To find the probabilities of a randomly selected U.S. adult using social media and being aged 18-29, we can use probability concepts and the given information.
Explanation:To solve this problem, we can use the concept of probability. Let's define the following events:
A: randomly selected U.S. adult uses social mediaB: randomly selected U.S. adult is aged 18-29From the information given, we have P(A' ∩ B') = 35%, P(A' ∩ B) = 0, P(A ∩ B') = 67.2%. We need to find the probabilities P(A) and P(B).
Using the formula for the probability of the complement of an event, we have P(A') = 35%.Using the given value of the percentage of U.S. young adults age 30 and older (78%), we have P(B') = 100% - 78% = 22%.Using the formula P(A ∩ B') = P(A) - P(A ∩ B), we have P(A) = P(A ∩ B') + P(A' ∩ B') = 67.2% + 35% = 102.2%.Using the formula P(B) = P(B') - P(A' ∩ B'), we have P(B) = 22% - 0% = 22%.Therefore, the probability that a randomly selected U.S. adult uses social media is 102.2%, the probability that a randomly selected U.S. adult is aged 18-29 is 22%, and the probability that a randomly selected U.S. adult is 18-29 and a user of social media is 67.2%.
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a. Probability of a randomly selected U.S. adult using social media is 45.2%.
b. Probability of a randomly selected U.S. adult being aged 18-29 is 65%.
c. Probability of a randomly selected U.S. adult being 18-29 and a user of social media is approximately 29.38%.
a. Probability a randomly selected U.S. adult uses social media:
Probability of being 18-29 or use social media = 67.2%
Percentage of US adults aged 30 and older = 78%
Thus, percentage of US adults aged 18-29 = 100% - 78% = 22%
Probability of using social media = 67.2% - 22% = 45.2%
b. Probability of a randomly selected U.S. adult being aged 18-29:
Percentage of adults not using social media = 35%
Thus, percentage of adults aged 18-29 = 100% - 35% = 65%
c. Probability of a randomly selected U.S. adult being 18-29 and a user of social media:
Probability of using social media = 45.2%
Probability of being 18-29 = 65%
Probability of being 18-29 and using social media = 0.452 * 0.65 = 29.38%
Use the empirical rule to solve the problem. The annual precipitation for one city is normally distributed with a mean of 288 inches and a standard deviation of 3.7 inches. Fill in the blanks. In 95.44% of the years, the precipitation in this city is between ___ and ___ inches. Round your answers to the nearest tenth as needed.
Answer:
[tex]z=-1.99<\frac{a-288}{3.7}[/tex]
[tex]z=1.99<\frac{a-288}{3.7}[/tex]
And if we solve for a we got
[tex]a=288 -1.99*3.7=214.4[/tex]
[tex]a=288 +1.99*3.7=295.4[/tex]
And the limits for this case are: (214.4; 295.4)
Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Solution to the problem
Let X the random variable that represent the annual precipitation of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(288,3.7)[/tex]
Where [tex]\mu=288[/tex] and [tex]\sigma=3.7[/tex]
The confidence level is 95.44 and the signficance is [tex] 1-0.9544=0.0456[/tex] and the value of [tex]\alpha/2 =0.0228[/tex]. And the critical value for this case is [tex]z = \pm 1.99[/tex]
Using this condition we can find the limits
[tex]z=-1.99<\frac{a-288}{3.7}[/tex]
[tex]z=1.99<\frac{a-288}{3.7}[/tex]
And if we solve for a we got
[tex]a=288 -1.99*3.7=214.4[/tex]
[tex]a=288 +1.99*3.7=295.4[/tex]
And the limits for this case are: (214.4; 295.4)
Which solid does the net form?
c. traingular prism
Answer:
Triangular Prism
I'm guessing this means what 3D shape it produces, which would lead my conclusion to it being a triangular prism.
Which events have a probability of 25 percent? Select three options.
Answer:
The answer is A, C, E
A: choosing a green jelly bean out of a bag that contains 2 green jelly beans, 1 red jelly bean, and 5 yellow jelly beans
C: spinning a number less than 2 on a spinner that has four equal sections numbered from 1 to 4
D: choosing a spade out of a standard deck of cards that contains 13 hearts, 13 clubs, 13 diamonds, and 13 spades
Therefore the answer is A,C,E
Edg : )
Somebody please help me with this i have no clue how to do any of this
Ms. Vargas recorded the hourly wage of each employee at her business.
Which of the following statements are true?
Chose all correct answers.
A. The mean of the data describes how the hourly wages vary.
B. The mean absolute deviation of the data describes how the hourly wages vary. C. The interquartile range of the data summarizes the hourly wages.
D. The median of the data summarizes the hourly wages.
E.The maximum value of the data summarizes the hourly wages.
Answer:
I think it is A and B. Not one hundred percent sure but,50 percent sure.
Step-by-step explanation:
Answer:
B.) and D.)
Step-by-step explanation:
Why am I not getting the right answer
Answer:
You could’ve doing the math wrong... what’s the question?
Step-by-step explanation:
Find the volume
A.I0 cubic units
B.20 cubic units
C.22 cubic units
D.10 cubic units
Answer:
its b
Step-by-step explanation:
FOR THE LENGTH YOU SEE HOW LONG IT IS THEN FINALLY YOU SEE THE WIDTH ITS ONLY ONE AND THEN THE HIGHT HOW TALL IT IS AND THEN YOU MULTIPLY TO GET Your answer
4. Determine which of the following are equivalence relations and/or partial CHAPTER 6. RELATIONS 121 ordering relations for the given sets: (a) A = { lines in the plane}, and r defined by xry if and only if x is parallel to y. Assume every line is parallel to itself. (b) A = R and r defined by xry if and only if |x − y| ≤ 7.
Answer:
Determine which of the following are equivalence relations and/or partial CHAPTER 6. RELATIONS 121 ordering relations for the given sets: (a) A = { lines in the plane}, and r defined by xry if and only if x is parallel to y. Assume every line is parallel to itself. (b) A = R and r defined by xry if and only if |x − y| ≤ 7.
A is an Equivalence Relation.
Step-by-step explanation:
For a reflexive case : x is parallel to itself => x R x
For a symmetric : x is parallel to y => y is parallel to x.
Thus, x R y = y R x
For a transitive: x is parallel to y, and y is parallel to z then x, y, z are parallel to each other.
=> x R y and y R z = x R z
Therefore, A is an Equivalence Relation.
The results of a medical test show that of 39 people selected at random who were given the test, 33 tested negative and 6 tested positive. Determine the odds against a person selected at random from these 39 people testing negative on the test.
Given Information:
Total number of people = 39
Number of people who tested negative = 33
Number of people who tested positive = 6
Required Information:
Probability of finding a person who tested negative = ?
Answer:
P( x = 1 ) = 0.000315 = 0.0315%
Step-by-step explanation:
Let the probability of success is denoted by those who test negative,
p = 33/39
p = 0.85
Then the probability of failure is denoted by those who test positive,
q = 1 - p = 1 - 0.15 = 0.85
We want to find out the probability of finding a random person who test negative.
P( x = 1 ) = (p³⁹⁻¹)(1 - p)¹
P( x = 1 ) = (0.85³⁹⁻¹)(1 - 0.15)¹
P( x = 1 ) = 0.0021*0.15
P( x = 1 ) = 0.000315
Therefore, the Probability of finding a person who tested negative is 0.0315%
Suppose 20 donors come to a blood drive. Assume that the blood donors are not related in any way, so that we can consider them to be independent. The probability that a donor has type "O" blood is 0.06. What is the probability that 1 or more donors have type O blood? A. 0.370 B. 0.290 C. 0.630 D. 0.710
Given Information:
Probability of success = p = 0.06
Number of trials = n = 20
Required Information:
Probability of 1 or more donors of "O" blood group = ?
Answer:
P( x ≥ 1 ) = 0.710
Step-by-step explanation:
We are given the probability that a donor has type "O" blood group.
We want to find out the probability of having 1 or more donors who has type "O" blood group out of 20 donors.
P( x ≥ 1 ) = 1 - P( x = 0)
So we will first find the probability that none of the donors has type "O" blood group then we will subtract that from 1 to get the probability of having 1 or more donors with "O" blood group.
P( x = 0) = (p⁰)(1 - p)²⁰
P( x = 0) = (0.06⁰)(1 - 0.06)²⁰
P( x = 0) = (1)(0.94)²⁰
P( x = 0) = 0.290
So the probability of having 1 or more donors with "O" blood group is
P( x ≥ 1 ) = 1 - P( x = 0)
P( x ≥ 1 ) = 1 - 0.290
P( x ≥ 1 ) = 0.710
P( x ≥ 1 ) = 71%
Therefore, the correct answer is D. 0.710
Final answer:
The probability that 1 or more donors have type O blood is approximately 0.370.
Explanation:
The probability that 1 or more donors have type O blood can be found using the complement rule. The complement of the event that 0 donors have type O blood is the event that 1 or more donors have type O blood.
The probability that a donor has type O blood is 0.06. The probability that a donor does not have type O blood is 1 - 0.06 = 0.94. Let's calculate the complement probability:
P(1 or more donors have type O blood) = 1 - P(0 donors have type O blood)
= 1 - (0.94)^{20} (since the donors are independent)
Using a calculator, we get P(1 or more donors have type O blood) ≈ 0.370
Latisha determined the approximate amount of time each student in her homeroom class spent outside on a sunny day and on a rainy day. The dot plots below show her results.
Which measures of center and variability can be used to most accurately compare the two data sets?
A.mean and MAD
B.mean and IQR
C.median and MAD
D.median and IQR
Answer:
THE ANSWER IS A. I HAD THE SAME QUESTION
Step-by-step explanation:
Answer:
A
Step-by-step explanation:
What 4c=12
4c/4=12.4
Answer:
c =3
Step-by-step explanation:
4c = 12
Divide each side by 4
4c/4 =12/4
c = 3
Use the convolution theorem to find the inverse Laplace transform of the given function. StartFraction 5 Over s cubed (s squared plus 25 )EndFraction 5 s3s2+25 laplace transform Superscript negative 1 Baseline StartSet StartFraction 5 Over s cubed (s squared plus 25 )EndFraction EndSet (t )ℒ−1 5 s3s2+25(t)
Answer:
[tex]\frac{1}{2}[/tex][tex]t^{2}Sin5t[/tex]
Step-by-step explanation:
using the Convolution theorem to find the inverse of :
[tex]\frac{5}{s^{3}(s^{2}+25 ) }[/tex]
[tex]L^{-1}[/tex] [tex]\frac{5}{s^{3}(s^{2}+25 ) }[/tex] = [tex]\frac{1}{s^{3} }[/tex] × [tex]\frac{5}{s^{2}+25}[/tex]
we know from derivation that
Sin(at) = [tex]\frac{a}{s^{2}+a^{2} }[/tex]
Hence: [tex]\frac{5}{s^{2}+25}[/tex] = Sin5t
Also: [tex]L^{-1}[/tex] [tex]\frac{n!}{s^{n+1} }[/tex] = [tex]t^{n}[/tex]
[tex]L^{-1}[/tex] [tex]\frac{1}{s^{3} }[/tex] = [tex]\frac{1}{2}[/tex] [tex]L^{-1}[/tex] ([tex]\frac{2!}{s^{3} }[/tex])
= [tex]\frac{1}{2}[/tex][tex]t^{2}[/tex]
therefore [tex]L^{-1}[/tex] [tex]\frac{5}{s^{3}(s^{2}+25 ) }[/tex] = [tex]\frac{1}{2}[/tex][tex]t^{2}Sin5t[/tex]
Joey read in his biology book that fish activity increases with water temperature, and he decided to investigate this issue by conducting an experiment. On nine successive days, he measured fish activity and water temperature in his aquarium. He found the equation of the LSRL to be y t ˆ = + 152.1 37.65 , where yˆ is the predicted amount of activity. If one of Joey’s days had a water temperature of 69.0 degrees Fahrenheit and the fish activity on that day was 265, what is the residual activity for this particular temperature?
The residual value is the difference between the actual and predicted values. Hence, the residual activity at a temperature of 69°F will be −2484.95
The equation of the Least Square Regression Line is given as :
y = 152.1 + 37.65tThe predicted amount of activity, y at a temperature of 69°F will be :
y = 152.1 + 37.65(69)
y = 2749.95
The actual amount of activity at a temperature of 69°F is 265
The residual activity at 69°F temperature can be calculated thus :
Residual = Actual - Predicted
Residual = 265 - 2749.95 = −2484.95
Therefore, the residual activity at 69°F is −2484.95
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kayla did 243 jumping jacks every day for 30 days and jean did 289 every day for 30 days how many mor jumping jacks didi jean do
Answer:
1380 more jumping jacks
Step-by-step explanation:
Kayla did 243 jumping jacks each day for 30 days, so we multiply 243 by 30 to get the total number of jumping jacks: 243 * 30 = 7290 jumping jacks.
Jean did 289 every day for 30 days, so we multiply 289 by 30 to get the total number of jumping jacks: 289 * 30 = 8670 jumping jacks.
Now subtract 7290 from 8670: 8670 - 7290 = 1380 more jumping jacks.
Hope this helps!
Answer: 1380
Step-by-step explanation : jean did 8760 jumping jacks. kayla did 7920. subtract 7920 from 8760 to get 1380
Davis performed an experiment in which he spun a 4-color spinner 20 times. He recorded his results in the frequency table.
Davis predicts that if he spins the spinner another 20 times, the frequency for the red and green segments will still be equal. Is Davis correct?
Yes. Another experiment will give the same numbers as the first experiment.
Yes. The experimental probabilities remain the same for all experiments.
No. The frequencies will be multiplied by 20.
No. The frequencies and probabilities for red and green segments might change in the next experiment.
Answer:
No. The frequencies and probabilities for red and green segments might change in the next experiment.
Step-by-step explanation:
I just did it and it said my answer was right
Answer:
Step-by-step explanation:
No. The frequencies and probabilities for red and green segments might change in the next experiment.
i just did it and got it right so i hope you do too
James is 5 feet 4 inches tall, how tall is James in inches
Answer:
64 in.
Step-by-step explanation:
One foot is 12 inches, so you would have to multiply 12 and 5 to get 60. You then add the 60 to the 4 to get 64 inches. Hope this helped!
Recipes for the same type of cookies can vary in terms of ingredients and baking times. From a collection of chocolate chip cookie recipes, a baker randomly selected 5 recipes. From a collection of oatmeal raisin cookie recipes, the baker randomly selected 4 recipes. The mean baking times, in minutes, for each sample were recorded as x¯C and x¯O, respectively. What is the correct unit of measure for the standard deviation of the sampling distribution of x¯C−x¯O?
Answer:
answer should be minutes.
Step-by-step explanation:
in the problem, it is stated that the mean is in "baking times." this can lead one to assume that the standard deviation between the two xC-xO can be the variation in minutes that the two cookies types have.
1/3=n+3/4 what dies n equal??
Find the Jacobian ∂(x, y, z) ∂(u, v, w) for the indicated change of variables. If x = f(u, v, w), y = g(u, v, w), and z = h(u, v, w), then the Jacobian of x, y, and z with respect to u, v, and w is ∂(x, y, z) ∂(u, v, w) = ∂x ∂u ∂x ∂v ∂x ∂w ∂y ∂u ∂y ∂v ∂y ∂w ∂z ∂u ∂z ∂v ∂z ∂w . x = 1 6 (u + v), y = 1 6 (u − v), z = 6uvw
Answer:
The Jacobian ∂(x, y, z) ∂(u, v, w) for the indicated change of variables
= -3072uv
Step-by-step explanation:
Step :-(i)
Given x = 1 6 (u + v) …(i)
Differentiating equation (i) partially with respective to 'u'
[tex]\frac{∂x}{∂u} = 16(1)+16(0)=16[/tex]
Differentiating equation (i) partially with respective to 'v'
[tex]\frac{∂x}{∂v} = 16(0)+16(1)=16[/tex]
Differentiating equation (i) partially with respective to 'w'
[tex]\frac{∂x}{∂w} = 0[/tex]
Given y = 1 6 (u − v) …(ii)
Differentiating equation (ii) partially with respective to 'u'
[tex]\frac{∂y}{∂u} = 16(1) - 16(0)=16[/tex]
Differentiating equation (ii) partially with respective to 'v'
[tex]\frac{∂y}{∂v} = 16(0) - 16(1)= - 16[/tex]
Differentiating equation (ii) partially with respective to 'w'
[tex]\frac{∂y}{∂w} = 0[/tex]
Given z = 6uvw ..(iii)
Differentiating equation (iii) partially with respective to 'u'
[tex]\frac{∂z}{∂u} = 6vw[/tex]
Differentiating equation (iii) partially with respective to 'v'
[tex]\frac{∂z}{∂v} =6 u (1)w=6uw[/tex]
Differentiating equation (iii) partially with respective to 'w'
[tex]\frac{∂z}{∂w} =6 uv(1)=6uv[/tex]
Step :-(ii)
The Jacobian ∂(x, y, z)/ ∂(u, v, w) =
[tex]\left|\begin{array}{ccc}16&16&0\\16&-16&0\\6vw&6uw&6uv\end{array}\right|[/tex]
Determinant 16(-16×6uv-0)-16(16×6uv)+0(0) = - 1536uv-1536uv
= -3072uv
Final answer:-
The Jacobian ∂(x, y, z)/ ∂(u, v, w) = -3072uv
A national magazine took a poll to determine company's opinions about outsourcing. Outsourcing is defined as the assignment of critical, but non-core, business functions to outside specialists. The magazine found that 26 percent of business executives felt that the benefits of outsourcing were less than what they expected. A sample of 100 business executives was polled. In this sample 19% of the business executives felt the benefits of outsourcing were less than what they expected. You want to determine whether the proportion of business executives in the sample differs from the national magazine poll. Which statistical test should you use?
Answer:
Hypothesis test on a proportion.
There is no enough evidence to support the claim that the proportion of business executives in the sample differs from the national magazine poll.
Step-by-step explanation:
We should test the hypothesis of the proportion, with a z-statistic.
The claim is that the proportion of business executives in the sample differs from the national magazine poll.
Then, the null and alternative hypothesis are:
[tex]H_0: \pi=0.26\\\\H_a:\pi<0.26[/tex]
The significance level is assumed to be 0.05.
The sample, of size n=100, has a proportion p=0.19.
The standard error for the proportion is:
[tex]\sigma_p=\sqrt{\dfrac{\pi(1-\pi)}{n}}=\sqrt{\dfrac{0.26*0.74}{100}}=\sqrt{0.001924}=0.044[/tex]
Then, the z-statistic can be calculated as:
[tex]z=\dfrac{p-\pi+0.5/n}{\sigma_p}=\dfrac{0.19-0.26+0.5/100}{0.044}=\dfrac{-0.065}{0.044}=-1.477[/tex]
The P-value for this one-tailed test is:
[tex]P-value=P(z<-1.477)=0.07[/tex]
As the P-value is bigger than the significance level, the effect is not significant. The null hypothesis failed to be rejected.
There is no enough evidence to support the claim that the proportion of business executives in the sample differs from the national magazine poll.
Most hard drives are divided into sectors of 512 bytes each. Our disk has a size of 16 GB. Calculate how many sectors the disk has.
Answer:
33,554,432 sectors
Step-by-step explanation:
-We first convert both sizes to a uniform measure(both to be in bytes or gigabytes).
-We convert the drive size into Bytes:
[tex]1GB=1,073,741,824 \ Bytes\\\\\therefore 16GB=16\times1,073,741,824=1.717986918\times10^{10}\ Bytes[/tex]
#We then divide the Drive size by the sector size to get the number of sectors.
-Let X be the number of sectors:
[tex]X=\frac{Drive \ Size}{Sector \ Size}\\\\=\frac{1.717986918\times10^{10}}{512}\\\\=33,554,432[/tex]
Hence, the disk has 33,554,432 sectors.
To calculate the number of sectors in a 16GB disk, you first convert the disk size from GB to bytes. Then, you divide the total bytes by the size in bytes of one sector. So, a 16GB disk has 33,554,432 sectors.
Explanation:The disk size is provided in gigabytes (GB), but first, we need to convert it into bytes, as sectors are measured in bytes. There are 1,073,741,824 bytes in 1 GB. Therefore, a 16 GB disk has 16 * 1,073,741,824 = 17,179,869,184 bytes.
Since each sector is 512 bytes, we can find the number of sectors by dividing the total number of bytes by the size of one sector: 17,179,869,184 / 512 = 33,554,432 sectors.
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An ordinary (fair) coin is tossed 3 times. Outcomes are thus triples of "heads" (An ordinary (fair) coin is tossed 3 times. Ou) and "tails" () which we write , , etc. For each outcome, let be the random variable counting the number of tails in each outcome. For example, if the outcome is , then . Suppose that the random variable is defined in terms of as follows: . The values of are thus:Outcome Value of Calculate the probability distribution function of , i.e. the function . First, fill in the first row with the values of . Then fill in thevalue x of X Px(X)
The probability distribution function of X, filled in:
X P(X)
0 1/8
1 3/8
2 3/8
3 1/8
Possible outcomes: There are 2^3 = 8 possible outcomes when a fair coin is tossed 3 times (HHH, HHT, HTH, THH, HTT, THT, TTH, TTT).
Counting tails: We count the number of tails in each outcome to determine the values of X.
Probability of each X: We calculate the probability of each value of X by counting the outcomes that correspond to that value and dividing by the total number of outcomes (8).
Detailed breakdown:
X = 0: Only one outcome has 0 tails (HHH), so P(X=0) = 1/8.
X = 1: Three outcomes have 1 tail (HHT, HTH, THH), so P(X=1) = 3/8.
X = 2: Three outcomes have 2 tails (HTT, THT, TTH), so P(X=2) = 3/8.
X = 3: Only one outcome has 3 tails (TTT), so P(X=3) = 1/8.
6x A = 24 show your work
Answer:
4
Step-by-step explanation:
6 × A = 24
6 × A ÷ 6 = 24 ÷ 6
A = 4
Answer:
A = 4
Step-by-step explanation:
6×A = 24
6A = 24
A = 24 : 6
A = 4
Orange M&M’s: The M&M’s web site says that 20% of milk chocolate M&M’s are orange. Let’s assume this is true and set up a simulation to mimic buying 200 small bags of milk chocolate M&M’s. Each bag contains 55 candies. We made this dotplot of the results. Normal sampling distribution of proportion of orange candies Now suppose that we buy a small bag of M&M’s. We find that 25.5% (14 of the 55) of the M&M’s are orange. What can we conclude? Group of answer choices This result is not surprising because we expect to see many samples with 14 or more orange candies. This result is surprising because we expect the orange candies to make up no more than 20% of the candies in a packet. This result is surprising because it is unlikely that we will select a random sample with 25.5% or more orange candies if 20% of milk chocolate M&M’s are orange.
The initial expectation is that 20% of the M&Ms in a bag would be orange. However, the occurrence of 25.5% orange M&Ms in a bag can be considered surprising but doesn't negate the initial percentage due to statistical sampling variability.
Explanation:This question is dealing with the concept of statistical sampling and proportions. Given that the M&M’s website states that 20% of milk chocolate M&M’s are orange, we would expect, on average, that 20% or 11 (0.20 * 55) of the M&M’s in a small bag of 55 M&M’s to be orange. However, in this instance, we observe that 25.5% (or 14 of the 55) of the M&M’s are orange.
This result might be surprising because it is more than the average expectation of 20%. But it's important to remember the concept of sampling variability. Even though we expect 20%, the actual results can vary from this expectation due to random chance.
In conclusion, this result can be considered surprising because we have got a higher percentage of orange candies than expected, but it does not necessarily prove that the initial assertion of 20% orange M&M’s is incorrect due to the inherent variability in statistical sampling.
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The observed result of 25.5% orange candies per bag is surprising given the stated 20% proportion. However, due to sampling variability, it falls within one standard deviation from the mean, indicating it's a possible result in a sample.
Explanation:The conclusion based on your provided information would be that this result is surprising because it is unlikely to select a sample with 25.5% or more orange candies if only 20% of milk chocolate M&M's are orange. This observation is compared to the expected proportion of 20%, i.e., the stated percentage of orange M&M's as per the website information.
To calculate how surprising this outcome is, we can use the concept of standard deviation. Based on the binominal distribution, the standard deviation (σ) is calculated as √npq, where n is the number of trials (in this case, the number of candies per bag, 55), p is the probability of success (the expected proportion of orange candies, 0.20), and q is the probability of failure (1-p).
Applying these values, σ = √(55)(0.20)(0.80), we get σ ≈ 4.7. Therefore, observing 14 orange candies is about one standard deviation above the expected value (11 candies, or 20% of 55). Although that might seem like a significant deviation, in a normal distribution about 68% of the data falls within one standard deviation of the mean. So while it is surprising, it is still a possible variation to have 25.5% or more orange candies in a single bag due to sampling variability.
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A company offers collision insurance and liability insurance. The claim amount under collision insurance is normally distributed with mean 10, 000 and standard deviation 2, 000; the claim amount under liability insurance is normally distributed with mean 9, 000 and standard deviation 2, 000. Assuming independence, find the probability that the claim amount under liability insurance exceeds the claim amount under collision insurance.
Answer:
36.32% probability that the claim amount under liability insurance exceeds the claim amount under collision insurance.
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
Two variables:
X1 and X2, probability X1 exceeds X2
[tex]\mu(X_{1} - X_{2}) = \mu_{X_{1}} - \mu_{X_{2}}[/tex]
[tex]\sigma^{2}(X_{1} - X_{2}) = \sigma_{X_{1}}^{2} + \sigma_{X_{2}}^{2}[/tex]
In this problem, we have:
X1: Liability insurance:
So [tex]\mu_{X_{1}} = 9000, \sigma_{X_{1}} = 2000[/tex]
X2: Collusion insurance:
So [tex]\mu_{X_{2}} = 10000, \sigma_{X_{2}} = 2000[/tex]
Then
[tex]\mu(X_{1} - X_{2}) = \mu_{X_{1}} - \mu_{X_{2}} = 9000 - 10000 = -1000[/tex]
[tex]\sigma^{2}(X_{1} - X_{2}) = 2000^{2} + 2000^{2} = 8000000[/tex]
[tex]\sigma = \sqrt{8000000} = 2828.43[/tex]
Assuming independence, find the probability that the claim amount under liability insurance exceeds the claim amount under collision insurance.
This is [tex]P(X_{1} - X_{2} > 0)[/tex], which is 1 subtracted by the pvalue of Z when X = 0. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{0 - (-1000)}{2828.43}[/tex]
[tex]Z = 0.35[/tex]
[tex]Z = 0.35[/tex] has a pvalue of 0.6368
1 - 0.6368 = 0.3632
36.32% probability that the claim amount under liability insurance exceeds the claim amount under collision insurance.
What is the area of this tile ?
Two planes are 1530 mi apart and are traveling toward each other. One plane is traveling 100 mph faster than the other plane. The planes meet in 1.5 h. Find the speed of each plane.
Answer: The speed of one plane is 560 mph and the speed of the other is 460 mph.
Step-by-step explanation: Please see the attachments below
the quoteint of 2 and a number x,times 3
Answer:
just look it up on m..a.t.h. w.a.y.!!
Step-by-step explanation:
it works for me
Find the given percent of 7% of 98
Answer:
Hello! 7% of 98 is 6.86.
Step-by-step explanation:
98 x 0.07 = 6.86
If you move the percentage mark of the 7% two times to the left, you get 7% as a decimal, which is 0.07. You then take 0.07 and multiply it by 98 to get 6.86.
A note organizer is not important if you know how to take good notes.
Please select the best answer from the choices provided
OT
OF
Answer:
Yes it is true.
A note organizer is not important if you know how to take good notes is false.
What is a note organizer?A note organizer is a tool or system used to help keep notes organized and easy to access.
There are many different types of note organizers, ranging from physical tools like binders or folders to digital tools like note-taking apps or software.
The purpose of a note organizer is to help individuals keep track of important information, categorize and label their notes, and quickly find the information they need when they need it.
We have,
While it is true that taking good notes is essential, having a note organizer can also be important in keeping your notes organized and easily accessible.
A note organizer can help you categorize and label your notes, making it easier to find the information you need when you need it.
Thus,
A note organizer is not important if you know how to take good notes is false.
Learn more about note organizers here:
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