Answer:
The answer is False
Step-by-step explanation:
The count does no suggest that the normal model is a good fit for sampling the distribution because the questions used for the test and survey is the one of many other question about cheating which implies that if other questions about college life are being used as the survey, the response would probably be that more of the student would not have admitted to cheating. This concept therefore disobeys the normal distribution model which is a bell shaped model and therefore assumes that at an average, the number of students that admitted to cheating in the major courses should be equal.
A website advertises job openings on its website, but job seekers have to pay to access the list of job openings. The website recently completed a survey to estimate the number of days it takes to find a new job using its service. It took the last 30 customers an average of 60 days to find a job. Assume the population standard deviation is 10 days. Construct a 90% confidence interval of the population mean number of days it takes to find a job.
Answer:
The 90% confidence interval would be given by (57.006;62.994)
Step-by-step explanation:
Previous concepts
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
[tex]\bar X=60[/tex] represent the sample mean
[tex]\mu[/tex] population mean (variable of interest)
[tex]\sigma=10[/tex] represent the population standard deviation
n=30 represent the sample size
Assuming the X follows a normal distribution
[tex]X \sim N(\mu, \sigma=10)[/tex]
The sample mean [tex]\bar X[/tex] is distributed on this way:
[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]
The confidence interval on this case is given by:
[tex]\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex] (1)
The next step would be find the value of [tex]\z_{\alpha/2}[/tex], [tex]\alpha=1-0.90=0.1[/tex],[tex]\alpha/2 =0.05[/tex] and [tex]z_\alpha/2=1.64[/tex]
Using the normal standard table, excel or a calculator we see that:
[tex]z_{\alpha/2}=1.64[/tex]
Since we have all the values we can replace:
[tex]60 - 1.64\frac{10}{\sqrt{30}}=57.006[/tex]
[tex]60 + 1.64\frac{10}{\sqrt{30}}=62.994[/tex]
So on this case the 90% confidence interval would be given by (57.006;62.994)
Final answer:
To construct the 90% confidence interval for the average number of days it takes to find a job using the website's service, the formula for a confidence interval is applied using the given sample mean of 60 days, population standard deviation of 10 days, and a sample size of 30 customers. The calculated interval is between 57 and 63 days.
Explanation:
To construct a 90% confidence interval for the population mean number of days it takes to find a job using the website's service, we need to use the formula:
CI = \(\bar{x} \pm z*\frac{\sigma}{\sqrt{n}}\)
Where:
\(\bar{x}\) is the sample mean (60 days)
\(z\) is the z-score that corresponds to the desired confidence level (For 90%, \(z=1.645\) from the z-table)
\(\sigma\) is the population standard deviation (10 days)
\(n\) is the sample size (30 customers)
Plugging the values into the formula gives:
CI = 60 \pm 1.645 * (\frac{10}{\sqrt{30}})
Calculating the margin of error:
ME = 1.645 * (\frac{10}{\sqrt{30}}) \approx 3.00
Now compute the confidence interval:
CI = [60 - 3.00, 60 + 3.00]
CI = [57.00, 63.00]
So, we are 90% confident that the population mean number of days it takes to find a job using the website's service is between 57 and 63 days.
Suppose the interval [4,6] is partitioned into n=4 subintervals. What is the subinterval length Δx? List the grid points x0, x1, x2, x3, x4. Which points are used for theleft, right, and midpoint Riemann sums?
If a generic interval [tex][a,b][/tex] is partitioned into [tex]n[/tex] subintervals, each one has a length:
[tex]\Delta x = \dfrac{b-a}{n}.[/tex]
In this case, [tex]a = 4[/tex], [tex]b=6[/tex] and [tex]n=4[/tex], so:
[tex]\Delta x = \dfrac{6-4}{4} = \dfrac{2}{4} = \dfrac{1}{2} = 0.5.[/tex]
The grid points are given by:
[tex]x_k = a + k\Delta x, \quad\textrm{with } k \in \{0,1,2,3,4\}.[/tex]
Since [tex]a = 4[/tex] and [tex]\Delta x = 0.5[/tex], we have:
[tex]x_0 = 4 + 0 \times 0.5 = 4 \\x_1 = 4 + 1 \times 0.5 = 4 + 0.5 = 4.5\\x_2 = 4 + 2 \times 0.5 = 4 + 1 = 5\\x_3 = 4 + 3 \times 0.5 = 4 + 1.5 = 5.5\\x_4 = 4 + 4 \times 0.5 = 4 + 2 = 6[/tex]
The [tex]4[/tex] subintervals are of the form [tex]I_k = [x_{k-1}, x_k], \quad\textrm{with } k \in \{1, 2, 3,4\}[/tex]:
[tex]I_1 = [x_0, x_1] = [4,4.5]\\I_2 = [x_1, x_2] = [4.5,5]\\I_3 = [x_2, x_3] = [5,5.5]\\I_4 = [x_3, x_4] = [5.5, 6][/tex]
For the left Riemann sums we will use the left-handed points, namely:
[tex]\{x_0, x_1, x_2, x_3\} = \{4,4.5,5,5.5\}.[/tex]
For the right Riemann sums we will use the right-handed points, namely:
[tex]\{x_1, x_2, x_3, x_4\} = \{4.5,5,5.5,6\}.[/tex]
For the midpoint Riemann sums we will use the average between the two extrema of each subinterval, given by
[tex]I_k \to \tilde{x}_k = \dfrac{x_{k-1}-x_k}{2}, \quad \textrm{with } k \in\{1,2,3,4\}.[/tex]
This gives the midpoints:
[tex]I_1 = [x_0, x_1] = [4,4.5] \to \tilde{x}_1 = \dfrac{x_0 + x_1}{2} = \dfrac{4+4.5}{2} = \dfrac{8.5}{2} = 4.25\\\\I_2 = [x_1, x_2] = [4.5,5] \to \tilde{x}_2 = \dfrac{x_1 + x_2}{2} = \dfrac{4.5+5}{2} = \dfrac{9.5}{2} = 4.75\\\\I_3 = [x_2, x_3] = [5,5.5] \to \tilde{x}_3 = \dfrac{x_2 + x_3}{2} = \dfrac{5+5.5}{2} = \dfrac{10.5}{2} = 5.25\\\\I_4 = [x_3, x_4] = [5.5,6] \to \tilde{x}_4 = \dfrac{x_3 + x_4}{2} = \dfrac{5.5+6}{2} = \dfrac{11.5}{2} = 5.75[/tex]
The points used for the midpoint Riemann sums are therefore:
[tex]\{\tilde{x}_1,\tilde{x}_2,\tilde{x}_3,\tilde{x}_4\} =\{4.25,4.75,5.25,5.75\}.[/tex]
The interval and sub intervals are illustrations of Riemann sums
The length of the sub-intervals is 0.5The grid points are 4, 4.5, 5, 5.5 and 6The midpoints are 4.25, 4.75, 5.25 and 5.75The given parameters are:
[tex]\mathbf{Interval = [4,6]}[/tex]
[tex]\mathbf{n =4}[/tex] --- sub intervals
(a) The sub interval length
This is calculated as:
[tex]\mathbf{\Delta x = \frac{b - a}{n}}[/tex]
Where:
[tex]\mathbf{[a,b] =[4,6]}[/tex]
So, we have:
[tex]\mathbf{\Delta x = \frac{6 - 4}{4}}[/tex]
[tex]\mathbf{\Delta x = \frac{2}{4}}[/tex]
[tex]\mathbf{\Delta x = 0.5}[/tex]
Hence, the length of the sub-intervals is 0.5
(b) The grid points
This is calculated as:
[tex]\mathbf{x_k = a + k\Delta x}[/tex]
So, we have:
[tex]\mathbf{x_0 = 4 + 0 \times 0.5 = 4}[/tex]
[tex]\mathbf{x_1 = 4 + 1 \times 0.5 = 4.5}[/tex]
[tex]\mathbf{x_2 = 4 + 2 \times 0.5 = 5}[/tex]
[tex]\mathbf{x_3 = 4 + 3 \times 0.5 = 5.5}[/tex]
[tex]\mathbf{x_4 = 4 + 4 \times 0.5 = 6}[/tex]
So, the grid points are 4, 4.5, 5, 5.5 and 6
(c) The left, right and midpoint Riemann sums
The left points are 4, 4.5, 5 and 5.5The right points are 4.5, 5, 5.5 and 6The midpoint is the average of the left and right points.
So, we have:
[tex]\mathbf{x_0 = 0.5 \times (4 + 4.5) = 4.25}[/tex]
[tex]\mathbf{x_1 = 0.5 \times (4.5 + 5) = 4.75}[/tex]
[tex]\mathbf{x_2 = 0.5 \times (5 + 5.5) = 5.25}[/tex]
[tex]\mathbf{x_3 = 0.5 \times (5.5 + 6) = 5.75}[/tex]
So, the midpoints are 4.25, 4.75, 5.25 and 5.75
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One study reports that 34% of newly hired MBAs are confronted with unethical business practices during their first year of employment. One business school dean wondered if her MBA graduates had similar experiences. She surveyed recent graduates from her school's MBA program to find that 28% of the 116 graduates from the previous year claim to have encountered unethical business practices in the workplace. Can she conclude that her graduates' experiences are different?
Answer:
[tex]z=\frac{0.28 -0.34}{\sqrt{\frac{0.34(1-0.34)}{116}}}=-1.364[/tex]
[tex]p_v =2*P(Z<-1.364)=0.173[/tex]
The p value obtained was a very high value and using the significance level assumed [tex]\alpha=0.05[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIl to reject the null hypothesis, and we can said that at 5% of significance the proportion of graduates from the previous year claim to have encountered unethical business practices in the workplace is not significant different from 0.34.
Step-by-step explanation:
1) Data given and notation
n=116 represent the random sample taken
X represent the number graduates from the previous year claim to have encountered unethical business practices in the workplace
[tex]\hat p=0.28[/tex] estimated proportion of graduates from the previous year claim to have encountered unethical business practices in the workplace
[tex]p_o=0.34[/tex] is the value that we want to test
[tex]\alpha=0.05[/tex] represent the significance level
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value (variable of interest)
2) Concepts and formulas to use
We need to conduct a hypothesis in order to test the claim that the proportion is 0.34 or no.:
Null hypothesis:[tex]p=0.34[/tex]
Alternative hypothesis:[tex]p \neq 0.34[/tex]
When we conduct a proportion test we need to use the z statistic, and the is given by:
[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)
The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].
3) Calculate the statistic
Since we have all the info requires we can replace in formula (1) like this:
[tex]z=\frac{0.28 -0.34}{\sqrt{\frac{0.34(1-0.34)}{116}}}=-1.364[/tex]
4) Statistical decision
It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.
The next step would be calculate the p value for this test.
Since is a bilateral test the p value would be:
[tex]p_v =2*P(Z<-1.364)=0.173[/tex]
The p value obtained was a very high value and using the significance level assumed [tex]\alpha=0.05[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIl to reject the null hypothesis, and we can said that at 5% of significance the proportion of graduates from the previous year claim to have encountered unethical business practices in the workplace is not significant different from 0.34.
From the top of a vertical tower, 331 feet above the surface of the earth, the angle of depression to a doghouse is 28 degrees 8'. How far is it from the doghouse to the foot of the tower?
Answer:
619.13 feet
Step-by-step explanation:
Please find the attachment.
Let x represent the distance between doghouse to the foot of the tower.
We have been given that from the top of a vertical tower, 331 feet above the surface of the earth, the angle of depression to a doghouse is 28 degrees 8'. We are asked to find the distance between doghouse to the foot of the tower.
First of all, we will convert our given angle into degrees as it is given in degrees and minutes.
We will divide 8 by 60 to convert 8 minutes into degrees as:
[tex]\frac{8}{60}=0.13333[/tex]
The doghouse, tower and angle of depression forms a right triangle with respect to ground, where, 331 feet is opposite side and x is adjacent side to angle 28.13 degrees.
[tex]\text{tan}=\frac{\text{Opposite}}{\text{Adjacent}}[/tex]
[tex]\text{tan}(28.13^{\circ})=\frac{331}{x}[/tex]
[tex]x=\frac{331}{\text{tan}(28.13^{\circ})}[/tex]
[tex]x=\frac{331}{0.534623339864}[/tex]
[tex]x=619.12747783[/tex]
Upon rounding to nearest hundredth, we will get:
[tex]x\approx 619.13[/tex]
Therefore, the doghouse is 619.13 feet far from the foot of the tower.
A leprechaun places a magic penny under a girl's pillow. The next night there are 2 magic pennies under her pillow. Each night the number of magic pennies doubles. How much money will the girl have after 25 nights? PLEASE HURRY
Answer:
The girl will have $335,544.32
Step-by-step explanation:
2^25 = 33,554,432
Divide by 100 to turn the amount of pennies into dollars:
33,554,432/100
$335,544.32
Help please!!! 20 points will mark brainliest!! :)
Answer:
The equation has zeroes at -4, 0 and 4 and is a minimum cubic degree or a 3 degree equation.
Step-by-step explanation:
The polynomial has roots at places where f(x) cuts the x axis.
The function cuts the x-axis at 3 points: 4, -4, and 0.
It cuts the axis at the point, x=0; x=4; x=-4
Therefore, the points equation has to be of the form, k*x*(x-4)*(x+4)*p(x)
where k is any arbitrary constant and p(x) is a polynomial of any degree depending on what the equation does in the region and (5,∞) and (-∞,-5).
Therefore, the equation has zeroes at -4, 0 and 4 and is a minimum cubic degree or a 3 degree equation.
Complete the two column proof Given: 22= 24,m_2 = 110°
Prove: m 23 = 70°
Statement
Proof
22 24, mZ2=1109
Given
m 2 2 = m 24
Definition of congruent angles
m 24 = 1100
m 23 and m 24 are a linear pair
Definition of a linear pair (shown in
diagram)
Step-by-step explanation:
I took 3 indicated as 5 and its adjacent angle to be 3
<2 = <4
As <2 =<3( corresponding angle)
And <3 = <4 ( Vertically opp.angle)
hence <2 = <4
<2 =>110
so , <4 = 110
So, <4 =>110
As <4 and <5 form linear pair
So <4 + <5 =>180
<5 = 180 -110 =>70
As i took <5 as replacing angle to <3
So According to Question fig
<3 =>70
Hence proved
Average expenditure on different items such as food, clothes, fuel comes under ____. Select one: a. Descriptive statistics b. Inferential statistics c. Applied statistics d. Business statistics e. Industrial statistics
Answer:
Descriptive
Step-by-step explanation:
Statistics can be broadly classified into two main branches
i) Descriptive and ii) inferential
Descriptive statistics deal with values such as mean, standard deviation from the data.
Inferential statistics is used to predict unknown values from the observed values.
Applied statistics mainly analyses the data according to the needs of business or industries.
Hence the average expenditure on different items such as food, clothes, fuel comes under
Descriptive Statistics
Final answer:
Average expenditure on different items like food, clothes, and fuel falls under descriptive statistics, essential for summarizing trends in expenditure data. The correct option is a.
Explanation:
Descriptive statistics involve summarizing trends in data, such as calculating averages and standard deviations, making them suitable for analyzing average expenditures on items like food and clothes.
Inferential statistics are used to make inferences about populations based on sample data, aiming to find cause and effect relationships or correlations, which is essential when studying average expenditures across different categories.
Therefore, analyzing average expenditures on various items falls under the domain of descriptive statistics as it involves summarizing and interpreting trends in expenditure data.
Assume that the weights of Chinook Salmon in the Columbia River are normally distributed. You randomly catch and weigh 40 such salmon. The mean weight from your sample is 23.6 pounds with a standard deviation of 3.5 pounds. Test the claim that the mean weight of Columbia River salmon is greater than 23 pounds. Test this claim at the 0.10 significance level.
(a) What type of test is this?i) This is a right-tailed test.ii) This is a two-tailed test. iii) This is a left-tailed test.(b) What is the test statistic? Round your answer to 2 decimal places.
Answer:
a) i) This is a right-tailed test.
b)
[tex]\text{Test statistic} = 1.08[/tex]
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = 23 pounds
Sample mean, [tex]\bar{x}[/tex] = 23.6 pounds
Sample size, n = 40
Alpha, α = 0.10
Sample standard deviation, s = 3.5 pounds
First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu = 23\text{ pounds}\\H_A: \mu > 23\text{ pounds}[/tex]
This is a one tailed(right).
Formula:
[tex]\text{Test statistic} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }[/tex]
Putting all the values, we have
[tex]\text{Test statistic} = \displaystyle\frac{23.6 - 23}{\frac{3.5}{\sqrt{40}} } = 1.08[/tex]
A continuous random variable may assume :
a. any numerical value in an interval or collection of intervals.b. finite number of values in a collection of intervals.c. an infinite sequence of values.d. only the positive integer values in an interval.
Answer:
Option A) any numerical value in an interval or collection of intervals
Step-by-step explanation:
Continuous Random Variable:
A continuous random variable can take any value within an interval.Thus, it can take infinite values since there are infinite numbers in an interval.A continuous variable is a variable whose value is obtained by measuring.Examples: height of students in class , weight of students in class, time it takes to get to school, distance traveled between classes.Thus, the correct meaning of continuous random variable is explained by Option A)Option A) any numerical value in an interval or collection of intervals
A continuous random variable can take on any numerical value within a given range or collection of ranges, and it's a characteristic feature of it to take an infinite number of values in any interval. Some examples of this can be a person's height, time, temperature, and weight in physics.
Explanation:A continuous random variable is a type of random variable that can assume any numerical value in a given interval or collection of intervals, making option a correct. This is in contrast to a discrete random variable, which can only take on a finite number of values. A key characteristic of continuous random variables is that they can take on an infinite number of values in any interval.
For example, the height of a person can be treated as a continuous variable, since it can theoretically take any value within a certain range, not just whole number values. The same applies to variables such as time, temperature, and weight in physics.
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The random variable x is the number of occurrences of an event over an interval of ten minutes. It can be assumed that the probability of an occurrence is the same in any two time periods of an equal length. It is known that the mean number of occurrences in ten minutes is 5.3. The probability that there are 8 occurrences in ten minutes is
A. .0241
B. .0771
C. .1126
D. .9107
Answer:
The probability that there are 8 occurrences in ten minutes is
option B. 0 .0771
Step-by-step explanation:
Given:
Random Variable = x
Mean number of occurrences in ten minutes is 5.3.
The probability of an occurrence is the same in any two time periods of an equal length
To Find:
The probability that there are 8 occurrences in ten minutes = ?
Solution:
Let X be the number of occurrences of the event X
[tex]X \sim {Pois} (\lambda)[/tex]
[tex]\lambda = E(X) = 5.3[/tex]
Possion of distribution is given by ,
[tex]P(X=x) = \frac{e^{- \lambda} \lambda^{x}}{x!}[/tex]
Substituting the values,
[tex]P(X=8) = \frac{e^{- 5.3} 5.3^{8}}{8!}[/tex]
[tex]P(X=8) = \frac{(0.004994) ( 622596.904)}{40320}[/tex]
[tex]P(X=8) = \frac{(3109.24894)}{40320}[/tex]
P(X=8) = 0.0771
To find the probability of 8 occurrences in ten minutes with a mean of 5.3, use the Poisson distribution formula to calculate the probability, resulting in approximately 0.0771.
Given: Mean number of occurrences in ten minutes = 5.3
Formula: Probability mass function of a Poisson distribution is given by: P(x) = (e^-λ * λ^x) / x!
Calculations: Plugging in the values, P(X=8) = (e^(-5.3) * 5.3^8) / 8! ≈ 0.0771
A disadvantage of using an arithmetic mean to summarize a set of data is that __________. Select one: a. The arithmetic mean sometimes has two values b. It can be used for interval and ratio data c. It is always different from the median d. It can be biased by one or two extremely small or large values
Answer:
d. It can be biased by one or two extremely small or large values
Step-by-step explanation:
Specially in a small sample, a single measure can cause a large difference.
For example, you are selling yourself as a tutor, you have five students. 4 of them got good grades, but the other one got 0. The arithmetic mean is not going to be kind to your averages, in virtue of the outlier.
So the correct answer is:
d. It can be biased by one or two extremely small or large values
The arithmetic mean's disadvantage is that extreme values or outliers in the data set can significantly skew the mean. This makes the mean a less accurate reflection of the central tendency or 'average' of the data.
Explanation:The correct answer to your question is "d. It can be biased by one or two extremely small or large values." The arithmetic mean, often simply called the 'mean', is a type of average most commonly used. It is calculated by adding all numbers in a set and then dividing by the quantity of numbers. While useful, it has a significant disadvantage in its sensitivity to extreme values, also referred to as 'outliers'. If there are one or two extremely large or small numbers in the data set, these can drastically affect the calculated mean, thus not accurately representing the central tendency of the overall data.
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There are 3 alternative routes by which you may drive to work: Alabaster Street, Brillantine Street, and Clancy Street. It is the beginning of rush hour, and by experience Alabaster street will be closed by a car crash on average in 25 minutes, Brillantine street in 50 minutes, and Clancy Street in 40 minutes. Accident times are completely unpredictable. If you leave for work in one hour (60 minutes), what is the probability that (at the moment you leave) a route to work will still be open?
The probability that at least one route to work will still be open is [tex]\( \frac{209}{216} \)[/tex] when leaving.
To find the probability that at least one route to work will still be open when you leave for work, we need to find the probability that none of the streets will be closed within the next hour.
Let [tex]\( A \)[/tex], [tex]\( B \)[/tex], and [tex]\( C \)[/tex] represent the events that Alabaster, Brillantine, and Clancy Streets are closed due to accidents within the next hour, respectively.
The probability of each street being closed within the next hour is:
[tex]\[ P(A) = \frac{25}{60} = \frac{5}{12} \][/tex]
[tex]\[ P(B) = \frac{50}{60} = \frac{5}{6} \][/tex]
[tex]\[ P(C) = \frac{40}{60} = \frac{2}{3} \][/tex]
Since the accidents on each street are independent events, the probability that all three streets remain open is the product of the probabilities that each street remains open:
[tex]\[ P(\text{})[/tex] = [tex]P(\neg A \cap \neg B \cap \neg C) = P(\neg A) \times P(\neg B) \times P(\neg C) ][/tex]
where [tex]\( \neg A \)[/tex], [tex]\( \neg B \)[/tex], and[tex]\( \neg C \)[/tex] represent the complementary events that the streets are not closed.
So, we have:
[tex]\[ P(\text{})[/tex] = [tex](1 - P(A)) \times (1 - P(B)) \times (1 - P(C)) ][/tex]
[tex]\[ = \left(1 - \frac{5}{12}\right) \times \left(1 - \frac{5}{6}\right) \times \left(1 - \frac{2}{3}\right) \][/tex]
[tex]\[ = \left(\frac{7}{12}\right) \times \left(\frac{1}{6}\right) \times \left(\frac{1}{3}\right) \][/tex]
[tex]\[ = \frac{7}{12} \times \frac{1}{6} \times \frac{1}{3} \]\[ = \frac{7}{216} \][/tex]
Therefore, the probability that at least one route to work will still be open when you leave for work is [tex]\( 1 - \frac{7}{216} = \frac{209}{216} \)[/tex].
On the average, 1.6 customers per minute arrive at any one of the checkout counters of Sunshine food market.
What type of probability distribution can be used to find out the probability that there will be no customers arriving at a checkout counter in 10 minutes?
-Poisson distribution
-Normal distribution
-Binomial distribution
-None of these choices.
Use technology to find the P-value for the hypothesis test described below. The claim is that for 12 AM body temperatures, the mean is mu μ greater than > 98.6 98.6 degrees °F. The sample size is n equals = 9 9 and the test statistic is t equals = 2.132 2.132.
Answer:
P-value = 0.032794
Step-by-step explanation:
We are given the following information in the question:
Population mean, μ = 98.6 degrees
Sample size, n = 9
Alpha, α = 0.05
Test t-statistic = 2.132
The null and the alternate hypothesis :
[tex]H_{0}: \mu = 98.6\text{ degrees}\\H_A: \mu > 98.6\text{ degrees}[/tex]
We have to find the p-value for degree of freedom 8 and significance level 0.05
The calculated p-value is 0.032794
Shane receives an hourly wage of $30.40 an hour as an emergency room nurse. When called in at night, he receives time and half. How much does he make if he works 15 hours at night?
Mutliply his hourly rate by 1.5 to find his night pay:
30.40 x 1.5 = $45.60 per hour at night.
Multiply his rate by number of hours:
45.60 x 15 = $684
A random sample of the amounts for 22 purchases was taken. The mean was $42.97, the standard deviation was $22.82, and the margin of error for a 95% confidence interval was $10.12. Assume that t Subscript n minus 1 Superscript starequals2.0 for the 95% confidence intervals. a) To reduce the margin of error to about $5, how large would the sample size have to be? b) How large would the sample size have to be to reduce the margin of error to $1.0?
Answer: a) 84 and b ) 2084
Step-by-step explanation:
Given : Sample standard deviation : s= $22.82
(Population standard deviation is unknown ) , so we use t-test.
Critical value or the 95% confidence intervals :[tex]t_{n-1}*=2.0[/tex]
Formula to find the sample size :
[tex]n=(\dfrac{t^*\cdot s}{E})^2[/tex]
a) E = 5
[tex]n=(\dfrac{(2)\cdot 22.82}{5})^2[/tex]
[tex]n=(9.128)^2=83.320384\approx84[/tex]
i.e. Required sample size : n= 84
b) E = 1
[tex]n=(\dfrac{(2)\cdot 22.82}{1})^2[/tex]
[tex]n=(45.64)^2=2083.0096\approx2084[/tex]
i.e. Required sample size : n= 2084
The radioactive isotope carbon-14 is present in small quantities in all life forms, and it is constantly replenished until the organism dies, after which it decays to stable carbon-12 at a rate proportional to the amount of carbon-14 present, with a half-life of 5595 years. Suppose C(t) is the amount of carbon-14 present at time t.(a) Find the value of the constant k in the differential equation.(b) In 1988 three teams of scientists found that the Shroud of Turin, which was reputed to be the burial cloth of Jesus, contained about 91 percent of the amount of carbon-14 contained in freshly made cloth of the same material. How old is the Shroud of Turin, according to these data?
Answer:
a) [tex]k = 0.000124[/tex]
b) According to these data, the Shroud of Turin has around 760 years.
Step-by-step explanation:
The amount of carbon-14 is modeled by the following equation:
[tex]C(t) = C_{0}e^{-kt}[/tex]
In which [tex]C_{0}[/tex] is the initial amount and k is the rate of decrease.
(a) Find the value of the constant k in the differential equation.
Half-life of 5595 years.
So [tex]C(5595) = 0.5C_{0}[/tex]
[tex]C(t) = C_{0}e^{-kt}[/tex]
[tex]0.5C_{0} = C_{0}e^{-5595k}[/tex]
[tex]e^{-5595k} = 0.5[/tex]
Applying ln to both sides
[tex]-5595k = -0.69[/tex]
[tex]k = 0.000124[/tex]
b) In 1988 three teams of scientists found that the Shroud of Turin, which was reputed to be the burial cloth of Jesus, contained about 91 percent of the amount of carbon-14 contained in freshly made cloth of the same material. How old is the Shroud of Turin, according to these data?
This is t when [tex]C(t) = 0.91C_{0}[/tex]
[tex]C(t) = C_{0}e^{-kt}[/tex]
[tex]0.91C_{0} = C_{0}e^{-0.000124t}[/tex]
[tex]e^{-0.000124t} = 0.91[/tex]
Applying ln to both sides
[tex]-0.000124t = -0.094[/tex]
[tex]t = 760.57[/tex]
According to these data, the Shroud of Turin has around 760 years.
What percentage of youth sport participants has experienced a sport related injury?
A.50%100%
B. 75%
C. 25%
D. 95%
Answer:
B.75%
Step-by-step explanation:
Answer:
A. 50%
Step-by-step explanation:
about 50% b/c over half the people who play sports all get injuries
The side of the base of a square prism is increasing at a rate of 5 meters per second and the height of the prism as decreasing at a rate of 2 meters per second. At a certain instant, the base's side is 6 meters and the height is 7 meters. What is the rate of change of the volume of the prism at that instant fin cubic meters per second?
A. -348
B. 492
C. -492
D. 348
Answer:
D. 348
Step-by-step explanation:
The volume of the square prisma is given by the following formula:
[tex]V = s^{2}h[/tex]
In which h is the height, and s is the side of the base.
Let's use implicit derivatives to solve this problem:
[tex]\frac{dV}{dt} = 2sh\frac{ds}{dt} + s^{2}\frac{dh}{dt}[/tex]
In this problem, we have that:
[tex]\frac{ds}{dt} = 5, \frac{dh}{dt} = -2, h = 7, s = 6[/tex]
So
[tex]\frac{dV}{dt} = 2sh\frac{ds}{dt} + s^{2}\frac{dh}{dt}[/tex]
[tex]\frac{dV}{dt} = 2*6*7*5 + (6)^{2}*(-2) = 348[/tex]
So the correct answer is:
D. 348
We guess, based on historical data, that 30% of graduating high-school seniors in a large city will have completed a first-year calculus course. What's the minimum sample size needed to construct a 95% confidence interval for a proportion with a margin of error of 2.5%?
Answer:
[tex]n=\frac{0.3(1-0.3)}{(\frac{0.025}{1.96})^2}=1290.78[/tex]
And rounded up we have that n=1291
Step-by-step explanation:
1) Previous concepts
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The population proportion have the following distribution
[tex]p \sim N(p,\sqrt{\frac{p(1-p)}{n}})[/tex]
2) Solution to the problem
In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2 =0.025[/tex]. And the critical value would be given by:
[tex]z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96[/tex]
The margin of error for the proportion interval is given by this formula:
[tex] ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex] (a)
And on this case we have that [tex]ME =\pm 0.025[/tex] and we are interested in order to find the value of n, if we solve n from equation (a) we got:
[tex]n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}[/tex] (b)
And replacing into equation (b) the values from part a we got:
[tex]n=\frac{0.3(1-0.3)}{(\frac{0.025}{1.96})^2}=1290.78[/tex]
And rounded up we have that n=1291
This exercise uses the population growth model. It is observed that a certain bacteria culture has a relative growth rate of 15% per hour, but in the presence of an antibiotic the relative growth rate is reduced to 8% per hour. The initial number of bacteria in the culture is 28. Find the projected population after 24 hours for the following conditions. (Round your answers to the nearest whole number.) (a) No antibiotic is present, so the relative growth rate is 15%. (b) An antibiotic is present in the culture, so the relative growth rate is reduced to 8%.
Answer:
a) P(24) = 1025.
b) P(24) = 191.
Step-by-step explanation:
This population can be modeled by the following exponential model.
[tex]P(t) = P_{0}e^{rt}[/tex]
In which P(t) is the population after t hours, [tex]P_{0}[/tex] is the initial population and r is the decimal growth rate.
The initial number of bacteria in the culture is 28. This means that [tex]P_{0} = 28[/tex].
Population after 24 hours.
(a) No antibiotic is present, so the relative growth rate is 15%.
So r = 0.15.
[tex]P(t) = P_{0}e^{rt}[/tex]
[tex]P(24) = 28e^{0.15*24} = 1024.75[/tex]
(b) An antibiotic is present in the culture, so the relative growth rate is reduced to 8%.
So r = 0.08.
[tex]P(t) = P_{0}e^{rt}[/tex]
[tex]P(24) = 28e^{0.08*24} = 190.99[/tex]
Audrey, an astronomer is searching for extra-solar planets using the technique of relativistic lensing. Though there are believed to be a very large number of planets that can be found this way, actually finding one takes time and luck; and finding one planet does not help at all with finding planets of other stars in the same part of the sky. Audrey is good at it, and finds one planet at a time, on average once every three months.
a.) Find the expected value and standard deviation of the number of planets she will find in the next two years.
b.) When she finds her sixth new planet, she will be eligible for a prize. Find the expected value and standard deviation of the amount of time until she is eligible for that prize.
c.) Find the probability that she will become eligible for that prize within one year.
Answer:
Step-by-step explanation:
The model [tex]N (t)[/tex], the number of planets found up to time [tex]t[/tex], as a Poisson process. So, the [tex]N (t)[/tex] has distribution of Poison distribution with parameter [tex](\lambda t)[/tex]
a)
The mean for a month is, [tex]\lambda = \frac{1}{3}[/tex] per month
[tex]E[N(t)]= \lambda t\\\\=\frac{1}{3}
(24)\\\\=8[/tex]
(Here. t = 24)
For Poisson process mean and variance are same,
[tex]Var[N (t)]= Var[N(24)]\\= E [N (24)]\\=8[/tex]
(Poisson distribution mean and variance equal)
The standard deviation of the number of planets is,
[tex]\sigma( 24 )] =\sqrt{Var[ N(24)]}=\sqrt{8}= 2.828
[/tex]
b)
For the Poisson process the intervals between events(finding a new planet) have independent exponential distribution with parameter [tex]\lambda[/tex]. The sum of [tex]K[/tex] of these independent exponential has distribution Gamma [tex](K, \lambda)[/tex].
From the given information, [tex]k = 6[/tex] and [tex]\lambda =\frac{1}{3}
[/tex]
Calculate the expected value.
[tex]E(x)=\frac{\alpha}{\beta}\\\\=\frac{K}{\lambda}
\\\\=\frac{6}{\frac{1}{3}}\\\\=18[/tex]
(Here, [tex]\alpha =k[/tex] and [tex]\beta=\lambda[/tex])
C)
Calculate the probability that she will become eligible for the prize within one year.
Here, 1 year is equal to 12 months.
P(X ≤ 12) = (1/Г (k)λ^k)(x)^(k-1).(e)^(-x/λ)
[tex]=\frac{1}{Г (6)(\frac{1}{3})^6}(12)^{6-1}e^{-36}\\\\=0.2148696\\=0.2419\\=21.49%[/tex]
Hence, the required probability is 0.2149 or 21.49%
We can calculate the expected value and standard deviation of the number of planets Audrey will find in the next two years using probability and statistics. Using geometric distribution, we can also calculate the expected value and standard deviation of the amount of time until Audrey is eligible for a prize after finding her sixth new planet. Finally, we can calculate the probability of Audrey becoming eligible for the prize within one year using the binomial distribution formula.
Explanation:To find the expected value and standard deviation of the number of planets Audrey will find in the next two years, we can use the concepts of probability and statistics. Since she finds one planet every three months on average, the expected value can be calculated by multiplying the average number of planets found per month (1/3) by the number of months in two years (24). The standard deviation can be calculated using the formula sqrt(n * p * q), where n is the number of trials, p is the probability of success, and q is the probability of failure.
To find the expected value and standard deviation of the amount of time until Audrey is eligible for the prize after finding her sixth new planet, we can use the concept of geometric distribution. The expected value can be calculated by taking the reciprocal of the probability of success (1/6) and the standard deviation can be calculated using the formula sqrt((1-p) / (p^2)).
To find the probability that Audrey will become eligible for the prize within one year, we can calculate the cumulative probability of finding six or more new planets in one year using the binomial distribution formula.
Learn more about Expected value and standard deviation here:https://brainly.com/question/32256698
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Suppose that for all t, a particle moving with constant speed is parameterized by r(t). Given that the length of the path from t = 5 to t = 7 is equal to 8, find the value of the speed, llv(t)ll.
Answer: value of the speed, llv(t)ll = 4.0
Step-by-step explanation:
Given;
time interval t1 =5, t2= 7
Length of path ( distance) L = 8
Speed = distance travelled/ time taken
Speed = dL/dt
Speed = 8/(t2-t1) = 8/(7-5)
Speed = 8/2
Speed = 4.0
Since it's moving at constant speed the speed = 4.0
Considered safe for agricultural use. A well in Texas is used to water crops. This well is tested on a regular basis for arsenic. A random sample of 36 tests gave a sample mean of = 7.3 ppb arsenic, with s = 1.9 ppb. Does this information indicate that the mean level of arsenic in this well is less than 8 ppb? Use a 0.01 level of signifcance..
Answer:
This information indicates that the mean level of arsenic in this well is less than 8 ppbat 0.01 level of signifcance..
Step-by-step explanation:
Given that a well in Texas is used to water crops.
This well is tested on a regular basis for arsenic.
A random sample of 36 tests gave a sample mean of = 7.3 ppb arsenic, with s = 1.9 ppb
H0: Sample mean = 8
Ha: Sample mean <8
(left tailed test at 1% level)
Mean difference =-0.70
Std error of mean = [tex]\frac{1.9}{\sqrt{36} } \\=0.3167[/tex]
Test statistic t = -2.204
df = 35
p value = 0.0171
Since p >0.01 we accept H0
This information indicates that the mean level of arsenic in this well is less than 8 ppbat 0.01 level of signifcance..
A scientist was interested in studying if students political beliefs change as they go through college. Two hundred randomly selected students were asked before they entered college if they would consider themselves liberal or conservative. Four years later, the same two hundred students were asked if they would consider themselves, liberal or conservative. The scientist decided to perform McNemar's test. The data is below. What is the null hypothesis? After College Before College Liberal Conservative Liberal 80 15 Conservative 20 85
A. -0.85 or 0.85
B. -0.39 or 0.39
C. -9.75 or -9.75
D. 1.96 or -1.96
The null hypothesis in this considered experiment is: There is no change in their political beliefs as they go through college.
How to form the hypotheses?There are two hypotheses. First one is called null hypothesis and it is chosen such that it predicts nullity or no change in a thing. It is usually the hypothesis against which we do the test. The hypothesis which we put against null hypothesis is alternate hypothesis.
Null hypothesis is the one which researchers try to disprove.
Here, it is specified that the scientist wants to study if students political beliefs change as they go through college. He wants to test if there are changes in the proportions of people who are liberal( or conservative).
Given that:
100 students before and after were asked their policial belief, as shown in table:
Liberal Conservative
After college 80 20
Before college 85 15
Proportion of liberal = 1 - proportion of conservatives
So we will symbolize the hypotheses in terms of one of them, let it be proportions of liberals.
Sample size = 100
Favorable cases = Number of people from sample who consider themselves liberal. = X (say)
After college:Sample proportion is: [tex]\hat{p}_1 = \dfrac{X}{N} = \dfrac{80}{100} = 0.8[/tex]
Before college:Sample proportion is: [tex]\hat{p}_2 = \dfrac{X}{N} = \dfrac{85}{100} = 0.85[/tex]
Let p1 and p2 be the population proportion of people believing themself liberal after and before college respectively.
Then, the null hypothesis will assume that the claim of difference in belief the scientist wants to test is false, and therfore,
[tex]H_0: p_1 = p_2[/tex] or [tex]H_0: p_1 - p_2 = 0[/tex] (no difference, proportions are same, indicating that belief of students doesn't change much, as d)
Thus, the null hypothesis in this considered experiment is: There is no change in their political beliefs as they go through college.
Learn more about null and alternative hypothesis here:
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ANYTHING WILL HELP!!!! ASAP!!!!
Washington High School won the meet.
Johnson High School came in second with 159 points.
Difference between first and second was 3 points.
Step-by-step explanation:
At a particular swim meet, the details of the points which were awarded to the first three places finish are given.
First place finish got 10 points.Second place finish got 8 points.Third place finish got 7 points.To find the total number of points scored by each school is equal to sum of multiplying the respective points which were awarded to number of respective places finish.
i.e. Washington High School had 10 first places finish, so total points for first place finish [tex]= 10 \times10=100[/tex] points.
Similarly, for second places finish = [tex]6\times 8=48[/tex]
and for third places finish = [tex]=2\times 7=14[/tex]
Therefore, Washington High School had total points = 100 + 48 + 14 = 162 points
By using this method, we can make the matrices (Please refer the below attachment).
From the matrices,
Total number of points:
Washington High School = 162 pointsJohnson High School = 159 pointsRoosevelt High School = 108 pointsLewis High School = 96 pointsAmong these scores of four schools, 162 is the highest score. So, Washington High School won the meet.
Johnson High School came in second with 159 points.
Difference between first and second = 162 - 159 = 3 points.
What is 6 to the power of 8 in exponential form?
Answer: 1.7 x 10^6
Step-by-step explanation:
6^8 = 1,679,616
1,679,616 = 1.7 x 10^6
Verify that P = Ce^t /1 + Ce^t is a one-parameter family of solutions to the differential equation dP dt = P(1 − P).
Answer:
See verification below
Step-by-step explanation:
We can differentiate P(t) respect to t with usual rules (quotient, exponential, and sum) and rearrange the result. First, note that
[tex]1-P=1-\frac{ce^t}{1+ce^t}=\frac{1+ce^t-ce^t}{1+ce^t}=\frac{1}{1+ce^t}[/tex]
Now, differentiate to obtain
[tex]\frac{dP}{dt}=(\frac{ce^t}{1+ce^t})'=\frac{(ce^t)'(1+ce^t)-(ce^t)(1+ce^t)'}{(1+ce^t)^2}[/tex]
[tex]=\frac{(ce^t)(1+ce^t)-(ce^t)(ce^t)}{(1+ce^t)^2}=\frac{ce^t+ce^{2t}-ce^{2t}}{(1+ce^t)^2}=\frac{ce^t}{(1+ce^t)^2}[/tex]
To obtain the required form, extract a factor in both the numerator and denominator:
[tex]\frac{dP}{dt}=\frac{ce^t}{1+ce^t}\frac{1}{1+ce^t}=P(1-P)[/tex]
Management at a seaside resort is publishing a brochure and wants to include a statement about the proportion of clear days during their peak season. Out of a random sample of 150 days from over the last two peak seasons, 117 days were recorded as clear. They want to estimate the proportion of clear days to within a 5% margin of error with a 95% confidence interval. What's the sample size necessary to construct this interval?A. 384B. 264C. 383D. 385E. 263
Answer: B. 264
Step-by-step explanation:
Formula to calculate the sample size 'n' , if the prior estimate of the population proportion (p) is available:
[tex]n= p(1-p)(\dfrac{z}{E})^2[/tex]
, where z = Critical z-value corresponds to the given confidence interval
E= margin of error
Let p be the population proportion of clear days.
As per given , we have
Prior sample size : n= 150
Number of clear days in that sample = 117
Prior estimate of the population proportion of clear days = [tex]p=\dfrac{117}{150}[/tex]
E= 0.05
The critical z-value corresponding to 95% confidence interval = z*= 1.95 (By z-table)
Then, the required sample size will be :
[tex]n= \dfrac{117}{150}(1-\dfrac{117}{150})(\dfrac{1.96}{0.05})^2[/tex]
Simplify ,
[tex]n= (0.1716)(39.2)^2[/tex]
[tex]n= 263.687424\approx264[/tex]
Hence, the sample size necessary to construct this interval =264
Thus the correct option is B. 264