Answer:
Present value = $4,122.4
Accumulated amount = $4,742
Step-by-step explanation:
Data provided in the question:
Amount at the Start of money flow = $1,000
Increase in amount is exponentially at the rate of 5% per year
Time = 4 years
Interest rate = 3.5% compounded continuously
Now,
Accumulated Value of the money flow = [tex]1000e^{0.05t}[/tex]
The present value of the money flow = [tex]\int\limits^4_0 {1000e^{0.05t}(e^{-0.035t})} \, dt[/tex]
= [tex]1000\int\limits^4_0 {e^{0.015t}} \, dt[/tex]
= [tex]1000\left [\frac{e^{0.015t}}{0.015} \right ]_0^4[/tex]
= [tex]1000\times\left [\frac{e^{0.015(4)}}{0.015} -\frac{e^{0.015(0)}}{0.015} \right][/tex]
= 1000 × [70.7891 - 66.6667]
= $4,122.4
Accumulated interest = [tex]e^{rt}\int\limits^4_0 {1000e^{0.05t}(e^{-0.035t}} \, dt[/tex]
= [tex]e^{0.035\times4}\times4,122.4[/tex]
= $4,742
The present value and interest accumulated would be as follows:
Present Value = $ 4,122.4
Interest Accumulated = $ 4742
Given that,
Principal at the beginning of money flow = $1,000
Exponential interest rate = 5% per year
Time Period = 4 years
So,
The accumulated money flow's worth = [tex]1000e^{0.05t}[/tex]
The current value of the money can be determined by [tex]\int\limits^4_0 1000e^{0.05t}(e^{-0.035t}) {} \, dt[/tex]
On solving, we get
The present value = $ 4,122.4
Interest Accumulated = $4,742
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We select two distinct numbers (a, b) in the range 1 to 99 (inclusive). How many ways can we pick a and b such that their sum is even and a is a multiple of 9?
Answer:
The possible number of ways to select distinct (a, b) such that (a + b) is even is 534.
Step-by-step explanation:
The range 1 - 99 has 99 numbers, since 1 and 99 are inclusive.
Of these 50 numbers are odd and 49 are even.
The two distinct numbers a and b must have an even sum and a should be a multiple of 9.
The sum of two numbers is even only when both are odd or both are even.
The possible values that a can assume are,
a = {9, 18, 27, 36, 45, 54, 63, 72, 81, 90 and 99}
Thus, a can assume 6 odd values and 5 even values.
If a = odd number, then b can be any of the 49 out of 50 odd numbers.Total number of ways to select a and b such that both are odd and their sum is even is:
[tex]n(Odd\ a\ and\ b)=n(Odd\ value\ of\ a)\times n(Odd\ value\ of\ b)=6\times49=294[/tex]
If a = even number, then b can be any of the 48 out of 50 even numbers.Total number of ways to select a and b such that both are even and their sum is even is
[tex]n(Even\ a\ and\ b)=n(E\ value\ of\ a)\times n(Even\ value\ of\ b)=5\times48=240[/tex]
Total number of ways to select distinct (a, b) such that (a + b) is even is =
[tex]=n(Odd\ a\ and\ b)+n(Even\ a\ and\ b)=294+240=534[/tex]
Thus, the possible number of ways to select distinct (a, b) such that (a + b) is even is 534.
Approximately 10% of all people are left-handed. Consider a grouping of fifteen people. a.)State the random variable. b.)Write the probability distribution. c.)Draw a histogram. d.)Describe the shape of the histogram. e.)Find the mean. f.)Find the variance. g.)Find the standard deviation.
Answer:
a) left handed people
b) Binomial probability distribution with pdf
[tex]P(X=x)=15Cx0.1^{x} 0.9^{15-x}[/tex]
where x=0,1,2,...,15.
c) Histogram is attached
d) The shape of histogram depicts that distribution is rightly skewed.
e) 1.5
f) 1.35
g) 1.16
Step-by-step explanation:
a)
The random variable in the given scenario is " left handed people"
b)
The scenario represents the binomial probability distribution as the outcome is divided into one of two categories and experiment is repeated fixed number of times i.e. 15 and trails are independent. The pdf of binomial distribution is
[tex]P(X=x)=nCxp^{x} q^{n-x}[/tex]
Here n=15, p=0.1 and q=1-p=0.9.
So, the pdf would be
[tex]P(X=x)=15Cx0.1^{x} 0.9^{n-x}[/tex]
where x=0,1,2,...,15.
c)
Histogram is constructed by first computing probabilities on all x points i.e. x=0, x=1 , .... ,x=15 and then plotting all probabilities with respective x values. Histogram is in attached image.
d)
The tail of histogram is to the right side and thus the histogram depicts that given probability distribution is rightly skewed.
e)
The mean of binomial probability distribution is computed by multiplying number of trails and probability of success.
mean=np=15*0.1=1.5
f)
The variance of binomial probability distribution is computed by multiplying number of trails and probability of success and probability of failure.
variance=npq=15*0.1*0.9=1.35
g)
The standard deviation can be calculated by simply taking square root of variance
S.D=√npq=√1.35=1.16
The proportion of left-handed people follows a binomial distribution
The random variable is left-handed peopleThe probability distribution function is [tex]\mathbf{P(x) = ^nC_x 0.1^x 0.9^{n -x}}[/tex]The mean is 1.5The variance is 1.35The standard deviation is 1.16The given parameters are:
[tex]\mathbf{n = 15}[/tex] -- the sample size
[tex]\mathbf{p = 10\%}[/tex] --- the proportion of left-handed people
(a) The random variable
The distribution is about left-handed people.
Hence, the random variable is left-handed people
(b) The probability distribution
If the proportion of left-handed people is 10%, then the proportion of right-handed people is 90%.
So, the probability distribution function is:
[tex]\mathbf{P(x) = ^nC_x p^x (1 - p)^{n -x}}[/tex]
This gives
[tex]\mathbf{P(x) = ^nC_x (10\%)^x (1 - 10\%)^{n -x}}[/tex]
[tex]\mathbf{P(x) = ^nC_x 0.1^x 0.9^{n -x}}[/tex]
Hence, the probability distribution function is [tex]\mathbf{P(x) = ^nC_x 0.1^x 0.9^{n -x}}[/tex]
(c) The histogram
To do this, we calculate P(x) for x = 0 to 15
[tex]\mathbf{P(0) = ^{15}C_0 \times 0.1^0 \times 0.9^{15 -0} = 0.206}[/tex]
[tex]\mathbf{P(1) = ^{15}C_1 \times 0.1^1 \times 0.9^{15 -1} = 0.343}[/tex]
.....
..
[tex]\mathbf{P(15) = ^{15}C_{15} \times 0.1^{15} \times 0.9^{15 -15} = 10^{-15}}[/tex]
See attachment for the histogram
(d) The mean
This is calculated as:
[tex]\mathbf{\bar x = np}[/tex]
So, we have:
[tex]\mathbf{\bar x = 15 \times 10\% }[/tex]
[tex]\mathbf{\bar x= 1.5}[/tex]
Hence, the mean is 1.5
(e) The variance
This is calculated as:
[tex]\mathbf{Var = np(1 - p)}[/tex]
So, we have:
[tex]\mathbf{Var = 15 \times 10\% \times (1 - 10\%)}[/tex]
[tex]\mathbf{Var = 1.35}[/tex]
Hence, the variance is 1.35
(f) The standard deviation
This is calculated as:
[tex]\mathbf{\sigma = \sqrt{Var}}[/tex]
So, we have:
[tex]\mathbf{\sigma = \sqrt{1.35}}[/tex]
[tex]\mathbf{\sigma =1.16}[/tex]
Hence, the standard deviation is 1.16
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The branch of statistical studies called inferential statistics refers to drawing conclusions about sample data by analyzing the corresponding population. True or False
Answer: False
Step-by-step explanation:
In Inferential statistics the sample data is analysed rather than analysing the whole population, analysing the whole population may sometimes be impossible, like analysing the population of a whole country. Therefore, for inferential statistics, conclusions about the whole population are drawn by analyzing the corresponding sample data. The sample data are selected from the population and then analysed, the results can then be used to conclude on the whole population.
Consider the demand curve Qd = 150 - 2P and the supply curve Qs = 50 + 3P. What is total expenditure at equilibrium? Make sure to round your answers to the nearest 100th percentage point. Also, do not write any symbol. For example, write 123.34 for $123.34. Let me know if you have questions. Good luck.
Answer:
20
Step-by-step explanation:
At equilibrium quantity demanded Qd equals quantity supply Qs
Qd = Qs (At equilibrium)
⇒ 150 - 2P = 50 + 3P
solving for P in the equation,
3P + 2P = 150 - 50
5P = 100
[tex]P = \frac{100}{5}[/tex]
P = 20
Una familia dedica dos tercios de sus ingresos a cubrir gastos de funcionamiento, ahorra la cuarta parte del total y gadta el resto en ocio.¿Qué fraccion de los ingresos invierte en ocio?
Answer:
la familia invierte 8.33% de los ingresos totales en ocio
Step-by-step explanation:
Representando los ingresos totales por I:
- Gastos de funcionamiento = 2/3*I
- Ahorro = 1/4*I
- En ocio : lo que resta = I - 2/3*I - 1/4*I = I - 11/12*I = 1/12*I (8.33% de I)
por lo tanto la familia invierte 8.33% de los ingresos totales en ocio
A history class is comprised of 7 female and 10 male students. If the instructor of the class randomly chooses 7 students from the class for an oral exam, what is the probability that 5 female students and 2 male students will be selected? Round your answer to 3 decimal places.
Answer:
The probability of selecting 5 female and 2 male students is 0.052.
Step-by-step explanation:
The class comprises of 7 female students and 10 male students.
Total number of students: 17.
Number of female students, 7.
Number of male students, 10.
The probability of an event E is:
[tex]P(E)=\frac{Favorable\ outcomes}{Total\ number\ of] outcomes}[/tex]
The number of ways to select 7 students from 17 is:
[tex]N ={17\choose 7}=\frac{17!}{7!(17-7)!}= 19448[/tex]
The number of ways to select 5 female students of 7 females is:
[tex]n(F) ={7\choose 5}=\frac{7!}{5!(7-5)!}= 21[/tex]
The number of ways to select 2 male students of 10 males is:
[tex]n(M) ={10\choose 2}=\frac{10!}{2!(10-2)!}= 45[/tex]
Compute the probability of selecting 5 female and 2 male students as follows:
P (5 F and 2 M) = [n (F) × n (M)] ÷ N
[tex]=\frac{21\times45}{19448} \\=0.05183\\\approx0.052[/tex]
Thus, the probability of selecting 5 female and 2 male students is 0.052.
Positive-sequence components consist of three phasors with _____ magnitudes and _____ phase displacement in positive sequence; negativesequence components consist of three phasors with _____ magnitudes and _____ phase displacement in negative sequence; and zero-sequence components consist of three phasors with _____ magnitudes and _____ phase displacement.
Answer: The answers in order are: Equal, 120°, Equal, 120°, Equal, no
Step-by-step explanation:
The positive sequence components have equal magnitudes and 120° phase displacements in positive sequence. Their phase sequence is same as that of the system one. Also their phase rotations is same like the system.
The negative sequence components have equal magnitudes with same 120° phase displacements. Their phase sequence is opposite to that of system but their phase rotation is same like the system.
Zero sequence components have equal magnitude with no phase displacement. They behave like negative sequence in terms of phase sequence and phase rotation.
Answer:
Equal, 120°, Equal, 120°, Equal, no
Step-by-step explanation:
An urn has 3 blue balls 4 red balls. 3 balls are drawn without replacement. find the probability of drawing 2 blue and 1 red given at least 1 blue is drawn
Answer:
Frist case: P=12/35
Second case: P=31/35
Step-by-step explanation:
An urn has 3 blue balls 4 red balls. 3 balls are drawn without replacement.
Frist case:
We calculate the number of possible combinations
{7}_C_{3}=\frac{7!}{3! · (7-3)!}=35
We calculate the number of favorable combinations
{3}_C_{2} · {4}_C_{1} =
=\frac{3!}{2! · (3-2)!} · \frac{4!}{1! · (4-1)!}
=3 · 4 = 12
Therefore, the probability is
P=12/35
Second case:
When we count on at least one ball to be blue, we go over the probability complement.
We calculate the probability that all the balls are red, then subtract this from 1.
We calculate the number of possible combinations
{7}_C_{3}=\frac{7!}{3! · (7-3)!}=35
We calculate the number of favorable combinations
{4}_C_{3} = \frac{4!}{3! · (4-3)!=4
The probability is
P=4/35.
Therefore the probability on at least one ball to be blue
P=1-4/35
P=31/35
what percentage of eligibible american vote? In 2008, a random sample of 500 american adults was take and we found that 68% of them voted. How many of the 500 adults in the sample voted? Now construct a 95% confidence interval estimate of the population percent of Americans that vote and write a sentence to explain the confidence interval.
Answer:
340 of the adults in the sample voted.
The 95% confidence interval estimate of the population percent of Americans that vote is (0.6391, 0.7209). This means that we are 95% sure that the true proportion of Americans that vote is between 0.6391 and 0.7209.
Step-by-step explanation:
In 2008, a random sample of 500 american adults was take and we found that 68% of them voted. How many of the 500 adults in the sample voted?
This is 68% of 500.
So 0.68*500 = 340.
340 of the adults in the sample voted.
Now construct a 95% confidence interval estimate of the population percent of Americans that vote and write a sentence to explain the confidence interval.
In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence interval [tex]1-\alpha[/tex], we have the following confidence interval of proportions.
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which
Z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].
For this problem, we have that:
[tex]n = 500, p = 0.68[/tex]
95% confidence interval
So [tex]\alpha = 0.05[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.05}{2} = 0.975[/tex], so [tex]Z = 1.96[/tex].
The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.68 - 1.96\sqrt{\frac{0.68*0.32}{500}} = 0.6391[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.68 + 1.96\sqrt{\frac{0.68*0.32}{500}} = 0.7209[/tex]
The 95% confidence interval estimate of the population percent of Americans that vote is (0.6391, 0.7209). This means that we are 95% sure that the true proportion of Americans that vote is between 0.6391 and 0.7209.
In a recent poll^1 of 1000 American adults, the number saying that exercise is an important part of daily life was 753. Use strategy or other technology to find a 90% confidence interval for the proportion of American adults who think exercise is an important part of daily life. The 90% confidence interval is
Answer:
[tex]0.753 - 1.64\sqrt{\frac{0.753(1-0.753)}{1000}}=0.731[/tex]
[tex]0.753 + 1.64\sqrt{\frac{0.753(1-0.753)}{1000}}=0.775[/tex]
The 95% confidence interval would be given by (0.731;0.775)
Step-by-step explanation:
Previous concepts
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The population proportion have the following distribution
[tex]p \sim N(p,\sqrt{\frac{p(1-p)}{n}})[/tex]
Solution to the problem
In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 90% of confidence, our significance level would be given by [tex]\alpha=1-0.90=0.1[/tex] and [tex]\alpha/2 =0.005[/tex]. And the critical value would be given by:
[tex]z_{\alpha/2}=-1.64, z_{1-\alpha/2}=1.64[/tex]
The confidence interval for the mean is given by the following formula:
[tex]\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex]
The estimated proportion for this case is:
[tex] \hat p = \frac{X}{n}= \frac{753}{1000}=0.753[/tex]
If we replace the values obtained we got:
[tex]0.753 - 1.64\sqrt{\frac{0.753(1-0.753)}{1000}}=0.731[/tex]
[tex]0.753 + 1.64\sqrt{\frac{0.753(1-0.753)}{1000}}=0.775[/tex]
The 95% confidence interval would be given by (0.731;0.775)
The length of side AB is ....
Answer:
ccvnnnrxfh koo HD fyu
Step-by-step explanation:
yyyhjkuz do Lakewooddrink instill k is s
Quota sampling is most commonly used in a. descriptive research. b. collecting primary data. c. surveys. d. population research. e. exploratory studies.
Quota sampling is most commonly used in surveys. It is a non-probability sampling technique where researchers select individuals who meet certain criteria to be included in the sample.
Explanation:Quota sampling is most commonly used in surveys. It is a non-probability sampling technique where researchers select individuals who meet certain criteria to be included in the sample.
In quota sampling, the researcher sets quotas or targets for each subgroup they want to include in the sample based on certain characteristics. For example, if the researcher wants equal representation of males and females in the sample, they would set quotas for each gender and continue selecting individuals until the quotas are met.
Quota sampling is often used when it is not possible or practical to obtain a random sample, but the researcher still wants to ensure representation of different subgroups within the sample.
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With your typical convenience store customer, there is a 0.23 probability of buying gasoline. The probability of buying groceries is 0.76 and the conditional probability of buying groceries given that they buy gasoline is 0.85. a. Find the probability that a typical customer buys both gasoline and groceries. b. Find the probability that a typical customer buys gasoline or groceries. c. Find the conditional probability of buying gasoline given that the customer buys groceries. d. Find the conditional probability of buying groceries given that the customer did not buy gasoline. e. Are these two events (groceries, gasoline) mutually exclusive? f. Are these two events independent?
Answer:
a) P ( A & B ) = 0.1995
b) P (A U B ) = 0.7905
c) P (A/B) = 0.2625
d) P(B/A') = 0.194805
e) NOT mutually exclusive
f) NOT Independent
Step-by-step explanation:
Declare Events:
- buying gasoline = Event A
- buying groceries = Event B
Given:
- P(A) = 0.23
- P(B) = 0.76
- P(B/A) = 0.85
Find:
- a. Find the probability that a typical customer buys both gasoline and groceries.
- b. Find the probability that a typical customer buys gasoline or groceries.
- c. Find the conditional probability of buying gasoline given that the customer buys groceries.
- d. Find the conditional probability of buying groceries given that the customer did not buy gasoline.
- e Are these two events (groceries, gasoline) mutually exclusive?
- f Are these two events independent?
Solution:
- a) P ( A & B ) ?
P ( A & B ) = P(B/A) * P(A) = 0.85*0.23 = 0.1995
- b) P (A U B ) ?
P (A U B ) = P(A) + P(B) - P(A&B)
P (A U B ) = 0.23 + 0.76 - 0.1995
P (A U B ) = 0.7905
- c) P ( A / B )?
P ( A / B ) = P(A&B) / P(B)
= 0.1995 / 0.76
= 0.2625
- d) P( B / A') ?
P( B / A') = P ( B & A') / P(A')
P ( B & A' ) = 1 - P( A / B) = 1 - 0.85 = 0.15
P ( B / A' ) = 0.15 / (1 - 0.23)
= 0.194805
- e) Are the mutually exclusive ?
The condition for mutually exclusive events is as follows:
P ( A & B ) = 0 for mutually exclusive events.
In our case P ( A & B ) = 0.1995 is not zero.
Hence, NOT MUTUALLY EXCLUSIVE
- f) Are the two events independent?
The condition for independent events is as follows:
P ( A & B ) = P (A) * P(B) for mutually exclusive events.
In our case,
0.1995 = 0.23*0.76
0.1995 = 0.1748 (NOT EQUAL)
Hence, NOT INDEPENDENT
a. The probability a customer buys both gasoline and groceries is 0.23 * 0.85 = 0.1955.
b. The probability a customer buys either gasoline or groceries is 0.23 + 0.76 - 0.1955 = 0.7945.
c. Conditional probability of buying gasoline given groceries: 0.1955 / 0.76 ≈ 0.2572.
d. Conditional probability of buying groceries given no gasoline: 0.76 - 0.1955 / 0.77 ≈ 0.7331.
e. No, these events are not mutually exclusive.
f. No, these events are not independent.
Let's calculate the probabilities and answer each part of the question:
a. To find the probability that a typical customer buys both gasoline and groceries, you can use the formula for conditional probability:
P(Gasoline and Groceries) = P(Groceries | Gasoline) * P(Gasoline)
P(Gasoline and Groceries) = 0.85 * 0.23 = 0.1955
So, the probability that a typical customer buys both gasoline and groceries is 0.1955.
b. To find the probability that a typical customer buys gasoline or groceries, you can use the addition rule for probabilities:
P(Gasoline or Groceries) = P(Gasoline) + P(Groceries) - P(Gasoline and Groceries)
P(Gasoline or Groceries) = 0.23 + 0.76 - 0.1955 = 0.7945
So, the probability that a typical customer buys gasoline or groceries is 0.79
c. To find the conditional probability of buying gasoline given that the customer buys groceries, you can use the formula for conditional probability:
P(Gasoline | Groceries) = P(Gasoline and Groceries) / P(Groceries)
P(Gasoline | Groceries) = 0.1955 / 0.76 ≈ 0.2572
So, the conditional probability of buying gasoline given that the customer buys groceries is approximately 0.2572.
d. To find the conditional probability of buying groceries given that the customer did not buy gasoline, you can use the formula for conditional probability:
P(Groceries | No Gasoline) = P(Groceries and No Gasoline) / P(No Gasoline)
First, calculate P(No Gasoline):
P(No Gasoline) = 1 - P(Gasoline) = 1 - 0.23 = 0.77
Now, calculate P(Groceries and No Gasoline):
P(Groceries and No Gasoline) = P(Groceries) - P(Gasoline and Groceries) = 0.76 - 0.1955 = 0.5645
Now, find P(Groceries | No Gasoline):
P(Groceries | No Gasoline) = 0.5645 / 0.77 ≈ 0.7331
So, the conditional probability of buying groceries given that the customer did not buy gasoline is approximately 0.7331.
e. These two events (buying groceries and buying gasoline) are not mutually exclusive because it's possible for a customer to buy both groceries and gasoline, as we calculated in part (a).
f. To determine whether these two events are independent, we need to check if the conditional probabilities match the unconditional probabilities:
P(Gasoline | Groceries) = P(Gasoline) and P(Groceries | No Gasoline) = P(Groceries)
Let's check:
P(Gasoline | Groceries) ≈ 0.2572
P(Gasoline) = 0.23
P(Groceries | No Gasoline) ≈ 0.7331
P(Groceries) = 0.76
These conditional probabilities are not equal to the unconditional probabilities, so the events are not independent. In an independent event scenario, the conditional probabilities would be equal to the unconditional probabilities.
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Shureka Washbum has scores of 74, 88, 61, and 83 on her algebra tests.
a. Use an inequality to find the scores she must make on the final exam to pass the course with an average of 76 or higher, given that the final exam counts as two
tests
b. Explain the meaning of the answer to part (a).
The solution set is {x{ }
(Type an inequality.)
Step-by-step explanation:
If x is her score on the final exam, then the average is:
(74 + 88 + 61 + 83 + 2x) / 6
(306 + 2x) / 6
51 + ⅓x
We want this to be greater than or equal to 76.
51 + ⅓x ≥ 76
⅓x ≥ 25
x ≥ 75
In order to get an average of 76 or higher, Shureka's score on the final exam must be greater than or equal to 75.
A process is normally distributed with a mean of 10.2 hits per minute and a standard deviation of 1.04 hits. If a randomly selected minute has 13.9 hits, would the process be considered in control or out of control?
Answer:
This process is out of control.
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.
A probability is said to be unusual if it's z-score has a pvalue of 0.05 or lower, or a pvalue of 0.95 or higher.
In this problem, we have that:
[tex]\mu = 10.2 \sigma = 1.04[/tex]
If a randomly selected minute has 13.9 hits, would the process be considered in control or out of control?
The process will be considered out of control if it's z-score(Z when X = 13.9) has a pvalue of 0.95 or higher. Otherwise(it will be positive, since 13.9 is above the mean), it will be considered in control.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{13.9 - 10.2}{1.04}[/tex]
[tex]Z = 3.56[/tex]
[tex]Z = 3.56[/tex] has a pvalue of 0.9999. So there is only a 1-0.9999 = 0.0001 = 0.01% probability of getting 13.9 hits a minute.
So this process is out of control.
If using the empirical rule, since 13.9 hits is more than two standard deviations above the mean of 10.2 hits, the process could be considered out of control. However, establishing specific control limits is necessary for a definitive answer.
Explanation:To determine whether a process is in control or out of control, we assess whether a given measurement is within the expected range of a normal distribution, often using the empirical rule or control limits. Given that the process has a mean of 10.2 hits per minute and a standard deviation of 1.04 hits, under the empirical rule, approximately 95% of the data should fall within two standard deviations of the mean (that is, between roughly 8.12 and 12.28 hits).
With 13.9 hits in a randomly selected minute, this count is significantly more than two standard deviations above the mean, suggesting that the process might be out of control. However, to make a definitive statement about control status, specific control limits must be established, often based on the particular specifications of the process being monitored.
Consider an airfoil in a wind tunnel (i.e., a wing that spans the entire test section). Prove that the lift per unit span can be obtained from the pressure distributions on the top and bottom walls of the wind tunnel (i.e., from the pressure distributions on the walls above and below the airfoil).
Answer:
The solution proved are in the attached file below. Also the explanation is in the attached file
Step-by-step explanation:
The area of a parking lot is calculated to be 5,474 ft2 with an estimated standard deviation of 2 ft2 . What is the Maximum Anticipated Error? There is a 90% chance that the error range will be what?
Answer: There is a 90% chance that the error range will be with in 3.29 ft² .
Step-by-step explanation:
Given : The area of a parking lot is calculated to be 5,474 ft² .
Estimated standard deviation = 2 ft²
The critical z-value for 90% confidence interval is 1.645 (from z-table)
Then, the Maximum Anticipated Error = ( critical z-value ) x ( standard deviation )
= 1.645 (2) =3.29 ft²
i.e. Maximum Anticipated Error = 3.29 ft²
Hence, there is a 90% chance that the error range will be with in 3.29 ft² .
Of the air conditioner repair shops listed in a particular phone book, 87% are competent. A competent repair shop can repair an air conditioner 85% of the time; an incompetent shop can repair an air conditioner 60% of the time. Suppose the air conditioner was repaired correctly.
A. Find the probability that it was repaired by a competent shop, given that it was repaired correctly.
Answer:
There is a 90.46% probability that it was repaired by a competent shop, given that it was repaired correctly.
Step-by-step explanation:
We have these following probabilities:
An 87% probability that an air conditioner repair shop is competent.
A 13% probability that an air conditioner repair shop is incompetent.
An 85% probability that an compotent shop can repair the air.
A 60% probability than an incompetent shop can repair the air.
This can be formulated as the following problem:
What is the probability of B happening, knowing that A has happened?
It can be calculated by the following formula
[tex]P = \frac{P(B).P(A/B)}{P(A)}[/tex]
Where P(B) is the probability of B happening, P(A/B) is the probability of A happening knowing that B happened and P(A) is the probability of A happening.
In this problem we have that:
Probability that it was repaired by a competent shop, given that it was repaired correctly.
P(B) is the probability that it was repaired by a competent shop. 87% of the shops are competent, so [tex]P(B) = 0.87[/tex]
P(A/B) is the probability that it was repaired correctly, given that it was repaired by a competent shop. There is an 85% probability that an compotent shop can repair the air. So [tex]P(A/B) = 0.85[/tex]
P(A) is the probability that an air was repaired correctly.
This is 85% of 87% and 60% of 13%. So
[tex]P(A) = 0.85*0.87 + 0.60*0.13 = 0.8175[/tex]
Finally
[tex]P = \frac{P(B).P(A/B)}{P(A)} = \frac{0.87*0.85}{0.8175} = 0.9046[/tex]
There is a 90.46% probability that it was repaired by a competent shop, given that it was repaired correctly.
The mean cost of a meal for two in a mid-range restaurant in Tokyo is $40 (Numbeo.com website, December 14, 2014). How do prices for comparable meals in Hong Kong compare? The DATAfile HongKongMeals contains the costs for a sample of 42 recent meals for two in Hong Kong mid-range restaurants.
Data:
Meal Price ($)
22.78
33.89
22.77
18.04
23.29
35.28
42.38
36.88
38.55
41.68
25.73
34.19
31.75
25.24
26.32
19.57
36.57
32.97
36.83
30.17
37.29
25.37
24.71
28.79
32.83
43.00
35.23
34.76
33.06
27.73
31.89
38.47
39.42
40.72
43.92
36.51
45.25
33.51
29.17
30.54
26.74
37.93
a. With 95% confidence, what is the margin of error (to 2 decimals)?
b. What is the 95% confidence interval estimate of the population mean (to 2 decimals)?
c. How do prices for meals for two in mid-range restaurants in Hong Kong compare to prices for comparable meals in Tokyo restaurants?
a. To find the margin of error, we first calculate the standard error of the mean using the formula [tex]\(SE = \frac{s}{\sqrt{n}}\)[/tex], where s is the sample standard deviation and n is the sample size. Then, we use the formula for the margin of error [tex]\(ME = Z \times SE\)[/tex], where Z is the z-score corresponding to the desired level of confidence.
b. Once we have the margin of error, we can construct the confidence interval estimate of the population mean by adding and subtracting the margin of error from the sample mean.
c. To compare the prices for meals for two in mid-range restaurants in Hong Kong to those in Tokyo, we can use the confidence interval estimate of the population mean. If the confidence interval includes the mean price for Tokyo ($40), it suggests that there may not be a significant difference in prices between the two cities. However, if the confidence interval does not include $40, it suggests that there may be a significant difference in prices between the two cities.
Explanation:
a. Margin of error:
1. Calculate the sample mean [tex](\( \bar{x} \))[/tex] and sample standard deviation s from the given data.
2. Determine the sample size n.
3. Find the standard error of the mean SE using the formula [tex]\( SE = \frac{s}{\sqrt{n}} \)[/tex].
4. Look up the z-score corresponding to the desired level of confidence (e.g., 95%) from the standard normal distribution table.
5. Multiply the z-score by the standard error to find the margin of error [tex](\( ME \))[/tex].
b. Confidence interval estimate:
1. Calculate the margin of error ME.
2. Subtract the margin of error from the sample mean to find the lower bound of the confidence interval.
3. Add the margin of error to the sample mean to find the upper bound of the confidence interval.
c. Price comparison:
1. Check if the confidence interval estimate of the population mean includes the mean price for Tokyo ($40).
2. If the confidence interval includes $40, it suggests that there may not be a significant difference in prices between the two cities.
3. If the confidence interval does not include $40, it suggests that there may be a significant difference in prices between the two cities.
English and American spelling are rigour and rigor, respectively. A man staying at Al Rashid hotel writes this word, and a letter taken at random from his spelling is found to be a vowel. If 40 percent of the English-speaking men at the hotel are English and 60 percent are American, what is the probability that the writer is an Englishman
Answer:
If 40 percent of the English-speaking men at the hotel are English and 60 percent are American, the the probability that the writer is an Englishman
is 40% or 0.4.
Step-by-step explanation:
i) If 40 percent of the English-speaking men at the hotel are English and 60 percent are American, the the probability that the writer is an Englishman
is 40% or 0.4.
The probability that the writer is an Englishman given that a vowel was chosen from his spelling is approximately 45.45%. This was calculated using the respective probabilities of selecting a vowel from the words 'rigour' and 'rigor' and Bayes' Theorem.
Given the scenario where a man writes a word, and a letter taken at random is a vowel, we aim to determine the probability that the writer is an Englishman.
The words are:
British: rigourAmerican: rigorLetters 'i', 'o', and 'u' are the vowels. Let's calculate the likelihood of selecting a vowel from each spelling:
Rigour contains 3 vowels (i, o, u) out of 6 letters, P(vowel) = 3/6 = 0.5Rigor contains 2 vowels (i, o) out of 5 letters, P(vowel) = 2/5 = 0.4Using Bayes' Theorem, let's find the probability the writer is English (E), given a vowel (V) was selected:
P(E|V) = [P(V|E) * P(E)] / [P(V|E) * P(E) + P(V|A) * P(A)]
Where:
P(E) = 0.4 (probability of being English)P(A) = 0.6 (probability of being American)P(V|E) = 0.5 (probability of picking a vowel if English)P(V|A) = 0.4 (probability of picking a vowel if American)So:
P(E|V) = [0.5 * 0.4] / [0.5 * 0.4 + 0.4 * 0.6] = 0.2 / (0.2 + 0.24) = 0.2 / 0.44 ≈ 0.4545
Therefore, the probability that the writer is an Englishman is approximately 0.4545 or 45.45%.