Answer:
[tex]P(X<67.5)=P(\frac{X-\mu}{\sigma}<\frac{67.5-\mu}{\sigma})=P(Z<\frac{67.5-75}{6})=P(z<-1.25)[/tex]
And we can find this probability using the normal standard table or excel:
[tex]P(z<-1.25)=0.106[/tex]
Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Solution to the problem
Let X the random variable that represent the number of days before cartridge runs out of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(75,6)[/tex]
Where [tex]\mu=75[/tex] and [tex]\sigma=6[/tex]
We are interested on this probability
[tex]P(X<67.5)[/tex]
And the best way to solve this problem is using the normal standard distribution and the z score given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
If we apply this formula to our probability we got this:
[tex]P(X<67.5)=P(\frac{X-\mu}{\sigma}<\frac{67.5-\mu}{\sigma})=P(Z<\frac{67.5-75}{6})=P(z<-1.25)[/tex]
And we can find this probability using the normal standard table or excel:
[tex]P(z<-1.25)=0.106[/tex]
a) What percentage of the area under the normal curve lies to the left of μ? % (b) What percentage of the area under the normal curve lies between μ − σ and μ + σ? % (c) What percentage of the area under the normal curve lies between μ − 3σ and μ + 3σ? %
Answer:
a) 50%
b) 68%
c) 99%
Step-by-step explanation:
for a standard normal curve ,
a) since the standard normal curve is symmetric and centred around μ , 50% of the curve lies at the left of μ and 50% lies to the right
b) according to the 68-95-99 rule, 68% of the standard normal curve lies from μ − σ and μ + σ
c) from the same rule , 99% of the standard normal curve lies from μ − 3σ and μ + 3σ
Answer:
a) 50%
b) 68%
c) 99.7%
Step-by-step explanation:
The Empirical Rule states that, for a normally distributed random variable:
68% of the measures are within 1 standard deviation of the mean.
95% of the measures are within 2 standard deviation of the mean.
99.7% of the measures are within 3 standard deviations of the mean.
The normal distribution is also symmetric, which means that 50% of the measures are below the mean and 50% are above.
In this problem, we have that:
Mean μ
Standard deviation σ
Area under the normal curve = percentage
a) What percentage of the area under the normal curve lies to the left of μ?
Normal distribution is symmetric, so the answer is 50%.
(b) What percentage of the area under the normal curve lies between μ − σ and μ + σ?
Within 1 standard deviation of the mean, so 68%.
(c) What percentage of the area under the normal curve lies between μ − 3σ and μ + 3σ?
Within 3 standard deviation of the mean, so 99.7%.
There are 98 balls in a box. 25 of them are red, 19 of them are green, 30 of them are purple,and 24 of them are blue. Suppose Hao draws 22 balls from the box with replacement (hedraws the ball, records its color, and then puts it back into the box). Find the probabilitythat he draws 2 red balls, 5 green balls, 10 purple balls, and 5 blue balls.
Answer:
Since the balls are drawn with replacement, it means the individual probability remain constant,
I have solved this problem on paper (Figures Attached).
Thanks.
Answer:
Since this problem related to the replacement problem thus
Pr(2 red)=(25/98)(25/98)=0.0651
Pr(5 green)=(19/98)(19/98)(19/98)(19/98)(19/98)
Pr(5 green)=2.73*10^-4
Pr(10 purple)=((30/98))^10=7.22*10^-6
Pr(5 blue)=((24/98))^5=8.8*10^-4
A student is asked to find the derivative of y = x sin2 (x) with respect to variable x, given x, y > 0. They provide the following answer: dy dx = sin2 (x) · x sin2 (x)−1 · cos2 (x) Is the student correct? If the student is correct, then explain how they used derivative rules correctly to find this derivative. If the student is incorrect, then give the correct answer and provide an explanation that you would use to correct the student’s thinking.
Answer:
[tex]\frac{dy}{dx} = 2x ( 2cos2x) +sin2x[/tex]
[tex]\frac{dy}{dx} = 4x( cos2x)+sin2x[/tex]
Step-by-step explanation:
Given y = x sin2x .....(1)
Applying UV formula [tex]\frac{d(UV)}{dx} = u \frac{dv}{dx} + v\frac{du}{dx}[/tex]
Differentiating with respective to 'x' we get
[tex]\frac{dy}{dx} = x ( 2cos2x)\frac{d(2x)}{dx} +sin2x (1)[/tex]
[tex]\frac{dy}{dx} = x ( 2cos2x)(2) +sin2x (1)[/tex]
Final answer:-
[tex]\frac{dy}{dx} = 4x( cos2x)+sin2x[/tex]
Mike brought a new car and financed $25,000 to make the purchase. He financed the car for 60 months with an APR of 6.5%. Determine each of the following:
A. Mike's monthly payment
B. Total cost if Mike's car
C. Total interest Mike pays over the life of the loan
Final answer:
Mike's monthly car payment is approximately $489.99. The total cost of the car including interest over 60 months is $29,399.40, and the total interest Mike pays over the loan's lifetime is $4,399.40.
Explanation:
To determine Mike's monthly car payment, total cost of the car, and total interest paid over the life of the loan, we need to use the formula for the monthly payment on an installment loan, which can be found using the formula:
P = (Pv * r) / (1 - (1 + r)-n)
Where:
P is the monthly paymentPv is the present value or the amount of the loan, which is $25,000r is the monthly interest rate (annual rate divided by 12), which is 0.065/12n is the total number of payments (months), which is 60Lets calculate the monthly payment (P):
P = ($25,000 * 0.065/12) / (1 - (1+0.065/12)-60)
= $489.99 approximately
Mike's total cost of the car is the monthly payment multiplied by the number of payments:
$489.99 * 60 = $29,399.40
The total interest Mike pays is the total cost minus the loan amount:
$29,399.40 - $25,000 = $4,399.40
Thus, Mike's monthly payment is approximately $489.99, the total cost of his car will be $29,399.40, and the total interest paid over the life of the loan is $4,399.40.
Suppose that a die is rolled twice. What are the possible values that the following random variables can take on: a. the maximum value to appear in the two rolls; b. the minimum value to appear in the two rolls; c. the sum of the two rolls; d. the value of the first roll minus the value of the second roll
Answer:
a. A = {1, 2, 3, 4, 5, 6}
b. B = {1, 2, 3, 4, 5, 6}
c. C = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}
d. D = {-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5}
Step-by-step explanation:
a. the maximum value to appear in the two rolls
Since only the maximum value is computed, the variable can assume any integer from 1 to 6:
A = {1, 2, 3, 4, 5, 6}
b. the minimum value to appear in the two rolls;
Since only the minimum value is computed, the variable can assume any integer from 1 to 6:
B = {1, 2, 3, 4, 5, 6}
c. the sum of the two rolls;
The minimum value would be from rolling two ones (sum is 2) and the maximum value would be from rolling two sixes (sum is 12). Every integer in-between is possible:
C = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}
d. the value of the first roll minus the value of the second roll
The minimum value would be from rolling a one and a six (result is -5) and the maximum value would be from rolling a six and a one (result is 5). Every integer in-between is possible:
D = {-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5}
Final answer:
Explanation of possible values for maximum, minimum, sum, and difference in rolling two dice.
Explanation:
a. The possible values for the maximum value:
1, if the same number appears twice on both rolls
2 to 6, for individual numbers in different scenarios
b. The possible values for the minimum value:
1 to 5, as the minimum value will exclude the maximum value
c. The possible values for the sum:
2 to 12, representing all possible sums from rolling two dice
d. The possible values for the difference between two rolls:
-5 to 5, as the difference can range from -5 to 5
based on the graph below, what is yhe total number of solutions to the equation f(x)= g(x)?
1
2
3
4
Answer:
Based on the graph below, what is the total number of solutions to the equation f(x)= g(x) will be 3.
Step-by-step explanation:
The intersection points of both graphs would be the total number of solutions to the equation f(x)= g(x).
From the given diagram, it is clear that both the graphs intersect at three locations points or intersection points. The approximations locations of The intersection points of both graphs are
(1.5, 4.125)(-1, 0), and (-2.5, -3.5)Therefore, based on the graph below, what is the total number of solutions to the equation f(x)= g(x) will be 3.
A company has learned that the relationship between its advertising and sales shows diminishing marginal returns. That is, as it saturates consumers with ads, the benefits of increased advertising diminish. The company should expect to find linear association between its advertising and sales.
Answer:
A company has learned that the relationship between its advertising and sales shows diminishing marginal returns. That is, as it saturates consumers with ads, the benefits of increased advertising diminish. The company should expect to find a linear association between its advertising and sales - This statement is false
Step-by-step explanation:
According to the scenario given for the company, it was said that the marginal return diminished after a saturation point, therefore, the company should rather expect a non-linear pattern and not a linear pattern.
Therefore, the statement expressed in the question is false.
In class we derived the MOM and MLE for an exponential distribution with parameter ????. Conduct a Bootstrap simulation to compare the estimation of λ with sample sizes of n = 10, n = 100, and n = 500. Choose true value λ = 0.2 and use B = 1000. Calculate and compare the mean and standard error for each set of simulations to each other as well as their theoretical values.
Answer:
Below is the R code for the bootstrapping in exponential distribution. The result is attached below.
####################################
rm(list=ls(all=TRUE))
set.seed(12345)
N=c(10,100,500)
Rate=0.2
B=1000
MN=SE=rep()
for(i in 1:length(N))
{
n=N[i]
X=rexp(n,rate=Rate)
EST=1/mean(X)
ESTh=rep()
for(j in 1:B)
{
Xh=rexp(n,rate=EST)
ESTh[j]=1/mean(Xh)
}
MN[i]=mean(ESTh)
SE[i]=sd(ESTh)
}
cbind(N,Rate,MN,SE)
Step-by-step explanation:
Suppose that MX=V is a linear system, for some matrix M and some vector V. Let the vector P be a particular solution to the system and the vector H a homogeneous solution to the system. Which of the following vectors must be a particular solution to the system? Select all that apply.
A. 2H-P
B. 2P+2H
C. 3H-P
D. H
E. 2P+H
F. P+3H
G. P
H. P+H
I. P-3H
J. 3P+H
K. 2H-P
L. P+2H
M. 3P+3H
N. P-2H
O. H-P
P. P-H
From what I understand, the general solution = particular+homogeneous.
I found that possible solutions could be P+H, a scalar * P, or a scalar * P + scalar*H. I tried these options and didn't get it right. Any help would be appreciated
Answer:
The Answer is "G. P"
Step-by-step explanation:
They ask for "Which of the following vectors must be a PARTICULAR solution to the system?" so the answer simply G. which is P
Let p1 be the proportion of successes in the first population and let p2 be the proportion of successes in the second population. Suppose that you are testing the hypotheses: H0:p1−p2=0Ha:p1−p2≠0 Futhermore suppose that z∗=1.73, find and input the p-value for this test. Round your answer to 4 decimal places.
Answer:
0.0836
Step-by-step explanation:
Given that p1 be the proportion of successes in the first population and let p2 be the proportion of successes in the second population
[tex]H_0:p_1-p_2=0\\H_a:p_1-p_2\neq 0[/tex]
is the hypotheses created
This is two tailed
Test statistic Z = 1.73
corresponding p value
= 0.08363
=0.0836 (after rounding off to 4 decimals)
The p-value for this test is 0.0802.
The p-value for this two-tailed test can be calculated by doubling the probability of observing a test statistic at least as extreme as the observed z-score of 1.73, under the assumption that the null hypothesis is true.
Given that the z-score is 1.73, we need to find the area under the standard normal distribution curve that lies beyond this z-score. This area represents the probability of observing a value at least as extreme as 1.73 standard deviations from the mean under the null hypothesis.
Using a standard normal distribution table or a calculator, we can find the probability corresponding to a z-score of 1.73. Let's denote this probability as P(Z > 1.73).
P(Z > 1.73) = 1 - P(Z < 1.73)
Using a standard normal distribution table or calculator, we find that P(Z < 1.73) is approximately 0.9599. Therefore:
P(Z > 1.73) = 1 - 0.9599 = 0.0401
Since this is a two-tailed test, we need to consider the probability of observing a z-score at least as extreme in either direction (both tails of the distribution). Thus, we multiply the single-tail probability by 2:
p-value = 2 * P(Z > 1.73) = 2 * 0.0401 = 0.0802
Rounded to four decimal places, the p-value is 0.0802.
Therefore, the p-value for this test is 0.0802.
The answer is: 0.0802.
The mean investment that employees put into their companies 401k per year is $10,000 with standard deviation of $500 Assuming the investment follow a normal distribution, determine the following a. what proportion of employees put between $9, 500 and $11,000 into the 401 k per year b. What proportion of employee put more than $11, 500 into the 401 k per year? c. What proportional of employees put less than $11,000 into the 401k per year?d. What proportional of employees put more than $9,000 into the 401k per year?e. What proportional of employees put between than $11,000 and $11, 500 into the 401k per year?f. How much would an employees need to put into his or her 401 K to be in the upper 10% of investors?
Answer:
a) 81.85% of employees put between $9, 500 and $11,000 into the 401 k per year
b) 0.13% of employee put more than $11, 500 into the 401 k per year
c) 97.72% of employees put less than $11,000 into the 401k per year.
d) 97.72% of employees put more than $9,000 into the 401k per year
e) 2.15% of employees put between than $11,000 and $11, 500 into the 401k per year
f) An employee would need to put $10,640 into his or her 401 K to be in the upper 10% of investors
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 = 10000, \sigma = 500[/tex]
a. what proportion of employees put between $9, 500 and $11,000 into the 401 k per year
This is the pvalue of Z when X = 11000 subtracted by the pvalue of Z when X = 9500. So
X = 11000
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{11000 - 10000}{500}[/tex]
[tex]Z = 2[/tex]
[tex]Z = 2[/tex] has a pvalue of 0.9772.
X = 9500
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{9500 - 10000}{500}[/tex]
[tex]Z = -1[/tex]
[tex]Z = -1[/tex] has a pvalue of 0.1587
0.9772 - 0.1587 = 0.8185
81.85% of employees put between $9, 500 and $11,000 into the 401 k per year
b. What proportion of employee put more than $11, 500 into the 401 k per year?
This is 1 subtracted by the pvalue of Z when X = 11500. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{11500 - 10000}{500}[/tex]
[tex]Z = 3[/tex]
[tex]Z = 3[/tex] has a pvalue of 0.9987
1 - 0.9987 = 0.0013
0.13% of employee put more than $11, 500 into the 401 k per year
c. What proportional of employees put less than $11,000 into the 401k per year?
This is the pvalue of Z when X = 11000. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{11000 - 10000}{500}[/tex]
[tex]Z = 2[/tex]
[tex]Z = 2[/tex] has a pvalue of 0.9772.
97.72% of employees put less than $11,000 into the 401k per year.
d. What proportional of employees put more than $9,000 into the 401k per year?
This is 1 subtracted by the pvalue of Z when X = 9000. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{9000 - 10000}{500}[/tex]
[tex]Z = -2[/tex]
[tex]Z = -2[/tex] has a pvalue of 0.0228.
1 - 0.0228 = 0.9772
97.72% of employees put more than $9,000 into the 401k per year
e. What proportional of employees put between than $11,000 and $11, 500 into the 401k per year?
This is the pvalue of Z when X = 11500 subtracted by the pvalue of Z when X = 11000. So
X = 11500
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{11500 - 10000}{500}[/tex]
[tex]Z = 3[/tex]
[tex]Z = 3[/tex] has a pvalue of 0.9987
X = 11000
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{11000 - 10000}{500}[/tex]
[tex]Z = 2[/tex]
[tex]Z = 2[/tex] has a pvalue of 0.9772.
0.9987 - 0.0972 = 0.0215
2.15% of employees put between than $11,000 and $11, 500 into the 401k per year
f. How much would an employees need to put into his or her 401 K to be in the upper 10% of investors?
This is the value of Z when X has a pvalue of 1-0.1 = 0.9. So it is X when Z = 1.28.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]1.28 = \frac{X - 10000}{500}[/tex]
[tex]X - 10000 = 500*1.28[/tex]
[tex]X = 10640[/tex]
An employee would need to put $10,640 into his or her 401 K to be in the upper 10% of investors
You estimate that you can save $3,800 by selling your home yourself rather than using a real estate agent. What would be the future value of that amount if invested for five years at 5 percent? Use Exhibit_1-A. (Round time value factor to 3 decimal places and final answer to 2 decimal places.)
Answer:
$4848.8
Step-by-step explanation:
(1 + 0.05)⁵ = 1.276
FV = 3800 × 1.276
= 4848.8
Final answer:
The future value of $3,800 invested for five years at a 5% interest rate, using the formula [tex]FV = PV[/tex] x [tex](1 + r)^t[/tex], is approximately $4,849.07.
Explanation:
You are asking how to calculate the future value of an amount of money when invested at a certain interest rate over a set period of time. Specifically, you want to know the future value of $3,800 invested for five years at a 5% interest rate.
To calculate the future value (FV) of money we use the formula:
[tex]FV = PV X (1 + r)^t[/tex]
Where:
PV is the present value or initial amount ($3,800)
r is the annual interest rate (5%, or as a decimal, 0.05)
t is the time in years the money is invested (5 years)
Using the formula, we get:
FV = $3,800 x (1 + 0.05)5
FV = $3,800 x (1.276281)
FV = $3,800 x 1.276 (rounded to three decimal places)
FV = $4,849.07 (rounded to two decimal places)
Hence, the future value of $3,800 invested for five years at a 5% interest rate would be approximately $4,849.07.
If a procedure meets all of the conditions of a binomial distribution except the number of trials is not fixed, then the geometric distribution can be used. The probability of getting the first success on the xth trial is given by P(x)equalsp (1 minus p )Superscript x minus 1, where p is the probability of success on any one trial. Subjects are randomly selected for a health survey. The probability that someone is a universal donor (with group O and type Rh negative blood) is 0.07. Find the probability that the first subject to be a universal blood donor is the seventh person selected. The probability is nothing. (Round to four decimal places as needed.)
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
Multiply.
(x - 6)(x - 4)
Answer:
[tex]x^{2} -10x+24[/tex]
Step-by-step explanation:
Given that (-4,9) is on the graph of f(x), find the corresponding point for the function. f(4x).
Answer:
(-1,9)
Step-by-step explanation:
In a certain very large city, the Department of Transportation (D.O.T.) has organized a complex system of bus transportation. In an advertising campaign, citizens are encouraged to use the new "GO-D.O.T!" system and head for the nearest bus stop to be transported to and from the central city. Suppose that at one of the bus stops the amount of time (in minutes) that a commuter must wait for a bus is a uniformly distributed random variable, T.
The possible values of T run from 0 minutes to 20 minutes.
(a) What is the probability that a randomly selected commuter will spend more than 7 minutes waiting for GO-D.O.T?
(b) What is the standard deviation?
(c) What is the probability that a randomly selected commuter will spend longer than 10 minutes but no more than 17 minutes waiting for the GO-D.O.T?
(d) What is the average waiting time?
Answer:
a) Probability that a randomly selected commuter will spend more than 7 minutes waiting for GO-D.O.T = P(7 < x ≤ 20) = 0.65
b) Standard deviation of the uniform distribution = 5.77 minutes
c) Probability that a randomly selected commuter will spend longer than 10 minutes but no more than 17 minutes waiting for the GO-D.O.T = P(10 < x < 17) = 0.35
d) average waiting time for the uniform distribution = 10 minutes.
Step-by-step explanation:
This is a uniform distribution problem with lower limit of 0 minute and upper limit of 20 minutes.
a = 0, b = 20
Probability = f(x) = [1/(b-a)] ∫ dx (with the definite integral evaluated between the two intervals whose probability is required.
a) Probability that a randomly selected commuter will spend more than 7 minutes waiting for GO-D.O.T
P(7 < x ≤ 20) = f(x) = [1/(b-a)] ∫²⁰₇ dx
P(7 < x ≤ 20) = (20-7)/(20-0) = (13/20) = 0.65
b) Standard deviation of the uniform distribution
Standard deviation of a uniform distribution is given as
σ = √[(b-a)²/12]
σ = √[(20-0)²/12]
σ = √[20²/12]
σ = 5.77 minutes
c) Probability that a randomly selected commuter will spend longer than 10 minutes but no more than 17 minutes waiting for the GO-D.O.T = P(10 < x < 17)
P(10 < x < 17) = (17-10)/(20-0)
P(10 < x < 17) = (7/20) = 0.35
d) The average waiting time.
The average of a uniform distribution = (b+a)/2
Average waiting time = (20+0)/2
Average waiting time = 10 minutes
Hope this Helps!!!
The dimensions of a closed rectangular box are measured x, y and z as 100 cm, 70 cm, and 30 cm, respectively, with a possible error of 0.2 cm in each dimension. The surface area and the volume of the box is given by the equations S(x, y, z) = 2xy + 2xz + 2yz, V(x, y, z) = xyz Find the linear approximation of S at the point (96, 69, 29). b. Suppose the box has been measured with a ruler that has one centimeter gradation, find the actual maximum error in measuring the surface of the box. c. Find L(101,71,31) -L(100,70,30) d. Use differentials to estimate the error in the measurement of the surface area of the box. e. Compare the answers of parts c to d and the d to b. What do you conclude? f. A coat of paint of thickness 0.0002 cm is applied to the exterior surface of the box. Use differentials to estimate the amount of the paint needed.
Answer:see the pictures attached
Step-by-step explanation:
Suppose that there are six prospective jurors, four men and two women, who might be impaneled to sit on the jury in a criminal case. Two jurors are randomly selected from these six to fill the two remaining jury seats.
a. List the simple events in the experiment
b. What is the probability that both impaneled jurors are women?
Exercise:
Jury Duty Three people are randomly selected from voter registration and driving records to report for jury duty. The gender of each person is noted by the county clerk.
a. Define the experiment.
b. List the simple events in S.
c. If each person is just as likely to be a man as a woman, what probability do you assign to each simple event?
d. What is the probability that only one of the three is a man?
e. What is the probability that all three are women?
Answer:
(1)
(a) Shown below.
(b) The probability that both impaneled jurors are women is 0.0667.
(2)
(a) Sampling 3 people and noting their gender.
(b) Shown below.
(c) The probability of each simple event is 0.125.
(d) The probability of selecting only one male is 0.375.
(e) The probability of selecting all 3 females is 0.125.
Step-by-step explanation:
(1)
Let the 4 men be denoted as: M₁, M₂, M₃ and M₄.
And the 2 women be denoted as: W₁ and W₂.
(a)
A jury of two is to be selected.
The simple events in this experiment are:
(M₁, M₂), (M₁, M₃), (M₁, M₄), (M₁, W₁), (M₁, W₂)
(M₂, M₃), (M₂, M₄), (M₂, W₁), (M₂, W₂)
(M₃, M₄), (M₃, W₁), (M₃, W₂)
(M₄, W₁), (M₄, W₂)
(W₁, W₂)
(b)
The total possible number of jury selections is, N = 15.
The possible combination such that both the jurors are woman is, n = 1.
Compute the probability of selecting two women jurors as follows:
[tex]P(2\ juror\ are\ women)=\frac{n}{N} =\frac{1}{15}=0.0667[/tex]
Thus, the probability that both impaneled jurors are women is 0.0667.
(2)
(a)
The experiment consists of sampling 3 people from the voter registration and driving records and noting the gender of each person.
(b)
The simple events are:
S = {(M, M, M), (M, M, F), (M, F, M), (F, M, M), (M, F, F), (F, F, M), (F, M, F), (F, F, F)}
Total number of simple events = 8.
(c)
If the probability of selecting a male is same as the probability of selecting a female, i.e. P (M) = P (F) = [tex]\frac{1}{2}[/tex] then,
The probability of each simple event is:
[tex]\frac{1}{2} \times\frac{1}{2} \times\frac{1}{2}=\frac{1}{8}=0.125[/tex]
(d)
The number of simple events with only 1 male is n = 3.
Compute the probability that only one of the three is a man as follows:
[tex]P(1\ male)=\frac{n}{N} =\frac{3}{8} =0.375[/tex]
Thus, the probability of selecting only one male is 0.375.
(e)
The number of simple events with all 3 females is n = 1.
Compute the probability that all 3 females are selected as follows:
[tex]P(3\ female)=\frac{n}{N} =\frac{1}{8} =0.125[/tex]
Thus, the probability of selecting all 3 females is 0.125.
Jack inherited a perpetuity-due, with annual payments of 15,000. He immediately exchanged the perpetuity for a 25-year annuity-due having the same present value. The annuity-due has annual payments of X. All the present values are based on an annual effective interest rate of 10% for the first 10 years and 8% thereafter. Calculate X.
Jack inherited a perpetuity-due, which he exchanged for a 25-year annuity-due. The present value of both are equal and based on different interest rates over the 25 years. The annual payment X of the annuity can be calculated using the formula for the present value of an annuity and the present value of the inherited perpetuity-due.
Explanation:This problem is about financial mathematics, specifically involving perpetuities and annuities. In a perpetuity-due, the payments are made at the beginning of each period indefinitely. An annuity-due is similar, but the payments only last a specified number of years.
The present value (PV) of a perpetuity-due with annual payments P and annual interest rate r is calculated by PV = P / r. Given that P is $15,000 and r is 10%, the present value of the perpetuity-due that Jack inherits is $150,000.
When Jack swaps this for a 25-year annuity-due, with the first 10 years at 10% interest and the next 15 years at 8% interest, we calculate the annual payment X using the formula for the present value of an annuity. This formula involves dividing the total present value by the sum of the present value factors for each year at the respective interest rates.
Consequently, X can be calculated as follows: X = (PV of perpetuity-due) / ∑ (discount factors for each year). A detailed calculation will give the exact value of X.
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Suppose that 20% of the adult women in the United States dye or highlight their hair. We would like to know the probability that a SRS of size 200 would come within plus or minus 3 percentage points of this true value. In other words, find probability that pˆ takes a value between 0.17 and 0.23.
Answer:
71.08% probability that pˆ takes a value between 0.17 and 0.23.
Step-by-step explanation:
We use the binomial approxiation to the normal to solve this question.
Binomial probability distribution
Probability of exactly x sucesses on n repeated trials, with p probability.
Can be approximated to a normal distribution, using the expected value and the standard deviation.
The expected value of the binomial distribution is:
[tex]E(X) = np[/tex]
The standard deviation of the binomial distribution is:
[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]
Normal probability distribution
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
When we are approximating a binomial distribution to a normal one, we have that [tex]\mu = E(X)[/tex], [tex]\sigma = \sqrt{V(X)}[/tex].
In this problem, we have that:
[tex]p = 0.2, n = 200[/tex]. So
[tex]\mu = E(X) = np = 200*0.2 = 40[/tex]
[tex]\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{200*0.2*0.8} = 5.66[/tex]
In other words, find probability that pˆ takes a value between 0.17 and 0.23.
This probability is the pvalue of Z when X = 200*0.23 = 46 subtracted by the pvalue of Z when X = 200*0.17 = 34. So
X = 46
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{46 - 40}{5.66}[/tex]
[tex]Z = 1.06[/tex]
[tex]Z = 1.06[/tex] has a pvalue of 0.8554
X = 34
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{34 - 40}{5.66}[/tex]
[tex]Z = -1.06[/tex]
[tex]Z = -1.06[/tex] has a pvalue of 0.1446
0.8554 - 0.1446 = 0.7108
71.08% probability that pˆ takes a value between 0.17 and 0.23.
The probability that [tex]\( \hat{p} \)[/tex] takes a value between 0.17 and 0.23 is approximately 0.7108.
1. Given that 20% of adult women dye or highlight their hair, the true population proportion [tex]\( p \)[/tex] is 0.20.
2. We want to find the probability that a sample proportion [tex]\( \hat{p} \)[/tex] from a simple random sample (SRS) of size 200 falls within plus or minus 3 percentage points of this true value. In other words, we want to find [tex]\( P(0.17 < \hat{p} < 0.23) \)[/tex].
3. The standard error of [tex]\( \hat{p} \)[/tex] is given by:
[tex]\[ SE = \sqrt{\frac{p(1-p)}{n}} \][/tex]
where [tex]\( p = 0.20 \)[/tex] (the true population proportion) and [tex]\( n = 200 \)[/tex] (the sample size).
4. Calculate the standard error:
[tex]\[ SE = \sqrt{\frac{0.20 \times (1-0.20)}{200}} = \sqrt{\frac{0.20 \times 0.80}{200}} \][/tex]
[tex]\[ SE = \sqrt{\frac{0.16}{200}} = \sqrt{0.0008} \approx 0.0283 \][/tex]
5. Next, we find the z-scores corresponding to the values 0.17 and 0.23 using the standard normal distribution table:
[tex]\[ z_{0.17} = \frac{0.17 - 0.20}{0.0283} \approx -1.0601 \][/tex]
[tex]\[ z_{0.23} = \frac{0.23 - 0.20}{0.0283} \approx 1.0601 \][/tex]
6. Using the z-scores, we find the corresponding probabilities from the standard normal distribution table:
[tex]\[ P(\hat{p} < 0.17) \approx P(Z < -1.0601) \approx 0.1446 \][/tex]
[tex]\[ P(\hat{p} < 0.23) \approx P(Z < 1.0601) \approx 0.8554 \][/tex]
7. Therefore, the probability that [tex]\( \hat{p} \)[/tex] takes a value between 0.17 and 0.23 is approximately:
[tex]\[ P(0.17 < \hat{p} < 0.23) = P(\hat{p} < 0.23) - P(\hat{p} < 0.17) \][/tex]
[tex]\[ \approx 0.8554 - 0.1446 = 0.7108 \][/tex]
8. Alternatively, we can find this probability directly using the cumulative distribution function (CDF) of the standard normal distribution:
[tex]\[ P(0.17 < \hat{p} < 0.23) = P(-1.0601 < Z < 1.0601) \][/tex]
[tex]\[ \approx \Phi(1.0601) - \Phi(-1.0601) \][/tex]
[tex]\[ \approx 0.8554 - 0.1446 = 0.7108 \][/tex]
9. Therefore, the probability that[tex]\( \hat{p} \)[/tex] takes a value between 0.17 and 0.23 is approximately 0.7108.
Determine if the described set is a subspace. The subset of Rn (n even) consisting of vectors of the form v = v1 vn , such that v1 − v2 + v3 − v4 + v5 − − vn = 0. The set is a subspace. The set is not a subspace.
Answer:
The set is a subspace
Step-by-step explanation:
We need to check 3 things: whether the 0 vector is in the set, whether the sum of 2 elements of the set is an element of the set and whether the product of an element of the set for a real scalar is an element of the set.
0 is in the setYes: the 0 vector (0, 0, ..., 0) satysfies the set property: 0-0+0-0........-0 = 0.
Given 2 elements v = (v1, ..., vn), w = (w1, ..., wn), is the sum v+2 = (v1+w1, v2+w2, ..., vn+wn) an element of the set?Yes: Note that (v1+w1)-(v2+w2)+(v3+w3)- ..... - (vn+wn) = v1-v1+v3 - ... - vn + w1 - w2 + w3 - ... - wn = 0+0 = 0.
Given an element of the set v = (v1, ... ,vn), and a real number r, is rv = (rv1, ..., rvn) an element of the set?Yes: By taking r as common factor, we have rv1 - rv2 + rv3 - ... - rvn = r * (v1-v2+v3 - ... - vn) = r*0 = 0.
Thus, the described set is effectively a subspace.
The described set is a subspace of Rn (n even). It satisfies all three conditions of a subspace: containing the zero vector, closed under addition, and closed under scalar multiplication.
Explanation:The set described is a subspace of ℝn (where n is even).
To determine if the set is a subspace, we need to check if it satisfies three conditions:
It contains the zero vector: The zero vector satisfies v1 - v2 + v3 - v4 + v5 - ... - vn = 0, so it is in the set.
Since the set satisfies all three conditions, it is a subspace of ℝn (where n is even).
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Conditional distribution. Give (in percents) the conditional distribution of income level among single men. Should your percents add to 100% (up to roundoff error)? Explain your reasoning.
Answer:
note:
solution is attached due to error in mathematical equation. please find the attachment
The conditional distribution of income level among single men should add up to 100%, since when calculating a conditional distribution we treat the subgroup as the total population. Thus, income levels among single men are all the possible outcomes for this group and should collectively correspond to 100% of cases.
Explanation:The conditional distribution in this scenario is the distribution of income level among single men, being key to understanding he relative frequency of various income levels within this specific group. Since we are looking at a subset of the total population (in this case, single men), the percentages should indeed add up to 100%. This is because when we calculate the conditional distribution, we treat this subgroup as if it were the total population. Therefore, the total should be 100%, reflecting the entire subgroup, not compared to the entire population. An example could be: If 10% of single men are in low income, 30% are in middle income, and 60% are in high income, these percentages represent the income level distribution among single men, and add up to 100%.
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Speeding on the I-5, Part I. The distribution of passenger vehicle speeds traveling on the Interstate 5 Freeway (I-5) in California is nearly normal with a mean of 72.6 miles/hour and a standard deviation of 4.78 miles/hour.47 (a) What percent of passenger vehicles travel slower than 80 miles/hour?(b) What percent of passenger vehicles travel between 60 and 80 miles/hour?(c) How fast to do the fastest 5% of passenger vehicles travel?(d) The speed limit on this stretch of the I-5 is 70 miles/hour. Approximate what percentage of the passenger vehicles travel above the speed limit on this stretch of the I-5.
Answer:
a) [tex]P(X<80)=P(\frac{X-\mu}{\sigma}<\frac{80-\mu}{\sigma})=P(Z<\frac{80-72.6}{4.78})=P(z<1.548)[/tex]
And we can find this probability using the normal standard distirbution or excel and we got:
[tex]P(z<1.548)=0.939[/tex]
And that correspond to 93.9 %
b) [tex]P(60<X<80)=P(\frac{60-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{80-\mu}{\sigma})=P(\frac{60-72.6}{4.78}<Z<\frac{80-72.6}{4.78})=P(-2.636<z<1.548)[/tex]
And we can find this probability with this difference:
[tex]P(-2.636<z<1.548)=P(z<1.548)-P(z<-2.636)[/tex]
And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.
[tex]P(-2.636<z<1.548)=P(z<1.548)-P(z<-2.636)=0.939-0.0042=0.935 [/tex]
So we have approximately 93.5%
c) [tex]z=1.64<\frac{a-72.6}{4.78}[/tex]
And if we solve for a we got
[tex]a=72.6 +1.64*4.78=80.439[/tex]
So the value of velocity that separates the bottom 95% of data from the top 5% is 80.439.
d) [tex]P(X>70)=P(\frac{X-\mu}{\sigma}>\frac{70-\mu}{\sigma})=P(Z>\frac{70-72.6}{4.78})=P(z>-0.544)[/tex]
And we can find this probability using the complement rule, normal standard distirbution or excel and we got:
[tex]P(z>-0.544)=1-P(z<-0.544) = 1-0.293=0.707 [/tex]
And that correspond to 70.7 %
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".
Part a
Let X the random variable that represent the vehicles speeds of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(72.6,4.78)[/tex]
Where [tex]\mu=72.6[/tex] and [tex]\sigma=4.78[/tex]
We are interested on this probability
[tex]P(X<80)[/tex]
And the best way to solve this problem is using the normal standard distribution and the z score given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
If we apply this formula to our probability we got this:
[tex]P(X<80)=P(\frac{X-\mu}{\sigma}<\frac{80-\mu}{\sigma})=P(Z<\frac{80-72.6}{4.78})=P(z<1.548)[/tex]
And we can find this probability using the normal standard distirbution or excel and we got:
[tex]P(z<1.548)=0.939[/tex]
And that correspond to 93.9 %
Part b
We want this probability
[tex]P(60<X<80)=P(\frac{60-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{80-\mu}{\sigma})=P(\frac{60-72.6}{4.78}<Z<\frac{80-72.6}{4.78})=P(-2.636<z<1.548)[/tex]
And we can find this probability with this difference:
[tex]P(-2.636<z<1.548)=P(z<1.548)-P(z<-2.636)[/tex]
And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.
[tex]P(-2.636<z<1.548)=P(z<1.548)-P(z<-2.636)=0.939-0.0042=0.935 [/tex]
So we have approximately 93.5%
Part c
For this part we want to find a value a, such that we satisfy this condition:
[tex]P(X>a)=0.05[/tex] (a)
[tex]P(X<a)=0.95[/tex] (b)
Both conditions are equivalent on this case. We can use the z score again in order to find the value a.
As we can see on the figure attached the z value that satisfy the condition with 0.95 of the area on the left and 0.05 of the area on the right it's z=1.64. On this case P(Z<1.64)=0.95 and P(z>1.64)=0.05
If we use condition (b) from previous we have this:
[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.95[/tex]
[tex]P(z<\frac{a-\mu}{\sigma})=0.95[/tex]
But we know which value of z satisfy the previous equation so then we can do this:
[tex]z=1.64<\frac{a-72.6}{4.78}[/tex]
And if we solve for a we got
[tex]a=72.6 +1.64*4.78=80.439[/tex]
So the value of velocity that separates the bottom 95% of data from the top 5% is 80.439.
Part d
[tex]P(X>70)[/tex]
And the best way to solve this problem is using the normal standard distribution and the z score given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
If we apply this formula to our probability we got this:
[tex]P(X>70)=P(\frac{X-\mu}{\sigma}>\frac{70-\mu}{\sigma})=P(Z>\frac{70-72.6}{4.78})=P(z>-0.544)[/tex]
And we can find this probability using the complement rule, normal standard distirbution or excel and we got:
[tex]P(z>-0.544)=1-P(z<-0.544) = 1-0.293=0.707 [/tex]
And that correspond to 70.7 %
Find the length of arc. DB. Leave your answer in terms of pi
Answer: Length of arc DB is 14π feet.
Step-by-step explanation:
The sum of the angles on a straight line is 180 degrees. This means that
m∠DAB + m∠CAB = 180
m∠DAB + 40 = 180
m∠DAB = 180 - 40
m∠DAB = 140°
The formula for determining the length of an arc is expressed as
Length of arc = θ/360 × 2πr
Where
θ represents the central angle.
r represents the radius of the circle.
π is a constant whose value is 3.14
From the information given,
r = 18 feet
θ = 140°
Therefore,
Length of arc DB = 140/360 × 2 × π × 18
Length of arc DB = 14π feet
In the final round of a TV game show, contestants have a chance to increase their current winnings of $1 million to $2 million. If they are wrong, their prize is decreased to $500,000. A contestant thinks his guess will be right 50 percent of the time. Should he play
Answer:
The contestant should not play.Step-by-step explanation:
As per the given question, their current winning is $1 million.
The probability of the guessing to be true is 50% = [tex]\frac{50}{100} = \frac{1}{2}[/tex].
There is also a possibility of 50% to be wrong, which can reduced the winning amount to $500,000 that is the half of the current amount.
Hence, the contestant should not play.
A 15-inch candle is lit and burns at a constant rate of 1.1 inches per hour. Let t represent the number of hours since the candle was lit, and suppose f is a function such that f ( t ) represents the remaining length of the candle (in inches) t hours after it was lit. Write a function formula for f . f ( t )
Answer: f(t) = 15 - 1.1t
Step-by-step explanation:
Let t represent the number of hours since the candle was lit.
A 15-inch candle is lit and burns at a constant rate of 1.1 inches per hour. This means that in t hours, the candle that would have burnt is 1.1t
The length of the candle that would be left after t hours is expressed as
15 - 1.1t
suppose f is a function such that f(t) represents the remaining length of the candle (in inches) t hours after it was lit, then
f(t) = 15 - 1.1t
A humanities professor assigns letter grades on a test according to the following scheme. A: Top 7% of scores B: Scores below the top 7% and above the bottom 56% C: Scores below the top 44% and above the bottom 19% D: Scores below the top 81% and above the bottom 6% F: Bottom 6% of scores Scores on the test are normally distributed with a mean of 72.1 and a standard deviation of 9.5. Find the minimum score required for an A grade. Round your answer to the nearest whole number, if necessary.
Answer:
The minimum score required for an A grade is 86.
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
[tex]\mu = 72.1, \sigma = 9.5[/tex]
Find the minimum score required for an A grade.
Top 7%, which is the value of X when Z has a pvalue of 1-0.07 = 0.93. So it is X when Z = 1.475. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]1.475 = \frac{X - 72.1}{9.5}[/tex]
[tex]X - 72.1 = 1.475*9.5[/tex]
[tex]X = 86[/tex]
The minimum score required for an A grade is 86.
The number and frequency of Atlantic hurricanes annually from 1940 through 2007 is shown here:
a. Find the probabilities of 0-8 hurricanes each season using these data
b. Assuming a Poisson distribution and using the mean number of hurricanes per season from the empirical data, compute the probabilities of experiencing 0-8 hurricanes in a season.
Compare these to your answer to part (a).
How good does a Poisson distribution model this phenomenon?
Number Frequency
0 5
1 16
2 19
3 13
4 3
5 5
6 4
7 2
8 1
Total 68
Answer:
The probability table is shown below.
A Poisson distribution can be used to approximate the model of the number of hurricanes each season.
Step-by-step explanation:
(a)
The formula to compute the probability of an event E is:
[tex]P(E)=\frac{Favorable\ no.\ of\ frequencies}{Total\ NO.\ of\ frequencies}[/tex]
Use this formula to compute the probabilities of 0 - 8 hurricanes each season.
The table for the probabilities is shown below.
(b)
Compute the mean number of hurricanes per season as follows:
[tex]E(X)=\frac{\sum x f_{x}}{\sum f_{x}}=\frac{176}{68}= 2.5882\approx2.59[/tex]
If the variable X follows a Poisson distribution with parameter λ = 7.56 then the probability function is:
[tex]P(X=x)=\frac{e^{-2.59}(2.59)^{x}}{x!} ;\ x=0, 1, 2,...[/tex]
Compute the probability of X = 0 as follows:
[tex]P(X=0)=\frac{e^{-2.59}(2.59)^{0}}{0!} =\frac{0.075\times1}{1}=0.075[/tex]
Compute the probability of X = 1 as follows:
[tex]\neq P(X=1)=\frac{e^{-2.59}(2.59)^{1}}{1!} =\frac{0.075\times7.56}{1}=0.1943[/tex]
Compute the probabilities for the rest of the values of X in the similar way.
The probabilities are shown in the table.
On comparing the two probability tables, it can be seen that the Poisson distribution can be used to approximate the distribution of the number of hurricanes each season. This is because for every value of X the Poisson probability is approximately equal to the empirical probability.
To address this question, probabilities from the given data are calculated first, after which Poisson distribution is applied to compute the probabilities using the mean number of hurricanes per season. A comparison of both results offers an insight into the accuracy of the Poisson distribution in modeling this phenomenon.
Explanation:To answer this question, we first have to calculate probabilities based on the empirical data and then compare them to probabilities computed under the assumption of a Poisson distribution.
First, we count total seasons from 1940 through 2007, which is 68. Using these frequencies, we can calculate the probability of having 0-8 hurricanes each season as follows: the number of seasons with a certain number of hurricanes divided by the total number of seasons.
For the Poisson distribution, we first need to calculate the average (mean) number of hurricanes per season, which is the sum of the product of the number of hurricanes and its frequency divided by the total number of seasons. After finding the mean, we can compute the probability of experiencing 0-8 hurricanes using the Poisson formula: e^(-mean) * (mean^n) / n!. After that, we can compare these probabilities with the ones derived from the empirical data.
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The table in the shape of a circle has a diameter of 6 feet. How much fabric is needed to make a table cloth if it hangs 1 foot off the table all the way around?jdjdkndjcjjfnnfm
Answer: area of fabric needed is 50.24 ft²
Step-by-step explanation:
The table in the shape of a circle has a diameter of 6 feet. This means that the diameter of the fabric that would just fit the table is 6 feet. Therefore, the diameter of the fabric needed to make a table cloth if it hangs 1 foot off the table would be 6 + 1 + 1 = 8 feet
The formula for determining the area of a circle is expressed as
Area = πr²
Where
r represents radius of the circle.
π is a constant whose value is 3.14
Radius = diameter/2. Therefore
r = 8/2 = 4
Area of fabric = 3.14 × 4²
= 50.24 ft²
Final answer:
To make a tablecloth for a circular table with a diameter of 6 feet, and an overhang of 1 foot, the student would need approximately 50.27 square feet of fabric.
Explanation:
The student is asking about finding the amount of fabric needed to create a tablecloth for a circular table with specific dimensions. Given that the table has a diameter of 6 feet, and the tablecloth needs to hang 1 foot off the table all the way around, we need to calculate the diameter of the fabric required.
To solve this, we need to add the overhang to the diameter of the table, considering that the overhang occurs on both sides:
Diameter of the table: 6 feetOverhang on one side: 1 footOverhang on the other side: 1 footTotal diameter needed: 6 feet + 1 foot + 1 foot = 8 feetNow, to find the area of fabric needed, we apply the formula for the area of a circle which is π × radius². First, we find the radius by halving the diameter:
Radius of the fabric: 8 feet / 2 = 4 feetThen, we calculate the area:
Area of fabric: π × (4 feet)² = π × 16 feet²Finally, we can approximate π as 3.1416 to get an approximate area of:
Area of fabric ≈ 3.1416 × 16 feet² ≈ 50.2656 square feetTherefore, the student would need approximately 50.27 square feet of fabric to make the tablecloth.
A student takes an exam containing 1414 multiple choice questions. The probability of choosing a correct answer by knowledgeable guessing is 0.30.3. At least 99 correct answers are required to pass. If the student makes knowledgeable guesses, what is the probability that he will pass? Round your answer to four decimal places.
Answer:
0.0082 = 0.82% probability that he will pass
Step-by-step explanation:
For each question, there are only two possible outcomes. Either the students guesses the correct answer, or he guesses the wrong answer. The probability of guessing the correct answer for a question is independent of other questions. So we use the binomial probability distribution to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
In this problem we have that:
[tex]n = 14, p = 0.3[/tex].
If the student makes knowledgeable guesses, what is the probability that he will pass?
He needs to guess at least 9 answers correctly. So
[tex]P(X \geq 9) = P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14)[/tex]
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 9) = C_{14,9}.(0.3)^{9}.(0.7)^{5} = 0.0066[/tex]
[tex]P(X = 10) = C_{14,10}.(0.3)^{10}.(0.7)^{4} = 0.0014[/tex]
[tex]P(X = 11) = C_{14,11}.(0.3)^{11}.(0.7)^{3} = 0.0002[/tex]
[tex]P(X = 12) = C_{14,12}.(0.3)^{12}.(0.7)^{2} = 0.000024[/tex]
[tex]P(X = 13) = C_{14,13}.(0.3)^{13}.(0.7)^{1} = 0.000002[/tex]
[tex]P(X = 14) = C_{14,14}.(0.3)^{14}.(0.7)^{0} \cong 0 [/tex]
[tex]P(X \geq 9) = P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) = 0.0066 + 0.0014 + 0.0002 + 0.000024 + 0.000002 = 0.0082[/tex]
0.0082 = 0.82% probability that he will pass