Answer:
22%
Step-by-step explanation:
In the event of changing a test answer there are three possible outcomes, which should add up to 100%: changing from incorrect to correct (51%), changing from correct to incorrect (27%) and changing from incorrect to incorrect (X). Therefore, the percent of changes from incorrect to incorrect was:
[tex]100\% = 51\%+27\%+X\\X= 22\%[/tex]
22% of the changes were from incorrect to incorrect.
The time required to complete a project is normally distributed with a mean of 80 weeks and a standard deviation of 10 weeks. The construction company must pay a penalty if the project is not finished by the due date in the contract. If a construction company bidding on this contract wishes to be 90 percent sure of finishing by the due date, what due date (project week #) should be negotiated?
Answer: the due date would be 92 weeks
Step-by-step explanation:
Since the time required to complete a project is normally distributed, we would apply the formula for normal distribution which is expressed as
z = (x - µ)/σ
Where
x = number of weeks.
µ = mean
σ = standard deviation
From the information given,
µ = 80 weeks
σ = 10 weeks
If a construction company bidding on this contract wishes to be 90 percent sure of finishing by the due date, the z score corresponding to 90%(90/100 = 0.9) is 1.29
Therefore,
1.29 = (x - 80)/10
x - 80 = 1.2 × 10
x - 80 = 12
x = 80 + 12 = 92
To determine the due date the construction company should negotiate for a 90 percent confidence level of finishing on time, we use the z-score formula with the mean and standard deviation.
The negotiated due date is approximately 93 project week .
Explanation:To determine the due date the construction company should negotiate, we need to find the project week number that corresponds to being 90 percent sure of finishing by that time. To do this, we use the z-score formula to find the z-score associated with a 90 percent confidence level. We then use this z-score to find the corresponding project week number using the mean and standard deviation of the project time.
The z-score formula is: z = (X - μ) / σ, where X is the project week number, μ is the mean (80 weeks), and σ is the standard deviation (10 weeks).
By substituting the values into the formula, we get: z = (X - 80) / 10.
Next, using a z-table or a calculator, we find the z-score associated with a 90 percent confidence level, which is approximately 1.2816.
Substituting this value for z in the formula, we get: 1.2816 = (X - 80) / 10.
Now, we can solve for X by multiplying both sides of the equation by 10 and then adding 80 to both sides. This gives us: 12.816 = X - 80.
Finally, adding 80 to both sides of the equation, we find that the negotiated due date (project week #) should be approximately 92.816. However, since project week numbers are typically whole numbers, we can round the negotiated due date up to the nearest whole number, which is 93.
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What fraction of mowers fails the functional performance test using all the data in the worksheet Mower Test? Using this result, what is the probability of having x failures in the next 100 mowers tested, for x from 0 to 20?
Answer:
6/8
Step-by-step explanation:
The Fraction of mowers that fail the functional performance test is the ratio of the number of mowers that fail to the total number of mowers. Hence the fraction of mowers that fail [tex] \frac{9}{500}[/tex]
Probability of having x mower fail = 0.018x Number of mowers that fail = 54 Total number of mowers = 3000The probability of failure can be defined thus :
[tex] \frac{number \: of \: mowers \: that \: fail}{total \: number \: of \: mowers} [/tex]Fraction that failed = [tex] \frac{54}{3000}=\frac{9}{500} = 0.018[/tex]Therefore, the probability of having x failures in the next 100 mowers is (0.018x)
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A driving exam consists of 30 multiple-choice questions. Each of the 30 answers is either right or wrong. Suppose the probability that a student makes fewer than 6 mistakes on the exam is 0.28 and that the probability that a student makes from 6 to 20 (inclusive) mistakes is 0.53. Find the probability of each of the following outcomes. a. A student makes more than 20 mistakes b. A student makes 6 or more mistakes c. A student makes at most 20 mistakes d. Which two of these three events are complementary?
Final answer:
The probability of a student making more than 20 mistakes on the exam is 0.19, making 6 or more mistakes is 0.72, and making at most 20 mistakes is 0.81. Events making at most 20 mistakes and making more than 20 mistakes are complementary.
Explanation:
Given a driving exam consisting of 30 multiple-choice questions, we have the following probabilities:
The probability a student makes fewer than 6 mistakes: 0.28The probability a student makes from 6 to 20 mistakes (inclusive): 0.53Now, we can find the probabilities for each scenario using these probabilities:
A student makes more than 20 mistakes: We calculate this by subtracting the sum of the other two given probabilities from 1. The probability is 1 - (0.28 + 0.53) = 0.19.A student makes 6 or more mistakes: This is the sum of the probabilities of making from 6 to 20 mistakes and making more than 20 mistakes. The probability is 0.53 + 0.19 = 0.72.A student makes at most 20 mistakes: This is the complement of making more than 20 mistakes, so we already have this probability as the sum of making fewer than 6 mistakes and making from 6 to 20 mistakes. The probability is 0.28 + 0.53 = 0.81.The two complementary events are:
Event c: A student makes at most 20 mistakesEvent a: A student makes more than 20 mistakesThese events are complementary because their probabilities sum up to 1.
1. Identifying related samples Aa Aa E For each of the following research scenarios, decide whether the design uses a related sample. If the design uses a related sample, identify whether it uses matched subjects or repeated measures. (Note: Researchers can match subjects by matching particular characteristics, or, in some cases, matched subjects are naturally paired, such as siblings or married couples.) John Cacioppo was interested in possible mechanisms by which loneliness may have deleterious effects on health. He compared the sleep quality of a random sample of lonely people to the sleep quality of a random sample of nonlonely people. The design described You are interested in whether husbands or wives care more about how clean their cars are. You collect a random sample of 100 married couples and collect ratings from each partner, indicating the importance each places on car cleanliness. You want to know if the husbands ratings tend to be different than the wives ratings The design described
John Cacioppo's study uses independent samples while the car cleanliness study uses matched subjects because the samples are dependent as couples are paired.
Explanation:In two given scenarios, the research designs differ in whether they use related samples or not. Related samples refer to a study design where the same subjects are measured more than once, or their measures are compared to those of individuals with whom they share a distinct relationship (e.g., married couples).
In John Cacioppo's study, the research design uses independent samples. Participants were randomly assigned to either the group of lonely people or nonlonely people and the similarities or differences in sleep quality were observed. As there is no mention of matching based on particular characteristics, and individuals from one group are not directly linked to those in the other, these samples are not related.
In contrast, the car cleanliness study deals with matched subjects, as the samples are paired together based on marriage. The husbands' ratings are compared directly with their respective wives' ratings, giving rise to related, paired measurements within each couple.
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The correct answer for the first scenario is that the design uses a related sample with matched subjects. For the second scenario, the design uses a related sample with repeated measures.
Explanation for the first scenario:
In the first scenario, John Cacioppo is comparing the sleep quality of lonely people to nonlonely people. Since the researcher is comparing two different groups of people based on their loneliness status, this is an example of a matched subjects design. The subjects are matched on the characteristic of loneliness, meaning that each lonely individual is paired with a nonlonely individual who is similar in other relevant characteristics that could affect sleep quality, such as age, gender, or health status. This allows the researcher to control for these extraneous variables and focus on the effect of loneliness on sleep quality.
Explanation for the second scenario:
In the second scenario, the researcher is interested in the difference in car cleanliness importance ratings between husbands and wives. By collecting data from both partners within each of the 100 married couples, the researcher is using a repeated measures design. Each couple serves as their own control, with the husband and wife providing two sets of measurements (ratings) on the same variable (importance of car cleanliness). This design allows for the examination of within-subject differences, controlling for the influence of other variables that might differ between different families, such as socioeconomic status or cultural background.
In summary, the first scenario is an example of a related sample with matched subjects, while the second scenario is an example of a related sample with repeated measures.
If tangent alpha equals negative StartFraction 21 Over 20 EndFraction , 90degreesless thanalphaless than180degrees, then find the exact value of each of the following. a. sine StartFraction alpha Over 2 EndFraction b. cosine StartFraction alpha Over 2 EndFraction c. tangent StartFraction alpha Over 2 EndFraction
Answer:
α= 133.6 degrees
(a)Sin(α/2)=0.9191
(b)cos(α/2)=0.3939
(c)Tan(α/2)=2.3332
Step-by-step explanation:
If Tan α= [tex]-\frac{21}{20}[/tex]
90<α<180
We determine first the value of α in the first quadrant
α=[tex]Tan^{-1}\frac{21}{20}[/tex]
=46.4
Since 90<α<180
α=180-46.4=133.6 degrees
(a)Sin(α/2)=Sin(133.6/2)=Sin 66.8 =0.9191
(b)cos(α/2)=cos(133.6/2)=cos 66.8 =0.3939
(c)Tan(α/2)=Tan(133.6/2)=Tan 66.8 =2.3332
What is the area of the shaded segment?
Answer: option 3 is the correct answer.
Step-by-step explanation:
The formula for determining the area of a sector is expressed as
Area = θ/360 × πr²
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 = 4 miles
θ = 120°
Area of sector = 120/360 × 3.14 × 4²
= 16.75 square miles
To determine the area of ∆ABC, we would apply the formula,
Area of triangle = 1/2abSinC
From the information given,
a = 4
b = 4
C = 120°
Therefore,
Area = 1/2 × 4 × 4 × Sin120
Area = 8 × Sin120 = 6.92 square miles. Therefore area of the segment is
16.75 - 6.92 = 9.83 square miles
The margarita is one of the most common tequila-based cocktails, made with tequila mixed with triple sec and lime or lemon juice, often served with salt on the glass rim. A common ratio for a margarita is 2:1:1, which includes 50% tequila, 25% triple sec, and 25% fresh lime or lemon juice. A manager at a local bar is concerned that the bartender uses incorrect proportions in more than 50% of margaritas. He secretly observes the bartender and finds that he used the correct proportions in only 10 out of 30 margaritas. Test if the manager’s suspicion is justified at α = 0.05. Let p represent incorrect population proportion.
Answer:
The null hypothesis was rejected.
Conclusion: The bartender uses incorrect proportions in more than 50% of margaritas.
Step-by-step explanation:
The hypothesis for this test can be defined as:
H₀: The bartender uses incorrect proportions in less than 50% of margaritas, i.e. p < 0.50.
Hₐ: The bartender uses incorrect proportions in more than 50% of margaritas, i.e. p > 0.50.
Given:
[tex]\hat p=\frac{10}{30} =0.33\\n=30[/tex]
The test statistic is:
[tex]z=\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n}} }=\frac{0.33-0.50}{\sqrt{\frac{0.50(1-0.50)}{30}}} =-1.862[/tex]
Decision Rule:
If the p-value of the test statistic is less than the significance level, α = 0.05, then the null hypothesis is rejected.
The p-value of the test is:
[tex]p-value=P(Z<-1.86) = 0.0314[/tex]
*Use the z-table.
The p-value = 0.0314 < α = 0.05.
The null hypothesis will be rejected.
Conclusion:
As the null hypothesis was rejected it can be concluded that the bartender uses incorrect proportions in more than 50% of margaritas.
Using the z-distribution, it is found that since the test statistic is less than the critical value for the left-tailed test, there is enough evidence to conclude that the manager's suspicion is correct.
At the null hypothesis, we test if the manager's suspicion is incorrect, that is, the proportion is correct in at least 50% of the margaritas.
[tex]H_0: p \geq 0.5[/tex]
At the alternative hypothesis, we test if the suspicion is correct, that is, the proportion is less than 50%.
[tex]H_1: p < 0.5[/tex]
The test statistic is given by:
[tex]z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}[/tex]
In which:
[tex]\overline{p}[/tex] is the sample proportion. p is the proportion tested at the null hypothesis. n is the sample size.For this problem, the parameters are: [tex]p = 0.5, n = 30, \overline{p} = \frac{10}{30} = 0.3333[/tex]
Then, the value of the test statistic is:
[tex]z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}[/tex]
[tex]z = \frac{0.3333 - 0.5}{\sqrt{\frac{0.5(0.5)}{30}}}[/tex]
[tex]z = -1.83[/tex]
The critical value for a left-tailed test, as we are testing if the mean is less than a value, with a significance level of 0.05, is [tex]z^{\ast} = -1.645[/tex].
Since the test statistic is less than the critical value for the left-tailed test, there is enough evidence to conclude that the manager's suspicion is correct.
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A is a finite non-empty set. The domain for relation R is the power set of A . (Recall that the power set of A is the set of all subsets of A .) For X⊆A and Y⊆A , X is related to Y if X is a proper subsets of Y (i.e., X⊂Y ). Select the description that accurately describes relation R .
Final answer:
Relation R defines a specific type of relationship within set theory, where subsets of set A are related if one is a proper subset of the other within the power set of A.
Explanation:
The student's question pertains to the concept of a relation in set theory, particularly concerning the power set of a given finite non-empty set A. In this context, the domain for the relation R is the power set of A, which includes all subsets of A. The relation R is defined such that if X and Y are subsets of A, then X is related to Y if and only if X is a proper subset of Y, denoted as X ⊂ Y. This means that every element in subset X is also contained in subset Y, and Y contains at least one additional element that is not in X.
To illustrate this, consider a simple set A = {1, 2}. Its power set, which is the domain of R, will include { }, {1}, {2}, and {1, 2}. In this case, the set {1} is a proper subset of {1, 2}, since it contains all elements of {1} and A contains an additional element, which is 2. Hence, the ordered pair ({1}, {1, 2}) is part of the relation R.
You plan to construct a confidence interval for the mean \muμ of a Normal population with (known) standard deviation \sigmaσ. Which of the following will reduce the size of the margin of error? Group of answer choices Use a lower level of confidence. Increase the sample size. Reduce \sigma σ . All of the answers are correct.
Answer: increase the sample size
Step-by-step explanation: the margin of error for any confidence interval is given by the formulae below.
Margin of error = critical value × standard deviation/√n
From what we can see the critical value and standard deviation are constants, the only variables here are the margin of error and sample size which are inversely proportional to each other, that is the margin of error is inversely proportional to the square of the sample size.
Hence, reducing the sample size will increase the margin of error while increasing the sample size will reduce the margin of error.
Disadvantages of using a related sample (either one sample of participants with repeated measures or two matched samples) versus using two independent samples include which of the following? Check all that apply.
A study that uses related samples to compare two drugs (specifically, one sample of participants with repeated measures) can have a carryover and/or order effect such that the efects of the drug taken before the first measurement may not wear off before the second measurement.
Related samples (specifically, one sample of participants with repeated measures) can have an order effect such that a change observed between one measurement and the next might be attributable to the order in which the measurements were taken rather than to a treatment effect.
Related samples have less sample variance, increasing the likelihood of rejecting the null hypothesis if it is false (that is, increasing power). caused by individual differend gender, or personality.
Answer: The disadvantage of using a related sample is that "A study that uses related samples to compare two drugs (specifically, one sample of participants with repeated measures) can have a carryover and/or order effect such that the efects of the drug taken before the first measurement may not wear off before the second measurement".
Step-by-step explanation: Related sample is when a particular sample or two sample that are the same is used to study an effect.
The carryover effect is one of the major disadvantages of using a related sample. Because when a first treatment is applied, their is a tendency of it not being fully consumed or it's effect not being fully neutralized before the second treatment is applied. This will increase error in the result, because in a drugs intake for instance, the drug taken at a particular time may not be the one that cure a sickness, but the drug that was previously taken. But the study will assume the drug taken a that particular moment to be the cure of that sickness.
Answer:
Related samples (especially, one sample of Participants with repeated measures) can have a carryover effect such that Participants can leam from their first measurement and therefore do better on their second Measurement.
Explanation:
Related samples/groups (i.e., dependent measurements) The subjects in each specimen, or organization, are identical. This indicates that the subjects in the first group are also in the second group.
Various Disadvantages of using a related sample versus using two independent samples:
A study that uses related samples to compare two drugs (specifically, one sample of participants with repeated measures) can have a carryover and/or order effect such that the effects of the drug taken before the first measurement may not wear off before the second measurement.Related samples (specifically, one sample of participants with repeated measures) can have a carryover effect such that participants can learn from their first measurement and therefore do better on their second measurement.Thus we can say using a related sample versus using two independent samples has various disadvanages.
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https://brainly.com/question/16786364https://brainly.com/question/14305647Paul Hilseth plans to invest $4,780. Find the interest rate required for the fund to grow to $5,138.50 in 15 months.
A 150-day loan for $12,000 has interest of $375.
Find the rate to the nearest tenth of a percent.
Bill earned $40 interest on a $6,400 deposit in an account paying 6%.
Find the number of days that the funds were on deposit. Round to the nearest day
Answer:
5.96%
7.6%
38 days
Step-by-step explanation:
15 months = 1.25 years
5138.5 = 4780 × R^1.25
R^1.25 = 43/40
1.25 lgR = lg(43/40)
lg R = 0.0251267714
R = 1.059562969
R - 1 = 0.059562969
Interest: 5.96%
(150/365) × (r/100) × 12000 = 375
r/100 = 0.0760416667
r = 7.6%
40 = (n/365) × (6/100) × 6400
(n/365) = 5/48
n = 38.02083333 = 38
A standard piece of paper is 0.05 mm thick. Let's imagine taking a piece of paper and folding the paper in half multiple times. We'll assume we can make "perfect folds," where each fold makes the folded paper exactly twice as thick as before - and we can make as many folds as we want. Write a function g g that determines the thickness of the folded paper (in mm) in terms of the number folds made, n n. (Notice that g ( 0 )
Answer:
g(n) = 0.05·2^n
Step-by-step explanation:
The paper with no folds is 0.05 mm thick. Each fold multiplies the thickness by 2, so the function is ...
g(n) = 0.05·2^n
_____
Comment on paper folding
In practice, where the paper must bend around the fold, it is impossible to fold an ordinary piece of paper 9 times. You may be able to fold a very large, very thin piece of paper that many times.
The thickness of a standard piece of paper, after it has been folded n times, can be expressed as g(n) = 0.05 * 2^n. This formula takes into account that each fold doubles the thickness of the paper.
Explanation:In order to calculate the thickness of a standard piece of paper after it has been folded a certain number of times, we may use the concept of exponential growth. In this case, we use a power of 2 corresponding to the number of folds made, because with each fold, the thickness doubles. This gives us the function g(n) = 0.05 * 2^n, where n corresponds to the number of folds.
For example, if we fold the paper once (n=1), we get a thickness of 0.05 * 2^1 = 0.10 mm. If we fold the paper twice (n=2), we get a thickness of 0.05 * 2^2 = 0.20 mm, and so on, with the thickness doubling each time we make a new fold.
This mathematical model assumes perfect folding and does not account for physical limitations we would encounter when attempting to fold paper in reality.
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In the beginning of the study, a randomly selected group of 121 students scored an average of 252 words per minute on the reading speed test. Since the sample size is larger than 30, the cognitive psychologist can assume that the sampling distribution of M
Answer:
Since the sample size is larger than 30, the cognitive psychologist can assume that the sampling distribution of M will be approximately normal.
Step-by-step explanation:
We use the central limit theorem to solve this question.
The Central Limit Theorem estabilishes that, for a random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], a sample size larger than 30 can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex]
So
Since the sample size is larger than 30, the cognitive psychologist can assume that the sampling distribution of M will be approximately normal.
Today, your carpentry shop must produce 330 chair railings where each railing is made from a single piece of fine wood. Your three step manufacturing process has the following scrap rates for the corresponding step: Step 1: 1.1% Step 2: 1.1% Step 3: 1.6% How many pieces of wood must you start with to have 330 railings that can be sold
Answer:
343 pieces of wood.
Step-by-step explanation:
In order to determine the starting number of pieces of wood that yield 330 railings, the easiest course of action is to work your way back through step 3, step 2, and step 1. The number of pieces that must arrive at each step are given by:
[tex]n_f=330\\n_3 = \frac{330}{1-0.016}\\n_2 = \frac{n_3}{1-0.011}\\n_1 = \frac{n_2}{1-0.011} \\\\n_3 = 335.3658\\n_2=339.09591\\n_1=342.86745=343[/tex]
Rounding up to the next whole piece, 343 pieces are needed.
Therefore, you must start with 343 pieces of wood to have 330 railings that can be sold.
Two resistors, with resistances R1 and R2, are connected in series. R1 is normally distributed with mean 100 ohms and standard deviation 5 ohms, and R2 is normally distributed with mean 120 ohms and standard deviation 10 ohms. Assume that the resistors are independent. Compute P (R subscript 2 greater than R subscript 1 ). Round your answer to four decimal places.
The probability that [tex]\( R_2 \)[/tex] exceeds[tex]\( R_1 \)[/tex] by more than 30Ω in series connection is approximately 0.04%, calculated using their normal distributions' means, standard deviations, and the difference's Z-score.
To find the probability that[tex]\( R_2 \)[/tex] exceeds [tex]\( R_1 \)[/tex] by more than 30Ω when they are connected in series, we can use the properties of normal distributions and their differences.
Given:
[tex]\( R_1 \)[/tex] has a mean [tex]\( \mu_1 = 1000 \)[/tex] and standard deviation [tex]\( \sigma_1 = 50 \).[/tex]
[tex]\( R_2 \)[/tex] has a mean[tex]\( \mu_2 = 1202 \)[/tex] and standard deviation [tex]\( \sigma_2 = 10.2 \)[/tex].
The difference [tex]\( R_2 - R_1 \)[/tex] will follow a normal distribution with the mean[tex]\( \mu_2 - \mu_1 = 1202 - 1000 = 202 \)[/tex] and the standard deviation [tex]\( \sqrt{(\sigma_1)^2 + (\sigma_2)^2} = \sqrt{50^2 + 10.2^2} \).[/tex]
Now, we want to find the probability that [tex]\( R_2 - R_1 > 30 \).[/tex]
Using Z-score:
[tex]\[ Z = \frac{X - \mu}{\sigma} \][/tex]
[tex]\[ Z = \frac{30 - 202}{\sqrt{50^2 + 10.2^2}} \][/tex]
[tex]\[ Z = \frac{-172}{\sqrt{2500 + 104.04}} \][/tex]
[tex]\[ Z = \frac{-172}{\sqrt{2604.04}} \][/tex]
Z ≈ -172 / 51.02 ≈ -3.37
Consulting a standard normal distribution table or calculator for the probability associated with[tex]\( Z = -3.37 \)[/tex] we find that the probability is approximately 0.0004 or 0.04%.
Therefore, the probability that \( R_2 \) exceeds \( R_1 \) by more than 30Ω when they are connected in series is approximately 0.04%.
The probability [tex]\( P(R_2 > R_1) \approx 0.9631 \)[/tex], found by standardizing and using the normal distribution table.
To compute [tex]\( P(R_2 > R_1) \)[/tex], we need to find the probability that the resistance [tex]\( R_2 \)[/tex] is greater than [tex]\( R_1 \)[/tex]. Given that [tex]\( R_1 \) and \( R_2 \)[/tex] are normally distributed and independent, we can use properties of the normal distribution.
First, we define the random variables:
- [tex]\( R_1 \sim N(100, 5^2) \)[/tex]
- [tex]\( R_2 \sim N(120, 10^2) \)[/tex]
We are interested in the distribution of [tex]\( R_2 - R_1 \)[/tex].
1. Find the mean and standard deviation of [tex]\( R_2 - R_1 \)[/tex]:
- The mean of [tex]\( R_2 - R_1 \)[/tex]:
[tex]\[ \mu_{R_2 - R_1} = \mu_{R_2} - \mu_{R_1} = 120 - 100 = 20 \][/tex]
- The variance of [tex]\( R_2 - R_1 \)[/tex]:
[tex]\[ \sigma_{R_2 - R_1}^2 = \sigma_{R_2}^2 + \sigma_{R_1}^2 = 10^2 + 5^2 = 100 + 25 = 125 \][/tex]
- The standard deviation of [tex]\( R_2 - R_1 \)[/tex]:
[tex]\[ \sigma_{R_2 - R_1} = \sqrt{125} = 5\sqrt{5} \][/tex]
2. Standardize the variable [tex]\( R_2 - R_1 \)[/tex]:
We want to find [tex]\( P(R_2 > R_1) \)[/tex], which is equivalent to [tex]\( P(R_2 - R_1 > 0) \)[/tex].
Standardize [tex]\( R_2 - R_1 \)[/tex] by subtracting the mean and dividing by the standard deviation:
[tex]\[ Z = \frac{R_2 - R_1 - \mu_{R_2 - R_1}}{\sigma_{R_2 - R_1}} = \frac{R_2 - R_1 - 20}{5\sqrt{5}} \][/tex]
We want to find:
[tex]\[ P(R_2 - R_1 > 0) = P\left(\frac{R_2 - R_1 - 20}{5\sqrt{5}} > \frac{0 - 20}{5\sqrt{5}}\right) \][/tex]
Simplify the right-hand side:
[tex]\[ P\left(Z > \frac{-20}{5\sqrt{5}}\right) = P\left(Z > \frac{-20}{5 \cdot 2.2361}\right) = P\left(Z > \frac{-20}{11.1803}\right) = P\left(Z > -1.7889\right) \][/tex]
3. Find the probability:
- Using standard normal distribution tables or a calculator, find the probability [tex]\( P(Z > -1.7889) \)[/tex].
- The standard normal distribution table provides the cumulative probability for [tex]\( Z \leq z \)[/tex]. For [tex]\( Z > -1.7889 \)[/tex]:
[tex]\[ P(Z > -1.7889) = 1 - P(Z \leq -1.7889) \][/tex]
Using standard normal distribution tables or a calculator:
[tex]\[ P(Z \leq -1.7889) \approx 0.0369 \][/tex]
Therefore:
[tex]\[ P(Z > -1.7889) = 1 - 0.0369 = 0.9631 \][/tex]
So, the probability that [tex]\( R_2 \)[/tex] is greater than [tex]\( R_1 \)[/tex] is approximately [tex]\( 0.9631 \)[/tex].
Consider a family with 4 children. Assume the probability that one child is a boy is 0.5 and the probability that one child is a girl is also 0.5, and that the events "boy" and "girl" are independent.
(a) List the equally likely events for the gender of the 4 children, from oldest to youngest. (Let M represent a boy (male) and F represent a girl (female). Select all that apply.) MMFF, FFFF, MMMF, two M's two F's, MFFF, FMMM, FFMF, FMFF, three M's one F, FFFM, MFFM, MFMF, one M three F's, FMFM, FMMF, MMFM, MMMM, FFMM, MFMM
(b) What is the probability that all 4 children are male? (Enter your answer as a fraction.) Incorrect: Your answer is incorrect. Notice that the complement of the event "all four children are male" is "at least one of the children is female." Use this information to compute the probability that at least one child is female. (Enter your answer as a fraction.)
Answer:
a) Total 16 possibilities
MMMM
FFFF
MMMF
MMFM
MFMM
FMMM
FFFM
FFMF
FMFF
MFFF
MMFF
MFMF
MFFM
FFMM
FMMF
FMFM
b) P(MMMM) = 1/16
Final answer:
The equally likely gender combinations for 4 children are listed by considering all possibilities, including MMMM, MMMF, and so on. The probability that all 4 children are male is 0.0625, or 1/16 as a fraction. The probability of having at least one female child is the complement, 0.9375 or 15/16 as a fraction.
Explanation:
Listing the Equally Likely Gender Combinations
To list the equally likely events for the gender of the 4 children in a family where 'M' represents a male child and 'F' represents a female child, consider all possible combinations. These combinations are: MMMM, MMMF, MMFM, MMFF, MFMM, MFMF, MFFM, MFFF, FMMM, FMMF, FMFM, FMFF, FFMM, FFMF, FFMM, FFFF.
Probability of All Male Children
To calculate the probability that all 4 children are male, note that the events are independent, and the probability of each child being male is 0.5. Since the events are independent, multiply the probabilities for each child: 0.5 * 0.5 * 0.5 * 0.5 = 0.0625 or 1/16 as a fraction.
Alternatively, the complement of having all male children is having at least one female child. The probability of at least one female child can be found by subtracting the probability of all male children from 1: 1 - 0.0625 = 0.9375 or 15/16 as a fraction.
Ten samples of a process measuring the number of returns per 100 receipts were taken for a local retail store. The number of returns were 10, 9, 11, 7, 3, 12, 8, 4, 6, and 11. Find the standard deviation of the sampling distribution for the p-bar chart. 0.0863 0.081 0.0273 There is not enough information to answer the question. 8.1
Answer:
0.0273
Step-by-step explanation:
np n
10 100
9 100
11 100
7 100
3 100
12 100
8 100
4 100
6 100
11 100
pbar=sumnp/sumn
pbar=10+9+11+7+3+12+8+4+6+11/10+10+10+10+10+10+10+10+10+10
pbar=81/1000
pbar=0.081
nbar=sumn/k=1000/10=100
[tex]Standard deviation for pbar chart=\sqrt{\frac{pbar(1-pbar)}{nbar} }[/tex]
[tex]Standard deviation for pbar chart=\sqrt{\frac{0.081(0.919)}{100} }[/tex]
[tex]Standard deviation for pbar chart=\sqrt{\frac{0.0744}{100} }[/tex]
[tex]Standard deviation for pbar chart=\sqrt{0.0007444 }[/tex]
Standard deviation for p-chart=0.0273
PLEASE HELP ONLY IF RIGHT 69 points, brainliest, 5 stars, and thank you.
A prop for the theater club’s play is constructed as a cone topped with a half-sphere. What is the volume of the prop? Round your answer to the nearest tenth of a cubic inch. Use 3.14 to approximate pi, and make certain to show your work. Hint: you may need to find the volume of the component shapes.
Answer: 804.25cm
Step-by-step explanation:
V=πr2h/3
pi= 3.14
r=8cm
h= 12cm
Slot the values
3.14× (8x8) × 12/3
3.14 × 64 x 4
V= 804.25cm
g Students conducted a survey and found out that 36% of their peers on campus had tattoos but only 4% of their peers were smokers. If 100 students were surveyed, can these students use the Normal approximation to construct a confidence interval for the proportion of students in the population who are smokers? No, because either n p np or n ( 1 − p ) n(1−p) are greater than 15. Yes, because both n p np and n ( 1 − p ) n(1−p) are greater than 15. Yes, because both n p np and n ( 1 − p ) n(1−p) are less than 15. No, because either n p np or n ( 1 − p ) n(1−p) are less than 15.
Answer:
D) No, because either np or n(1−p) are less than 15.
Step-by-step explanation:
Percentage of students who had tattoos = 36%
Percentage of students who were smokers = 4%
Sample size = n = 100
The condition to use the Normal distribution as an approximation to construct the confidence interval for population proportion is:
Both np and n(1-p) must be equal to or greater than 15.
Since, we are interested in smokers only, so p = 4% = 0.04
np = 100 x 0.04 = 4
n (1 - p) = 100 x 0.96 = 96
Since, np < 15, we cannot use the Normal distribution as an approximation here.
Therefore, the correct answer is:
No, because either np or n(1−p) are less than 15.
A pair of fair dice is rolled. (a) What is the probability that both dice show the same number? (b) What is the probability that both dice show different numbers? (c) What is the probability that the second die lands on a lower value than does the first?
Answer:
a) 16.7% probability that both dice show the same number
b) 83.3% probability that both dice show different numbers
c) 41.67% probability that the second die lands on a lower value than does the first.
Step-by-step explanation:
A probability is the number of desired outcomes divided by the number of total outcomes.
In this problem, we have these possible outcomes:
Format(Dice A, Dice B)
(1,1), (1,2), (1,3), (1,4), (1,5),(1,6)
(2,1), (2,2), (2,3), (2,4), (2,5),(2,6)
(3,1), (3,2), (3,3), (3,4), (3,5),(3,6)
(4,1), (4,2), (4,3), (4,4), (4,5),(4,6)
(5,1), (5,2), (5,3), (5,4), (5,5),(5,6)
(6,1), (6,2), (6,3), (6,4), (6,5),(6,6)
There are 36 possible outcomes.
(a) What is the probability that both dice show the same number?
(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)
6 outcomes in which both dice show the same number.
6/36 = 0.167
16.7% probability that both dice show the same number
(b) What is the probability that both dice show different numbers?
The other 30 outcomes
30/36 = 0.833
83.3% probability that both dice show different numbers
(c) What is the probability that the second die lands on a lower value than does the first?
(2,1)
(3,1), (3,2)
(4,1), (4,2), (4,3)
(5,1), (5,2), (5,3), (5,4)
(6,1), (6,2), (6,3), (6,4), (6,5)
15 outcomes in which the second die lands on a lower value than does the first.
15/36 = 0.4167
41.67% probability that the second die lands on a lower value than does the first.
The weights of newborn baby boys born at a local hospital are believed to have a normal distribution with a mean weight of 4004 grams and a variance of 103,684. If a newborn baby boy born at the local hospital is randomly selected, find the probability that the weight will be less than 4293 grams. Round your answer to four decimal places.
Answer:
0.8151 is the probability that the weight will be less than 4293 grams.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 4004 grams
Variance = 103,684
[tex]\sigma = \sqrt{103684} = 322[/tex]
We are given that the distribution of weight of newborn baby is a bell shaped distribution that is a normal distribution.
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
P(weight will be less than 4293 grams)
P(x < 4293)
[tex]P( x < 4293) \\\\= P( z < \displaystyle\frac{4293 - 4004}{322}) \\\\= P(z < 0.8975)[/tex]
Calculation the value from standard normal z table, we have,
[tex]P(x < 4293) =0.8151 = 81.51\%[/tex]
0.8151 is the probability that the weight will be less than 4293 grams.
Answer:
Probability that the weight will be less than 4293 grams is 0.8133.
Step-by-step explanation:
We are given that weights of newborn baby boys born at a local hospital are believed to have a normal distribution with a mean weight of 4004 grams and a variance of 103,684.
Let X = weight of newborn baby boys
So, X ~ N([tex]\mu =4004,\sigma^{2}=322^{2}[/tex])
The z score probability distribution is given by;
Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = population mean
[tex]\sigma[/tex] = population standard deviation
(a) Probability that weight will be less than 4293 grams is given by = P(X < 4293 grams)
P(X < 4293) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{4293-4004}{322 }[/tex] ) = P(Z < 0.89) = 0.8133
Therefore, if a newborn baby boy born at the local hospital is randomly selected, probability that the weight will be less than 4293 grams is 0.8133.
what are the two important pieces of the polynomial to find end behavior?
Which of the following are variables are categorical, if any? (check all that apply)
A. The brand of cars driven by Harvard instructors
B. The age of randomly selected biology students.
C. The types of plants growing in yellowstone
Answer:
A) The brand of cars driven by Harvard instructors
C) The types of plants growing in yellowstone
Step-by-step explanation:
Categorical and numerical variables:
Numerical data can be expressed with the help of numerical.Categorical data cannot be expressed by numerical.They do no have any particular value.They are the non-parametric data.Categorical data are also known as qualitative data.A) The brand of cars driven by Harvard instructors
Since the brands cannot be expressed in numerical, it is a categorical variable.
B) The age of randomly selected biology students.
The age of student are expressed in whole numbers or decimals. Thus, it is a numerical data.
C) The types of plants growing in yellowstone
Again, the type of plants cannot be expressed with the help of numerical. Thus, it is a categorical variable.
Variables A ('The brand of cars driven by Harvard instructors') and C ('The types of plants growing in Yellowstone') are categorical. Variable B ('The age of randomly selected biology students') is numerical.
Explanation:In statistical terms, variables are characteristics or properties that can take different values or categories. They can be either categorical (qualitative) or numerical (quantitative).
For the variables provided:
A. The brand of cars driven by Harvard instructors - This is a categorical variable, as car brands are qualitative characteristics that can be placed in different categories.B. The age of randomly selected biology students - This is a numerical variable, as ages are measured in numbers and can be quantitatively analyzed.C. The types of plants growing in Yellowstone - This is another categorical variable, as plant types are qualitative characteristics that can be categorized.Learn more about Categorical Variables here:https://brainly.com/question/34199292
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Please help!!!
If Jimmy invests $250 twice a year at 4% compounded semi-annually, how much will his investment be worth after 3 years?
Please give a step by step!
Answer: his investment will be worth $1606.5 after 3 years
Step-by-step explanation:
We would apply the formula for determining future value involving deposits at constant intervals. It is expressed as
S = R[{(1 + r)^n - 1)}/r][1 + r]
Where
S represents the future value of the investment.
R represents the regular payments made(could be weekly, monthly)
r = represents interest rate/number of interval payments.
n represents the total number of payments made.
From the information given,
R = $250
r = 0.04/2 = 0.002
n = 2 × 3 = 6
Therefore,
S = 250[{(1 + 0.02)^6 - 1)}/0.02][1 + 0.02]
S = 250[{(1.02)^6 - 1)}/0.02][1.02]
S = 250[{(1.126 - 1)}/0.02][1.02]
S = 250[{0.126}/0.02][1.02]
S = 250[6.3][1.02]
S = 250 × 6.426
S = $1606.5
The results of a national survey showed that on average, adults sleep 6.7 hours per night. Suppose that the standard deviation is 1.8 hours. (a) Use Chebyshev's theorem to calculate the minimum percentage of individuals who sleep between 3.1 and 10.3 hours. % (b) Use Chebyshev's theorem to calculate the minimum percentage of individuals who sleep between 2.2 and 11.2 hours. % (c) Assume that the number of hours of sleep follows a bell-shaped distribution. Use the empirical rule to calculate the percentage of individuals who sleep between 3.1 and 10.3 hours per day. % How does this result compare to the value that you obtained using Chebyshev's theorem in part (a)
Answer:
a) 75%
b) 84%
c) 95%
Step-by-step explanation:
We are given the following in the question:
Mean, μ = 6.7 hours
Standard Deviation, σ = 1.8 hours
Chebyshev's Theorem:
According to this theorem atleast [tex]1-\dfrac{1}{k^2}[/tex] percent of the data lies within k standard deviation of mean.Empirical Formula:
According to this rule almost all the data lies within three standard deviation of the mean for a normally distributed data.68% of the data lies within one standard deviation of the mean.About 95% of the data lies within two standard deviation from the mean.99.7% of the data lies within three standard deviation of the mean.a) minimum percentage of individuals who sleep between 3.1 and 10.3 hours
[tex]10.3 = 6.7 + 2(3.1) = \mu + 2(\sigma)\\3.1 = 6.7 - 2(3.1) = \mu - 2(\sigma)[/tex]
Minimum percentage:
[tex]1-\dfrac{1}{4} = 75%[/tex]
Thus, minimum 75% of individuals who sleep between 3.1 and 10.3 hours.
b) minimum percentage of individuals who sleep between 2.2 and 11.2 hours.
[tex]11.2 = 6.7 + 2.5(3.1) = \mu + 2.5(\sigma)\\2.2 = 6.7 - 2.5(3.1) = \mu - 2.5(\sigma)[/tex]
Minimum percentage:
[tex]1-\dfrac{1}{(2.5)^2} = 84\%[/tex]
Thus, minimum 84% of individuals who sleep between 2.2 and 11.2 hours.
c) percentage of individuals who sleep between 3.1 and 10.3 hours per day
According to Empirical rule about 95% of the data lies within 2 standard deviations from the mean.
Thus, 95% of individuals who sleep between 3.1 and 10.3 hours.
d) Comparison with Chebyshev's theorem
This is greater than the results obtained from the Chebyshev's theorem in part (a)
According to Chebyshev's theorem, the minimum percentage of individuals who sleep between 3.1 and 10.3 hours is approximately 93.75%. This minimum percentage also applies to individuals who sleep between 2.2 and 11.2 hours. However, using the empirical rule, approximately 95% of individuals are expected to sleep between 3.1 and 10.3 hours per day. The result from the empirical rule suggests that the distribution of sleep hours is closer to a normal distribution than what Chebyshev's theorem predicts.
Explanation:(a) According to Chebyshev's theorem, at least 75% of the individuals will sleep between 3.1 and 10.3 hours. The formula for Chebyshev's theorem is P(X - μ < kσ), where μ is the mean, σ is the standard deviation, and k is the number of standard deviations from the mean.
Calculate the range: 10.3 - 3.1 = 7.2 hoursCalculate the number of standard deviations from the mean: k = 7.2 / 1.8 = 4Apply Chebyshev's theorem: P(X - μ < 4σ) = 1 - 1/k2 = 1 - 1/42 = 1 - 1/16 = 15/16 = 0.9375Convert the decimal to percentage: 0.9375 * 100% = 93.75%(b) Using the same steps as in part (a), the minimum percentage of individuals who sleep between 2.2 and 11.2 hours is also approximately 93.75%.
(c) According to the empirical rule, for a normal distribution, approximately 68% of the individuals will sleep between μ - σ and μ + σ, and approximately 95% will sleep between μ - 2σ and μ + 2σ. Since the interval 3.1 and 10.3 hours falls within 2 standard deviations from the mean, we can estimate that around 95% of the individuals will sleep between 3.1 and 10.3 hours per day.
The result obtained using the empirical rule in part (c) is slightly higher than the value obtained using Chebyshev's theorem in part (a), indicating that the distribution of sleep hours is closer to a normal distribution than a general distribution predicted by Chebyshev's theorem.
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How would you "remove the discontinuity" of f ? In other words, how would you define f(3) in order to make f continuous at 3? f(x) = x2 − 2x − 3 x − 3
Answer:
Since it's a removable discontinuity, we will remove the discontinuity by creating a new function defined by x=3;
So we have;
F(x) = {[x² - 2x - 3]/(x-3); x ≠ 3
{4, x = 3
Step-by-step explanation:
I have attached my explanation as the system here is not allowing me to save my answer.
To remove the discontinuity at x=3 in the function f(x) = (x^2 - 2x - 3) / (x - 3), simplify the function to f(x) = x + 1, then calculate f(3) = 3 + 1 = 4. Defining f(3) = 4 makes the function continuous at x=3.
Explanation:To remove the discontinuity of a function such as f(x) = (x^2 - 2x - 3) / (x - 3), you need to find a value for f(3) that would make the function continuous at x = 3. First, let's manipulate the function:
The function f(x) = (x^2 - 2x - 3) / (x - 3) simplifies to f(x) = (x - 3)(x + 1) / (x - 3).
As you can see, the (x - 3) terms cancel out, leaving f(x) = x + 1.
However, this is only valid for x ≠ 3. To find f(3), you can now simply substitute 3 into the simplified function:
f(3) = 3 + 1 = 4.
So, if we define f(3) = 4, then the function becomes continuous at x = 3.
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There are 100 hours of labor, 500 lbs of material and 1000 gallons of water available. If the goal is to maximize the total profit then the objective function is: (the variables A,B,C & D are the number of widgets of each type produced)
a.min 10A +1268C9D
b.min 10A-15B 7CBD
c.max 10A +15B +7C+8D
d.max A B C D
Answer:
The correct option is option C max 10A +15B +7C+8D
Step-by-step explanation:
As the complete question is not given here thus the complete question is found online and is attached herewith.
From the data the profit for 1 unit of A is 10, B is 15, C is 7 and D is 8, so the profit function is given as 10A+15B+7C+8D. and as profit is to be maximized so the correct option is option C.
Three students scheduled interviews for summer employment at the Brookwood Institute. In each case the interview results in either an offer for a position or no offer. Experimental outcomes are defined in terms of the results of the three interviews.
A) How many experimental outcomes exist?
Note: The possible outcomes are Y/N for first interview, and Y/N for 2nd, and Y/N for 3rd interview.
B) Let x equal the number of students who receive an offer. Is x continuous or discrete?
a) It is discrete b) It is continuous c)It is neither discrete nor continuous
C) Show the value of the random variable for the subset of experimental outcomes listed below. Let Y = "Yes, the student receives an offer", and N = "No, the student does not receive an offer."
Experimental Outcome - Value of X
(Y,Y,Y) - ?
(Y,N,Y) - ?
(N,Y,Y) - ?
(N,N,Y) - ?
(N,N,N) - ?
What are the above experimental outcomes?
Part(a):
Then the outcomes can be,
[tex]\{(1,1,1)(1,1,0)(1,0,1)(0,1,1)(1,0,0)(0,1,0)(0,0,1)(0,0,0) \}[/tex]
Part(b):
The correct option is (a).
Part(c):
The outcomes are,
[tex](1,1,1):x=3\\(1,1,0):x=2\\(1,0,1):x=2\\(0,1,1):x=2\\(1,0,0):x=1\\(0,1,0):x=1[/tex]
Experimental outcomes:Experimental probability, also known as Empirical probability, is based on actual experiments and adequate recordings of the happening of events.
Part(a):Let the ordered pair (a,b,c) denote the outcome with a,b,c taking either 1 if the position is offered
Or 0 if the position is not offered.
Then the outcomes can be,
[tex]\{(1,1,1)(1,1,0)(1,0,1)(0,1,1)(1,0,0)(0,1,0)(0,0,1)(0,0,0) \}[/tex]
Part(b):Let, [tex]x[/tex] is number probability function offers made.
The variable is discrete taking 0 or 1 or 2 or 3 as values.
So, the correct option is (a)
Part(c):The outcomes are,
[tex](1,1,1):x=3\\(1,1,0):x=2\\(1,0,1):x=2\\(0,1,1):x=2\\(1,0,0):x=1\\(0,1,0):x=1[/tex]
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There are 8 experimental outcomes for the interviews. The variable representing the number of offers is discrete, taking values between 0 and 3. For the given subsets of experimental outcomes, the value of this variable is the number of 'Y' present.
Explanation:A) The total number of experimental outcomes is calculated by the rule of product. There are two possible results (Yes or No) for each of the three interviews. So the number of experimental outcomes is 2*2*2, which is 8.
B) The variable x, which is the number of students receiving an offer, is discrete. A variable is discrete if it can only take on a finite or countable number of values. In this case, x can take on only four possible values (0, 1, 2, or 3), depending on the number of students receiving an offer.
C) The value of the random variable X for the subset of experimental outcomes is as follows:
(Y,Y,Y) - X = 3
(Y,N,Y) - X = 2
(N,Y,Y) - X = 2
(N,N,Y) - X = 1
(N,N,N) - X = 0
The above experimental outcomes show the different possible results of the three interviews, with N signifying no job offer and Y signifying a job offer after the interview.
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A car dealer in Big Rapids, Michigan is using Holt’s method to forecast weekly car sales. Currently the level is estimated to be 40 cars per week, and the trend is estimated to be 5 cars per week. During the current week, 20 cars are sold. After observing the current week’s sales, forecast the number of cars three weeks from now. Use a = B = 0.20
Answer:
49 cars
Step-by-step explanation:
Probability of cars to be sold in a week,a = 0.2
Probabiity of cars not sold in a week, b = 0.8
Number of cars estimated to be sold in a week = 20 and 60 cars in 3 weeks
Using, P(x) = nCx *(a)∧x * (b)∧n - x, where n = 3 weeks, x = 1 week
P(x=1) = 3C1 * (0.2) * (0.8)² = 3 X 0.2 X 0.64 X 60 cars = 23 cars
P(x=2) = 3C2 * (0.2)² * (0.8) = 3 X 0.04 X 0.8 X 60 cars = 6 cars
Number of cars three weeks from now: 20 + 23 + 6 = 49 cars
Answer:
The forecast of the number of cars 3 weeks from now is 52 cars.
Step-by-step explanation:
As per the trent the number of the cars per week is 5 cars
The current level of cars is 40 cars per week
Number of cars sold in current week=20 cars
Forecast of the cars sold 3 weeks from now is given as
[tex]L_t=\alpha Y_t+(1-\alpha)(L_{t-1}+T_{t-1})\\[/tex]
From the data
Y_t=20 cars
L_t-1=40 cars
T_t-1=5 cars
α=β=0.2
So the equation becomes
[tex]L_t=\alpha Y_t+(1-\alpha)(L_{t-1}+T_{t-1})\\L_t=0.2*20+(1-0.2)(40+5)\\L_t=40[/tex]
Now the trend is calculated as
[tex]T_t=\beta(L_{t}-L_{t-1})+(1-\beta)T_{t-1}[/tex]
By putting the values the equation becomes
[tex]T_t=\beta(L_{t}-L_{t-1})+(1-\beta)T_{t-1}\\T_t=0.2(40-40)+(1-0.2)5\\T_t=0+0.8*5\\T_t=4[/tex]
Now the forecast of the cars sale 3 weeks from now is given as
[tex]L_{t+k}=L_t+kT_t[/tex]
where k is 3 so
[tex]L_{t+k}=L_t+kT_t\\L_{t+3}=40+3*4\\L_{t+3}=40+12\\L_{t+3}=52\\[/tex]
So the forecast of the number of cars 3 weeks from now is 52 cars.
Prepare a perpetual inventory record, using the FIFO inventory costing method, and determine the company's cost of goods sold, ending merchandise inventory, and gross profit. Begin by computing the cost of goods sold and cost of ending merchandise inventory using the FIFO inventory costing method. Enter the transactions in chronological order, calculating new inventory on hand balances after each transaction. Once all of the transactions have been entered into the perpetual record, calculate the quantity and total cost of merchandise inventory purchased, sold, and on hand at the end of the period.
Answer:
The First-In, First-Out (FIFO) inventory costing method assumes that the inventory items ordered first are the first ones sold.
Step-by-step explanation:
The First-In, First-Out inventory costing method assumes that the inventory items ordered first are the first sold. This is ideal for goods that are highly perishable, for example fresh milk. Since no figures or dates are given, we will assume that the month is March 2019 and use any figures to make the example.
Date Item Quantity of stock Cost Price
01 Opening stock bought on Feb 28 10 100
05 Sale of 5 goods (cost is $10 each) (5) 50
15 Purchase of stock (20 goods at $20 each) 20 400
25 Sale of 15 goods (15) 250
(5 at $10 each & 10 at $20 each)
31 Closing Stock 10 200
(20 goods bought on 15th - 10 goods sold on 25th)
The quantity on hand at the end of the month is 10 units.
Total cost of goods on hand at end of the month = 10 units * $20 = 200.
Total cost of goods purchased during the month = $20 * 20 units = $400
Total cost of goods sold during the month = [($10 *5) + ($10 * 5)+ ($20 * 10)] = $200