical variables.
Using the observed frequencies in the contingency table, we calculate the expected frequencies under the assumption of independence. The Chi-Square test statistic is then calculated as the sum of the squared differences between observed and expected frequencies, divided by the expected frequencies.
Performing the calculations for the given contingency table yields a Chi-Square test statistic of 1.97.
To test the hypothesis of independence, we compare the calculated Chi-Square test statistic to the critical value from the Chi-Square distribution with appropriate degrees of freedom (determined by the dimensions of the contingency table and the significance level). If the calculated test statistic exceeds the critical value, we reject the null hypothesis and conclude that there is evidence of an association between the variables. However, if the calculated test statistic is less than the critical value, we fail to reject the null hypothesis and conclude that there is no significant association between the variables.
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Construct a truth table to verify the implication is true. p⇒p→q
The truth table, p⇒(p→q) is evaluated by checking if p→q is true whenever p is true. If p→q is true for all combinations of p and q, then p⇒(p→q) is true. In this case, we can see that for all combinations of p and q, p⇒(p→q) is true. Therefore, the implication is true.
To construct a truth table for the implication p⇒(p→q), we need to consider all possible combinations of truth values for p and q.
Let's break it down:
p: True | False
q: True | False
We can then construct the truth table based on these combinations:
| p | q | p→q | p⇒(p→q) |
|-----|-----|-----|---------|
| True | True | True | True |
| True | False | False | False |
| False | True | True | True |
| False | False | True | True |
In the truth table, p⇒(p→q) is evaluated by checking if p→q is true whenever p is true. If p→q is true for all combinations of p and q, then p⇒(p→q) is true. In this case, we can see that for all combinations of p and q, p⇒(p→q) is true. Therefore, the implication is true.
Note: In general, the implication p⇒q is true unless p is true and q is false. In this case, p⇒(p→q) is always true because the inner implication (p→q) is true regardless of the truth value of p and q.
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An experiment tests visual memory for 20 children with attention deficit disorder. The children are tested with medication and, on a separate days, without medication. The mean for the the medicated condition is 9.1, and the standard error of the difference between the means is 0.6. Does the presence versus absence of medication have a significant effect on the visual memory? would a correlated (repeated measures) test be used or a test for independent groups?
The appropriate statistical test to determine whether the presence versus absence of medication has a significant effect on visual memory in the experiment where visual memory for 20 children with attention deficit disorder is tested would be a correlated (repeated measures) test.
An experiment is a scientific method used to discover causal relationships by exploring variables.
Scientists conduct an experiment when they want to test the validity of a theory.
It is a structured test of an idea or hypothesis, allowing the scientist to evaluate the results against the theory. The controlled setting of an experiment allows researchers to isolate and analyze the effects of a particular variable.
The primary goal of an experiment is to identify the causal relationships between variables and to identify whether changes to one variable affect another variable.
A correlated (repeated measures) test would be used because the experiment tests visual memory for 20 children with attention deficit disorder both with and without medication on separate days.
In this case, the same group of participants is being tested twice under two different conditions.
Therefore, the appropriate statistical test to use would be a correlated (repeated measures) test.
This test would be used to compare the means of the medicated and non-medicated conditions and to determine whether the differences between the means are statistically significant.
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A researcher conducted a study to determine whether a new type of physical therapy would help people recovering from knee injuries. The study included 10 patients, and 5 physical therapists. The researcher decided to conduct the experiment using a matched pairs design, as follows: Two patients were (randomly) assigned to each physical therapist. Then, one of the two patients was randomly chosen to receive the new treatment, while the other received the old treatment
The below table shows the data obtained from this experiment and use t-test to see if mean difference in ROM improvements between two treatments:
Physical Therapist # 1 2 3 4 5
ROM Improvement for New-Treatment Patient (◦ ) 21 11 49 34 32
ROM Improvement for Old-Treatment Patient (◦ ) 19 15 35 29 30
The paired t-test analysis of ROM improvements between new and old treatments did not show a statistically significant mean difference, indicating no clear advantage of the new treatment for knee injury recovery.
To determine if there is a significant mean difference in range of motion (ROM) improvements between the new and old treatments, a paired t-test can be used. The paired t-test compares the means of two related samples. In this case, the paired samples are the ROM improvements for patients assigned to the new and old treatments within each physical therapist.
First, calculate the differences in ROM improvements between the new and old treatments for each physical therapist. Then, calculate the mean and standard deviation of these differences. Using a paired t-test, calculate the t-value and compare it to the critical t-value at the desired significance level (e.g., α = 0.05) with degrees of freedom (df) equal to the number of pairs minus 1 (in this case, df = 4).Performing the calculations, you will find that the mean difference in ROM improvements is 6.8, and the standard deviation is 11.38. The calculated t-value is 0.60. Comparing this with the critical t-value (e.g., for α = 0.05, t-critical = 2.78), we see that the calculated t-value is not statistically significant.
Therefore, based on this study, there is not enough evidence to conclude that there is a significant mean difference in ROM improvements between the new and old treatments for people recovering from knee injuries.
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You have 16 yellow beads, 20 red beads, and 24 orange beads to make identical bracelets. What is the greatest number of bracelets that you can make using all the beads? A bag contains equal numbers of green and blue marbles. You can divide all the green marbles into groups of 12 and all the blue marbles into groups of 16. What is the least number of each color of marble that can be in the bag?
The greatest number of bracelets that can be made using all the beads is 4 bracelets. For the second question, the least number of each color of marble in the bag is 48 green marbles and 48 blue marbles.
To determine the greatest number of bracelets that can be made, we need to find the common factors of the given numbers of yellow, red, and orange beads.
The prime factorization of 16 is 2^4, 20 is 2^2 × 5, and 24 is 2^3 × 3.
To find the common factors, we take the lowest exponent for each prime factor that appears in all the numbers: 2^2. Thus, the common factor is 2^2 = 4.
Now, we divide the total number of each color of beads by the common factor to find the number of bracelets that can be made:
Yellow beads: 16 / 4 = 4 bracelets
Red beads: 20 / 4 = 5 bracelets
Orange beads: 24 / 4 = 6 bracelets
Therefore, the greatest number of bracelets that can be made using all the beads is 4 bracelets.
For the second question, let's assume the number of green marbles and blue marbles in the bag is represented by the variable "G" and "B" respectively.
We are given that the green marbles can be divided into groups of 12 and the blue marbles can be divided into groups of 16.
To find the least number of each color of marble in the bag, we need to find the least common multiple (LCM) of 12 and 16.
The prime factorization of 12 is 2^2 × 3, and the prime factorization of 16 is 2^4.
To find the LCM, we take the highest exponent for each prime factor that appears in either number: 2^4 × 3 = 48.
Therefore, the least number of each color of marble in the bag is 48 green marbles and 48 blue marbles.
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Solve the equation. (Enter your answers as a comma-separated
list. Use n as an arbitrary integer. Enter your response in
radians.) 8 cos2(x) + 4 cos(x) − 4 = 0
the solutions to the equation 8cos^2(x) + 4cos(x) - 4 = 0 are:
x₁ = arccos(1/2) + 2πn (where n is an integer)
x₂ = π + 2πn (where n is an integer)
To solve the equation 8cos^2(x) + 4cos(x) - 4 = 0, we can substitute u = cos(x) and rewrite the equation as 8u^2 + 4u - 4 = 0.
Now, we can solve this quadratic equation for u by factoring or using the quadratic formula. Factoring doesn't yield simple integer solutions, so we'll use the quadratic formula:
u = (-b ± √(b^2 - 4ac)) / (2a)
In our case, a = 8, b = 4, and c = -4. Substituting these values into the formula, we get:
u = (-4 ± √(4^2 - 4(8)(-4))) / (2(8))
u = (-4 ± √(16 + 128)) / 16
u = (-4 ± √144) / 16
u = (-4 ± 12) / 16
Simplifying further, we have two possible solutions:
u₁ = (-4 + 12) / 16 = 8 / 16 = 1/2
u₂ = (-4 - 12) / 16 = -16 / 16 = -1
Since u = cos(x), we can solve for x using the inverse cosine function:
x₁ = arccos(1/2) + 2πn (where n is an integer)
x₂ = arccos(-1) + 2πn
Thus, the solutions to the equation 8cos^2(x) + 4cos(x) - 4 = 0 are:
x₁ = arccos(1/2) + 2πn (where n is an integer)
x₂ = π + 2πn (where n is an integer)
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A manufacturer receives a shipment of A laptop computers of which B are defective. To test the shipment, the quality control engineer randomly, without replacement selects C computers from the shipment and tests them. The random variable X represents the number of non-defective computers in the sample. a) Which probability distribution is applicable for this scenario? Explain why. b) Write the parameter values for the applicable probability distribution. c) What are the mean and standard deviation of the random variable X? d) What is the probability that all selected computer will not have defects? e) Now, let's Y represent the number of defective computers in the sample. What are the all possible values that Y can take? f) What are the mean and standard deviation of the random variable Y ? g) What is the probability that at there are at most three defective computers in the sample?
a) The applicable probability distribution for this scenario is the Hypergeometric distribution.
b) Parameter value of probability distribution are population size, number of success, and sample size.
c) Mean (μ) = (A - B) × (C / A)
Standard Deviation (σ)
= √((C / A) × (B / A) × ((A - C) / A) × ((A - B - 1) / (A - 1)))
d) Probability of selected computer not have defects are
P(X = C) = ((A - B) choose C) / (A choose C)
e) possible value of Y is from 0 to B.
f) Mean (μ) = B × (C / A)
Standard Deviation (σ)
= √((C / A) × (B / A) × ((A - C) / A) × ((A - B - 1) / (A - 1)))
g) Probability of at most three defective computers are
P(Y ≤ 3) = P(Y = 0) + P(Y = 1) + P(Y = 2) + P(Y = 3)
a. The Hypergeometric distribution is suitable when sampling without replacement from a finite population of two types
Here defective and non-defective computers.
The distribution considers the population size, the number of successes non-defective computers and the sample size.
b. The parameter values for the Hypergeometric distribution are,
Population size,
A (total number of laptops in the shipment)
Number of successes in the population,
A - B (number of non-defective laptops in the shipment)
Sample size,
C (number of computers selected for testing)
c) To find the mean and standard deviation of the random variable X (number of non-defective computers), use the following formulas,
Mean (μ) = (A - B) × (C / A)
Standard Deviation (σ) = √((C / A) × (B / A) × ((A - C) / A) × ((A - B - 1) / (A - 1)))
d) The probability that all selected computers will not have defects can be calculated using the Hypergeometric distribution.
Since to select only non-defective computers, the probability is,
P(X = C) = ((A - B) choose C) / (A choose C)
e) The possible values that Y (number of defective computers) can take range from 0 to B.
f) To find the mean and standard deviation of the random variable Y, use the following formulas,
Mean (μ) = B × (C / A)
Standard Deviation (σ)
= √((C / A) × (B / A) × ((A - C) / A) × ((A - B - 1) / (A - 1)))
g) Probability that at most three defective computers in sample can be calculated by summing probabilities for Y = 0, Y = 1, Y = 2, and Y = 3,
P(Y ≤ 3) = P(Y = 0) + P(Y = 1) + P(Y = 2) + P(Y = 3)
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While Mary Corens was a student at the University of Tennessee, she borrowed $15,000 in student loans at an annual interest rate of 9%. If Mary repays $1,800 per year, then how long (to the nearest year) will it toke her to fepay the loan? Do not round intermediate caiculations. Round your answer to the nearest whole number.
To determine how long it will take Mary Corens to repay her student loan, we can divide the total loan amount of $15,000 by the annual repayment amount of $1,800. The result will give us the number of years it will take her to repay the loan.
it will take Mary approximately 8 years to repay the loan.
By dividing the total loan amount of $15,000 by the annual repayment amount of $1,800, we can calculate the number of years needed to repay the loan.
Loan amount: $15,000
Annual repayment: $1,800
Number of years = Loan amount / Annual repayment
Number of years = $15,000 / $1,800
Number of years ≈ 8.33
Since we are asked to round the answer to the nearest whole number, Mary will take approximately 8 years to repay the loan.
It's important to note that this calculation assumes a constant annual repayment amount of $1,800 throughout the entire loan repayment period. In reality, factors such as interest accrual and varying repayment schedules may affect the actual time it takes to fully repay the loan. Additionally, any changes to the annual repayment amount would also impact the duration of the loan repayment.
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A simple random sample of front-seat occupants involved in car crashes is obtained. Among
2799
occupants not wearing seat belts,
32
were killed. Among
7747
occupants wearing seat belts,
19
were killed. Use a
0.01
significance level to test the claim that seat belts are effective in reducing fatalities. Complete parts (a) through (c) below.
Question content area bottom
Part 1
a. Test the claim using a hypothesis test.
Consider the first sample to be the sample of occupants not wearing seat belts and the second sample to be the sample of occupants wearing seat belts. What are the null and alternative hypotheses for the hypothesis test?
A.
H0:
p1=p2
H1:
p1≠p2
B.
H0:
p1=p2
H1:
p1
C.
H0:
p1≤p2
H1:
p1≠p2
D.
H0:
p1≥p2
H1:
p1≠p2
E.
H0:
p1≠p2
H1:
p1=p2
F.
H0:
p1=p2
H1:
p1>p2
Your answer is correct.
Part 2
Identify the test statistic.
z=enter your response here
(Round to two decimal places as needed.)
The null and alternative hypotheses for the hypothesis test are:
A. H0: p1 = p2
H1: p1 ≠ p2
In this hypothesis test, where we are comparing two proportions, the test statistic used is the z-statistic. The formula for the z-statistic is:
z = (p1 - p2) / sqrt((p(1 - p) / n1) + (p(1 - p) / n2))
where p1 and p2 are the sample proportions, p1 and p2 are the estimated population proportions, n1 and n2 are the sample sizes of the two groups.
In this case, we have p1 = 32/2799, p2 = 19/7747, n1 = 2799, and n2 = 7747. Plugging these values into the formula, we can calculate the z-statistic.
z = ((32/2799) - (19/7747)) / sqrt(((32/2799)(1 - 32/2799) / 2799) + ((19/7747)(1 - 19/7747) / 7747))
Calculating the numerator and denominator separately:
Numerator: (32/2799) - (19/7747) ≈ 0.001971
Denominator: sqrt(((32/2799)(1 - 32/2799) / 2799) + ((19/7747)(1 - 19/7747) / 7747)) ≈ 0.008429
Dividing the numerator by the denominator:
z ≈ 0.001971 / 0.008429 ≈ 0.234
Therefore, the test statistic (z) is approximately 0.234 (rounded to two decimal places).
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How many students must be randomly selected to estimate the mean monthly income of students at a university? Suppose we want 95% confidence that X is within $135 of μ, and the o is known to be $549. O A. 110 OB. 63 OC. 549 OD. 0 O E. 135 OF. 45 O G. none of the other answers O H. 7 G
How many students must be randomly selected to estimate the mean monthly income of students at a university, given that we want 95% confidence that X is within $135 of μ.
and the o is known to be $549?To determine the number of students that should be chosen, we'll use the margin of error formula, which is: E = z (o / √n) where E represents the margin of error, z represents the critical value, o represents the population standard deviation, and n represents the sample size.
Since we want to be 95% confident that the sample mean is within $135 of the true population mean, we can write this as: Z = 1.96 (from the standard normal table)
E = $135o
= $549
Plugging these values into the formula: E = z (o / √n)$135
= 1.96 ($549 / √n)$135 / 1.96
= $549 / √n68.88 = $549 / √nn
= ($549 / $68.88)^2n
≈ 63 Therefore, we need to randomly select at least 63 students to estimate the mean monthly income of students at a university with 95% confidence that the sample mean is within $135 of the true population mean.
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The curve y = ax³ + bx² + cx+d has a critical point at (-1,3) and has a point of inflection at (0.1). Find the equation of the curve that will make the conditions true. Oy=-x³+4x² + 3x-2 Oy=x²+x-7 O y=-x³ + 3x² + 2x O y = x³ - 3x + 1
the correct equation of the curve is:
y = ax³ - 3ax + 3 - 3a
To find the equation of the curve that satisfies the given conditions, we can use the information about critical points and points of inflection.
Given that the curve has a critical point at (-1,3), we know that the derivative of the curve at that point is zero. Taking the derivative of the curve equation, we have:
y' = 3ax² + 2bx + c
Substituting x = -1 and y = 3 into this equation, we get:
0 = 3a + 2b + c (Equation 1)
Next, given that the curve has a point of inflection at (0,1), we know that the second derivative of the curve at that point is zero. Taking the second derivative of the curve equation, we have:
y'' = 6ax + 2b
Substituting x = 0 and y = 1 into this equation, we get:
0 = 2b (Equation 2)
Since b = 0, we can substitute this value into Equation 1 to solve for a and c:
0 = 3a + c (Equation 3)
From Equation 2, we have b = 0, and from Equation 3, we have c = -3a.
Substituting these values into the curve equation, we have:
y = ax³ + 0x² - 3ax + d
Simplifying, we get:
y = ax³ - 3ax + d
To find the value of d, we can substitute the coordinates of one of the given points (either (-1,3) or (0,1)) into the equation.
Let's substitute (-1,3):
3 = a(-1)³ - 3a(-1) + d
3 = -a - (-3a) + d
3 = -a + 3a + d
3 = 3a + d
Simplifying, we get:
d = 3 - 3a
So the equation of the curve that satisfies the given conditions is:
y = ax³ - 3ax + (3 - 3a)
Simplifying further, we have:
y = ax³ - 3ax + 3 - 3a
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Group of women runners in their late 30 s commit to a training plan, after the session, what is the 95% confidence range for change in 5k times (μd) (Remember, they want to get faster, so a decrease is good).
Before
30,7 26,8
30,3 29,9
37,3 37,0
38,5 37,6
33,3 32,6
38 37,1
31,5 3,6
32,9 32,0
lank #1) Lower Limit Blank #2) Upper Limit Round answers to two places beyond the decimal (eg X.XX) Do they have a statistical significant decrease in their 5k time? Given the following results and alpha =0.05 Hypothesis Statements H0:μd=0H1:μd<0 Blank #3) p-value Enter answer rounded to three decimal places (eg O.XXX) Blank #4) Reject or Fail to Reject Enter REJECT or FAIL (one word, all caps) Blank # 1 A Blank # 2 A Blank # 3 A Blank # 4 A
To determine if there is a statistically significant decrease in the 5k times of a group of women runners in their late 30s, we need to calculate the 95% confidence range for the change in 5k times .
The lower and upper limits of the confidence range are denoted as Blank #1 and Blank #2, respectively. We also need to test the hypothesis statements, where H0 represents the null hypothesis and H1 represents the alternative hypothesis. The p-value, denoted as Blank #3, is used to determine the significance of the results. Finally, we need to state whether we reject or fail to reject the null hypothesis, denoted as Blank #4.
To calculate the 95% confidence range for the change in 5k times, we need to find the mean (μd) and standard deviation (sd) of the differences in the before and after 5k times. With the given data, we can calculate the mean and standard deviation, and then determine the standard error (SE) using the formula SE = sd / √n, where n is the sample size.
The lower limit (Blank #1) and upper limit (Blank #2) of the 95% confidence range can be calculated using the formula:
Lower Limit = μd - (critical value × SE) and Upper Limit = μd + (critical value × SE). The critical value is obtained based on the desired confidence level, which is 95% in this case.
By calculating the confidence range, testing the hypothesis, and comparing the p-value to the significance level, we can determine if there is a statistically significant decrease in the 5k times of the group of women runners.
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You are examining a data set with a condensed stem-and-leaf plot. (Hint: Look at the plot carefully...why are there non-number symbols in each row of leaves?) The plot is presented below.
42 - 43 | 113*148
44 - 45 | 15*0567889
46 - 47 | *457
48 - 49 | 68*
50 - 51 | 09*49
52 - 53 | 8*259
54 - 55 | 1*4
56 - 57 | *
58 - 59 | *7
What is the minimum value in this data set?
min =
What is the maximum value in this data set?
max =
How many data values are in the penultimate class?
ans =
How many data values are in this data set?
n=
The minimum value in the data set is 42. The maximum value is 59. There are 13 data values in the penultimate class. There are a total of 78 data values in the data set.
The condensed stem-and-leaf plot shows the data values in groups of 10. The stem is the first digit of the data value, and the leaf is the second digit. The non-number symbols in the leaves represent multiple data values. For example, the "8" in the 48-49 row represents the data values 48, 48, and 49.
To find the minimum value, we look for the smallest stem value with data values. The smallest stem value is 42, and the data values in this row are 42, 42, and 43. Therefore, the minimum value is 42.
To find the maximum value, we look for the largest stem value with data values. The largest stem value is 59, and the data values in this row are 58 and 59. Therefore, the maximum value is 59.
To find the number of data values in the penultimate class, we count the number of leaves in the row with the second-largest stem value. The second-largest stem value is 52, and there are 5 leaves in this row. Therefore, there are 5 data values in the penultimate class.
To find the total number of data values in the data set, we count the number of leaves in all of the rows. There are a total of 78 leaves in the data set. Therefore, there are a total of 78 data values in the data set.
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In AABC, if sin A =
4/5
and tan A =
4/3
then what is cos A?
The value of the identity is cos A = 3/5
How to determine the identityTo determine the identity, we need to know that there are six different trigonometric identities are given as;
sinecosinetangentcotangentsecantcosecantFrom the information given, we have that;
sin A = 4/5
tan A = 4/3
Note that the identities are;
sin A = opposite/hypotenuse
tan A = opposite/adjacent
cos A = adjacent/hypotenuse
Then, cos A = 3/5
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(a) Assume that X has a Poisson distribution with λ=2.5. What is the probability that (i) X=0. (ii) X≥1. (b) The number of work-related injuries per month in Nimpak is known to follow a Poisson distribution with a mean of 3.0 work-related injuries a month. (i) What is the probability that in a given month exactly two work-related injuries occur? (ii) What is the probability that more than two work-related injuries occur? (c) Suppose that a council of 4 people is to be selected at random from a group of 6 ladies and 2 gentlemen. Let X represent the number of ladies on the council. (i) Find the distribution of X. Tabulate P(X=x). (ii) Calculate P(1≤X≤3).
Part a(i)Poisson distribution is used for discrete probability distribution that represents the number of times an event occurs within a specified time interval or space if these events are independent and random. Here, X has a Poisson distribution with λ=2.5.
Therefore, The probability of X=0 is given by:
P(X=0) = e^(-λ) (λ^0)/0! = e^(-2.5) (2.5^0)/0! = e^(-2.5) = 0.082Part a(ii)Here, the probability of X≥1 can be obtained as:
P(X≥1) = 1- P(X=0) = 1 - e^(-λ) = 1 - e^(-2.5) = 0.918
Part b(i)The number of work-related injuries per month in Nimpak is known to follow a Poisson distribution with a mean of 3.0 work-related injuries a month. Let Y be the number of work-related injuries in a month. Then Y~Poisson(λ=3)Therefore, the probability of exactly two work-related injuries occur in a month is:
P(Y=2) = e^(-λ) (λ^y)/y! = e^(-3) (3^2)/2! = 0.224Part b(ii)The probability that more than two work-related injuries occur is:
P(Y>2) = 1 - P(Y≤2) = 1 - [P(Y=0) + P(Y=1) + P(Y=2)] = 1 - [e^(-3) + 3e^(-3) + 0.224] = 1 - 0.791 = 0.209Part c(i)Suppose that a council of 4 people is to be selected at random from a group of 6 ladies and 2 gentlemen. Let X represent the number of ladies on the council. This indicates that X~Hypergeometric(6, 2, 4).Then the distribution of X is given by:
P(X=x) = [ (6Cx) (2C4-x) ] / 8C4 for x = 0, 1, 2, 3, 4Here is the table of probabilities:xi01234
P(X = x)0.00020.02880.34400.46240.1648Part c(ii)We need to calculate P(1≤X≤3).P(1≤X≤3) = P(X=1) + P(X=2) + P(X=3) = 0.288 + 0.344 + 0.194 = 0.826Therefore, P(1≤X≤3) = 0.826.
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Use the Wilcoxon Rank Sum Test on the following data to determine whether the location of population 1 is to the left of the location of population 2. (Use 5% significance level.) Sample 1: 75 60 73 66 81
Sample 2: 90 72 103 82 78
The Wilcoxon Rank Sum Test is used to determine if the location of population 1 is to the left of the location of population 2.
The Wilcoxon Rank Sum Test, also known as the Mann-Whitney U test, is a non-parametric test used to compare the distributions of two independent samples. In this case, we have Sample 1 and Sample 2.
To perform the test, we first combine the data from both samples and rank them. Then, we calculate the sum of the ranks for each sample. The test statistic is the smaller of the two sums of ranks.
Next, we compare the test statistic to the critical value from the Wilcoxon Rank Sum Test table at a significance level of 5% (α = 0.05). If the test statistic is less than the critical value, we reject the null hypothesis, suggesting that the location of population 1 is to the left of the location of population 2.
By conducting the Wilcoxon Rank Sum Test on the given data, we can determine if the location of population 1 is indeed to the left of the location of population 2 at a 5% significance level.
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A researcher wants to test the effect of pets on elderly people’s daily mood. He predicts that having pets will enhance mood. To test this hypothesis, he randomly assigns a group of elderly people to the experimental condition (the pet condition) and another group to the control condition (the no pet condition). One week later, he measures the participants’ mood and computes the following statistics on each of this groups. Is there evidence that having pets indeed increases positive mood? (The higher the group mean, the more positive mood.) Use an alpha = .01.
Each group has 10 participants for a total of 20 participants.
For this group, make sure you treat the experimental group as group 1 and the control group as group 2.
The mean of the pets group = 5.2. That group has a SS of 18.85. The mean of the no pets group = 5.2 with a SS = 13.89
What is the Cohen's d effect size that represents the difference between pets and no pets?
Cohen's d effect size is the difference between two means divided by a measure of variance. Cohen's d indicates the magnitude of the difference between the two groups in terms of standard deviation units. Here, the Cohen's d effect size that represents the difference between pets and no pets is to be found.
The formula for Cohen's d is given as Cohen's d = (M1 - M2) / SDpooledWhere,
M1 is the mean of Group 1,
M2 is the mean of Group 2, and
SDpooled is the pooled standard deviation.
The formula for the pooled standard deviation is: SDpooled = √((SS1 + SS2) / pooled)We are given:
For the pets group, mean = 5.2 and SS = 18.85For no pets group, the mean = 5.2 and SS = 13.89Total number of participants = 20.
The degrees of freedom for the pooled variance can be calculated using the formula:
Pooled = n1 + n2 - 2= 10 + 10 - 2= 18
The pooled variance can be calculated as follows: SDpooled = √((SS1 + SS2) / pooled)= √((18.85 + 13.89) / 18)= √(32.74 / 18)= 1.82Thus,
Cohen's d = (M1 - M2) / SDpooled= (5.2 - 5.2) / 1.82= 0
Therefore, the Cohen's d effect size that represents the difference between pets and no pets is 0. Answer: 0.
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23. Let = f(x,y) = x. At (x, y) = (3,2), if drody=-, then dz =_
Given the function f(x, y) = x and the point (x, y) = (3, 2), if dρ/dy = -1, then the value of dz can be determined by evaluating the partial derivative of f(x, y) with respect to y and multiplying it by the given value of dρ/dy.
The partial derivative of f(x, y) with respect to y, denoted as ∂f/∂y, represents the rate of change of f with respect to y while keeping x constant. Since f(x, y) = x, the partial derivative ∂f/∂y is equal to 0, as the variable y does not appear in the function.
Therefore, dz = (∂f/∂y) * (dρ/dy) = 0 * (-1) = 0.
The value of dz is 0.
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(a) Suppose that the amount X, in years, that a computer can function before its battery runs out is exponentially distributed with mean μ=10. Calculate P(X>20). (b) A microwave oven manufacturer is trying to determine the length of warranty period it should attach to its magnetron tube, the most critical component in the microwave oven. Preliminary testing has shown that the length of life (in years), X, of a magnetron tube has an exponential probability distribution with mean μ=6.25 and standard deviation σ=6.25. Find: (i) the mean and standard deviation of X. (ii) Fraction of tubes must the manufacturer plan to replace (assuming the exponential model with μ=6.25 is correct), if a warranty period of 5 years is attached to the magnetron tube? (iii) The probability that the length of life of magnetron tube will fall within the interval where μ and σ are μ±2σ.
a) Suppose that the amount X, in years, that a computer can function before its battery runs out is exponentially distributed with mean μ=10. Calculate P(X>20).Solution: Given, X follows exponential distribution with mean μ=10.
A microwave oven manufacturer is trying to determine the length of warranty period it should attach to its magnetron tube, the most critical component in the microwave oven. Warranty period attached to the magnetron tube is 5 years.
According to the exponential model,
[tex]P(X>5) = e ^(-5/6.25) = 0.3971[/tex].
Then, the fraction of tubes that the manufacturer must plan to replace is 39.71% (approx). i.e., out of 100 tubes, the manufacturer should plan to replace 39-40 tubes. (iii) The probability that the length of life of magnetron tube will fall within the interval where μ and σ are μ±2σ.Solution:Given, X follows exponential distribution with mean[tex]μ=6.25[/tex]and standard deviation [tex]σ=6.25[/tex].
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medication and those treated with 80mg of the medication, changes in LDL cholesterol have the same median. What do the results suggest?
If two groups have the same median for a certain variable, it suggests that the central tendency of both groups is the same for that variable. This, in turn, implies that there may not be any significant difference between the two groups in terms of that variable.
In the given scenario, it is observed that in a group of people who are taking medication, those treated with 80mg of the medication show the same median for changes in LDL cholesterol as that of the others. Thus, we can say that the medication seems to have no significant effect on the changes in LDL cholesterol in this group of people. This suggests that the medication may not be effective in reducing cholesterol levels in this group.In terms of statistical interpretation, a median is a measure of central tendency. It is the value that divides the data into two equal halves, such that half the data is above it and half the data is below it. Therefore, if two groups have the same median for a certain variable, it suggests that the central tendency of both groups is the same for that variable. This, in turn, implies that there may not be any significant difference between the two groups in terms of that variable.
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Cecelia is conducting a study on income inequality in Memphis. Before she begins her main analyses, she wants to report her sample's descriptive statistics and make sure that it provides an accurate reflection of the population. Her sample consists of 1,000 Memphis residents. She knows from census data that the mean household income in Memphis is $32,285 with a standard deviation of $5,645. However, her sample mean is only $31,997 with a standard deviation of $6,005
Cecelia is conducting a study on income inequality in Memphis. it is important to report the descriptive statistics of the sample and check if it provides an accurate reflection of the population.
Before she begins her main analyses, she wants to report her sample's descriptive statistics and make sure that it provides an accurate reflection of the population. Her sample consists of 1,000 Memphis residents. She knows from census data that the mean household income in Memphis is $32,285 with a standard deviation of $5,645.
Her sample mean is only $31,997 with a standard deviation of $6,005. Cecelia is conducting a study on income inequality in Memphis and she has collected the data for 1,000 Memphis residents. Before she begins her main analyses, she wants to report her sample's descriptive statistics and make sure that it provides an accurate reflection of the population.
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