Answer:
Given p = 0.07 as the probability that someone is a universal donor
In case of Geometric Distribution, Probability of getting the first success on nth trial is given by
[tex]P (X=n) = p (1-p) ^ {n-1}[/tex]
where p is the probability of success on any one trial and (1-p) shows the probability of failure.
So the probability of the first subject to be a universal blood donor will be the seventh person is
[tex]P (X=7) = 0.07 (1-0.07) ^ {7-1} = 0.07 (0.93) ^ 6 = 0.07*0.647 = 0.0453[/tex]
So the final probability is 0.0453
what is the area of a triangle expressed as a monomial
Answer:
[tex] 28x^2 [/tex]
Step-by-step explanation:
[tex]area \:o f \: \triangle = \frac{1}{2} \times 8x \times 7x \\ = 4x \times 7x \\ = 28 {x}^{2} [/tex]
The histogram to the right represents the weights (in pounds) of members of a certain high-school math team. How many team members are included in the histogram?
Answer:
Total member = a+b+c+d+e+f+g=5+4+2+0+5+0+4=20 member
Step-by-step explanation:
Let suppose the attached histogram
a) Total members whose weight 110 = 5
b) Total members whose weight 120 = 4
c) Total members whose weight 130 = 2
d) Total members whose weight 140 = 0
e) Total members whose weight 150 = 5
f) Total members whose weight 160 = 0
g) Total members whose weight 170 = 4
Total member = a+b+c+d+e+f+g=5+4+2+0+5+0+4=20 member
A histogram is a statistical graphical representation of data with the help of bars with different heights.
In a recent survey, 10 percent of the participants rated Pepsi as being "concerned with my health." PepsiCo's response included a new "Smart Spot" symbol on its products that meet certain nutrition criteria, to help consumers who seek more healthful eating options. At α = .05, would a follow-up survey showing that 18 of 100 persons now rate Pepsi as being "concerned with my health" prove that the percentage has increased?
Answer:
[tex]z=\frac{0.18 -0.1}{\sqrt{\frac{0.1(1-0.1)}{100}}}=2.67[/tex]
[tex]p_v =P(z>2.67)=0.0038[/tex]
So the p value obtained was a very low value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of adults who rate Pepsi as being "concerned with my health" i significantly higher than 0.1
Step-by-step explanation:
Data given and notation
n=100 represent the random sample taken
X=18 represent the adults who rate Pepsi as being "concerned with my health"
[tex]\hat p=\frac{18}{100}=0.18[/tex] estimated proportion of adults who rate Pepsi as being "concerned with my health"
[tex]p_o=0.10[/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)
Concepts and formulas to use
We need to conduct a hypothesis in order to test the claim that the true proportion is higher than 0.1.:
Null hypothesis:[tex]p \leq 0.1[/tex]
Alternative hypothesis:[tex]p > 0.1[/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].
Calculate the statistic
Since we have all the info requires we can replace in formula (1) like this:
[tex]z=\frac{0.18 -0.1}{\sqrt{\frac{0.1(1-0.1)}{100}}}=2.67[/tex]
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 right tailed test the p value would be:
[tex]p_v =P(z>2.67)=0.0038[/tex]
So the p value obtained was a very low value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of adults who rate Pepsi as being "concerned with my health" i significantly higher than 0.1
The student is asked to perform a hypothesis test at the alpha = 0.05 level to determine if the percentage of participants rating Pepsi as 'concerned with my health' has statistically increased from 10% to 18% in a follow-up survey.
The question involves conducting a hypothesis test to determine if the percentage of participants rating Pepsi as being "concerned with my health" has increased after a new initiative. With an initial percentage of 10%, a follow-up survey showed that 18 out of 100 (or 18%) now rate Pepsi in this manner. To determine if this is a statistically significant increase, at the alpha = 0.05 significance level, a hypothesis test can be performed using the binomial or normal approximation to the binomial distribution.
Null hypothesis (H0): p = 0.10 (The percentage of those who believe Pepsi is concerned with their health has not changed)
Alternative hypothesis (Ha): p > 0.10 (The percentage has increased)
If the calculated p-value is less than the significance level (0.05), we can reject the null hypothesis in favor of the alternative hypothesis, thus concluding that the percentage has indeed increased.
2. The Hereford Cattle Society says that the mean weight of a one-year-old Hereford bull is 1135 pounds, with a standard deviation of 97 pounds. Suppose 40 bulls are randomly selected and loaded on a train car. Find the probability their combined weight exceeds 46000 pounds. (Hint: The combined weight exceeds 46000 pounds if the average weight exceeds 46000 40 = 1150 pounds.)
Answer:
16.35% probability their combined weight exceeds 46000 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], the sample means with size 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 = 1135, \sigma = 97, n = 40, s = \frac{97}{\sqrt{40}} = 15.34[/tex]
Find the probability their combined weight exceeds 46000 pounds.
This is 1 subtracted by the pvalue of Z when [tex]X = \frac{46400}{40} = 1150[/tex]. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{1150 - 1135}{15.34}[/tex]
[tex]Z = 0.98[/tex]
[tex]Z = 0.98[/tex] has a pvalue of 0.8365
1 - 0.8365 = 0.1635
16.35% probability their combined weight exceeds 46000 pounds.
Final answer:
The probability that the combined weight of 40 randomly selected one-year-old Hereford bulls exceeds 46,000 pounds is approximately 16.35%. This calculation uses the Z-score related to the sample mean weight exceeding 1,150 pounds.
Explanation:
To find the probability that the combined weight of 40 randomly selected one-year-old Hereford bulls exceeds 46,000 pounds, we first need to determine if the average weight per bull exceeds 1,150 pounds (since 46,000 ÷ 40 = 1,150). The mean weight of a Hereford bull is given as 1,135 pounds with a standard deviation of 97 pounds. Using the Central Limit Theorem, we can find the mean and standard deviation for the sample mean weight of 40 bulls.
The standard deviation of the sample mean (σ_x-bar) is σ ÷ √{n} = 97 pounds ÷ √{40}, which equals approximately 15.34 pounds. To find the Z-score, we use the formula Z = (X - μ) ÷ σ_x-bar, where X is 1,150 pounds, μ (the mean) is 1,135 pounds, and σ_x-bar is 15.34 pounds, yielding a Z-score of about 0.98.
Using the standard normal distribution table, a Z-score of 0.98 corresponds to a probability of approximately 0.8365. However, since we want the probability that the sample mean is greater than 1,150 pounds, we need to look at the upper tail, which is 1 - 0.8365 = 0.1635. Therefore, there's a 16.35% chance the combined weight of 40 randomly selected one-year-old Hereford bulls exceeds 46,000 pounds.
please help I’m desperate
Ian has decided to buy a new car for $22,000 and agreed to make monthly payments for three years at
6.5% compounded monthly.
a. How much is each payment?
b. How much total interest will he pay over the life of the loan?
Answer:
Step-by-step explanation:
a) We would apply the periodic interest rate formula which is expressed as
P = a/[{(1+r)^n]-1}/{r(1+r)^n}]
Where
P represents the monthly payments.
a represents the amount of the loan
r represents the annual rate.
n represents number of monthly payments. Therefore
a = $22000
r = 0.065/12 = 0.0054
n = 12 × 3 = 36
Therefore,
P = 22000/[{(1+0.0054)^36]-1}/{0.0054(1+0.0054)^36}]
22000/[{(1.0054)^36]-1}/{0.0054(1.0054)^36}]
P = 22000/{1.214 -1}/[0.0054(1.214)]
P = 22000/(0.214/0.0065556)
P = 22000/32.64
P = $674
b) The total amount that he would pay in 3 years is
674 × 36 = 24265
total interest paid over the life of the loan is
24265 - 22000 = $2265
Zenich Corp manufactures laptops. The waiting time is 72 minutes before the start of production and the manufacturing lead time is 160 minutes per laptop. What is its manufacturing cycle efficiency? (Round to the nearest whole percent.)
Answer:
The Manufacturing cycle efficiency = 69 %
Step-by-step explanation:
The waiting time is = 72 minute
The manufacturing lead time is ( [tex]T_{lead}[/tex] )= 160 minutes
Total cycle time ( [tex]T_{c}[/tex] ) = 160 + 72 = 232 minutes
Manufacturing cycle efficiency = [tex]\frac{T_{lead} }{T_{c} }[/tex]
⇒ Manufacturing cycle efficiency = [tex]\frac{160}{232}[/tex]
⇒ Manufacturing cycle efficiency = 69 %
Therefore, the Manufacturing cycle efficiency = 69 %
Executives at a large multinational company are being accused of spending too much time on email and not enough time with their employees whom they supervise. In order to see if this is true, the VP of Human Resources decides to do a comparison of the number of minutes per day that the top 100 executives spend on e-mail compared to the total minutes spent by everyone in the company. The average minutes spent per day by each employee was 96 minutes. The average for just the top 100 executives was 91 minutes with a standard deviation of 25 minutes. On average, are the Executives spending more time on e-mail than all the employees? If you wanted to use the the 1% level of significance to test your hypothesis, what would be the correct critical value?
Answer:
We conclude that there is not enough evidence to support the claim that company is spending too much time on email.
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = 96 minutes
Sample mean, [tex]\bar{x}[/tex] = 91 minutes
Sample size, n = 100
Alpha, α = 0.05
Sample standard deviation, s = 25 minutes
First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu \leq 96\text{ minutes}\\H_A: \mu > 96\text{ minutes}[/tex]
We use one-tailed t test to perform this hypothesis.
Formula:
[tex]t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }[/tex]
Putting all the values, we have
[tex]t_{stat} = \displaystyle\frac{91 - 96}{\frac{25}{\sqrt{100}} } = -2[/tex]
Now, [tex]t_{critical} \text{ at 0.05 level of significance, 99 degree of freedom } = 1.6603[/tex]
Since,
The calculated test statistic is less than the critical value, we fail to reject the null hypothesis and accept it.
Thus, we conclude that there is not enough evidence to support the claim that company is spending too much time on email.
Find the function r that satisfies the given condition. r'(t) = (e^t, sin t, sec^2 t): r(0) = (2, 2, 2) r(t) = ()
Answer:
r(t) = (e^t +1, -cos(t) + 3, tan(t) + 2)
Step-by-step explanation:
A primitive of e^t is e^t+c, since r(0) has 2 in its first cooridnate, then
e^0+c = 2
1+c = 2
c = 1
Thus, the first coordinate of r(t) is e^t + 1.
A primitive of sin(t) is -cos(t) + c (remember that the derivate of cos(t) is -sin(t)). SInce r(0) in its second coordinate is 2, then
-cos(0)+c = 2
-1+c = 2
c = 3
Therefore, in the second coordinate r(t) is equal to -cos(t)+3.
Now, lets see the last coordinate.
A primitive of sec²(t) is tan(t)+c (you can check this by derivating tan(t) = sin(t)/cos(t) using the divition rule and the property that cos²(t)+sin²(t) = 1 for all t). Since in its third coordinate r(0) is also 2, then we have that
2 = tan(0)+c = sin(0)/cos(0) + c = 0/1 + c = 0
Thus, c = 2
As a consecuence, the third coordinate of r(t) is tan(t) + 2.
As a result, r(t) = (e^t +1, -cos(t) + 3, tan(t) + 2).
To find the function r(t) that satisfies the given condition, integrate each component of r'(t) and solve for the constants using the initial condition.
Explanation:To find the function r(t) that satisfies the condition r'(t) = (e^t, sin t, sec^2 t) and r(0) = (2, 2, 2), we need to integrate each component of r'(t) with respect to t.
The integral of e^t with respect to t is e^t + C_1, where C_1 is a constant. The integral of sin t with respect to t is -cos t + C_2, where C_2 is a constant.The integral of sec^2 t with respect to t is tan t + C_3, where C_3 is a constant.Combining these integrals, we get r(t) = (e^t + C_1, -cos t + C_2, tan t + C_3). Substituting the initial condition r(0) = (2, 2, 2), we can solve for the constants and obtain the function r(t) = (e^t + 2, -cos t + 2, tan t + 2).
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A city park employee collected 1,850 cents in nickels, dimes, and quarters at the bottom of a wishing well. There were 30 nickels, and a combined total of 110 dimes and quarters. How many dimes and quarters were at the bottom of the wishing well?
dimes ??
quarters ???
can someone tell me how to solve this problem, ive been getting it wrong. I think I've missed a step maybe.
Answer: there are 70 dimes and 40 quarters.
Step-by-step explanation:
A dime is 10 cents
A nickel is 5 cents
A quarter is 25 cents
Let x represent the number of dimes at the bottom.
Let y represent the number of quarters at the bottom.
The city park employee collected 1,850 cents in nickels, dimes, and quarters at the bottom of a wishing well. If there were 30 nickels, it means that
10x + 25y + 30 × 5 = 1850
10x + 25y = 1850 - 150
10x + 25y = 1700- - - - - - - - -1
There is a combined total of 110 dimes and quarters. It means that
x + y = 110
Substituting x = 110 - y into equation 1, it becomes
10(110 - y) + 25y = 1700
1100 - 10y + 25y = 1700
1100 + 15y = 1700
15y = 1700 - 1100
15y = 600
y = 600/15
y = 40
x = 110 - y = 110 - 40
x = 70
Let X, the number of flaws on the surface of a randomly selected boiler of a certain type, have a Poisson distribution with parameter μ = 5. Use the cumulative Poisson probabilities from the Appendix Tables to compute the following probabilities. (Round your answers to three decimal places.)
(a) P(X ≤ 8)
(b) P(X = 8)
(c) P(9 ≤ X)
(d) P(5 ≤ X ≤ 8)
(e) P(5 < X < 8)
Answer:
[tex]a. \ P(X\leq 8)=0.932\\b. \ P(X=8)=0.065\\c. \ P(9\leq X)=0.068\\d. \ P(5\leq X \leq 8)=0.491\\[/tex]
e. P(5<X<8)=0.251
Step-by-step explanation:
Given that boiler follows a Poisson distribution,$\sim$
[tex]X $\sim$Poi(\mu=5)[/tex]
Poisson Distribution formula is given by the expression:-
[tex]p(x,\mu)=\frac{e^{-\mu}\mu^x}{x!}[/tex]
Our probabilities will be calculated as below:
a.
[tex]P(X=x)=\frac{5^xe^{-5}}{x!}\\P(X\leq 8)=P(X=0)+P(X=1)+...+P(X=8)\\=\frac{5^0e^{-5}}{1!}+...+\frac{5^8e^{-5}}{8!}\\=0.006738+...+0.065278\\=0.961[/tex]
b.
[tex]P(X=8)=\frac{5^8e^{-5}}{8!}\\=0.065[/tex]
c.
[tex]P(9\leq X)=1-P(X=0)-P(X=1)-...-P(X=8)\\=0.068[/tex]
d.
[tex]P(5\leq X\leq 8)=P(X=5)+P(X=6)+P(X=7)+P(X=8)\\=0.491[/tex]
e.P(5<X<8)
[tex]P(X=6)+P(X=7)\\=0.251[/tex]
Jane and Nancy were both awarded National Merit Scholarships (for academic excellence), and both love mathematics. Jane breezed through Dr. Morgan's calculus class. Thus, Nancy should do well in that class, too."
This argument is:
a. Deductive and Valid
b. Deductive and Invalid
c. Inductive and Strong
d. Inductive and Weak
Answer:
c. Inductive and Strong
Step-by-step explanation:
In inductive reasoning, provided data is analyzed in order to reach a conclusion. In this case, the argument provides data regarding Jane and Nancy's awards and their love for mathematics and then draws a conclusion regarding Nancy's performance in a particular class, this is an example of inductive reasoning.
As for the strength of the argument, it is plausible to infer that Jane and Nancy have similar mathematics skills since they both love calculus and excel academically. Therefore, if Jane does well in the calculus class, it is a strong argument to say that Nancy does as well.
The answer is :
c. Inductive and Strong
Final answer:
The argument is Deductive and Invalid as it does not guarantee Nancy's success in the calculus class based on Jane's performance.
Explanation:
The argument presented is Deductive and Invalid. This is because the conclusion that Nancy will do well in the calculus class based on Jane's performance does not necessarily follow logically. Just because Jane excelled does not guarantee the same result for Nancy.
In deductive reasoning, the conclusion must follow necessarily from the premises, which is not the case here. An example of a valid deductive argument would be 'All humans are mortal, Socrates is a human, therefore Socrates is mortal.'
Because the argument in question does not provide a guarantee of Nancy's success based on Jane's performance, it is categorized as Invalid in the realm of deductive reasoning.
A teen chewing bubble gum blows a huge bubble, the volume of which satisfies the differential equation:dV/dt=2(V)^(0.5)Where t is the time in seconds after the teen start to blow the bubble. If the bubble pops as soon as it reaches 529 cubic centimeters in volume, how many seconds does it take for the bubble to pop? Assume that the bubble had no volume when the teen first started blowing.
The ODE is separable:
[tex]\dfrac{\mathrm dV}{\mathrm dt}=2\sqrt V\implies\dfrac{\mathrm dV}{2\sqrt V}=\mathrm dt[/tex]
Integrate both sides to get
[tex]\sqrt V=t+C[/tex]
Assuming the bubble starts with zero volume, so that [tex]V(0)=0[/tex], we find
[tex]\sqrt0=0+C\implies C=0[/tex]
Then the volume of the bubble at time [tex]t[/tex] is
[tex]V(t)=t^2[/tex]
and so the time it take for the bubble to reach a volume of 629 cubic cm is
[tex]t^2=529\implies t=23[/tex]
or 23 seconds after the teen first starts blowing the bubble.
The materials and equipment you will have include 100% alcohol, test tubes (large diameter test tubes), fermentation tubes (smaller diameter tubes closed off at one end), yeast, and sugar. The yeast and sugar are mixed to start the fermentation process. The fermentation tube can be used to collect gases. State a testable hypothesis for your speculation from item 8 above, using the above equipment.
Answer:
A hypothesis is a proposed statement explaining the relationship between two or more variable which can be verifiable through experiment. In this case a testable hypothesis would be.
When yeast and sugar are mixed together during formation, it will cause the sugar to undergo a high rate of glycolysis and CO[tex]^{2}[/tex] will be present in the fermentation tube.
A section of an exam contains four True-False questions. A completed exam paper is selected at random, and the four answers are recorded. Assuming the outcomes to be equally likely, find the probability that all the answers are the same.
The probability that all the answers are the same in the section of the exam is 1/8 or 12.5%.
Explanation:To find the probability that all the answers are the same, we need to consider the number of ways in which all the answers can be the same, divided by the total number of possible outcomes. Since there are only two possible answers (True or False) for each question, there are a total of 2^4 = 16 possible outcomes.
Out of these 16 outcomes, there are only two ways in which all the answers can be the same: either all True or all False.
Therefore, the probability that all the answers are the same is 2/16 = 1/8 = 0.125, or 12.5%.
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The mean and the standard deviation of the sampled population are 225.8and 17.8 respectively. What is the mean and standard deviation of the sample mean when n = 25?
Answer:
Mean and standard deviation of the sample mean are 225.8 and 3.56 respectively.
Step-by-step explanation:
The mean (μₓ) and standard deviation of the sample mean (σₓ) are related to the mean (μ) and standard deviation of the population (σ) through the following relationship
μₓ = μ = 225.8
σₓ = σ/√n = 17.8/√25 = 17.8/5 = 3.56
find the measure of exterior angle tuv below .
Given:
m∠T = 61°
m∠S = 5x - 15
m ext∠TUV = 8x - 8
To find:
The measure of exterior angle TUV
Solution:
STU is a triangle.
By exterior angle of triangle property:
The measure of an exterior angle is equal to the sum of the opposite interior angles.
m ext∠TUV = m∠T + m∠S
8x - 8 = 61 + 5x - 15
8x - 8 = 46 + 5x
Add 8 on both sides.
8x - 8 + 8 = 46 + 5x + 8
8x = 54 + 5x
Subtract 5x from both sides.
8x - 5x = 54 + 5x - 5x
3x = 54
Divide by 3 on both sides.
[tex]$\frac{3x}{3} =\frac{54}{3}[/tex]
x = 18
Substitute x = 18 in m ext∠TUV.
m ext∠TUV = 8(18) - 8
= 144 - 8
= 136
Therefore measure of exterior angle ∠TUV is 136°.
Seventy-five (75) customers at Crummy Burger today were entered in a drawing. Three names will be drawn, and each will win a prize. Determine the number of ways that the prizes can be distributed if first prize is a $25 gift card, second prize is a free Crummy Combo Meal, and third prize is a free Jumbo Crummy Burger.
Answer:
405150
Step-by-step explanation:
3 people are to be selected from 75 people. This could be done using the combination function, [tex]\binom{75}{3}[/tex]. But thus function only gives the number of ways of choosing any 3 out of the 75 customers. For any single selection of any 3 customers, there are 3! ways of sharing the three prizes among them. Hence the total number of ways of sharing the prizes is
[tex]\binom{75}{3}\times3! = 405150[/tex]
In fact, this is the permutation function given by
[tex]{}^{75}P_3 = \dfrac{75!}{(75-3)! = 405150}[/tex]
Given the system of equations
(D+5)x+D2y=cost
D2x+(D+9)y=t2+3t
(a) Write the system as a differentialequation for x , in operator form, by filling thepolynomial operator in D (on the left-hand-sideof the equation) and the corresponding right-hand-side of theequation:
x=
(Write your expression for the right-hand-side as a function oft only---that is, with no derivativeoperators.)
(b) Note that we could equally well solve fory. Do this here, obtaining an equivalent differentialequation for y.
y=
(Write your expression for the right-hand-side as a function oft only---that is, with no derivativeoperators.)
The differential equations for x and y are:
x = (cost - D^2 (t2 + 3t)) / ((D+5) + D^2 (D+9))
y = (t2 + 3t - D2 cost) / (D2 / (D+5) + (D+9))
Here's how to solve the problem:
(a) Write the system as a differential equation for x:
Eliminate y: Substitute the second equation into the first equation to eliminate y.
(D+5)x + D2(t2 + 3t) = cost
Rearrange and define operator: Define P(D) = (D+5) + D^2 (D+9).
P(D) x = cost - D^2 (t2 + 3t)
Write as differential equation:
x = (cost - D^2 (t2 + 3t)) / P(D)
(b) Write the system as a differential equation for y:
Eliminate x: Substitute the first equation into the second equation to eliminate x.
D2((cost - D2y) / (D+5)) + (D+9)y = t2 + 3t
Rearrange and define operator: Define Q(D) = D2 / (D+5) + (D+9).
Q(D) y = t2 + 3t - D2 cost
Write as differential equation:
y = (t2 + 3t - D2 cost) / Q(D)
Therefore, the differential equations for x and y are:
x = (cost - D^2 (t2 + 3t)) / ((D+5) + D^2 (D+9))
y = (t2 + 3t - D2 cost) / (D2 / (D+5) + (D+9))
(a)[tex]\(x = \frac{1}{D^3 + 14D + 5}\cdot \left( \cos(t) \right)\)[/tex]
(b)[tex]\(y = \frac{1}{D^3 + 14D + 5}\cdot \left( t^2 + 3t \right)\)[/tex] by expressing (x) and (y) in terms of differential operators and given functions of (t), we derive their differential equations .
Explanation:The given system of equations can be transformed into differential equations using operator (D), representing differentiation with respect to a variable, often time.
(a) To derive the differential equation for (x), we isolate (x) in terms of the differential operator \(D\). Rearranging the first equation gives us [tex]\(x = \frac{\cos(t)}{D+5} - \frac{D^2y}{D+5}\).[/tex]Substituting the expression for (y) from the second equation into this yields[tex]\(x = \frac{\cos(t)}{D+5} - \frac{D^2}{D+5}\left(\frac{t^2 + 3t}{D+9}\right)\).[/tex] Simplifying further leads to [tex]\(x = \frac{1}{D^3 + 14D + 5}\cdot \left( \cos(t) \right)[/tex]\).
(b) Similarly, isolating \(y\) from the given equations leads to [tex]\(y = \frac{t^2 + 3t}{D+9} - \frac{D+5}{D+9}x\).[/tex]Substituting the expression for (x) from the first equation into this yields [tex]\(y = \frac{t^2 + 3t}{D+9} - \frac{D+5}{D+9}\left(\frac{\cos(t)}{D+5} - \frac{D^2y}{D+5}\right)\).[/tex]Further simplification results in [tex]\(y = \frac{1}{D^3 + 14D + 5}\cdot \left( t^2 + 3t \right)\).[/tex]
Therefore, by expressing (x) and (y) in terms of differential operators and given functions of (t), we derive their differential equations as specified in the final answer.
The proportion of baby boys born in the United States has historically been 0.508. You choose an SRS of 50 newborn babies and find that 45% are boys. Do ALL calculations to 5 decimal places before rounding.
Answer:
[tex]z=\frac{0.45 -0.508}{\sqrt{\frac{0.508(1-0.508)}{50}}}=-0.82[/tex]
[tex]p_v =P(z<-0.82)=0.206[/tex]
So the p value obtained was a very high value and using the significance level given [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 newborn boys babies is not significantly lower than 0.508
Step-by-step explanation:
Data given and notation
n=50 represent the random sample taken
[tex]\hat p=0.45[/tex] estimated proportion of newborn boys babies
[tex]p_o=0.508[/tex] is the value that we want to test
[tex]\alpha=0.05[/tex] represent the significance level
Confidence=95% or 0.95 (asumed)
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value (variable of interest)
Concepts and formulas to use
We assume that we want to check if the true proportion is less than 0.508.
Null hypothesis:[tex]p\geq 0.508[/tex]
Alternative hypothesis:[tex]p < 0.508[/tex]
When we conduct a proportion test we need to use the z statisitc, 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].
Calculate the statistic
Since we have all the info requires we can replace in formula (1) like this:
[tex]z=\frac{0.45 -0.508}{\sqrt{\frac{0.508(1-0.508)}{50}}}=-0.82[/tex]
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 significance level assumed is [tex]\alpha=0.05[/tex]. The next step would be calculate the p value for this test.
Since is a left tailed test the p value would be:
[tex]p_v =P(z<-0.82)=0.206[/tex]
So the p value obtained was a very high value and using the significance level given [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 newborn boys babies is not significantly lower than 0.508
It is concluded that the null hypothesis H₀ is not rejected. There is not enough evidence to claim that the population proportion p is less than 0.508, at the α =0.05 significance level.
What is the z test statistic for one sample proportion?Suppose that we have:
n = sample size[tex]\hat{p}[/tex] = sample proportionp₀ = population proportion (hypothesised)Then, the z test statistic for one sample proportion is:
[tex]Z = \dfrac{\hat{p} - p_0}{\sqrt{\dfrac{p_0(1-p_0)}{n}}}[/tex]
Making the hypothesis and performing the test, assuming that we want to test if the population mean proportion is < 0.508 at the level of significance 0.05.
(1) Null and Alternative HypothesesThe following null and alternative hypotheses for the population proportion needs to be tested:
[tex]H_0: p \geq 0.508 \\\\ H_a: p & < & 0.508 \end{array}[/tex]
This corresponds to a left-tailed test, for which a z-test for one population proportion will be used.
(2) Rejection RegionBased on the information provided, the significance level is [tex]\alpha = 0.05[/tex], and the critical value for a left-tailed test is [tex]z_c = -1.64[/tex]
The rejection region for this left-tailed test is [tex]R = \{z: z < -1.645\}[/tex]
(3) Test StatisticsThe z-statistic is computed as follows:
[tex]z & = & \displaystyle \frac{\hat p - p_0}{\sqrt{ \displaystyle\frac{p_0(1-p_0)}{n}}} \\\\& = & \displaystyle \frac{0.45 - 0.508}{\sqrt{ \displaystyle\frac{ 0.508(1 - 0.508)}{50}}} \\\\ & = & -0.82[/tex]
(4) Decision about the null hypothesisSince it is observed that [tex]z = -0.82 \ge z_c = -1.645[/tex], it is then concluded that the null hypothesis is not rejected.
Thus, it is concluded that the null hypothesis H₀ is not rejected. Therefore, there is not enough evidence to claim that the population proportion p is less than 0.508, at the α =0.05 significance level.
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Petroleum pollution in oceans stimulates the growth of certain bacteria. An assessment of this growth has been madew by counting the bacteria in each of 5 randomly chosen specimens of ocean water (of a fixed size). The 5 counts obtained were as follows.
41, 62, 45, 48, 69
Find the deviation of this sample of numbers. Round your answers to at least two decimal places.
Final answer:
The standard deviation of the sample of numbers is approximately 10.68, after calculating the mean, finding the deviations, squaring them, averaging the squared deviations to get the variance, and then taking the square root of the variance.
Explanation:
The student is asking how to calculate the standard deviation of a sample of numbers. The sample given consists of the counts of bacteria found in ocean water, which are 41, 62, 45, 48, 69. To calculate the standard deviation, follow these steps:
Calculate the mean (average) of the sample.Subtract the mean from each number in the sample to find their deviations.Square each deviation to get the squared deviations.Calculate the average of the squared deviations (this is the variance).Take the square root of the variance to get the standard deviation.Performing the calculations, you get:
Mean (average): (41 + 62 + 45 + 48 + 69) / 5 = 265 / 5 = 53Deviation for each number: -12, 9, -8, -5, 16Squared deviations: 144, 81, 64, 25, 256Variance: (144 + 81 + 64 + 25 + 256) / 5 = 570 / 5 = 114Standard deviation: √114 ≈ 10.68 (rounded to two decimal places)A principal of $4700 was invested at 4.75% interest, compounded annually.
Lett be the number of years since the start of the investment. Let y be the value of the investment, in dollars.
Write an exponential function showing the relationship between y and t.
Answer:
Step-by-step explanation:
We would apply the formula for determining compound interest which is expressed as
y = P(1 + r/n)^nt
Where
y = the value of the investment at the end of t years
r represents the interest rate.
n represents the periodic interval at which it was compounded.
P represents the principal or initial amount invested
From the information given,
P = $4700
r = 4.75% = 4.75/100 = 0.0475
n = 1 because it was compounded once in a year.
Therefore, the exponential function showing the relationship between y and t is
y = 4700(1 + 0.0475/1)^1 × t
y = 4700(1.0475)^t
Suppose that combined verbal and math SAT scores follow a normal distribution with a mean 896 and standard deviation 174. Suppose further that Peter finds out he scored in the top 2.5 percentile (97.5% of students scored below him). Determine how high Peter's score must have been.
Let x* denote Peter's score. Then
P(X > x*) = 0.025
P((X - 896)/174 > (x* - 896)/174) = 0.025
P(Z > z*) = 1 - P(Z < z*) = 0.025
P(Z < z*) = 0.975
where Z follows the standard normal distribution (mean 0 and std dev 1).
Using the inverse CDF, we find
P(Z < z*) = 0.975 ==> z* = 1.96
Then solve for x*:
(x* - 896)/174 = 1.96 ==> x* = 1237.04
so Peter's score is roughly 1237.
A company is considering producing some new Gameboy electronic games. Based on past records, management believes that there is a 70 percent chance that each of these will be successful and a 30 percent chance of failure. Market research may be used to revise these probabilities. In the past, the successful products were predicted to be successful based on market research 90 percent of the time. However, for products that failed, the market research predicted these would be successes 20 percent of the time. If market research is performed for a new product, what is the probability that the results indicate an unsuccessful market for the product and the product is actually unsuccessful
The probability that the market research predicts an unsuccessful market for the product and the product is actually unsuccessful is 24 percent. This is calculated using the concept of conditional probability and multiplying the probability of the product actually failing (30%) with the probability of market research correctly predicting a failure (80%).
Explanation:To calculate the probability that the market research indicates an unsuccessful market for the product and the product is actually unsuccessful, we first need to understand the concept of conditional probability, given certain conditions.
Here, there is a 30 percent chance that a product will fail. The market research incorrectly predicts success for failed products 20 percent of the time. Therefore, the market research predicts failure for failed products 80 percent of the time (100% - 20%).
By multiplying these probabilities together, we can find the joint probability that a product is unsuccessful and market research also predicts it to be unsuccessful.
The result is 0.30 x 0.80 = 0.24. Therefore, there is a 24 percent chance that the market research will predict a failure and the product will also actually be a failure.
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The probability that market research indicates an unsuccessful market and the product is unsuccessful is 24%.
To calculate the probability that the results indicate an unsuccessful market for the product and that the product is unsuccessful, we need to use Bayes' Theorem. Initially, we have the probability of a product being successful, P(Success) = 0.70, and the probability of a product being a failure, P(Failure) = 0.30. Furthermore, if the product is successful, market research predicts it successful 90% of the time. If the product is a failure, market research incorrectly predicts it as a success 20% of the time, meaning it correctly predicts failure 80% of the time (100% - 20%).
We are interested in P(Market Research Failure and Actual Failure). This is the probability that the market research indicates a failure and the product indeed fails. Using the probabilities provided:
P(Market Research Successful | Success) = 0.90P(Market Research Successful | Failure) = 0.20P(Market Research Failure | Success) = 0.10P(Market Research Failure | Failure) = 0.80Therefore, P(Market Research Failure and Actual Failure) can be calculated as:
P(Market Research Failure and Actual Failure) = P(Market Research Failure | Failure) x P(Failure) = 0.80 x 0.30 = 0.24 or 24%.
Find the slope of the line passing through each pair of points or state that the slope is undefined. Assume that all variables represent positive real numbers. Then indicate whether the line through the points rises, falls, is horizontal, or is vertical. (-a, 0) and (0, -b)
Answer:
-b/a and line is falling
Step-by-step explanation:
Let (-a,0) and (0,-b) be point 1 and 2, respectively. The slope passing through point (-a,0) and (0,-b) can be calculated using the following equation
[tex]m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-b - 0}{0 - (-a)} = -b/a[/tex]
As b and a are positive real number, then the slope m = -b/a is a negative real number. The line passing through these 2 points is falling.
The slope of the line passing through points (-a, 0) and (0, -b) is calculated using the formula for slope, resulting in a slope of -b/a, which indicates that the line falls as it moves from left to right.
To calculate the slope (m), we use the formula m = (y2 - y1) / (x2 - x1), with (x1, y1) = (-a, 0) and (x2, y2) = (0, -b). Substituting the values, we get m = (-b - 0) / (0 - (-a)) = -b/a. Since both a and b are positive real numbers, the sign of the slope depends on the negative numerator, indicating the line falls as it goes from left to right. If a and/or b were zero, revealing a vertical or horizontal line, a different methodology would apply, such as indicating an undefined slope for a vertical line or a zero slope for a horizontal line.
f(x) = x3 − 9x2 − 21x + 8
(a) Find the interval on which f is increasing.
Find the interval on which f is decreasing
(b) Find the local minimum and maximum values of f.
(c) Find the inflection point.
(d)Find the interval on which f is concave up.
(e)Find the interval on which f is concave down.
The function f(x) = x3 - 9x2 - 21x + 8 increases in the intervals (-∞, -1) and (7, ∞), and decreases in the interval (-1, 7). It achieves a local minimum at -190 and a local maximum at -23. The inflection point is at x = 3, with the function being concave up for x < 3 and concave down for x > 3.
Explanation:Given the function f(x) = x3 - 9x2 - 21x + 8, we need to find intervals where the function is increasing or decreasing, and determine its local minima/maxima, inflection points, and where it is concave up or down.
To find the intervals, we need to first take the derivative of f(x): f'(x) = 3x2 - 18x - 21 which becomes 3(x - 7)(x + 1). The critical points are then 7 and -1.
The function increases in the interval (-∞, -1) and decreases in the interval (-1, 7). The function again increases in the interval (7, ∞).
Local minima and maxima can be found by evaluating the function at the critical points. f(-1) = -23 and f(7) = -190, so the local maximum value of f is -23 and local minimum is -190.
To find the inflection point, we compute the second derivative f''(x) = 6x - 18, which simplifies to 6(x - 3), giving us an inflection point at x = 3.
The function is concave up for x < 3 and concave down for x > 3.
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a) f(x) is increasing for x in (- ∞, - 1) cup(7, infty)
f(x) is decreasing for x \in (- 1, 7)
b)Local minimum=-237 Local maximum=19
c)Point of inflection: (x, y) = (3, - 109)
d)Interval for concave upward curve: x \in (3, ∞)
e)Interval for concave upward curve: x \in (- ∞, 3)
The given function is [tex]f(x) = x ^ 3 - 9x ^ 2 - 21x + 8[/tex]
[tex]f' * (x) = 3x ^ 2 - 18x - 21[/tex]
f^ prime prime (x) = 6x - 18
a)f(x) is increasing for
f' * (x) > 0
[tex]3x ^ 2 - 18x - 21 > 0[/tex]
[tex]x ^ 2 - 6x - 7 > 0[/tex]
[tex]x ^ 2 - 7x + x - 7 > 0[/tex]
x(x - 7) + 1(x - 7) > 0
(x - 7)(x + 1) > 0
x > 7 and x > - 1or x < 7 and x < - 1
f(x) is decreasing for
f' * (x) < 0
3x ^ 2 - 18x - 21 < 0
x ^ 2 - 6x - 7 < 0
x ^ 2 - 7x + x - 7 < 0
x(x - 7) + 1(x - 7) < 0
(x - 7)(x + 1) < 0
x > 7 and x < - 1 or x < 7 and x > - 1
x=(-1,7)
b)for local extrema
f' * (x) = 0
3x ^ 2 - 18x - 21 = 0
x ^ 2 - 6x - 7 < 0
x ^ 2 - 7x + x - 7 = 0
x(x - 7) + 1(x - 7) = 0
(x - 7)(x + 1) = 0
x = 7 and x = - 1
At x=-1:
f(- 1) = (- 1) ^ 3 - 9 * (- 1) ^ 2 - 21(- 1) + 8 = 19
At x=7:
f(7) = (7) ^ 3 - 9 * (7) ^ 2 - 21(7) + 8 = - 237
Local minimum=-237
Local maximum=19
c) For point of inflection:
f^ prime prime (x) = 0
Rightarrow 6x - 18 = 0
Rightarrow x = 3
y = f(3) = (3) ^ 3 - 9 * (3) ^ 2 - 21(3) + 8 = - 109
(x, y) = (3, - 109)
d)Interval for concave upward curve:
f^ prime prime (x) > 0
6x - 18 > 0
x \in (3, ∞)
e)Interval for concave upward curve:
f^ prime prime (x) < 0
6x - 18 < 0
X€(−0,3)
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Football Strategies A particular football team is known to run 30% of its plays to the left and 70% to the right. A linebacker on an opposing team notes that the right guard shifts his stance most of the time (80%) when plays go to the right and that he uses a balanced stance the remainder of the time. When plays go to the left, the guard takes a balanced stance 90% of the time and the shift stance the remaining 10%. On a particular play, the linebacker notes that the guard takes a balanced stance.
a. What is the probability that the play will go to the left?
b. What is the probability that the play will go to the right?
c. If you were the linebacker, which direction would you prepare to defend if you saw the balanced stance?
Answer:
a) 0.659
b) 0.341
c) It'll be smarter to prepare to defend the left hand side as the probability of play going through that side when the right guard takes a balanced stance is almost double of the probability that the play would go through the right when the right guard takes a balanced stance.
Step-by-step explanation:
Considering the left first
Play goes through the left 30% of the time
- The right guard takes a balanced stance 90% of the time when play goes through the left, that is, 0.9 × 0.3 = 27% overall.
- He takes a shift stance 10% of the time that play goes through the left = 0.1 × 0.3 = 3% overall.
Play goes through the right, 70% of the time.
- He takes a balanced stance 20% of the time that play goes through the right, that is, 0.7 × 0.2 = 14% overall
- He takes a shift stance 80% of the time that play goes through the right, that is, 0.7 × 0.8 = 56% overall.
Then the right guard takes a balanced stance 27% + 14% of all the time = 41% overall.
a) Probability that play goes through the left with the right guard in a balanced stance = 27/41 = 0.659
b) Probability that play goes through the right with the right guard in a balanced stance = 14/41 = 0.341
c) It'll be smarter to prepare to defend the left hand side as the probability of play going through that side when the right guard takes a balanced stance is almost double of the probability that the play would go through the right when the right guard takes a balanced stance.
Using conditional probability, we can find the probabilities of the play going left or right. The probability of the play going left is 0.3, and the probability of the play going right is 0.7. If the linebacker sees a balanced stance, the probability of the play going left is approximately 0.658, so the linebacker should prepare to defend against a play going left.
Explanation:To find the probabilities of the play going left or right, we need to use conditional probability. Let's denote L as the event of the play going left, and R as the event of the play going right. We are given P(L) = 0.3 and P(R) = 0.7. We also have the following conditional probabilities: P(B|L) = 0.9 (balanced stance when play goes left) P(S|L) = 0.1 (shift stance when play goes left) P(B|R) = 0.2 (balanced stance when play goes right) P(S|R) = 0.8 (shift stance when play goes right)
a. To find the probability that the play will go left, we can use the Law of Total Probability. P(L) = P(L∩B) + P(L∩S) = P(L)P(B|L) + P(L)P(S|L) = 0.3 X 0.9 + 0.3 * 0.1 = 0.27 + 0.03 = 0.3.
b. Similarly, to find the probability that the play will go right, P(R) = P(R∩B) + P(R∩S) = P(R)P(B|R) + P(R)P(S|R) = 0.7 X 0.2 + 0.7 X 0.8 = 0.14 + 0.56 = 0.7.
c. If the linebacker sees a balanced stance, it means that P(B) = P(B|L)P(L) + P(B|R)P(R) = 0.9 X 0.3 + 0.2 X 0.7 = 0.27 + 0.14 = 0.41. Now, we can compare the conditional probability of the play going left given a balanced stance, P(L|B) = P(B|L)P(L)/P(B) = (0.9 X 0.3)/0.41 = 0.27/0.41 ≈ 0.658. Therefore, the linebacker should prepare to defend against a play going to the left.
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The gypsy moth is a serious threat to oak and aspen trees. A state agriculture department places traps throughout the state to detect the moths. When traps are checked periodically, the mean number of moths per trap is only 1.2, but some traps have several moths. The distribution of moth counts in traps is strongly right skewed, with standard deviation 1.4. A random sample of 60 traps has x = 1 and s = 2.4.
Let X = the number of moths in a randomly selectd trap
(a) For the population distribution, what is the ...
...mean? =
...standard deviation? =
(b) For the distribution of the sample data, what is the ...
...mean? =
...standard deviation? =
(c) What shape does the distribution of the sample data probably have?
Exactly NormalApproximately Normal Right skewedLeft skewed
(d) For the sampling distribution of the sample mean with n = 60, what is the ...
...mean? =
...standard deviation? = (Use 3 decimal places)
(e) What is the shape of the sampling distribution of the sample mean?
Left skewedApproximately Normal Right skewedExactly Normal
(f) If instead of a sample size of 60, suppose the sample size were 10 instead. What is the shape of the sampling distribution of the sample mean for samples of size 10?
Approximately normalSomewhat left skewed Exactly NormalSomewhat right skewed
(g) Can we use the Z table to calculate the probability a randomly selected sample of 10 traps has a sample mean less than 1?
No, because the sampling distribution of the sample mean is somewhat right skewedYes, by the Central Limit Theorem, we know the sampling distribution of the sample mean is normal
Answer:
Step-by-step explanation:
Hello!
X: number of gypsy moths in a randomly selected trap.
This variable is strongly right-skewed. with a standard deviation of 1.4 moths/trap.
The mean number is 1.2 moths/trap, but several have more.
a.
The population is the number of moths found in traps places by the agriculture departments.
The population mean μ= 1.2 moths per trap
The population standard deviation δ= 1.4 moths per trap
b.
There was a random sample of 60 traps,
The sample mean obtained is X[bar]= 1
And the sample standard deviation is S= 2.4
c.
As the text says, this variable is strongly right-skewed, if it is so, then you would expect that the data obtained from the population will also be right-skewed.
d. and e.
Because you have a sample size of 60, you can apply the Central Limit Theorem and approximate the distribution of the sampling mean to normal:
X[bar]≈N(μ;σ²/n)
The mean of the distribution is μ= 1.2
And the standard deviation is σ/√n= 1.4/50= 0.028
f. and g.
Normally the distribution of the sample mean has the same shape of the distribution of the original study variable. If the sample size is large enough, as a rule, a sample of size greater than or equal to 30 is considered sufficient you can apply the theorem and approximate the distribution of the sample mean to normal.
You have a sample size of n=10 so it is most likely that the sample mean will have a right-skewed distribution as the study variable. The sample size is too small to use the Central Limit Theorem, that is why you cannot use the Z table to calculate the asked probability.
I hope it helps!
Find the arc length of AB. Round your answer to the nearest hundredth.
The arc length of AB = 8.37 meters
Solution:
Degree of AB (θ) = 60°
Radius of the circle = 8 m
Let us find the arc length of AB.
Arc length formula:
[tex]$\text{Arc length}=2 \pi r\left(\frac{\theta}{360^\circ}\right)[/tex]
[tex]$=2 \times 3.14 \times 8 \left(\frac{60^\circ}{360^\circ}\right)[/tex]
[tex]$=2 \times 3.14 \times 8 \left(\frac{1}{6}\right)[/tex]
Arc length = 8.37 m
Hence the arc length of AB is 8.37 meters.
A population has mean 187 and standard deviation 32. If a random sample of 64 observations is selected at random from this population, what is the probability that the sample average will be less than 182
Answer:
11.51% probability that the sample average will be less than 182
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 = 187, \sigma = 32, n = 64, s = \frac{32}{\sqrt{64}} = 4[/tex]
What is the probability that the sample average will be less than 182
This is the pvalue of Z when X = 182. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{182 - 187}{4}[/tex]
[tex]Z = -1.2[/tex]
[tex]Z = -1.2[/tex] has a pvalue of 0.1151
11.51% probability that the sample average will be less than 182
3/5 of your class earned an A or a B last week. 1/4 of those teammates earned an A. what fraction of you class earned an A? Possible answers could be 4/9, 7/20, 2/1, 3/20 but which one is it?
Answer:
3/20
Step-by-step explanation:
3/5 earned an A or B = "3/5 of the class"1/4 of 3/5 earned an A = "3/5 of class divided by 4"[tex]\frac{3}{5}[/tex] ÷ 4 = [tex]\frac{3}{5}[/tex] [tex]* \frac{1}{4}[/tex] = [tex]\frac{3 * 1}{5 * 4}[/tex] = [tex]\frac{3}{20}[/tex]
To calculate the fraction of the class that earned an A, multiply 3/5 by 1/4, which equals 3/20. Thus, 3/20 of the class earned an A. This is the correct answer among the given options.
Explanation:Calculating the Fraction of a Class That Earned an A
To find out what fraction of the class earned an A, we start with the fact that 3/5 of the class earned an A or a B.
We also know that 1/4 of those students earned an A.
To determine the fraction of the entire class that earned an A, we multiply these two fractions together:
Fraction of class that earned an A = (3/5) × (1/4)
This multiplication gives us:
(3 × 1) / (5 × 4) = 3/20
Therefore, 3/20 of the class earned an A.
This fraction cannot be reduced further and it is also one of the options provided, confirming that it is the correct answer.