Full Question
Eighteen telephones have just been received at an authorized service center. Six of these telephones are cellular, six are cordless, and the other six are corded phones. Suppose that these components are randomly allocated the numbers 1, 2, . . . , 18 to establish the order in which they will be serviced.
What is the probability that after servicing twelve of these phones, phones of only two of the three types remain to be serviced?
What is the probability that two phones of each type are among the first six serviced?
Answer:
a. 0.149
b. 0.182
Step-by-step explanation:
Given
Number of telephone= 18
Number of cellular= 6
Number of cordless = 6
Number of corded = 6
a.
There are 18C6 ways of choosing 6 phones
18C6 = 18564
From the Question, there are 3 types of telephone (cordless, Corded and cellular)
There are 3C2 ways of choosing 2 out of 3 types of television
3C2 = 3
There are 12C6 ways of choosing last 6 phones from just 2 types (2 types = 6 + 6 = 12)
12C6 = 924
There are 2 * 6C6 * 6C0 ways of choosing none from any of these two types of phones
2 * 6C6 * 6C0 = 2 * 1 * 1 = 2.
So, the probability that after servicing twelve of these phones, phones of only two of the three types remain to be serviced is
3 * (924 - 2) / 18564
= 3 * 922/18564
= 2766/18564
= 0.149
b)
There are 6C2 * 6C2 * 6C2 ways of choosing 2 cellular, 2 cordless, 2 corded phones
= (6C2)³
= 3375
So, the probability that two phones of each type are among the first six serviced is
= 3375/18564
= 0.182
The length of side AB is ....
Answer:
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Step-by-step explanation:
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Money Flow The rate of a continuous money flow starts at $1000 and increases exponentially at 5% per year for 4 years. Find the present value and accumulated amount if interest earned is 3.5% compounded continuously.
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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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.
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.
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.
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.
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:
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
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 .
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 switchboard display in the store allows a customer to hook together any selection of components (consisting of one of each type). Use the product rules to answer the following questions: a. In how many ways can one component of each type be selected
Answer:
120 ways
Step-by-step explanation: If there are three different categories of components, that is, component A, comprising four types; component B comprising, five types and component C comprising, six types. Three types are to be selected, that is, one type from each category. The number of ways one component of each type be selected is:
Component A * Component B * Component C = 4 X 5 X 6 = 120
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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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%.
A baseball enthusiast believes pitchers who strike out a lot of batters also walk a lot of batters. He reached this conclusion by going to the library and examining the records of all major league pitchers between 1990 and 1995. What type of study is his decision based on? A) B) C) An observational study based on a sample survey D) An experiment. Anecdotal evidence. An observational study based on available data.
Answer:
D) An observational study based on available data.
Step-by-step explanation:
This is an observational study based on available data.
If it had been on a sample, he would take a sample of a few pitchers, and not studied the statistics of all major league pitchers during those seasons.
It is not an anecdotal evidence, because an anecdotal evidence is something without study, just an impression.
It is not an experiment, because he just studies(observes, that is why it is an observational study) the data, he does not change anything about the pitchers.
So the correct answer is:
D) An observational study based on available data.
The baseball enthusiast's decision is based on an observational study based on available data.
Explanation:The baseball enthusiast's decision is based on A) an observational study based on available data. In this case, the enthusiast examined the records of all major league pitchers between 1990 and 1995. This observational study involved collecting and analyzing data that was already available, without manipulating any variables or conducting an experiment.
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.
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)
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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Name the quadrant in which the angle below is located.
sine = 4 and cose is positive.
Answer:
first quadrant
Step-by-step explanation:
To know this we have to take into account that for the breast we have to look at the y-axis and in the cosine we will look at the x-axis
We have the positive and the negative part of the x-axis, now we notice that it says that the cosine is positive, then the quadrant will be with the positive x, these quadrants can be the first or the fourth.
Now if we look at sine = 4 is positive, this means that the axis y will also be positive, this means that it will be between the first and second quadrant.
Now if we look to meet the conditions of the sine and cosine, the angle has to be found in the first quadrant
Vectors are quantities that possess both magnitude and direction. In engineering problems, it is best to think of vectors as arrows, and usually it is best to manipulate vectors using components. In this tutorial, we consider the addition of two vectors using both of these techniques. Consider two vectors AAA_evec and BBB_evec that have lengths AAA and BBB, respectively. Vector BBB_evec make
Answer:
Check below
Step-by-step explanation:
Vectors are quantities that possess both magnitude and direction. In engineering problems, it is best to think of vectors as arrows, and usually it is best to manipulate vectors using components. In this tutorial, we consider the addition of two vectors using both of these techniques. Consider two vectors [tex]\vec{A}[/tex]and [tex]\vec{B}[/tex] that have lengths A and B, respectively. Vector [tex]\vec{B}[/tex] makes an angle?
1) Vector [tex]\vec{B}[/tex] makes an angle?
Yes, it does. Vector [tex]\vec{B}[/tex] makes an angle with [tex]\vec{A}[/tex], since both have the same origin and different direction.
2) From the direction of A.(Figure 1)In vector notation, the sum is represented by [tex]\vec{C}=\vec{A}+\vec{B}[/tex] where [tex]\vec{C}[/tex] is a new vector that is the sum of [tex]\vec{A}[/tex] and [tex]\vec{B}[/tex]. Find C, the length of C, which is the sum of A and B.
C is the resultant vector of this sum of vectors([tex](\vec{C}=\vec{A}+\vec{B})[/tex]
The length of c is found through the law of cosines, after projecting, vector a.
(Check 3rd picture)
2.2) The other technique to add vectors is to write them. As C is the resultant vector then we have
[tex]\vec{A}=\left \langle a_{1},a_{2} \right \rangle \vec{B}=\left \langle b_{1},b_{2} \right \rangle \vec{C}=\left \langle a_{1}+b_{1}, a_{2}+b_{2}\right \rangle[/tex]
In physics, vectors are quantities possessing both magnitude and direction. They can be represented as arrows on a graph, with their length and direction corresponding to the vector's magnitude and direction. Vectors are added geometrically using the head-to-tail method or analytically, where they are broken down into components, and these components are added separately.
Explanation:This can be visualized as an arrow, where its length corresponds to the magnitude, and its direction is represented by the way the arrow points. Some examples of vectors include displacement, velocity, and force.
When adding vectors, there are two methods to consider: geometric or component wise method. The geometric method involves representing the vectors as arrows on a graph and adding them using the head-to-tail method. The component-wise method involves breaking down the vector into its x and y components, and adding these components separately.
For instance, if you have two vectors A and B, vector A can be broken down into its x and y components: A_x and A_y. Likewise, vector B can be broken down into B_x and B_y. Simply add corresponding components to yield the resultant vector, R_x = (A_x + B_x) and R_y = (A_y + B_y)
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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
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
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:
What is the present value of a $50,000 decreasing perpetuity beginning in one year if the discount rate is 7% and the payments decline by 4% annually?
Answer:
Present Value = $1666666.67
Step-by-step explanation:
Present Value of a Growing Perpuity is calculated using the following formula
PV =D/(r - g)
Where D = Dividend
r = Discount Rate
g = Growth rate
D = $50,000
r = 7%
r = 7/100
r = 0.07
g = 4%
g = 4/100
g = 0.04
PV = D/(r-g)
Becomes
PV = $50,000/(0.07-0.04)
PV = $50,000/0.03
PV = $1,666,666.67
So the Present Value of the perpuity is $1,666,666.67
Preston and Joel are both solving the equation 2x=14. Preston is not sure what to do because he does not know a power of 2 that equals 14. Joel uses his calculator to graph y=2x and y=14 and find the point of intersection. Will Joel's method work?
Answer:
yes
Step-by-step explanation:
You can always separate an equation into two parts and see where those graphs intersect.
Joel's method works well.
_____
Additional comments
Preston should know that the invention of logarithms makes it easy to solve equations like this. x = log₂(14) = log(14)/log(2) ≈ 3.8073549.
As for Joel's method, I prefer to subtract the right side to get the equation ...
2^x -14 = 0
Then graphing y = 2^x -14, I look for the x-intercept. Most graphing calculators make it easy to find x- and y-intercepts. Not all make it easy to find points of intersection between different curves.
Answer:
Yes, the graph intersects around (3.807,14), so 3.807 is a good estimate of the solution to 2^x=14.
Step-by-step explanation:
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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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.
A small business makes 3-speed and 10-speed bicycles at two different factories. Factory A produces 16 3-speed and 20 10-speed bikes in one day while factory B produces 12 3-speed and 20 10- speed bikes daily. It costs $1000/day to operate factory A and $800/day to operate factory B. An order for 80 3-speed bikes and 140 10-speed bikes has just arrived How many days should each factory be operated in order to fill this order at a minimum cost? (Give your answers correct to two decimal places.) Factory A should be operated Factory B should be operated days. days. What is the minimum cost? (Give your answer correct to the nearest dollar.)
Answer:
Factory A should be operated 6.11 days
Factory B should be operated 6.88 days
The minimum cost for factory A is $6,110
The minimum cost for factory B is $5,504
Step-by-step explanation:
Factory A
Daily production: 16 3-speed and 20 10-speed bikes
Total daily production = 16+20 = 36 speed bikes
Order: 80 3-speed and 140 10-speed bikes
Total order = 80+140 = 220 speed bikes
Number of days = 220/36 = 6.11 days
Cost per day = $1,000
Minimum cost for 6.11 days = $1000 × 6.11 = $6,110
Factory B
Daily production: 12 3-speed and 20 10-speed bikes
Total daily production = 12+20 = 32 speed bikes
Total order = 220 speed bikes
Number of days = 220/36 = 6.88 days
Cost per day = $800
Cost for 6.88 days = $800 × 6.88 = $5,504
The business should operate Factory A for about 2.92 days and Factory B for about 4.17 days to fill the order at a minimum cost of $7148.
Explanation:This problem can be solved using Linear Programming, a mathematical model used in optimization problems. Suppose we let A be the number of days factory A operates and B be the number of days factory B operates. The cost to operate the factories is given by the equation C = 1000A + 800B.
The constraints are as follows:
Factory A produces 16 3-speed bikes per day and factory B produces 12, so 16A + 12B ≥ 80. Both factories produce 20 10-speed bikes per day, so 20A + 20B ≥ 140.By solving these equations, you find that Factory A should be operated for about 2.92 days and Factory B should be operated for 4.17 days, giving a minimum cost of about $7148.
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Determine whether the relation R defined below is reflexive, irreflexive, symmetric, asymmetric, antisymmetric, or transitive. For each property, either explain why R has that property or give an example showing why it does not.
a) Let A = {1, 2, 3, 4} and let R = { (2, 3) }
b) Let A = {1, 2, 3, 4} and let R = { (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 4), (3, 1), (3, 3), (4, 1), (4, 4) }.
Answer:
See below
Step-by-step explanation:
Remember some definitions about binary relations. If R⊆S×S then
R is reflexive if (a,a)∈R for all a∈SR is irreflexive if (a,a)∉R for all a∈SR is symmetric if (a,b)∈R implies (b,a)∈R for all a,b∈SR is asymmetric if (a,b)∈R implies (b,a)∉R for all a,b∈SR is antisymmetric if (a,b)∈R and (b,a)∈R imply that a=b, for all a,b∈SR is transitive if (a,b)∈R and (b,c)∈R imply (a,c)∈R for all a,b,c∈Sa) R is not reflexive since (1,1)∉R.
R is irreflexive, since (a,a)∉R for all a=1,2,3,4
R is asymmetric: (2,3)∈R and (3,2)∉R (thus R is not symmetric).
R is antisymmetric, there are no cases to check. R is transitive, there are no cases to check.
b) R is reflexive, checking case by case, (a,a)∈R for all a=1,2,3,4. Hence R is not irreflexive.
R is not asymmetric: (1,2)∈R but (2,1)∈R. R is not symmetric, since (4,1)∈R but (1,4)∉R
R is not antisymmetric: (1,2)∈R and (2,1)∈R but 1≠2.
R is not transitive: (1,2)∈R and (2,4)∈R but (1,4)∉R.
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.
A construction company needs to remove 2 1/6 tons of dirt from a construction site. They can remove 710 tons of dirt each hour. What is the total number of hours it will take to remove the dirt?
Answer:
13/4260 tons
Step-by-step explanation:
We have the rate at which they remove tons of dirt per hour. We also know that total that needs to be removed. We can determine the time by dividing the amount of tons that need to be removed by the rate:
[tex]=(13/6)/\cdot{710}=13/4260[/tex]
will take 13/4260 hours to remove the dirt