Answer:
Step-by-step explanation:
Given a rectangular tank with dimension (10ft by 20ft by 16ft)
Then the volume of the tank is
Volume =length × breadth ×height
Volume=10×20×16=3200ft³
V=3200ft³
Then,
∆F= weight density × volume
∆F= 62.4×3200
∆F= 199,680 lb
Then,
Let the 0-point on the x-axis be at the bottom of the tank, so the level of the water ranges from x = 0 to x = 16ft. (It would just as well to let 0 be ground level and let x range from x = −26ft to x − 0.) Then a slice of water at level x is raised (26− x)ft
Then ∆x=(26-x)ft
Work is given as
W= -∫F∆xdx. From 0 to 16
W= -∫199,680(26-x)dx From 0 to 16
W=-199,680∫26-x dx From 0 to 16
W=-199,680 (26x-x²/2). From 0 to 16
W=-199,680(26×16-0.5×16² -0-0)
W=-199,680(288)
W=-57,507,840J
The work done to empty the tank is 57,507,840J
Answer:
2 X 10∧6 ft.lb
Step-by-step explanation:
Volume of rectangular tank = Base area X Height = (10 X 20 X 16)ft³ = 3200ft³
Mass of tank filled with water = 3200ft³ X 62.4 lb/ft³ = 199680lb or 90706.37kg
Hence, work done due to gravity,Δg = m * g * h, where, m = mass of tank, g = gravity = 9.8m/s² and h = height = 10ft or 3.05 meter(m)
∴ Δg = 90706.37 * 9.8 * 3.05 = 2.711 X10∧6 J or 2 X 10∧6 ft.lb
Suppose that the current measurements in a strip of wire are assumed to follow a normal distribution with a mean of 10 millimeters and a standard deviation of 2 millimeters. What is the probability that a measurement exceeds 13 milliamperes
Answer: the probability that a measurement exceeds 13 milliamperes is 0.067
Step-by-step explanation:
Suppose that the current measurements in a strip of wire are assumed to follow a normal distribution, we would apply the formula for normal distribution which is expressed as
z = (x - µ)/σ
Where
x = current measurements in a strip.
µ = mean current
σ = standard deviation
From the information given,
µ = 10
σ = 2
We want to find the probability that a measurement exceeds 13 milliamperes. It is expressed as
P(x > 13) = 1 - P(x ≤ 13)
For x = 13,
z = (13 - 10)/2 = 1.5
Looking at the normal distribution table, the probability corresponding to the z score is 0.933
P(x > 13) = 1 - 0.933 = 0.067
Final answer:
To find the probability that a current measurement exceeds 13 milliamperes in a normally distributed set with mean 10 mA and SD 2 mA, calculate the Z-score and use a normal distribution table or software.
Explanation:
The student's question seems to mistakenly mix units (millimeters instead of milliamperes), but assuming the intent was to refer to electrical current and not physical measurements of wire, we will proceed on the basis that the actual question is about the probability of a current measurement exceeding 13 milliamperes.
To calculate the probability that a measurement exceeds 13 milliamperes when the current measurements in a strip of wire are normally distributed with a mean of 10 milliamperes and standard deviation of 2 milliamperes, we need to use the Z-score formula:
Z = (X - μ) / σ
where X is the value in question (13 milliamperes), μ is the mean (10 milliamperes), and σ is the standard deviation (2 milliamperes).
Plugging in the values:
Z = (13 - 10) / 2 = 1.5
After finding the Z-score, we would look up this value in a standard normal distribution table or use a statistical software to find the probability that Z exceeds 1.5, which gives us the probability that a measurement exceeds 13 milliamperes. For this Z-score, the probability is approximately 6.68%.
A rocket is launched upward from a launching pad and the function h determines the rocket's height above the launching pad (in miles) given a number of minutes t since the rocket was launched.
What does the equality h(4) = 516 convey about the rocket in this context? Select all that apply.
(A) The rocket travels 516 miles every 4 minutes
(B) 14 minutes after the rocket was launched, the rocket is 516 miles above the ground
(C) The rocket is currently 516 miles above the ground
(D) 516 minutes after the rocket was launched, the rocket is 4 miles above the ground
Answer:
(B) 4 minutes after the rocket was launched, the rocket is 516 miles above the ground
Step-by-step explanation:
The function h(t) represents the height of the rocket above the launchpad after a time t minutes.
h(4) = 516 means that when t = 4 minutes, the height h is 516 miles above the launch pad. Note that the time t is measured from when the rocket is launched.
The first option indicates a rate of change, which is a velocity. This is not indicated in the question because velocity is a time derivative of the the height function.
Option C implies that the rocket is 516 miles currently but we do not know what time currently is from the time of launch.
The fourth option has reversed the roles of the variable by implying the time is 516 minutes while the height is 4 miles which is not what the function means.
The weight of adobe bricks for construction is normally distributed with a mean of 3 pounds and a standard deviation of 0.25 pound. Assume that the weights of the bricks are independent and that a random sample of 25 bricks is chosen.
a) What is the probability that the mean weight of the sample is less than 3.10 pounds? Round your answer to four decimal places
b) What value will the mean weight exceed with probability 0.99? Round your answer to two decimal places.
Answer:
a) 0.9772 = 97.72% probability that the mean weight of the sample is less than 3.10 pounds
b) 2.88 pounds
Step-by-step explanation:
To solve this question, we have to understand the normal probability distribution and the central limit theorem.
Normal probability distribution:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
Central limit theorem:
The Central Limit Theorem estabilishes that, for a random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], a large sample size can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex]
In this problem, we have that:
[tex]\mu = 3, \sigma = 0.25, n = 25, s = \frac{0.25}{\sqrt{25}} = 0.05[/tex]
a) What is the probability that the mean weight of the sample is less than 3.10 pounds? Round your answer to four decimal places
This is the pvalue of Z when X = 3.10. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{3.1 - 3}{0.05}[/tex]
[tex]Z = 2[/tex]
[tex]Z = 2[/tex] has a pvalue of 0.9772
0.9772 = 97.72% probability that the mean weight of the sample is less than 3.10 pounds
b) What value will the mean weight exceed with probability 0.99? Round your answer to two decimal places.
This is the value of X when Z has a pvalue of 1-0.99 = 0.01. So X when Z = -2.33.
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]-2.33 = \frac{X - 3}{0.05}[/tex]
[tex]X - 3 = -2.33*0.05[/tex]
[tex]X = 2.88[/tex]
Isaac, a manager at a supermarket, is inspecting cans of pasta to make sure that the cans are neither dented nor have other defects. From past experience, he knows that 1 can in every 90 is defective. What is the probability that Isaac will find his first defective can among the first 50 cans
Answer:
0.0064 is the probability that Isaac will find his first defective can among the first 50 cans.
Step-by-step explanation:
W are given the following in the question:
Probability of defective can =
[tex]P(A) = \dfrac{1}{90}[/tex]
We have to find the probability that Isaac will find his first defective can among the first 50 cans.
Then the number of adults follows a geometric distribution, where the probability of success on each trial is p, then the probability that the kth trial (out of k trials) is the first success is
[tex]P(X=k) = (1-p)^{k-1}p[/tex]
We have to evaluate:
[tex]P(x = 50)\\= (1-\frac{1}{90})^{50-1}(\frac{1}{90})\\= 0.0064[/tex]
0.0064 is the probability that Isaac will find his first defective can among the first 50 cans.
The probability that an archer hits her target when it is windy is 0.4; when it is not windy, her probability of hitting the target is 0.7. On any shot, the probability of a gust of wind is 0.3. Find the probability that a. on a given shot there is a gust of wind and she hits her target.
b. she hits the target with her first shot.
c. she hits the target exactly once in two shots.
d. there was no gust of wind on an occasion when she missed.
Answer:
a) the probability is 0.12 (12%)
b) the probability is 0.61 (61%)
c) the probability is 0.476 (47.6%)
a) the probability is 0.538 (53.8%)
Step-by-step explanation:
a) denoting the event H= hits her target and G= a gust of wind appears hen
P(H∩G) = probability that a gust of wind appears * probability of hitting the target given that is windy = 0.3* 0.4 = 0.12 (12%)
b) for any given shot
P(H)= probability that a gust of wind appears*probability of hitting the target given that is windy + probability that a gust of wind does not appear*probability of hitting the target given that is not windy = 0.3*0.4+0.7*0.7 = 0.12+0.49 = 0.61 (61%)
c) denoting P₂ as the probability of hitting once in 2 shots and since the archer can hit in the first shot or the second , then
P₂ = P(H)*(1-P(H))+ (1-P(H))*P(H) = 2*P(H) *(1-P(H)) = 2*0.61*0.39= 0.476 (47.6%)
d) for conditional probability we can use the theorem of Bayes , where
M= the archer misses the shot → P(M) = 1- P(H) = 0.39
S= it is not windy when the archer shots → P(S) = 1- P(G) = 0.7
then
P(S/M) = P(S∩M)/P(M) = 0.7*(1-0.7)/0.39 = 0.538 (53.8%)
where P(S/M) is the probability that there was no wind when the archer missed the shot
A bottling company uses a filling machine to fill plastic bottles with a popular cola. The bottles are supposed to contain 300 ml. In fact, the contents vary according to a normal distribution with mean μ = 298 ml and standard deviation σ = 3ml.
What is the probability that a randomly selected bottle contains less than 295 ml?
Answer:
0.16
Step-by-step explanation:
For a normal distribution, the 68-95-99.7 rule says that 68% of the distribution lies within 1 standard deviation from the mean, 95% within two standard deviations and 99.7% within three standard deviations.
From the question,
Mean μ = 298 ml
Standard deviation σ = 3 ml
A value of 295 ml is within one standard deviation = 68%.
Since it's on the lower side, it's within 68% ÷ 2 = 34%.
The lower half below the mean is 50% of the distribution. Hence, for a selection less than 1 standard deviation, the probability is
50% - 34% = 16% = 0.16
Final answer:
The probability that a randomly selected bottle contains less than 295 ml of cola is calculated using the normal distribution. With a z-score of -1, ths probability is approximately 15.87%.
Explanation:
To determine the probability that a randomly selected bottle contains less than 295 ml of cola, we can use the properties of the normal distribution.
Given that the mean is 298 ml and the standard deviation is 3 ml, we must first find the z-score associated with 295 ml.
The z-score formula is z = (X - mu) / SD where X is the value for which we want to find the probability.
Substituting the given values into the formula, we get:
z = (295 - 298) / 3 = -1.
A z-score of -1 tells us that 295 ml is one standard deviation below the mean.
To find the probability corresponding to this z-score, we refer to a standard normal distribution table or use relevant statistical software. If we look up the probability for z = -1 in the z-table, we find that the area to the left of z is approximately 0.1587.
Therefore, the probability that a randomly selected bottle contains less than 295 ml of cola is about 15.87%.
A random variableX= {0, 1, 2, 3, ...} has cumulative distribution function.a) Calculate the probability that 3 ≤X≤ 5.b) Find the expected value of X, E(X), using the fact that. (Hint: You will have to evaluate an infinite sum, but that will be easy to do if you notice that
Answer:
a) P ( 3 ≤X≤ 5 ) = 0.02619
b) E(X) = 1
Step-by-step explanation:
Given:
- The CDF of a random variable X = { 0 , 1 , 2 , 3 , .... } is given as:
[tex]F(X) = P ( X =< x) = 1 - \frac{1}{(x+1)*(x+2)}[/tex]
Find:
a.Calculate the probability that 3 ≤X≤ 5
b) Find the expected value of X, E(X), using the fact that. (Hint: You will have to evaluate an infinite sum, but that will be easy to do if you notice that
Solution:
- The CDF gives the probability of (X < x) for any value of x. So to compute the P ( 3 ≤X≤ 5 ) we will set the limits.
[tex]F(X) = P ( 3=<X =< 5) = [1 - \frac{1}{(x+1)*(x+2)}]\limits^5_3\\\\F(X) = P ( 3=<X =< 5) = [-\frac{1}{(5+1)*(5+2)} + \frac{1}{(3+1)*(3+2)}}\\\\F(X) = P ( 3=<X =< 5) = [-\frac{1}{(42)} + \frac{1}{(20)}}]\\\\F(X) = P ( 3=<X =< 5) = 0.02619[/tex]
- The Expected Value can be determined by sum to infinity of CDF:
E(X) = Σ ( 1 - F(X) )
[tex]E(X) = \frac{1}{(x+1)*(x+2)} = \frac{1}{(x+1)} - \frac{1}{(x+2)} \\\\= \frac{1}{(1)} - \frac{1}{(2)}\\\\= \frac{1}{(2)} - \frac{1}{(3)} \\\\= \frac{1}{(3)} - \frac{1}{(4)}\\\\= ............................................\\\\= \frac{1}{(n)} - \frac{1}{(n+1)}\\\\= \frac{1}{(n+1)} - \frac{1}{(n+ 2)}[/tex]
E(X) = Limit n->∞ [1 - 1 / ( n + 2 ) ]
E(X) = 1
The null hypothesis in ANOVA is that all means of all groups are the same. The alternative is that at least one pair of means is different. We compute an F-statistic to explore sources of variability in our data to conduct the omnibus ANOVA. Question: what do you expect to happen when the null hypothesis is true?
A. More between group variability
B. Less between group variability
Answer:
Correct option: B. Less between group variability
Step-by-step explanation:
The Analysis of Variance (ANOVA) test is performed to determine whether there is a significant difference between the different group mean.
The hypothesis is defined as:
H₀: There is no difference between the group means, i.e. μ₁ = μ₂ = ... = μₙ
Hₐ: At least one of the mean is different from the others, i.e. μ[tex]_{i}[/tex] ≠ 0.
The test statistic is defined as:
[tex]F=\frac{SS_{between}}{SS_{within}}[/tex]
If the null hypothesis is true then the test statistic will be small and if it is false then the test statistic will be large.
In this case it is provided that the null hypothesis is true.
This implies that:
[tex]SS_{between}<SS_{within}[/tex]
Implying that the sum of squares for between group variability is less than within group variability.
Thus, if the null hypothesis is true there will be less between group variability.
A tortoise and a hare are competing in a 1600-meter race. The arrogant hare decides to let the tortoise have a 510-meter head start. When the start gun is fired the hare begins running at a constant speed of 9 meters per second and the tortoise begins crawling at a constant speed of 5 meters per second.
a. Define a function f to represent the tortoise's distance from the finish line (in meters) in terms of the number of seconds t since the start of the race.
b. Solve f(t)=0for t.
c. Define a function g to represent the hare's distance from the finish line (in meters) in terms of the number of seconds t since the start of the race.
d. Solve g(t)=0 for t
e. Who won the race?
Answer:
(a)f(t)=1090-5t
(b)f(0)=1090metres
(c)g(t)=1600-9t
(d)g(0)=1600metres
(e)The Hare
Step-by-step explanation:
Total Distance =1600 metres
(a)The tortoise has a 510m headstart and a speed of 5m/s
Distance=Speed X Time
Distance at 5m/s = 5t
Total Distance covered by the tortoise at any time t= 510+5t
Therefore, The tortoise's distance from the finish line (in meters) in terms of the number of seconds t since the start of the race is given as:
f(t)=1600-(510+5t)
f(t)=1090-5t
(b)f(0)=1090-(5X0)
=1090metres
(c)The hare has a speed of 9m/s
Distance=Speed X Time
Distance at 9m/s = 9t
Total Distance covered by the hare at any time t= 9t
Therefore, The hare's distance from the finish line (in meters) in terms of the number of seconds t since the start of the race is given as:
g(t)=1600-9t
(d)
g(0)=1600-(9X0)=1600metres
(e)The race is finished when the distance from the finish line=0
For the Tortoise
f(t)=1090-5t=0
1090=5t
t=218seconds
For the Hare
g(t)=1600-9t=0
9t=1600
t=177.8seconds
The hare takes a shorter time to reach the finish line so he won the race.
The time, in number of days, until maturity of a certain variety of tomato plant is Normally distributed, with mean μ and standard deviation s = 2.4. You select a simple random sample of four plants of this variety and measure the time until maturity. The sample yields ¯x = 65. You read on the package of seeds that these tomatoes reach maturity, on average, in 61 days. You want to test to see if your seeds are reaching maturity later than expected, which might indicate that your package of seeds is too old.
The appropriate hypotheses are:
a) H0 : μ = 61 , Hα : μ > 61 .
b) H0 : μ = 65 , Hα : μ < 65 .
c) H0 : μ = 61 , Hα : μ < 61 .
d) H0 : μ = 65 , Hα : μ > 65 .
Answer:
Option a) H0 : μ = 61 , Hα : μ > 61 .
Step-by-step explanation:
We are given that the time, in number of days, until maturity of a certain variety of tomato plant is Normally distributed, with mean μ and standard deviation s = 2.4.
You select a simple random sample of four plants of this variety and measure the time until maturity. The sample yields x bar = 65.
You read on the package of seeds that these tomatoes reach maturity, on average, in 61 days.
Now, since we know about;
sample size, n = 4
sample standard deviation, s = 2.4
sample, x bar = 65
The only line which represent information about the population mean is reading on the package of seeds that these tomatoes reach maturity, on average, in 61 days.
This shows that population mean , [tex]\mu[/tex] = 61 and in null and alternate hypothesis only population mean is tested.
So, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu[/tex] = 61
Alternate Hypothesis, [tex]H_1[/tex] : [tex]\mu[/tex] > 61 { we are asked for later than expected}
Long-run classical model from Chapter 3. You must provide properly labeled graphs to get full credit!!!!!!! 3) A) Suppose there is a permanent increase in the labor force (L). a) What will be the impact on the real wage (W/P) and the real rental price of capital (R/P)
Answer:
Below the classical model, economic growth is necessarily achieved because of stability in the wage level. For instance, one case of unemployment predominates at a real wage (W / P)1.
Currently, the excessive labor supply would lower the actual wage level before labor supply equals its demand. Eventually, real wage rates would decline to (W / P)F, whereby aggregate labor demand is perfectly matches by aggregate labor supply.
Only the supply side of the production market for products defines the quantity of output & jobs in the classical model.
As the classical method is supply-determined, it states that equiproportional increases (or declines) will not alter the supply of labor in both the rate of money wage and the price level.
Use false-position method to determine the drag coefficient needed so that an 95-kg bungee jumper has a velocity of 46 m/s after 9 s of free fall. Note: The acceleration of gravity is 9.81 m/s2. Start with initial guesses of xl = 0.2 and xu = 0.5 and iterate until the approximate relative error falls below 5%.
Answer:
solution attached below
Step-by-step explanation:
To determine the drag coefficient using the false-position method, start with initial guesses and iterate until the approximate relative error falls below 5%.
Explanation:To determine the drag coefficient needed for a bungee jumper to have a velocity of 46 m/s after 9 s of free fall using the false-position method, we can follow these steps:
Start with initial guesses of xl = 0.2 and xu = 0.5.Calculate the velocity at 9 s using the false-position method.If the calculated velocity is greater than 46 m/s, update xu with the calculated drag coefficient. If the calculated velocity is less than 46 m/s, update xl with the calculated drag coefficient.Repeat steps 2 and 3 until the approximate relative error falls below 5%.The final value of the drag coefficient will be the approximate solution.Learn more about Drag coefficient here:
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The lifetime of a certain transistor in a certain application has mean 900 hours and standard deviation 30 hours. Find the mean and standard deviation of the length of time that four bulbs will last
Answer:
[tex]E(T)=E(X1 + X2 + X3 + X4 =E(x1) + E(x2) + E(x3) + E(x4) [/tex]
[tex]E(T) =900+900+900+900 =3600[/tex]
And the mean would be:
[tex] \bar X = \frac{T}{4} = \frac{3600}{4}= 900[/tex]
And the standard deviation of total time would be:
[tex]SD(T)=\sqrt(Var(T)) = sqrt(Var(X1) + Var(x2) + Var(x3) + Var(x4))[/tex]
[tex]\sigma=sqrt(30^2+30^2+30^2+30^2) =\sqrt(3600) =60[/tex]
Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Solution to the problem
Let total life time of t four transistors T = X1 + X2 + X3 + X4 (where X1,X2,X3 and X4 are life time of individual transistors
For this case the mean length of time that four transistors will last
[tex]E(T)=E(X1 + X2 + X3 + X4 =E(x1) + E(x2) + E(x3) + E(x4) [/tex]
[tex]E(T) =900+900+900+900 =3600[/tex]
And the mean would be:
[tex] \bar X = \frac{T}{4} = \frac{3600}{4}= 900[/tex]
And the standard deviation of total time would be:
[tex]SD(T)=\sqrt(Var(T)) = sqrt(Var(X1) + Var(x2) + Var(x3) + Var(x4))[/tex]
[tex]\sigma=sqrt(30^2+30^2+30^2+30^2) =\sqrt(3600) =60[/tex]
The mean lifetime of four transistors is 3600 hours and the standard deviation of the lifetime is 60 hours.
Explanation:The mean and standard deviation are statistical concepts used to describe a dataset. The mean, or average, is the sum of all data points divided by the number of data points. In this case, the mean lifetime of a single transistor is 900 hours. If you have four transistors, their total lifetime would be four times the mean of a single transistor. So, the mean lifetime for four transistors is 4 * 900 = 3600 hours.
The standard deviation measures the amount of variation in a set of values. When you are considering the lifetime of multiple transistors, you add the variances - not the standard deviations - and then take the square root. The variance is the square of the standard deviation. So, the standard deviation for the lifetime of four transistors is sqrt(4) * 30 = 60 hours.
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The rate at which a professional tennis player used carbohydrates during a strenuous workout was found to be 1.7 grams per minute. If a line were graphed showing time (in minutes) on the horizontal axis and carbohydrates used (in grams) on the vertical axis, what would be the slope of the line?
How many carbohydrates (in grams) would the athlete use in 40 minutes?
Answer:
m=1.7
C=68 gr
Step-by-step explanation:
Function Modeling
We are given a relationship between the carbohydrates used by a professional tennis player during a strenuous workout and the time in minutes as 1.7 grams per minute. Being C the carbohydrates in grams and t the time in minutes, the model is
[tex]C=1.7t[/tex]
The slope m of the line is the coefficient of the independent variable, thus m=1.7
The graph of C vs t is shown in the image below.
To find how many carbohydrates the athlete would use in t=40 min, we plug in the value into the equation
[tex]C=1.7\cdot 40=68\ gr[/tex]
The slope of the line representing the rate of carbohydrate usage is 1.7. Multiply this rate (1.7 grams per minute) by the time (40 minutes) to find the total carbohydrates used, which is 68 grams.
Explanation:The rate at which the tennis player uses carbohydrates is 1.7 grams per minute. In the context of a graph, this rate would represent the slope of the line. So, the slope of the line would be 1.7. Slope, in mathematics, is defined as the change in the y-value (vertical axis) divided by the change in the x-value (horizontal axis). Here, the rate of carbohydrate usage (1.7 grams per minute) is the change in the y-value (carbohydrates used) per change in x-value (time).
Now, you also want to know how many carbohydrates the athlete would use in 40 minutes. We know that the rate of carbohydrate usage is 1.7 grams per minute. So, to find the total amount of carbohydrates used in 40 minutes, you'd simply multiply the rate by the time:
1.7 grams/minute * 40 minutes = 68 grams
So, the athlete would use 68 grams of carbohydrates in 40 minutes.
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Discuss the significance of the discontinuities of T to someone who uses the road. Because of the steady increases and decreases in the toll, drivers may want to avoid the highest rates at t = 7 and t = 24 if feasible. Because of the sudden jumps in the toll, drivers may want to avoid the higher rates between t = 0 and t = 7, between t = 10 and t = 16, and between t = 19 and t = 24 if feasible. The function is continuous, so there is no significance. Because of the sudden jumps in the toll, drivers may want to avoid the higher rates between t = 7 and t = 10 and between t = 16 and t = 19 if feasible
the answer is in the attachment
One hundred eight Americans were surveyed to determine the number of hours they spend watching television each month. It was revealed that they watched an average of 151 hours each month with a standard deviation of 32 hours. Assume that the underlying population distribution is normal.
Construct a 99% confidence interval for the population mean hours spent watching television per month.
Answer:
The 99% confidence interval for the population mean hours spent watching television per month is between 143.07 hours and 158.93 hours.
Step-by-step explanation:
We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:
[tex]\alpha = \frac{1-0.99}{2} = 0.005[/tex]
Now, we have to find z in the Ztable as such z has a pvalue of [tex]1-\alpha[/tex].
So it is z with a pvalue of [tex]1-0.005 = 0.995[/tex], so [tex]z = 2.575[/tex]
Now, find M as such
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.
[tex]M = 2.575*\frac{32}{\sqrt{108}} = 7.93[/tex]
The lower end of the interval is the mean subtracted by M. So it is 151 - 7.93 = 143.07 hours
The upper end of the interval is the mean added to M. So it is 151 + 7.93 = 158.93 hours
The 99% confidence interval for the population mean hours spent watching television per month is between 143.07 hours and 158.93 hours.
The Hyperbolic Functions and their Inverses: For our purposes, the hyperbolic functions, such as sinhx=ex−e−x2andcoshx=ex+e−x2 are simply extensions of the exponential, and any questions concerning them can be answered by using what we know about exponentials. They do have a host of properties that can become useful if you do extensive work in an area that involves hyperbolic functions, but their importance and significance is much more limited than that of exponential functions and logarithms. Let f(x)=sinhxcoshx.
d/dx f(x) =_________
Answer:
[tex]\dfrac{d}{dx}f(x)=\dfrac{e^{2x}+e^{-2x}}{2}[/tex]
Step-by-step explanation:
It is given that
[tex]\sinh x=\dfrac{e^x-e^{-x}}{2}[/tex]
[tex]\cosh x=\dfrac{e^x+e^{-x}}{2}[/tex]
[tex]f(x)=\sinh x\cosh x=[/tex]
Using the given hyperbolic functions, we get
[tex]f(x)=\dfrac{e^x-e^{-x}}{2}\times \dfrac{e^x+e^{-x}}{2}[/tex]
[tex]f(x)=\dfrac{(e^x)^2-(e^{-x})^2}{4}[/tex]
[tex]f(x)=\dfrac{e^{2x}-e^{-2x}}{4}[/tex]
Differentiate both sides with respect to x.
[tex]\dfrac{d}{dx}f(x)=\dfrac{d}{dx}\left(\dfrac{e^{2x}-e^{-2x}}{4}\right )[/tex]
[tex]\dfrac{d}{dx}f(x)=\left(\dfrac{2e^{2x}-(-2)e^{-2x}}{4}\right )[/tex]
[tex]\dfrac{d}{dx}f(x)=\left(\dfrac{2(e^{2x}+e^{-2x})}{4}\right )[/tex]
[tex]\dfrac{d}{dx}f(x)=\dfrac{e^{2x}+e^{-2x}}{2}[/tex]
Hence, [tex]\dfrac{d}{dx}f(x)=\dfrac{e^{2x}+e^{-2x}}{2}[/tex].
A survey of 8 adults employed full-time was taken. Here are their reported numbers of hours worked per week: 50, 53, 46, 46, 49, 43, 41, 41 (a) What is the mean of this data set? If your answer is not an integer, round your answer to one decimal place. (b) What is the median of this data set? If your answer is not an integer, round your answer to one decimal place. (c) How many modes does the data set have, and what are their values? Indicate the number of modes by clicking in the appropriate circle, and then indicate the value(s) of the mode(s), if applicable. zero modes one mode: two modes:
Answer:
a) 46.1
b) 46
c) Two modes: 46, 41
Step-by-step explanation:
We are given the following sample of hours per week:
50, 53, 46, 46, 49, 43, 41, 41
a) mean of this data set
[tex]Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}[/tex]
[tex]Mean =\displaystyle\frac{369}{8} = 46.1[/tex]
b) Median of data set
[tex]Median:\\\text{If n is odd, then}\\\\Median = \displaystyle\frac{n+1}{2}th ~term \\\\\text{If n is even, then}\\\\Median = \displaystyle\frac{\frac{n}{2}th~term + (\frac{n}{2}+1)th~term}{2}[/tex]
Sorted data:
41, 41, 43, 46, 46, 49, 50, 53
Median =
[tex]=\dfrac{4^{th}+5^{th}}{2} = \dfrac{46+46}{2} = 46[/tex]
The median of data is 46.
c) Mode of the data set
Mode is the most frequent observation in the data.
The mode of the data are 46 and 41 as they appeared two times.
Thus, there are two modes.
The mean of the data set is 46.1, the median is 46, and there are two modes: 41 and 46.
To answer the questions based on the provided data set of hours worked by eight adults, we need to perform some basic statistical calculations.
(a) Mean
The mean is calculated by summing all the data points and then dividing by the number of data points:
Mean = (50 + 53 + 46 + 46 + 49 + 43 + 41 + 41) / 8 = 369 / 8 = 46.1
(b) Median
First, we need to arrange the data in ascending order: 41, 41, 43, 46, 46, 49, 50, 53. As there are 8 data points (even number), the median is the average of the 4th and 5th values:
Median = (46 + 46) / 2 = 46
(c) Mode
The mode is the number that appears most frequently in the data set. Here, 41 and 46 both appear twice:
There are two modes: 41 and 46 and the mean of the data set is 46.1.
A sample of 30 is taken from a population of Inconel weldments. The average thickness is 1.87 inches at a key location on the part. The sample standard deviation is 0.125 inches and will be used as an estimate of population standard deviation.Calculate the 99% confidence interval. (Hint: Z(a/2) is 2.58 for a 99% CI)a. (1.811, 1.929)b. (1.611, 1.729)c. (1.711, 1.829)d. (1.511, 1.629)
Answer:
a. (1.811, 1.929)
Step-by-step explanation:
We have to find find M as such
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
In which, z is Z(a/2), [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample. So
[tex]M = 2.58*\frac{0.125}{\sqrt{30}} = 0.059[/tex]
The lower end of the interval is the mean subtracted by M. So it is 1.87 - 0.059 = 1.811.
The upper end of the interval is the mean added to M. So it is 1.87 + 0.059 = 1.929
So the correct answer is:
a. (1.811, 1.929)
The timekeeper for a particular mile race uses a stopwatch to determine the finishing times of the racers. He then calculates that the mean time for the first three finishers was 5.75 minutes. After checking his stopwatch, he notices to his horror that the stopwatch begins timing at 45 seconds rather than at 0, resulting in scores each of which is 45 seconds too long. What is the correct mean time for the first three finishers?
Final answer:
The correct mean time for the first three finishers is 5 minutes, after subtracting the 45-second error from the initially recorded mean time of 5.75 minutes.
Explanation:
The correct mean time for the first three finishers, after adjusting for the stopwatch error, can be calculated by subtracting the 45 seconds from the initially recorded meantime. Since the mean time was calculated to be 5.75 minutes (or 345 seconds), we must correct each time by subtracting 45 seconds and then recalculating the mean.
To find the correct mean time, we do the following steps:
First, convert the mean time from minutes to seconds, so 5.75 minutes = 5 minutes and 45 seconds or 345 seconds.Subtract 45 seconds from each racer's finish time to correct the error: 345 seconds - 45 seconds = 300 seconds.Then, convert the corrected total time back to minutes, so 300 seconds is equal to 5 minutes.The correct mean time for the first three finishers is therefore 5 minutes.
what is the area of the shaded sector ?
Step-by-step explanation:
Find the area using a proportion.
A / (πr²) = θ / 360°
A / (π (20 ft)²) = 160° / 360°
A ≈ 558.5 ft²
Researchers conducted a study of obesity in children. They measured body mass index (BMI), which is a measure of weight relative to height. High BMI is an indication of obesity. Data from a study published in the Journal of the American Dietetic Association shows a fairly strong positive linear association between mother’s BMI and daughter’s BMI (r = 0.506). This means that obese mothers tend to have obese daughters.
1. Based on this study, what proportion of the variation in the daughter BMI measurements is explained by the mother BMI measurements?
2. What are some of the other variables that explain the variability in the daughter BMI?
Answer:
Part a
For this case after do the operations we got a value for the correlation coeffcient of:
[tex] r =0.506[/tex]
With this value we can find the determination coefficient:
[tex] r^2 = 0.506^2 = 0.256[/tex]
And with this value we can analyze the proportion of variance explained by one variable and the other. So we can conclude that 25.6% of the mother's BMI variation is explained by the daugther's BMI.
Part a
Since the BMI is a relation between height and weight, other possible variables that can explain the variability are (weight , height, age)
Step-by-step explanation:
Previous concepts
The correlation coefficient is a "statistical measure that calculates the strength of the relationship between the relative movements of two variables". It's denoted by r and its always between -1 and 1.
And in order to calculate the correlation coefficient we can use this formula:
[tex]r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^2 -(\sum x)^2][n\sum y^2 -(\sum y)^2]}}[/tex]
Solution to the problem
Part a
For this case after do the operations we got a value for the correlation coeffcient of:
[tex] r =0.506[/tex]
With this value we can find the determination coefficient:
[tex] r^2 = 0.506^2 = 0.256[/tex]
And with this value we can analyze the proportion of variance explained by one variable and the other. So we can conclude that 25.6% of the mother's BMI variation is explained by the daugther's BMI.
Part a
Since the BMI is a relation between height and weight, other possible variables that can explain the variability are (weight , height, age)
It is often said that your chances of winning the lottery if you buy a ticket are just slightly higher than if you don't buy one! Suppose a Lotto game consists of picking 6 of 48 numbers.
What is the probability of winning with the very first Lotto ticket you purchase?
Answer:
1/48 % or 6/48 % chance
Step-by-step explanation:
The probability of winning with the very first Lotto ticket you purchase of the Lotto game consisting of picking 6 of 48 numbers is 1/12271512 or approximately 0.0000000815.
What is a permutation?A permutation is a process of calculating the number of ways to choose a set from a larger set in a particular order.
If we want to choose a set of r items from a set of n items in a particular order, we find the permutation nPr = n!/(n-r)!.
What is a combination?A combination is a process of calculating the number of ways to choose a set from a larger set in no particular order.
If we want to choose a set of r items from a set of n items in no particular order, we find the combination nCr = n!/{(r!)(n-r)!}.
How do we solve the given question?In the question, we are asked to determine the probability of winning a lottery by picking 6 numbers from 48 numbers with the first ticket we purchase.
First, we need to calculate the number of combinations of choosing 6 numbers from 48 numbers. As we need to consider no particular order, we will use combinations,
48C6 = 48!/{(6!)(48-6)!} = 48!/(6!*42!) = (43*44*45*46*47*48)/(1*2*3*4*5*6*) (As 48! = 42!*43*44*45*46*47*48, and 42! cancels itself from the numerator and the denominator).
or, 48C6 = 12,271,512.
So, we get the number of combinations = 12,271,512.
We know that we will choose only one particular set of 6 numbers.
∴ The probability of winning on the very first ticket = 1/12,271,512 ≈ 0.0000000815
∴ The probability of winning with the very first Lotto ticket you purchase of the Lotto game consisting of picking 6 of 48 numbers is 1/12271512 or approximately 0.0000000815.
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A 20% solution of fertilizer is to be mixed with a 50% solution of fertilizer in order to get 180 gallons of a 40% solution. How many gallons of the 20% solution and 50%
solution should be mixed?
Answer: 60 gallons of the 20% solution and 120 gallons of the 50% solution should be mixed.
Step-by-step explanation:
Let x represent the number of gallons of 20% solution that should be mixed.
Let y represent the number of gallons of 50% solution that should be mixed.
A 20% solution of fertilizer is to be mixed with a 50% solution of fertilizer in order to get 180 gallons of a 40% solution. This means that
0.2x + 0.5y = 0.4×180
0.2x + 0.5y = 72- - - - - - - - - - - -1
Since the total number of gallons is 180, it means that
x + y = 180
Substituting x = 180 - y into equation 1, it becomes
0.2(180 - y) + 0.5y = 72
36 - 0.2y + 0.5y = 72
- 0.2y + 0.5y = 72 - 36
0.3y = 36
y = 36/0.3
y = 120
x = 180 - y = 180 - 120
x = 60
To get a 40% solution by mixing a 20% and 50% solution, you need 60 gallons of the 20% solution and 120 gallons of the 50% solution.
Explanation:To solve this problem, we can use the concept of mixtures. Let's represent the amount of 20% solution as x gallons and the amount of 50% solution as y gallons. From the given information, we can set up the following equations:
x + y = 180 (equation 1)
0.2x + 0.5y = 0.4 * 180 (equation 2)
Simplifying equation 2 gives us 0.2x + 0.5y = 72. To eliminate decimals, we can multiply both sides by 10 to get 2x + 5y = 720.
Now we have a system of equations. We can solve it by substitution or elimination. Let's use elimination:
Multiplying equation 1 by 2 gives us 2x + 2y = 360. Subtracting this from equation 2 gives us 3y = 360. Dividing both sides by 3 gives us y = 120. Substituting this value into equation 1 gives us x + 120 = 180. Subtracting 120 from both sides gives us x = 60.
Therefore, we need 60 gallons of the 20% solution and 120 gallons of the 50% solution.
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The weights of newborn children in the United States vary according to the Normal distribution with mean 7.5 pounds and standard deviation 1.25 pounds. The government classifies a newborn as having low birth weight if the weight is less than 5.5 pounds. (a) What is the probability that a baby chosen at random weighs less than 5.5 pounds at birth?(b) You choose three babies at random. What is the probability that their average birth weight is less than 5.5 pounds?
Answer:
a) 5.48% probability that a baby chosen at random weighs less than 5.5 pounds at birth
b) 0.28% probability that their average birth weight is less than 5.5 pounds
Step-by-step explanation:
To solve this question, the normal probability distribution and the central limit theorem are used.
Normal probability distribution:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
Central limit theorem:
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sample means n of at least 30 can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex]
In this problem, we have that:
[tex]\mu = 7.5, \sigma = 1.25[/tex]
(a) What is the probability that a baby chosen at random weighs less than 5.5 pounds at birth?
This is the pvalue of Z when X = 5.5. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{5.5 - 7.5}{1.25}[/tex]
[tex]Z = -1.6[/tex]
[tex]Z = -1.6[/tex] has a pvalue of 0.0548
5.48% probability that a baby chosen at random weighs less than 5.5 pounds at birth
(b) You choose three babies at random. What is the probability that their average birth weight is less than 5.5 pounds?
[tex]n = 3, s = \frac{1.25}{\sqrt{3}} = 0.7217[/tex]
This is the pvalue of Z when X = 5.5. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
This is the pvalue of Z when X = 5.5. So
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{5.5 - 7.5}{0.7217}[/tex]
[tex]Z = -2.77[/tex]
[tex]Z = -2.77[/tex] has a pvalue of 0.0028
0.28% probability that their average birth weight is less than 5.5 pounds
Suppose that your statistics professor tells you that the scores on a midterm exam were approximately normally distributed with a mean of 79 and a standard deviation of 6. The top 15% of all scores have been designated As. Your score is 89. Did you earn an A
Answer:
You earned an A.
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
[tex]\mu = 79, \sigma = 6[/tex]
The top 15% of all scores have been designated As.
This means that if Z for the score has a pvalue of 1-0.15 = 0.85 or higher, the score is designated as A.
Your score is 89. Did you earn an A?
We have to find the pvalue of Z when X = 89. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{89 - 79}{6}[/tex]
[tex]Z = 1.67[/tex]
[tex]Z = 1.67[/tex] has a pvalue of 0.9525. So yes, you earned an A.
Find out for what value of the variable:
do the trinomial a2+7a+6 and the binomial a+1 have the same value?
do the trinomial 3x2−x+1 and the trinomial 2x2+5x−4 have the same value?
The Main Answer for:
1. The value of the variable for which the trinomial [tex]a^2+7a+6[/tex] and the binomial a+1 have the same value is [tex]a=-1[/tex] or [tex]a=-5[/tex].
2. The value of the variable for which the trinomial [tex]3x^2-x+1[/tex] and the trinomial [tex]2x^2+5x-4[/tex] have the same value is [tex]x=1[/tex] or [tex]x=5[/tex].
To find the value of the variable for which two polynomials have the same value, we set them equal to each other and solve for the variable.
1. For the trinomial [tex]a^2+7a+6[/tex] and the binomial [tex]a+1[/tex]:
[tex]a^2+7a+6=a+1[/tex]
First, let's rewrite the equation in standard form:
[tex]a^2+7a+6-(a+1)=0\\a^2+7a+6-a-1=0\\a^2+6a+5=0[/tex]
Now, we can solve this quadratic equation by factoring:
[tex]a^2+6a+5=0\\(a+5)(a+1)=0[/tex]
Setting each factor equal to zero:
[tex]a+5=0[/tex]⇒[tex]a=-5[/tex]
[tex]a+1=0[/tex]⇒[tex]a=-1[/tex]
So, the value of the variable for which the trinomial [tex]a^2+7a+6[/tex] and the binomial a+1 have the same value is [tex]a=-1[/tex] or [tex]a=-5[/tex].
2. For the trinomial [tex]3x^2-x+1[/tex] and the trinomial [tex]2x^2+5x-4[/tex]:
[tex]3x^2-x+1=2x^2+5x-4[/tex]
Again, let's rewrite the equation in standard form:
[tex]3x^2-x+1-(2x^2+5x-4)=0\\3x^2-x+1-2x^2-5x+4=0\\X^2-6x+5=0[/tex]
Now, let's solve this quadratic equation by factoring:
[tex]X^2-6x+5=0\\(x-5)(x-1)=0[/tex]
Setting each factor equal to zero:
[tex]x-5=0[/tex]⇒[tex]x=5[/tex]
[tex]x-1=0[/tex]⇒[tex]x=1[/tex]
So, the value of the variable for which the trinomial [tex]3x^2-x+1[/tex] and the trinomial [tex]2x^2+5x-4[/tex] have the same value is [tex]x=1[/tex] or [tex]x=5[/tex].
COMPLETE QUESTION:
Find out for what value of the variable:
1. Do the trinomial [tex]a^2+7a+6[/tex] and the binomial [tex]a+1[/tex] have the same value?
2. Do the trinomial [tex]3x^2-x+1[/tex] and the trinomial [tex]2x^2+5x-4[/tex] have the same value?
Suppose that the warranty cost of defective widgets is such that their proportion should not exceed 5% for the production to be profitable. Being very cautious, you set a goal of having 0.05 as the upper limit of a 90% confidence interval, when repeating the previous experiment. What should the maximum number of defective widgets be, out of 1024, for this goal to be reached.
Answer: 63 defective widgets
Step-by-step explanation:
Given that the proportion should not exceed 5%, that is:
p< or = 5%.
So we take p = 5% = 0.05
q = 1 - 0.05 = 0.095
Where q is the proportion of non-defective
We need to calculate the standard error (standard deviation)
S = √pq/n
Where n = 1024
S = √(0.05 × 0.095)/1024
S = 0.00681
Since production is to maximize profit(profitable), we need to minimize the number of defective items. So we find the limit of defective product to make this possible using the Upper Class Limit.
UCL = p + Za/2(n-1) × S
Where a is alpha of confidence interval = 100 -90 = 10%
a/2 = 5% = 0.05
UCL = p + Z (0.05) × 0.00681
Z(0.05) is read on the t-distribution table at (n-1) degree of freedom, which is at infinity since 1023 = n-1 is large.
Z a/2 = 1.64
UCL = 0.05 + 1.64 × 0.00681
UCL = 0.0612
Since the UCL in this case is a measure of proportion of defective widgets
Maximum defective widgets = 0.0612 ×1024 = 63
Alternatively
UCL = p + 3√pq/n
= 0.05 + 3(0.00681)
= 0.05 + 0.02043 = 0.07043
UCL =0.07043
Max. Number of widgets = 0.07043 × 1024
= 72
The Wall Street Journal reported that Walmart Stores Inc. is planning to lay off 2300 employees at its Sam's Club warehouse unit. Approximately half of the layoffs will be hourly employees (The Wall Street Journal, January 25-26, 2014). Suppose the following data represent the percentage of hourly employees laid off for 15 Sam's Club stores. 55 56 44 43 44 56 60 62 57 45 36 38 50 69 65 a. Compute the mean and median percentage of hourly employees being laid off at these stores. b. Compute the first and third quartiles. c. Compute the range and interquartile range. d. Compute the variance and standard deviation. e. Do the data contain any outliers? f. Based on the sample data, does it appear that Walmart is meeting its goal for reducing the number of hourly employees?
Answer:
(a) The mean is 52 and the median is 55.
(b) The first quartile is 44 and the third quartile is 60.
(c) The value of range is 33 and the inter-quartile range is 16.
(d) The variance is 100.143 and the standard deviation is 10.01.
(e) There are no outliers in the data set.
(f) Yes
Step-by-step explanation:
The data provided is:
S = {55, 56, 44, 43, 44, 56, 60, 62, 57, 45, 36, 38, 50, 69, 65}
(a)
Compute the mean of the data as follows:
[tex]\bar x=\frac{1}{n}\sum x\\=\frac{1}{15}[55+ 56+ 44+ 43+ 44+ 56+ 60+ 62+ 57+ 45 +36 +38 +50 +69+ 65]\\=\frac{780}{15}\\=52[/tex]
Thus, the mean is 52.
The median for odd set of values is the computed using the formula:
[tex]Median=(\frac{n+1}{2})^{th}\ obs.[/tex]
Arrange the data set in ascending order as follows:
36, 38, 43, 44, 44, 45, 50, 55, 56, 56, 57, 60, 62, 65, 69
There are 15 values in the set.
Compute the median value as follows:
[tex]Median=(\frac{15+1}{2})^{th}\ obs.=(\frac{16}{2})^{th}\ obs.=8^{th}\ observation[/tex]
The 8th observation is, 55.
Thus, the median is 55.
(b)
The first quartile is the middle value of the upper-half of the data set.
The upper-half of the data set is:
36, 38, 43, 44, 44, 45, 50
The middle value of the data set is 44.
Thus, the first quartile is 44.
The third quartile is the middle value of the lower-half of the data set.
The upper-half of the data set is:
56, 56, 57, 60, 62, 65, 69
The middle value of the data set is 60.
Thus, the third quartile is 60.
(c)
The range of a data set is the difference between the maximum and minimum value.
Maximum = 69
Minimum = 36
Compute the value of Range as follows:
[tex]Range =Maximum-Minimum\\=69-36\\=33[/tex]
Thus, the value of range is 33.
The inter-quartile range is the difference between the first and third quartile value.
Compute the value of IQR as follows:
[tex]IQR=Q_{3}-Q_{1}\\=60-44\\=16[/tex]
Thus, the inter-quartile range is 16.
(d)
Compute the variance of the data set as follows:
[tex]s^{2}=\frac{1}{n-1}\sum (x_{i}-\bar x)^{2}\\=\frac{1}{15-1}[(55-52)^{2}+(56-52)^{2}+...+(65-52)^{2}]\\=100.143[/tex]
Thus, the variance is 100.143.
Compute the value of standard deviation as follows:
[tex]s=\sqrt{s^{2}}=\sqrt{100.143}=10.01[/tex]
Thus, the standard deviation is 10.01.
(e)
An outlier is a data value that is different from the remaining values.
An outlier is a value that lies below 1.5 IQR of the first quartile or above 1.5 IQR of the third quartile.
Compute the value of Q₁ - 1.5 IQR as follows:
[tex]Q_{1}-1.5QR=44-1.5\times 16=20[/tex]
Compute the value of Q₃ + 1.5 IQR as follows:
[tex]Q_{3}+1.5QR=60-1.5\times 16=80[/tex]
The minimum value is 36 and the maximum is 69.
None of the values is less than 20 or more than 80.
Thus, there are no outliers in the data set.
(f)
Yes, the data provided indicates that the Walmart is meeting its goal for reducing the number of hourly employees
Answer:
Step-by-step explanation:
A researcher interested in weight control wondered whether normal and overweight individuals differ in their reaction to the availability of food. Thus, normal and overweight participants were told to eat as many peanuts as they desired while working on a questionnaire. One manipulation was the proximity of the peanut dish (close or far from the participant); the second manipulation was whether the peanuts were shelled or unshelled. After filling out the questionnaire, the peanut dish was weighed to determine the amount of peanuts consumed.1. Identify the design (e.g., 2 X 2 factorial).2. Identify the total number of conditions.3. Identify the manipulated variable(s).4. Is this an IV X PV design? If so, identify the participant variable(s).5. Is this a repeated measures design? If so, identify the repeated variable(s).6. Identify the dependent variable(s).
Answer:
Complete question for your reference is below.
A researcher interested in weight control wondered whether normal and overweight individuals differ in their reaction to the availability of food. Thus, normal and overweight participants were told to eat as many peanuts as they desired while working on a questionnaire. One manipulation was the proximity of the peanut dish (close or far from the participant); the second manipulation was whether the peanuts were shelled or unshelled. After filling out the questionnaire, the peanut dish was weighed to determine the amount of peanuts consumed.
1. Identify the design (e.g., 2 X 2 factorial).
2. Identify the total number of conditions.
3. Identify the manipulated variable(s).
4. Is this an IV X PV design? If so, identify the participant variable(s).
5. Is this a repeated measures design? If so, identify the repeated variable(s).
6. Identify the dependent variable(s).
Step-by-step explanation:
Please find attached file for complete answer solution and explanation.
The research design is a 2x2 factorial with the manipulated or independent variables being the proximity of the peanut dish and whether the peanuts are shelled, resulting in four conditions. It is not an IV x PV design, nor a repeated measures design. The dependent variable in this study is the amount of peanuts consumed.
Explanation:The research design posed in this question appears to be a 2x2 factorial design. This is because there are two independent variables (the proximity of the peanut dish and whether the peanuts are shelled or not), each with two levels (close or far, and shelled or unshelled respectively).
Given this, there would be a total of four conditions in this study (close-shelled, close-unshelled, far-shelled, far-unshelled). The manipulated variables, also referred to as independent variables, in this study are the 'proximity of the peanut dish' and 'whether the peanuts are shelled or unshelled'.
This does not appear to be an IV X PV (Independent Variable x Participant Variable) design because there is no information about a participant variable being used here. It also does not seem to be a repeated measures design as individuals are not exposed to all conditions; rather they experience one specific condition. The dependent variable is the amount of peanuts consumed, as determined by weighing the peanut dish after participants have finished eating.
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