Answer: The probability of drawing a red marble the sixth time is 1/2
Step-by-step explanation:
Here is the complete question:
A box contains 10 red marbles and 10 green marbles. Sampling at random from the box five times with replacement, you have drawn a red marble all five times. What is the probability of drawing a red marble the sixth time?
Explanation:
Since the sampling at random from the box containing the marbles is with replacement, that is, after picking a marble, it is replaced before picking another one, the probability of picking a red marble is the same for each sampling. Probability, P(A) is given by the ratio of the number of favourable outcome to the total number of favourable outcome.
From the question,
Number of favourable outcome = number of red marbles =10
Total number of favourable outcome = total number of marbles = 10+10= 20
Hence, probability of drawing a red marble P(R) = 10 ÷ 20
P(R) = 1/2
Since the probability of picking a red marble is the same for each sampling, the probability of picking a red marble the sixth time is 1/2
The probability is [tex]\frac{1}{2}[/tex]
First, we have to calculate the probability of drawing a red marble on any given try using the formula of probability:
Probability of an event = [tex]\frac{Number\ of\ favorable\ outcomes}{Total\ number\ of\ outcomes}[/tex]
P(red) = [tex]\frac{Number\ of\ red\ marbles}{Total\ number\ of\ marbles} =\frac{10}{20} = \frac{1}{2}[/tex]
Since we are replacing after every time we pick up a marble, each event of picking a marble is an independent event
Thus, every time a marble is to be picked, the probabilities remain same.
Hence the probability of drawing a red marble the sixth time is = [tex]P(red) = \frac{1}{2}[/tex]
The complete question is:
A box contains 10 red marbles and 10 green marbles. Sampling at random from the box five times with replacement, you have drawn a red marble all five times. What is the probability of drawing a red marble the sixth time?
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.
pat was among 170 college students particiapting in a quantitative skills assesment. I t has been determined that there were 34 score that were lower than pat's. Pat's score therfore was the _______ percentiles?
Answer:
Pat's score therfore was the 20th percentiles
Step-by-step explanation:
When a value V is said to be in the xth percentile of a set, x% of the values in the set are lower than V and (100-x)% of the values in the set are higher than V.
In this problem, we have that:
34 scores lower than Pat's.
[tex]\frac{34}{170} = 0.20[/tex]
So Patrick's score was the 20th percentile.
The probability is 0.4 that a traffic fatality involves an intoxicated or alcohol-impaired driver or nonoccupant. In 7 traffic fatalities, find the probability that the number, Y, which involve an intoxicated or alcohol-impaired driver or nonoccupant is
a. exactly three; at least three; at most three.
b. between two and four, inclusive.
c. Find and interpret the mean of the random variable Y.
d. Obtain the standard deviation of Y.
Answer:
a.
[tex]P(X=3)=0.2903\\\\P(X \geq 3)=0.5801\\\\P(X\leq 3)0.7102[/tex]
b.
[tex]P(2\leq x\leq 4 )=0.7451[/tex]
c. mean=2.8
d . standard deviation=1.2961
Step-by-step explanation:
We determine that the accident rates follow a binomial distribution. The rate of success p=0.4 and sample n=7:
[tex]P(x)={n\choose x}p^x(1-p)^{n-x}[/tex]
#the probability of exactly three;
[tex]P(x)={n\choose x}p^x(1-p)^{n-x}\\\\P(X=3)={7\choose 3}0.4^3(0.6)^{4}\\\\=0.2903[/tex]
#At least(more than 2)
[tex]P(x)={n\choose x}p^x(1-p)^{n-x}\\\\P(X\geq 3)=1-P(X\leq 2)\\\\=1-{7\choose 0}0.4^0(0.6)^{7}-{7\choose 1}0.4^1(0.6)^{6}-{7\choose 2}0.4^2(0.6)^{5}\\\\=1-0.0280-0.1306-0.2613\\\\=0.5801[/tex]
#At most 3;
[tex]P(x)={n\choose x}p^x(1-p)^{n-x}\\\\P(X \leq 3)={7\choose 0}0.4^0(0.6)^{7}+{7\choose 1}0.4^1(0.6)^{6}+{7\choose 2}0.4^2(0.6)^{5}+{7\choose 3}0.4^3(0.6)^{4}\\\\\\\=0.0280+0.1306+0.2613+0.2903\\\\=0.7102[/tex]
b. Between 2 and 4:
Using the binomial expression, this probability is calculated as:
[tex]P(x)={n\choose x}p^x(1-p)^{n-x}\\\\P(2\leq x\leq 4 )={7\choose 2}0.4^2(0.6)^{5}+{7\choose 3}0.4^3(0.6)^{4}+{7\choose 4}0.4^4(0.6)^{3}\\\\\\\\\=0.2613+0.2903+0.1935\\\\=0.7451[/tex]
Hence,the probability of between 2 and four is 0.7451
c. From a above, we have the values of n=7 and p=0.4.
-We substitute this values in the formula below to calculate the mean:
-The mean of a binomial distribution is calculated as the product of the probability of success by the sample size, mean=np:
[tex]\mu=np, n=7, p=0.4\\\\\mu=7\times 0.4\\\\=2.8[/tex]
Hence, the standard deviation of the sample is 2.8
d. From a above, we have the values of n=7 and p=0.4
--We substitute this values in the formula below to calculate the standard deviation
-The standard deviation a binomial distribution is given as:
[tex]\sigma={\sqrt {np(1-p)}\\\\=\sqrt{7\times 0.4\times 0.6}\\\\=1.2961[/tex]
Hence, the standard deviation of the sample is 1.2961
We can calculate the probabilities of having exactly 3 fatalities, at least 3 fatalities, and at most 3 fatalities using the formula. The mean and standard deviation of the random variable Y can be calculated using specific formulas for a binomial distribution.
Explanation:To solve this problem, we will use the binomial probability formula. The probability of exactly 3 traffic fatalities involving an intoxicated or alcohol-impaired driver or nonoccupant can be calculated by plugging in the values of n (total number of trials) and p (probability of success in a single trial) into the formula. The probability of at least 3, or more than 3 traffic fatalities involving an intoxicated or alcohol-impaired driver or nonoccupant can be calculated by summing the probabilities of 3, 4, 5, 6, and 7 fatalities. The probability of at most 3 traffic fatalities involving an intoxicated or alcohol-impaired driver or nonoccupant can be calculated by summing the probabilities of 0, 1, 2, and 3 fatalities. The mean of the random variable Y can be calculated by multiplying the number of trials (7) by the probability of success (0.4). The standard deviation of Y can be calculated using the formula for the standard deviation of a binomial distribution.
Learn more about binomial probability here:https://brainly.com/question/33993983
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A manufacturer is studying the effects of cooking temperature, cooking time, and type of cooking oilfor making potato chips. Three different temperatures,4 different cooking times, and 3 different oils are to be used.
a. What is the total number of combinations to be studied?
b. How many combinations will be used for each type of oil?
c. Discuss why permutations are not an issue in this exercise.
Answer:
(a) The total number of combinations that can be applied for making potato chips is 36.
(b) The number of combinations that will be used for each type of oil is 12.
(c) Permutations are not an issue because order does not matter.
Step-by-step explanation:
The effects of cooking temperature, cooking time and cooking oil is studied for making potato chips.
The number of different temperatures applied is, n (T) = 3.
The number of different times taken is, n (t) = 4.
The number of different oils used is, n (O) = 3.
If an assignment can be done in n₁ ways and if for this assignment another assignment can be done in n₂ ways then these two assignments can be performed in (n₁ × n₂) ways.
(a)
Compute the total number of combinations that can be applied for making potato chips as follows:
Total number of combinations for making chips = n (T) × n (t) × n (O)
[tex]=3\times4\times 3\\=36[/tex]
Thus, the total number of combinations that can be applied for making potato chips is 36.
(b)
To make potato chips 4 different temperatures are used and 3 different oils are used.
Each of the oil type is cooked in 4 different temperatures.
So the number of ways to select each oil type is,
n (T) × n (O) = [tex]4\times3=12[/tex]
Thus, the number of combinations that will be used for each type of oil is 12.
(c)
Permutation is the arrangement of objects in a specified order.
Since in this case ordering of the the three effects, i.e. temperature. time and oil type is not important, permutations are not an issue.
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
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.
A similar problem is given at https://brainly.com/question/23265899
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
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.
Normal (or average) body temperature of humans is often thought to be 98.6° F. Is that number really the average? To test this, we will use a data set obtained from 65 healthy female volunteers aged 18 to 40 that were participating in vaccine trials. We will assume this sample is representative of a population of all healthy females.
A. The mean body temperature for the 65 females in our sample is 98.39° F and the standard deviation is 0.743° F. The data are not strongly skewed. Use the Theory-Based Inference applet to find a 95% confidence interval for the population mean body temperature for healthy female
B. Based on your confidence interval, is 98.6° F a plausi- ble value for the population average body temperature or is the average significantly more or less than 98.6° F? Explain how you are determining this.
C. In the context of this study, was it valid to use the theory-based (t-distribution) approach to find a confi- dence interval? Explain your reasoning.
Answer:
(a) 95% confidence interval for the population mean body temperature for healthy female is between a lower limit of 98.21 °F and an upper limit of 98.57 °F.
(b) The average is less than 98.6 °F
(c) Yes
Step-by-step explanation:
(a) Confidence interval = mean + or - Margin of Error (E)
mean = 98.39 °F
sd = 0.743 °F
n = 65
degree of freedom = n - 1 = 65 - 1 = 64
confidence level = 95%
t- value corresponding to 64 degrees of freedom and 95% confidence level is 1.9976.
E = t×sd/√n = 1.9976×0.743/√65 = 0.18 °F
Lower limit = mean - E = 98.39 - 0.18 = 98.21 °F
Upper limit = mean + E = 98.39 + 0.18 = 98.57 °F
95% confidence interval is between 98.21 °F and 98.57 °F.
(b) The average is less than 98.6 °F. The lower limit 98.21 °F and the upper limit 98.57 °F are both less than 98.6 °F
(c) It was valid to use the t-distribution approach to find the confidence Interval beci it gives a range of values for the population mean body temperature for healthy female.
consider a bank with two tellers. Three people, Anne, Betty and Carol enter the bank at almost the same time and in that order. Anne and Betty go directly into service while Carol waits or the first available teller. Suppose that the service times for two servers are exponentially distributed with mean three and six minutes (or they have rates of 20 and 10 per hour).
(a)What is the expected total amount of time for Carol to complete her businesses?
(b) What is the expected total time until the last of the three customers leaves?
(c) What is the probability for Anne, Betty, and Carol to be the last one to leave?
Final answer:
Explanation of expected total times for Carol and the last customer, along with the probability of being last.
Explanation:
(a) Expected total time for Carol to complete her businesses:
Carol will go to the first available teller, who has a service time of 3 minutes (or service rate of 20 per hour).
Hence, the expected total time for Carol to complete her businesses is 3 minutes.
(b) Expected total time until the last customer leaves:
Calculating the expected total time for each customer to complete their businesses gives Anne: 3 minutes, Betty: 6 minutes.
Therefore, the expected total time until the last customer leaves is 6 minutes.
(c) Probability of being the last to leave:
The last to leave will be the customer with the highest expected service time, which is Betty (6 minutes).
The probability for Anne, Betty, and Carol to be the last one to leave is: Anne 0, Betty 1, Carol 0.
(a) Expected total amount of time for Carol to complete her business: the expected total amount of time for Carol to complete her business is 2 minutes.
(b) Expected total time until the last of the three customers leaves:the expected total time until the last of the three customers leaves is 8 minutes.
c) Probability for Anne, Betty, and Carol to be the last one to leave:the probability for Anne, Betty, and Carol to be the last one to leave is [tex]\( \frac{1}{3} \)[/tex] for each of them.
Let's solve each part of the problem:
(a) Expected total amount of time for Carol to complete her business:
Since Carol waits for the first available teller, we need to consider the minimum of the service times of the two tellers. Let's denote:
- ( X ) as the service time for the first teller (with mean three minutes).
- ( Y ) as the service time for the second teller (with mean six minutes).
The minimum of two exponentially distributed random variables with means [tex]\( \mu_1 \) and \( \mu_2 \)[/tex] is also exponentially distributed with mean \( \frac{1}[tex]\( \frac{1}[/tex][tex]{\lambda_1 + \lambda_2} \), where \( \lambda_1 \) and \( \lambda_2 \)[/tex] are the rates.
Given:
- The rate for the first teller is [tex]\( \lambda_1 = \frac{1}{3} \)[/tex] per minute.
- The rate for the second teller is [tex]\( \lambda_2 = \frac{1}{6} \)[/tex] per minute.
So, the rate for the minimum service time for Carol is [tex]\( \lambda = \lambda_1 + \lambda_2 = \frac{1}{3} + \frac{1}{6} = \frac{1}{2} \)[/tex] per minute.
The expected total amount of time for Carol to complete her business is the reciprocal of the rate, which is [tex]\( \frac{1}{\lambda} = \frac{1}{\frac{1}{2}} = 2 \) minutes.[/tex]
Therefore, the expected total amount of time for Carol to complete her business is 2 minutes.
(b) Expected total time until the last of the three customers leaves:
The total time until the last customer leaves is the maximum of the service times of all three customers. Since Anne and Betty go directly into service, their service times are independent of each other and of Carol's service time.
Let's denote:
- ( Z ) as the service time for Carol (which we found to be 2 minutes).
- ( W ) as the maximum of the service times for Anne and Betty.
Since ( W ) is the maximum of two exponentially distributed random variables with the same rate, it follows that ( W ) is also exponentially distributed with the same rate.
So, the expected total time until the last of the three customers leaves is the sum of the expected service time for Carol and the expected maximum service time for Anne and Betty, which is [tex]\( 2 + 6 = 8 \)[/tex]minutes.
Therefore, the expected total time until the last of the three customers leaves is 8 minutes.
(c) Probability for Anne, Betty, and Carol to be the last one to leave:
Since Anne and Betty go directly into service, their service times are independent of each other and of Carol's service time. Therefore, each of them has an equal chance of being the last one to leave.
So, the probability for Anne, Betty, and Carol to be the last one to leave is [tex]\( \frac{1}{3} \)[/tex] for each of them.
Therefore, the probability for Anne, Betty, and Carol to be the last one to leave is [tex]\( \frac{1}{3} \)[/tex] for each of them.
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
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].
An average light bulb manufactured by the Acme Corporation lasts 300 days with a standard deviation of 50 days. Assuming that bulb life is normally distributed:
1. What is the probability that an Acme light bulb will last more than 300 days?
2. What is the probability that an Acme light bulb will last less than 300 days?
3. What is the probability that an Acme light bulb will last exactly 300 days?
4. In order to obtain a scientific survey with 95 % confidence level of public opining on something without making more than 3% error in either direction, how much percentage of all American adults should we ask?
Answer:
1. 90% 2. 10% 3. 50%
Step-by-step explanation:
Standard Deviation (σ) = 50 days
Average/Mean (μ) = 300days
Probability that it would last more than 300 days = P(Bulb>300 days)
We will assume there are 365 days in a year.
P(Bulb>300 days) implies that the bulb would
Using the normal equation;
z = standard/normal score = (x-μ)/σ where x is the value to be standardized
P(Bulb>300 days) implies x = 365 days
Therefore z = (365-300)/50 = 1.3
Using the normal graph for z=1.3, probability = 90%
2. P(Bulb<300 days) = 1 - P(Bulb>300 days)\
P(Bulb<300 days) = 1 - 0.9
P(Bulb<300 days) = 10%
3. P(Bulb=300 days) implies z=0 since x=300
Using the normal graph for z=0, probability =50%
1. The probability that an Acme light bulb will last more than 300 days is 50%. 2. The probability that an Acme light bulb will last less than 300 days is 50%. 3. The probability that an Acme light bulb will last exactly 300 days is zero.
Explanation:1. To find the probability that an Acme light bulb will last more than 300 days, we need to determine the area under the normal distribution curve to the right of 300 days. We use the z-score formula: z = (x - μ) / σ, where x is the value we are interested in, μ is the mean, and σ is the standard deviation. Substituting the values, we get: z = (300 - 300) / 50 = 0. The area to the right of 300 days is equal to the area to the left of 0. Using a standard normal distribution table, we find that the area to the left of 0 is 0.5. Therefore, the probability that an Acme light bulb will last more than 300 days is 0.5 or 50%.
2. To find the probability that an Acme light bulb will last less than 300 days, we need to determine the area under the normal distribution curve to the left of 300 days. Using the same z-score formula, we get: z = (300 - 300) / 50 = 0. The area to the left of 0 is 0.5. Therefore, the probability that an Acme light bulb will last less than 300 days is 0.5 or 50%.
3. To find the probability that an Acme light bulb will last exactly 300 days, we need to determine the area under the normal distribution curve at 300 days. Since the normal distribution is continuous, the probability of any single value is zero. Therefore, the probability that an Acme light bulb will last exactly 300 days is zero.
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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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.
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.
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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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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which of the following statistical techniques can the psychologist use to determine the developmental score of a typical 4 year old child despite the fact that no 4 year old children participated in the study
A psychologist can use normative data, sequential design, or parent-report questionnaires to estimate the developmental score of a typical 4-year-old child without direct participation from children of that specific age.
Explanation:To determine the developmental score of a typical 4-year-old child without having 4-year-old children participate in the study, a psychologist may use several statistical techniques grounded in developmental science. One method is to use normative data, which refers to average developmental milestones established through extensive research on children at various ages. By comparing existing data on children older and younger than 4 years, the psychologist can interpolate or approximate the developmental score for the typical 4-year-old based on these norms.
Another approach involves a sequential design, where overlapping cohorts of different age groups are studied over time. This method can help infer the developmental score of 4-year-olds by comparing changes in development across the different cohorts that are adjacent in age. Additionally, parent-report questionnaires can be used to gather data from parents or guardians about typical developmental milestones, which can then be statistically analyzed to estimate the developmental score of children in the age group of interest.
An oceanic island has 15 species of birds on it. Nearby, a new island is formed by volcanic activity. Since this new island is somewhat smaller than the original island, biogeography theory that it can support 8 species of birds. If the colonizing birds must migrate from the oceanic island to the newly formed island, how many different communities of 8 species could be formed on the new island
Answer:
6435 different communities.
Step-by-step explanation:
This is a combination problem,
We need to know how many ways 8 species out of 15 can be selected for the new small island
15C8 = 15!/(15-8)!(8!) = 15!/(8!)(7!) = 6435 different communities.
n a major league baseball game, the average is 1.0 broken bat per game. (a) Find the probability of no broken bats in a game. (Round your answer to 4 decimal places.) Probability (b) Find the probability of at least 2 broken bats in a game. (Round your answer to 4 decimal places.) Probability
Answer:
a) Probability of no broken bats in a game P(X=0) = 0.3678
b) Probability of at-least broken bats in a game P(X≥2) = 0.2644
Step-by-step explanation:
we will use Poisson distribution
P(X=x) = e^(-λ) λ^ r/r!
a) Probability of no broken bats in a game P(X=0) = e^-1
= 0.3678
b) Probability of at-least two broken bats in a game
P(X≥2) = 1-([p(x=0)+P(x=1)]
= 1- (e^(-1)+e^(-1))
= 1-(0.3678+0.3678)
P(X≥2)= 1- 0.7356
P(X≥2)= 0.2644
A coding supervisor must determine the number of FTEs needed to code 600 discharges per week. If it takes an average of 20 minutes to code each record and each coding professional works 7.5 productive hours per day, how many FTEs will the coding supervisor need
Answer:
6 Full Time Employees
Step-by-step explanation:
It takes 20 minutes to code each record.
Therefore, in 1 hour, number of records per coder = 60/20 = 3 records
Since each coding professional works 7.5 productive hours per day.
Number of Records Per Day for each coder
=(3X7.5) = 22.5 records
Assume a 5-Week Day
Number of Records that must be treated = 600/5 = 120 records a day
Number of FTE Employees needed
= 120 / 22.5 = 5.33
The employer will require at least 6 FTE since Number of Employees is a discrete data.
Final answer:
The coding supervisor will need approximately 5.33 FTEs to code 600 discharges per week, considering the given average time to code each record and daily productive hours. Since a fraction of an FTE isn't practical, the supervisor will need to employ 6 full-time coding professionals.
Explanation:
To determine the number of Full-Time Equivalents (FTEs) needed to code 600 discharges per week, we first need to calculate the total time required to code all records and then divide that time by the productive hours each coding professional works per day.
First, calculate the total weekly coding time: 600 discharges × 20 minutes per discharge = 12,000 minutes per week.
Then convert the total weekly coding time to hours: 12,000 minutes / 60 = 200 hours per week needed for coding.
Considering each coding professional works 7.5 productive hours per day, let's find out the daily capacity for one FTE: 7.5 hours per day × 5 days per week = 37.5 hours per week.
Now, divide the total weekly hours needed by the weekly capacity of one FTE: 200 hours per week / 37.5 hours per FTE per week = approximately 5.33 FTEs.
Therefore, the coding supervisor will need to employ roughly 5.33 FTEs to code 600 discharges per week, given the average coding time and productive hours per day. Since you can't have a fraction of an FTE in reality, the supervisor would need to round up to employ 6 full-time coding professionals.
Pls help me ASAP!!!!!!!
Answer:
The standard deviation is 5.83 kg
Step-by-step explanation:
Variance and Standard Deviation
Given a data set of random values, the variance is defined as the average of the squared differences from the mean. We use this formula to calculate the variance:
[tex]\displaystyle \sigma^2=\frac{\sum(x_i-\mu)^2}{n}[/tex]
Where [tex]\mu[/tex] is the mean of the values xi (i=1 to n), n the total number of values.
The standard deviation is known by the symbol [tex]\sigma[/tex] and is the square root of the variance.
We are given the value of the variance:
[tex]\sigma^2=34\ kg^2[/tex]
We now compute the standard deviation
[tex]\sigma=\sqrt{34\ kg^2}=5.83\ kg[/tex]
The standard deviation is 5.83 kg
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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
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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Paul 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
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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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
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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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.