In the given sample, the parameter [tex]\mu[/tex] represents the average temperature of the coffee in the population whose value is not known.
In statistics, a parameter means a numerical attribute of a population. A population parameter describes a particular aspect of the entire population, which is the complete set of individuals, items, or units of interest. The mean, standard deviation, and variance are some of the population parameters.
A coffee company wants to make sure that their coffee is being served at the right temperature. If it is too hot, the customers could burn themselves. If it is too cold, the customers will be unsatisfied. The company has determined that they want the average coffee temperature to be 65 degrees C. They take a sample of 20 orders of coffee and find the sample mean to be equal to 70.2 C. [tex]\mu[/tex] represents the average temperature of the coffee in the population, and the value is not known for this problem.
Also, [tex]\bar{X}[/tex] represents the sample mean here which is known and equals 70.2 C.
b
Therefore, [tex]\mu[/tex] represents the population mean and is not known in this problem.
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The symbol mu in the coffee company's problem represents the population mean, the desired average coffee temperature, which should be 65 degrees C.
In the context of the coffee company's quality control problem, the symbol mu represents the population mean, which is the average temperature of all cups of coffee served by the company. The sample mean, denoted as x-bar, is 70.2°C, which is the average temperature calculated from the sample of 20 orders of coffee. The goal of the company is to have mu equal to 65°C, as they want this to be the average serving temperature to ensure customer satisfaction and safety. If they find that x-bar is significantly different from mu, it may suggest that corrective actions are needed to reach the desired temperature.
Geologists estimate the time since the most recent cooling of a mineral by counting the number of uranium fission tracks on the surface of the mineral. A certain mineral specimen is of such an age that there should be an average of 6 tracks per cm2 of surface area. Assume the number of tracks in an area follows a Poisson distribution. Let X represent the number of tracks counted in 1 cm2 of surface area.
a)Find P(X = 7).
b)Find P(X ≥ 3).
c)Find P(2 < X < 7).
d)Find μX.
e)Find σX
Answer:
(a) The value of P (X = 7) is 0.1388.
(b) The value of P (X ≥ 3) is 0.9380.
(c) The value of P (2 < X < 7) is 0.5433.
(d) [tex]\mu_{X}=6[/tex]
(e) [tex]\sigma_{X}=2.45[/tex]
Step-by-step explanation:
Let X = number of uranium fission tracks on per cm² surface area of the mineral.
The average number of track per cm² surface area is, λ = 6.
The random variable X follows a Poisson distribution with parameter λ = 6.
The probability mass function of a Poisson distribution is:
[tex]P(X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!};\ x=0, 1, 2, 3...[/tex]
(a)
Compute the value of P (X = 7) as follows:
[tex]P(X=6)=\frac{e^{-6}(6)^{7}}{7!}=\frac{0.0025\times 279936}{5040}=0.1388[/tex]
Thus, the value of P (X = 7) is 0.1388.
(b)
Compute the value of P (X ≥ 3) as follows:
P (X ≥ 3) = 1 - P (X < 3)
= 1 - P (X = 0) - P (X = 1) - P (X = 2)
[tex]=1-\frac{e^{-6}(6)^{0}}{0!}-\frac{e^{-6}(6)^{1}}{1!}-\frac{e^{-6}(6)^{2}}{2!}\\=1-0.00248-0.01487-0.04462\\=0.93803\\\approx0.9380[/tex]
Thus, the value of P (X ≥ 3) is 0.9380.
(c)
Compute the value of P (2 < X < 7) as follows:
P (2 < X < 7) = P (X = 3) + P (X = 4) + P (X = 5) + P (X = 6)
[tex]=\frac{e^{-6}(6)^{3}}{3!}+\frac{e^{-6}(6)^{4}}{4!}+\frac{e^{-6}(6)^{5}}{5!}+\frac{e^{-6}(6)^{6}}{6!}\\=0.08924+0.13385+0.16062+0.16062\\=0.54433\\\approx0.5443[/tex]
Thus, the value of P (2 < X < 7) is 0.5433.
(d)
The mean of the Poisson distribution is:
[tex]\mu_{X}=\lambda=6[/tex]
(e)
The standard deviation of the Poisson distribution is:
[tex]\sigma_{X}=\sqrt{\sigma^{2}_{X}}=\sqrt{\lambda}=\sqrt{6}=2.4495\approx2.45[/tex]
Blue Ribbon taxis offers shuttle service to the nearest airport. You look up the online reviews for Blue Ribbon taxis and find that there are 17 17 reviews, six of which report that the taxi never showed up. Is this a biased sampling method for obtaining customer opinion on the taxi service
Answer:
Yes, this a biased sampling method.
Step-by-step explanation:
The sampling done is a biased sampling method.
This is because usually people who are upset or dissatisfied with the service are the ones who write online reviews, as there are no other way to release their frustration with the service provided.
The bias is likely to be directed towards the proportion of people who are not satisfied with service.
When two standard dice are thrown, what is the probability that the sum of the dots on the two top faces will be 3? Express your answer using three significant digits.
The probability that the sum of the dots on two dice will be 3 is 0.056, determined by finding the number of favorable outcomes (2) and dividing by the total number of outcomes (36).
Explanation:The subject of this question is probability, specifically focusing on an experiment involving the tossing of two six-sided dice. To find the probability that the sum of the dots on the two top faces will be 3, we first need to determine the total number of possible outcomes when two dice are thrown. For one six-sided die, there are 6 possible outcomes. Therefore, when two dice are thrown, the total number of outcomes is 6 * 6 = 36.
Next, we need to find the number of outcomes where the sum of the dots would be 3. Looking at our dice, we can see that this can only happen in two ways: getting a 1 on the first die and a 2 on the second, or getting a 2 on the first die and a 1 on the second.
So, the number of favorable outcomes is 2.
Probability is defined as the number of favorable outcomes divided by the total number of outcomes. Hence, the probability of the sum of the dots being 3 when two dice are thrown is 2/36. To get this in three significant figures, we divide 2 by 36 which gives the decimal 0.0556. Therefore, the probability to three significant figures is 0.056.
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Suppose f:double-struck Rn → double-struck Rm and g:double-struck Rp → double-struck Rq. (a) What must be true about the numbers n, m, p, and q for f ∘ g to make sense? n = q m = p n = p m = q n = m (b) What must be true about the numbers n, m, p, and q for g ∘ f to make sense? n = q m = p n = p m = q n = m (c) When does f ∘ f make sense?
Answer:
See the attached picture.
Step-by-step explanation:
See the attached picture.
To ensure compositions f ∘ g, g ∘ f, and f ∘ f make sense, certain conditions must be met regarding the number of inputs and outputs of the functions.
Explanation:(a) For f ∘ g to make sense, the number of inputs of g must match the number of outputs of f. Therefore, n must equal p, and m must equal q.
(b) For g ∘ f to make sense, the number of inputs of f must match the number of outputs of g. Therefore, n must equal q, and m must equal p.
(c) For f ∘ f to make sense, the number of outputs of f must match the number of inputs of f. Therefore, n must equal m.
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3.3.1. An urn contains five balls numbered 1 to 5. Two balls are drawn simultaneously. (a) Let X be the larger of the two numbers drawn. Find pX (k). (b) Let V be the sum of the two numbers drawn. Find pV (k).
Answer:
Step-by-step explanation:
Given that an urn contains five balls numbered 1 to 5
Two balls are drawn simultaneously.
a) X = the larger of two numbers drawn
Assuming balls are drawn without replacement, (since simultaneously drawn)
the sample space would be (1,2) (1,3)(1,4)(1,5) (2,3)(2,4)(2,5) (3,4) (3,5) (4,5)
n(S) = 10
n(x=2) = 1,
n(x=3) =2
n(x=4) = 3
n(x=5) = 4
Larger value X can take values only as 2,3,4 or 5
X 2 3 4 5
p 0.1 0.2 0.3 0.4
-------------------------------------------
b) V = sum of numbers drawn
V can take values as 3, 4, 5, 6, 7, 8 or 9
V 3 4 5 6 7 8 9
p 0.1 0.1 0.2 0.2 0.2 0.1 0.1
The probability that the larger number drawn will be k follows a specific probability distribution, as does the sum of the two numbers drawn. Using the given information and principles of probability, we can determine the probabilities for each value of k for both X and V. For X, the probability distribution pX(k) is [1/5, 1/5, 3/10, 1/5, 1/5]. For V, the probability distribution pV(k) is [1/20, 1/10, 3/20, 1/10, 1/10, 1/10, 3/20, 1/10, 1/20].
(a) Let's consider each possible outcome to find pX(k), the probability that the larger number drawn will be k:
If k is 1, there is only one possible outcome: drawing (1, anything). The probability of this outcome is (1/5) * (4/4) = 1/5.
If k is 2, there are two possible outcomes: drawing (2, 1) or (2, 3). The probability of these outcomes is (1/5) * (1/4) + (1/5) * (1/4) = 1/10 + 1/10 = 1/5.
If k is 3, there are three possible outcomes: drawing (3, 1), (3, 2), or (3, 4). The probability of these outcomes is (1/5) * (1/4) + (1/5) * (1/4) + (1/5) * (2/4) = 1/10 + 1/10 + 1/10 = 3/10.
If k is 4, there are four possible outcomes: drawing (4, 1), (4, 2), (4, 3), or (4, 5). The probability of these outcomes is (1/5) * (1/4) + (1/5) * (1/4) + (1/5) * (1/4) + (1/5) * (3/4) = 1/10 + 1/10 + 1/10 + 3/20 = 1/5.
If k is 5, there are five possible outcomes: drawing (5, 1), (5, 2), (5, 3), (5, 4), or (5, 5). The probability of these outcomes is (1/5) * (1/4) + (1/5) * (1/4) + (1/5) * (1/4) + (1/5) * (1/4) + (1/5) * (4/4) = 1/10 + 1/10 + 1/10 + 1/10 + 1/5 = 1/5.
Therefore, the probability distribution pX(k) is:
pX(1) = 1/5
pX(2) = 1/5
pX(3) = 3/10
pX(4) = 1/5
pX(5) = 1/5
(b) Let's consider each possible outcome to find pV(k), the probability that the sum of the two numbers drawn will be k:
If k is 2, there is only one possible outcome: drawing (1, 1). The probability of this outcome is (1/5) * (1/4) = 1/20.
If k is 3, there are two possible outcomes: drawing (1, 2) or (2, 1). The probability of these outcomes is (1/5) * (1/4) + (1/5) * (1/4) = 1/20 + 1/20 = 1/10.
If k is 4, there are three possible outcomes: drawing (1, 3), (2, 2), or (3, 1). The probability of these outcomes is (1/5) * (1/4) + (1/5) * (1/4) + (1/5) * (1/4) = 1/20 + 1/20 + 1/20 = 3/20.
If k is 5, there are four possible outcomes: drawing (1, 4), (2, 3), (3, 2), or (4, 1). The probability of these outcomes is (1/5) * (1/4) + (1/5) * (1/4) + (1/5) * (1/4) + (1/5) * (1/4) = 1/20 + 1/20 + 1/20 + 1/20 = 1/10.
If k is 6, there are five possible outcomes: drawing (1, 5), (2, 4), (3, 3), (4, 2), or (5, 1). The probability of these outcomes is (1/5) * (1/4) + (1/5) * (1/4) + (1/5) * (1/4) + (1/5) * (1/4) + (1/5) * (1/4) = 1/20 + 1/20 + 1/20 + 1/20 + 1/20 = 1/10.
If k is 7, there are four possible outcomes: drawing (2, 5), (3, 4), (4, 3), or (5, 2). The probability of these outcomes is (1/5) * (1/4) + (1/5) * (1/4) + (1/5) * (1/4) + (1/5) * (1/4) = 1/20 + 1/20 + 1/20 + 1/20 = 1/10.
If k is 8, there are three possible outcomes: drawing (3, 5), (4, 4), or (5, 3). The probability of these outcomes is (1/5) * (1/4) + (1/5) * (1/4) + (1/5) * (1/4) = 1/20 + 1/20 + 1/20 = 3/20.
If k is 9, there are two possible outcomes: drawing (4, 5) or (5, 4). The probability of these outcomes is (1/5) * (1/4) + (1/5) * (1/4) = 1/20 + 1/20 = 1/10.
If k is 10, there is only one possible outcome: drawing (5, 5). The probability of this outcome is (1/5) * (1/4) = 1/20.
Therefore, the probability distribution pV(k) is:
pV(2) = 1/20
pV(3) = 1/10
pV(4) = 3/20
pV(5) = 1/10
pV(6) = 1/10
pV(7) = 1/10
pV(8) = 3/20
pV(9) = 1/10
pV(10) = 1/20
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Suppose that, during the next hour, 20 customers each purchase one cup of coffee. Assume that the customers make their decisions independently, and each customer has a 17% probability of ordering decaffeinated coffee. What is the probability that everyone will order a coffee with caffeine? g
Answer:
0.0241 or 2.41%
Step-by-step explanation:
Since each customer has a 17% probability of ordering decaffeinated coffee, the probability of a customer ordering a coffee with caffeine is:
[tex]P(C) = 1-0.17 = 0.83[/tex]
If customers make their decisions independently, the probability that all 20 order a coffee with caffeine is:
[tex]P(X=20)=P(C)^{20}\\P(X=20) = 0.83^{20}=0.0241=2.41\%[/tex]
The probability is 0.0241 or 2.41%.
The Christmas Bird Count (CBC) is an annual tradition in Lexington, Massachusetts. A group of volunteers counts the number of birds of different species over a 1-day period. Each year, there are approximately 30–35 hours of observation time split among multiple volunteers. The following counts were obtained for the Northern Cardinal (or cardinal, in brief) for the period 2005–2011.
Year Number
2005 76
2006 47
2007 63
2008 53
2009 62
2010 69
2011 62
5.126 What is the mean number of cardinal birds per year observed from 2005 to 2011?
5.127 What is the standard deviation (sd) of the number of cardinal birds observed?
5.128 What is the probability of observing at least 60 cardinal birds in 2012? (Hint: Apply a continuity correction where appropriate.)
The observers wish to identify a normal range for the number of cardinal birds observed per year. The normal range will be defined as the interval (L, U), where L is the largest integer ≤ 15th percentile and U is the smallest integer ≥ 85th percentile.
5.129 If we make the same assumptions as in Problem 5.128, then what is L? What is U?
5.130 What is the probability that the number of cardinal birds will be ≥ U at least once on Christmas day during the 10-year period 2012–2021? (Hint: Make the same assumptions as in Problem 5.128.)
Answer:
(5.216) mean = 61.71
(5.217) standard deviation = 8.88
(5.218) P(X>=60) = 0.9238
(5.129) L = 79, U = 69
(5.130) P(X> or = U) = 0.7058
Step-by-step explanation:
The table of the statistic is set up as shown in attachment.
(5.216) mean = summation of all X ÷ no of data.
mean = 432/7 = 61.71 birds
(5.217)Standard deviation = √ sum of the absolute value of difference of X from mean ÷ number of data
S = √ /X - mean/ ÷ 7
= √551.428/7
S = 8.88
(5.218) P (X> or = 60)
= P(Z> or =60 - 61.71/8.8 )
= P(Z>or= - 0.192)
= 1 - P(Z< or = 0.192)
= 1- 0.0762
= 0.9238
(5.219)the 15th percentile=15/100 × 7
15th percentile = 1.05
The value is the number in the first position and that is 79,
L= 79
85th percentile = 85/100 × 7 = 5.95
The value is the number in the 6th position, and that is 69
U = 69
5.130) P(X>or = 60)
= P(Z>or= 69 - 61.71/8.8)
= P(Z> or = 0.8208)
= 1 - P(Z< or = 0.8209)
= 1 - 0.2942
= 0.7058
It’s estimated that 52% of American adults have incurred credit card debt. The department of finance surveys 2500 adults for a report.
Determine the probability that between 1200 and 1450 of those surveyed incurred debt.
Answer:
100% probability that between 1200 and 1450 of those surveyed incurred debt.
Step-by-step explanation:
We use the binomial approximation to the normal to solve this question.
Binomial probability distribution
Probability of exactly x sucesses on n repeated trials, with p probability.
Can be approximated to a normal distribution, using the expected value and the standard deviation.
The expected value of the binomial distribution is:
[tex]E(X) = np[/tex]
The standard deviation of the binomial distribution is:
[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]
Normal probability distribution
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
When we are approximating a binomial distribution to a normal one, we have that [tex]\mu = E(X)[/tex], [tex]\sigma = \sqrt{V(X)}[/tex].
In this problem, we have that:
[tex]n = 2500, p = 0.52[/tex]
So
[tex]\mu = E(X) = 2500*0.52 = 1300[/tex]
[tex]\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{2500*0.52*0.48} = 24.98[/tex]
Determine the probability that between 1200 and 1450 of those surveyed incurred debt.
This is the pvalue of Z when X = 1450 subtracted by the pvalue of Z when X = 1200. So
X = 1450
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{1450 - 1300}{24.98}[/tex]
[tex]Z = 6[/tex]
[tex]Z = 6[/tex] has a pvalue of 1
X = 1200
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{1200 - 1300}{24.98}[/tex]
[tex]Z = -4[/tex]
[tex]Z = -4[/tex] has a pvalue of 0
1 - 0 = 1
100% probability that between 1200 and 1450 of those surveyed incurred debt.
PLEASE HELP ONLY IF RIGHT 50 POINTS AND BRAINLIEST PLUS THANK YOU AND 5 STARS. My cousin need help.
Anthony has a sink that is shaped like a half-sphere. The sink has a volume of 4000/3*π in^3. One day, his sink clogged. He has to use one of two cylindrical cups to scoop the water out of the sink. The sink is completely full when Anthony begins scooping. Hint: you may need to find the volume for both.
1.)One cup has a diameter of 4 in. and a height of 8 in. How many cups of water must Anthony scoop out of the sink with this cup to empty it? Round the number of scoops to the nearest whole number, and make certain to show your work.
2.One cup has a diameter of 8 in. and a height of 8 in. How many cups of water must he scoop out of the sink with this cup to empty it? Round the number of scoops to the nearest whole number, and make certain to show your work.
Answer:
1) Number of scoops = 42
2)Number of scoops = 10
Step-by-step explanation:
Volume of the sink = (4000/3)π in³
1)
Diameter = 4 in ∵ Radius = 4/2 = 2 in
Height = 8 in
Volume of the cylindrical cup = [tex]\pi r^{2}h = \pi 2^{2} * 8 = 32\pi[/tex] in³
Number of cups of water that must be scooped out = Volume of the sink/ Volume of the cylindrical cup
= (4000/3)π / 32π ==> 42
Number of scoops = 42
b)
Diameter = 8 in ∵ Radius = 8/2 = 4 in
Height = 8 in
Volume of the cylindrical cup = [tex]\pi r^{2}h = \pi 4^{2} * 8 = 128\pi[/tex] in³
Number of cups of water that must be scooped out = Volume of the sink/ Volume of the cylindrical cup
= (4000/3)π / 128π ==> 10
Number of scoops = 10
A mexican restaurant sells quesadillas in two sizes: a “large” 12 inch-round quesadilla and a “small” 6 inch-round quesadilla. Which is larger, half of the 12-inch quesadilla or the entire 6 inch quesadilla?
Answer:
"half of the 12-inch quesadilla" is larger
Step-by-step explanation:
The "inch size" is the diameter of the circular quesadilla.
The area of circle is given by [tex]\pi r^2[/tex]
We also know radius is half of diameter.
HALF OF 12-INCH QUESADILLA:
12in diameter = 6 inch radius, the area is:
[tex]\pi(6)^2=36\pi[/tex]
Half of this is:
[tex]\frac{36\pi}{2}=18\pi[/tex]
ENTIRE 6-INCH QUESADILLA:
6 inch diameter means 6/2 = 3 inch radius
The area of this is:
[tex]\pi(3)^2 = 9\pi[/tex]
So, we can say that "half of the 12-inch quesadilla" is larger.
A class survey in a large class for first-year college students asked, "About how many minutes do you study on a typical weeknight?" The mean response of the 255 students was x¯¯¯ = 148 minutes. Suppose that we know that the studey time follows a Normal distribution with standard deviation σ = 65 minutes in the population of all first-year students at this university. Regard these students as an SRS from the population of all first-year students at this university. Does the study give good evidence that students claim to study more than 2 hours per night on the average? (a) State null and alternative hypotheses in terms of the mean study time in minutes for the population. (b) What is the value of the test statistic z? (c) Can you conclude that students do claim to study more than two hours per weeknight on the average?
Answer:
There is enough evidence to support the claim that students study more than two hours per weeknight on the average.
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = 2 hours = 120 minutes
Sample mean, [tex]\bar{x}[/tex] = 148 minutes = 2.467 hours
Sample size, n = 255
Alpha, α = 0.05
Population standard deviation, σ = 65 minutes
First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu = 120\text{ minutes}\\H_A: \mu > 120\text{ minutes}[/tex]
We use one-tailed z test to perform this hypothesis.
Formula:
[tex]z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }[/tex]
Putting all the values, we have
[tex]z_{stat} = \displaystyle\frac{148 - 120}{\frac{65}{\sqrt{255}} } = 6.8788[/tex]
Now, we calculate the p-value from the normal standard table.
P-value = 0.00001
Since the p-value is smaller than the significance level, we fail accept the null hypothesis and reject it, We accept the alternate hypothesis.
Conclusion:
Thus, there is enough evidence to support the claim that students study more than two hours per weeknight on the average.
Answer:
There is enough evidence to support the claim that students study more than two hours per weeknight on the average.
Step-by-step explanation:
A publisher reports that 344% of their readers own a particular make of car. A marketing executive wants to test the claim that the percentage is actually different from the reported percentage. A random sample of 220220 found that 300% of the readers owned a particular make of car. Is there sufficient evidence at the 0.020.02 level to support the executive's claim
Answer:
No, there is not enough evidence at the 0.02 level to support the executive's claim.
Step-by-step explanation:
We are given that a publisher reports that 34% of their readers own a particular make of car. A random sample of 220 found that 30% of the readers owned a particular make of car.
And, a marketing executive wants to test the claim that the percentage is actually different from the reported percentage, i.e;
Null Hypothesis, [tex]H_0[/tex] : p = 0.34 {means that the percentage of readers who own a particular make of car is same as reported 34%}
Alternate Hypothesis, [tex]H_1[/tex] : p [tex]\neq[/tex] 0.34 {means that the percentage of readers who own a particular make of car is different from the reported 34%}
The test statistics we will use here is;
T.S. = [tex]\frac{\hat p -p}{\sqrt{\frac{\hat p(1- \hat p)}{n} } }[/tex] ~ N(0,1)
where, p = actual % of readers who own a particular make of car = 0.34
[tex]\hat p[/tex] = percentage of readers who own a particular make of car in a
sample of 220 = 0.30
n = sample size = 220
So, Test statistics = [tex]\frac{0.30 -0.34}{\sqrt{\frac{0.30(1- 0.30)}{220} } }[/tex]
= -1.30
Now, at 0.02 significance level, the z table gives critical value of -2.3263 to 2.3263. Since our test statistics lie in the range of critical values which means it doesn't lie in the rejection region, so we have insufficient evidence to reject null hypothesis.
Therefore, we conclude that the actual percentage of readers who own a particular make of car is same as reported percentage and the executive's claim that it is different is not supported.
Rona drove 56 milesto visit a friend she drove 42 miles before stopping for gas what percent of the drive did rona complete before stopping for gas
Answer:
75%
Step-by-step explanation:
Find the percentage corresponding to 42/56: (0.75)(100%) = 75%
Answer:
75%
Step-by-step explanation:
Becauuuuuuseeee if you divide 42 by 56 it would be 0.75 and yeah. so take out the 0 and its 75%
A Type II error is defined as which of the following?
A. rejecting a false null hypothesis
B. rejecting a true null hypothesis
C. failing to reject a false null hypothesis
D. failing to reject a true null hypothesis
Answer:
Option C. failing to reject a false null hypothesis
Step-by-step explanation:
Type II error states that Probability of accepting null hypothesis given the fact that null hypothesis is false.
This is considered to be the most important error.
So, from the option given to us option C matches that failing to reject a false null hypothesis.
Calculate the probability that a particle in a one-dimensional box of length aa is found between 0.29a0.29a and 0.34a0.34a when it is described by the following wave function: 2a−−√sin(πxa)2asin(πxa). Express your answer to two significant figures.
Answer:
P(0.29a < X < 0.34a) = -(4a/π) [(cos 0.34πa²) - (cos 0.29πa²)]
Assuming the box is of unit length, a = 1,
0.29a = 0.29 and 0.34a = 0.34
P(0.29 < X < 0.34) = (-4/π) [(cos 0.34π) - (cos 0.29π)] = 0.167 = 0.17 to 2 s.f
Step-by-step explanation:
ψ(x) = 2a(√sin(πxa)
The probability of finding a particle in a specific position for a given energy level in a one-dimensional box is related to the square of the wavefunction
The probability of finding a particle between two points 0.29a and 0.34a is given mathematically as
P(0.29a < X < 0.34a) = ∫⁰•³⁴ᵃ₀.₂₉ₐ ψ²(x) dx
That is, integrating from 0.29a to 0.34a
ψ²(x) = [2a(√sin(πxa)]² = 4a² sin(πxa)
∫⁰•³⁴ᵃ₀.₂₉ₐ ψ²(x) dx = ∫⁰•³⁴ᵃ₀.₂₉ₐ (4a² sin(πxa)) dx
∫⁰•³⁴ᵃ₀.₂₉ₐ (4a² sin(πxa))
= - [(4a²/πa)cos(πxa)]⁰•³⁴ᵃ₀.₂₉ₐ
- [(4a/π)cos(πxa)]⁰•³⁴ᵃ₀.₂₉ₐ = -(4a/π) [(cos 0.34πa²) - (cos 0.29πa²)]
P(0.29a < X < 0.34a) = -(4a/π) [(cos 0.34πa²) - (cos 0.29πa²)]
Assuming the box is of unit length, a = 1
P(0.29 < X < 0.34) = (-4/π) [(cos 0.34π) - (cos 0.29π)] = (-1.273) [0.482 - 0.613] = 0.167.
Assume the random variable X has a binomial distribution with the given probability of obtaining a success. Find the following probability, given the number of trials and the probability of obtaining a success. Round your answer to four decimal places. P(X=17), n=18, p=0.9
Answer:
P(X = 17) = 0.3002
Step-by-step explanation:
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
In this problem we have that:
[tex]n = 18, p = 0.9[/tex]
We want P(X = 17). So
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 17) = C_{18,17}.(0.9)^{17}.(0.1)^{1} = 0.3002[/tex]
P(X = 17) = 0.3002
Answer:
P(X=17) = 0.3002 .
Step-by-step explanation:
We are given that the random variable X has a binomial distribution with the given probability of obtaining a success.
The above situation can be represented through Binomial distribution;
[tex]P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....[/tex]
where, n = number of trials (samples) taken = 18
r = number of success = 17
p = probability of success which in our question is given as 0.9 .
So, X ~ [tex]Binom(n=18,p=0.9)[/tex]
We have to find the probability of P(X = 17);
P(X = 17) = [tex]\binom{18}{17}0.9^{17} (1-0.9)^{18-17}[/tex]
= [tex]18 \times 0.9^{17} \times 0.1^{1}[/tex] { [tex]\because \binom{n}{r} = \frac{n!}{r! \times (n-r)!}[/tex] }
= 0.3002
Therefore, P(X=17) = 0.3002 .
A fair die is rolled 8 times. What is the probability that the die comes up 6 exactly twice? What is the probability that the die comes up an odd number exactly five times? Find the mean number of times a 6 comes up. Find the mean number of times an odd number comes up. Find the standard deviation of the number of times a 6 comes up. Find the standard deviation of the number of times an odd number comes up.
Answer:
0.2605, 0.2188, 1.33, 4, 1.0540, 1.4142
Step-by-step explanation:
A fair die is rolled 8 times.
a. What is the probability that the die comes up 6 exactly twice?
b. What is the probability that the die comes up an odd number exactly five times?
c. Find the mean number of times a 6 comes up.
d. Find the mean number of times an odd number comes up.
e. Find the standard deviation of the number of times a 6 comes up.
f. Find the standard deviation of the number of times an odd number comes up.
a. A die is rolled 8 times. If A represent the number of times a 6 comes up. For a fair die the probability that the die comes up 6 is 1/6 - Thus A ~ Bin(8, 1/6)
The probability mass function of the random variable A is
[tex]p(A) = \left \{ {\frac{8!}{x!(8 - x)!}*(\frac{1}{6} )^{A}*(\frac{5}{6} )^{8-A} } \right. for A=0,1, ...8[/tex]
hence, p(6 twice) implies P(A=2)
that is P(2) substitute A = 2
[tex]p(2) = \left \{ {\frac{8!}{2!(8 - 2)!}*(\frac{1}{6} )^{2}*(\frac{5}{6} )^{8-2} } \right. for A=0,1, ...8[/tex]
[tex]p(2)=\frac{8!}{2!6!} *(\frac{1}{6} )^{2} *(\frac{5}{6} )^{6}[/tex]
p(2) = 0.2605
b. If B represent the number of times an odd number comes up. For the fair die the probability that an odd number comes up is 0.5.
Thus B ~ Bin(8, 1/2 )
The probability mass function of the random variable B is given by
[tex]p(B) = \left \{ {\frac{8!}{B!(8 - B)!}*(\frac{1}{2} )^{B\\}*(\frac{1}{2} )^{8-B} } \right. for B=0,1, ...8[/tex]
hence p(odd comes up 5 times) is
[tex]p(x=5) = p(2)=\frac{8!}{5!3!} *(\frac{1}{2} )^{5} *(\frac{1}{2} )^{3}[/tex]
p(5) = 0.2188
c. let the mean no of times a 6 comes up be μₐ
and let the total number of outcomes be n
using the formula μₐ = nρₐ
μₐ = 8 * 1/6
= 1.33
d. let the mean nos of times an odd nos comes up be μₓ
let the total outcomes be n = 8
let the probability odd be pb = 1/2
μₓ = npb
= 8 * (1/2)
= 4
e. the standard deviation of a random variable A is given as follows
σₐ [tex]= \sqrt{np(1-p)}[/tex]
where p = 1/6 (prob 6 outcome)
n = total outcomes = 8
[tex]= \sqrt{8*\frac{1}{6}*\frac{5}{6} }[/tex]
= 1.0540
f. the standard dev of the binomial random variable Y is given by
σ [tex]= \sqrt{np(1-p)}[/tex]
where p = 1/2 and n = 8
= [tex]\sqrt{8*\frac{1}{2} *\frac{1}{2} }[/tex]
= 1.4142
An observation from a normally distributed population is considered "unusual" if it is more than 2 standard deviations away from the mean. There are several contaminants that can harm a city's water supply. Nitrate concentrations above 10 ppm (parts per million) are considered a health risk for infants less than six month of age. The City of Rexburg reports that the nitrate concentration in the city's drinking water supply is between 1.59 and 2.52 ppm (parts per million,) and values outside of this range are unusual. We will assume 1.59 ppm is the value of mu - 2 sigma and mu + 2 sigma is equal to 2.52 ppm. It is reasonable to assume the measured nitrate concentration is normally distributed.
(Source: City of Rexburg)
Use this information to answer questions 20 and 22.
20. Estimate the mean of the measured nitrate concentration in Rexburg's drinking water. (Round your answer to three decimal places)
21. Estimate the standard deviation of the measured nitrate concentration in Rexburg's drinking water. (Round your answer to three decimal places)
22. Between what two measured nitrate concentrations do approximately 68% of the data values lie?
A.2.055 and 2.520 ppm
B.1.357 and 2.753 ppm
C.1.823 and 2.288 ppm
D.1.590 and 2.520 ppm
Answer:
Step-by-step explanation:
Hello!
X: nitrate concentration in the city's drinking water supply.
X~N(μ;δ²)
A concentration above 10 ppm is a health risk for infants less than 6 months old.
The reported nitrate concentration is between 1.59 and 2.52 ppm, values outside this range are considered unusual.
If any value of a normal distribution that is μ ± 2δ is considered unusual, it is determined that:
μ - 2δ= 1.59
μ + 2δ= 2.52
20) and 21)
I'll clear the values of the mean and the standard deviation using the given information:
a) μ - 2δ= 1.59 ⇒ μ = 1.59 + 2δ
b) μ + 2δ= 2.52 ⇒ Replace the value of Mu from "a" in "b" and clear the standard deviation:
(1.59 + 2δ) + 2δ= 2.52
1.59 + 4δ= 2.52
4δ= 2.52 - 1.59
δ= 0.93/4
δ= 0.2325
Then the value of the mean is μ = 1.59 + 2(0.2325)
μ = 1.59 + 0.465
μ = 2.055
22)
Acording to the empirical rule, 68% of a normal distribution is between μ±δ
Then you can expect 68% of the distribution between:
μ-δ= 2.055-0.2325= 1.8225
μ+δ= 2.055+0.2325= 2.2875
Correct option: C. 1.823 and 2.288 ppm
I hope it helps!
The distribution of actual weight of tomato soup in a 16 ounce can is thought to be be shaped with a ean equal to 16 ounces, and a standard deviation equal to 0.25 ounces. Based on this information, between what two values could we expect 95% of all cans to weigh?
(a) 15.75 to 16.25 ounces
(b) 15.50 to 16.50 ounces
(c) 15.25 to 16.75 ounces
(d) 15 to 17 ounces
Answer:
(b) 15.50 to 16.50 ounces
Step-by-step explanation:
The Empirical Rule states that, for a normally distributed(bell shaped) random variable:
68% of the measures are within 1 standard deviation of the mean.
95% of the measures are within 2 standard deviation of the mean.
99.7% of the measures are within 3 standard deviations of the mean.
In this problem, we have that:
Mean = 16
Standard deviation = 0.25
Based on this information, between what two values could we expect 95% of all cans to weigh?
By the Empirical Rule, within 2 standard deviations of the mean. So
16 - 2*0.25 = 15.50 ounces
16 + 2*0.25 = 16.50 ounces
So the corret answer is:
(b) 15.50 to 16.50 ounces
Final answer:
By calculating two standard deviations from the mean in both directions, we determine that 95% of cans are expected to weigh between 15.5 and 16.5 ounces. Thus, option b is correct.
Explanation:
The question involves finding between which two values 95% of all cans of tomato soup would weigh, given that the mean weight is 16 ounces with a standard deviation of 0.25 ounces. This question can be answered by applying the empirical rule (also known as the 68-95-99.7 rule) for a normal distribution, which states that approximately 95% of the data falls within two standard deviations of the mean.
To calculate the range:
1. Subtract two standard deviations from the mean to find the lower limit.
16 ounces - (2 * 0.25 ounces) = 15.5 ounces
2. Add two standard deviations to the mean to find the upper limit.
16 ounces + (2 * 0.25 ounces) = 16.5 ounces
Therefore, we can expect 95% of all cans to weigh between 15.5 and 16.5 ounces, which corresponds to option (b) 15.50 to 16.50 ounces.
You are testing the null hypothesis that the population proportion equals .45, using data you collected from a sample of 100 adults. You sample proportion equals .30. What does Z equal
Answer:
[tex]z=\frac{0.3 -0.45}{\sqrt{\frac{0.45(1-0.45)}{100}}}=-3.015[/tex]
Step-by-step explanation:
Data given and notation
n=100 represent the random sample taken
[tex]\hat p=0.3[/tex] estimated proportion of interest
[tex]p_o=0.45[/tex] is the value that we want to test
[tex]\alpha[/tex] represent the significance level
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value (variable of interest)
Concepts and formulas to use
We need to conduct a hypothesis in order to test the claim that the true proportion is 0.45.:
Null hypothesis:[tex]p=0.45[/tex]
Alternative hypothesis:[tex]p \neq 0.45[/tex]
When we conduct a proportion test we need to use the z statistic, and the is given by:
[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)
The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].
Calculate the statistic
Since we have all the info requires we can replace in formula (1) like this:
[tex]z=\frac{0.3 -0.45}{\sqrt{\frac{0.45(1-0.45)}{100}}}=-3.015[/tex]
The z-value is approximately -3.01.
To test the null hypothesis about a population proportion, you can use the z-test for proportions. The formula for the z-test statistic for proportions is:
[tex]z= \frac{p -p_0}{\frac{\sqrt{p_0(1 -p_0)}}{n} }[/tex]
where:
p is the sample proportion,
p_0 is the hypothesized population proportion under the null hypothesis, n is the sample size.
In your case:
p =0.30 (sample proportion),
p_0 =0.45 (hypothesized population proportion),
n=100 (sample size).
Now plug these values into the formula:
Calculate the values within the brackets first:
[tex]z= \frac{0.30 - 0.45}{\frac{\sqrt{0.45(1 -0.45)}}{100} }[/tex]
[tex]z= \frac{-0.15}{\frac{\sqrt{0.45(0.55)}}{100} }[/tex]
[tex]z= \frac{-0.15}{\frac{\sqrt{0.2475}}{100} }[/tex]
[tex]z= \frac{-0.15}{\sqrt{0.002475} }[/tex]
z≈−3.01
So, the z-value is approximately -3.01.
for such more question on null hypothesis
https://brainly.com/question/4436370
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One of four calculator batteries is bad. An experiment consists of testing each battery until the dead one is found.
(a) How many possible outcomes are there for this experiment?
(b) Is the outcome GBGG (Good, Bad, Good, Good) possible? Why or why not?
Answer:
(a) 4 possible outcomes
(b) Not possible, testing stops upon finding the bad battery,
Step-by-step explanation:
Let G denote a good battery and B denote a bad battery.
Whenever a bad battery is found, the batteries stop being tested.
(a) The outcomes for this experiment are:
B
GB
GGB
GGGB
There are 4 possible outcomes.
(b) The outcome GBGG is not possible since testing stops upon finding the bad battery, the outcome in this case would be GB.
Suppose that the number of calls coming per minute into an airline reservation center follows a Poisson distribution. Assume that the mean number of calls is 3 calls per minute. The probability that no calls are received in a given one-minute period is _______.
Answer:
0.0497 is the required probability.
Step-by-step explanation:
We are given the following information in the question:
Mean number of calls = 3 calls per minute
The number of calls coming per minute is following a Poisson distribution.
Formula:
[tex]P(X =k) = \displaystyle\frac{\lambda^k e^{-\lambda}}{k!}\\\\ \lambda \text{ is the mean of the distribution}[/tex]
We have to evaluate
[tex]P( x =0) = \displaystyle\frac{\lambda^k e^{-\lambda}}{k!}\\\\P(x = 0) = \frac{3^0e^{-3}}{0!} = 0.0497[/tex]
0.0497 is the probability that no calls are received in a given one-minute period.
Big chickens: The weights of broilers (commercially raised chickens) are approximately normally distributed with mean 1511 grams and standard deviation 198 grams. Use the TI-84 Plus calculator to answer the following. (a) What proportion of broilers weigh between 1143 and 1242 grams?
Answer:
5.55% of broilers weigh between 1143 and 1242 grams
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.
In this problem, we have that:
[tex]\mu = 1511, \sigma = 198[/tex]
What proportion of broilers weigh between 1143 and 1242 grams?
This is the pvalue of Z when X = 1242 subtracted by the pvalue of Z when X = 1143.
X = 1242
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{1242 - 1511}{198}[/tex]
[tex]Z = -1.36[/tex]
[tex]Z = -1.36[/tex] has a pvalue of 0.0869
X = 1143
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{1143 - 1511}{198}[/tex]
[tex]Z = -1.36[/tex]
[tex]Z = -1.86[/tex] has a pvalue of 0.0314
0.0869 - 0.0314 = 0.0555
5.55% of broilers weigh between 1143 and 1242 grams
Based on data from a large study of healthy infants in six countries, the World Health Organization produced growth charts that are part of every pediatrician’s toolkit for monitoring a child’s overall health. According to the WHO report, girls who are one month old have a mean head circumference of 36.6 centimeters with a standard deviation of 1.2 centimeters. As with most body measurements, head circumference has a normal probability distribution. Medscape defines microcephaly (small head syndrome) as a head circumference that is more than two standard deviations below the mean. What is the probability that a one-month old girl will be categorized as having microcephaly? Group of answer choices 68% 95% 5% 2.5%
Answer:
We are interested on this probability
[tex]P(X<\mu -2\sigma = 36.6 -2*1.2 =34.2)[/tex]
And the best way to solve this problem is using the normal standard distribution and the z score given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
If we apply this formula to our probability we got this:
[tex]P(X<34.2)=P(\frac{X-\mu}{\sigma}<\frac{34.2-\mu}{\sigma})=P(Z<\frac{34.2-36.6}{1.2})=P(z<-2)[/tex]
And we can find this probability using the normal standard table or excel and we got:
[tex]P(z<-2)=0.025[/tex]
And the best answer for this case would be:
2.5%
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 head circumference of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(36.6,1.2)[/tex]
Where [tex]\mu=36.6[/tex] and [tex]\sigma=1.2[/tex]
We are interested on this probability
[tex]P(X<\mu -2\sigma = 36.6 -2*1.2 =34.2)[/tex]
And the best way to solve this problem is using the normal standard distribution and the z score given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
If we apply this formula to our probability we got this:
[tex]P(X<34.2)=P(\frac{X-\mu}{\sigma}<\frac{34.2-\mu}{\sigma})=P(Z<\frac{34.2-36.6}{1.2})=P(z<-2)[/tex]
And we can find this probability using the normal standard table or excel and we got:
[tex]P(z<-2)=0.025[/tex]
And the best answer for this case would be:
2.5%
A regression model, y = 4 + 4x1 + 6x2, with the undernoted parameters. Parameters: R2= 0.88 Sig of F: 0.04 p-value of x1= 0.05 p-value of x2= 0.08 a. Should NOT be rejected because all the parameters are generally acceptable. b. Should be rejected because the Sig of F: 0.04 is not high enough to be generally acceptable (95% or more) as a measure of confidence in the R2 . (1 - sig of F = Confidence Level) c. Should be rejected because the p-value of x2= 0.08 is not high enough to be generally acceptable (95% or more) (1- p = confidence level) d. Should be rejected because the R2 is very low
Answer:
a. Should NOT be rejected because all the parameters are generally acceptable.
Step-by-step explanation:
A warehouse receives orders for a particular product on a regular basis. When an order is placed, customers can order 3, 4, 5, ..., 22 units of the product. Historical data suggest that the size of any given order is equally likely to be of any of these sizes. Let X denote the size of an order.
Find the probability that a customer orders at most five units?
Answer:
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Step-by-step explanation:
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An individual is hosting a cookout for the kick ball team. The individual wants to have two hot dogs for each guest, and 6 extra hot dogs in case some teammates bring friends. Which variable is dependent?
Answer:
6 extra hot dogs.
Step-by-step explanation:
A dependent variable is the variable being tested and measured in a scientific experiment. The dependent variable is 'dependent' on the independent variable
The 6 hot dogs are dependent on whether the teammates will bring friends. In other words, if there are no friends, there are no hot dogs. Note that the teammates are the independent variable that is they are not dependent on any variable.
The probabilities that stock A will rise in price is 0.59 and that stock B will rise in price is 0.41. Further, if stock B rises in price, the probability that stock A will also rise in price is 0.61. a. What is the probability that at least one of the stocks will rise in price.b. Are events A and B mutually exclusive? i. Yes because P(A | B) = P(A).ii. Yes because P(A ∩ B) = 0.iii. No because P(A | B) ≠ P(A).iv. No because P(A ∩ B) ≠ 0.c. Are events A and B independent? i. Yes because P(A | B) = P(A).ii.Yes because P(A ∩ B) = 0.iii.No because P(A | B) ≠ P(A).iv.No because P(A ∩ B) ≠ 0.
Answer:
1) 0.75
2) No, the 2 events are not mutually exclusive because P(A&B) ≠ 0
3)The 2 events are not independent because P(A | B) ≠ P(A).
Step-by-step explanation:
1) P(A) = 0.59 and P(B) = 0.41
Thus; to find the probability that at least one of the stocks will rise in price;
P(A or B) = P(A) + P(B) - P(A&B)
Now, we know that if B rises in price, the probability that A will also rise is 0.61 i.e P(A|B) = 0.61
Hence;P(A & B) = P(A|B) x P(B) =0.61 x 0. 41 = 0.25
So P(A or B) = 0.59 + 0.41 - 0.25 = 0.75
B) we want to find out if events A and B are mutually exclusive ;
P(A|B) = P(A&B)/P(B), For the events to be mutually exclusive, P(A&B) has to be zero. Since P(A&B) ≠ 0,the 2 events are not mutually exclusive.
C) Since P(A|B) = 0.61 and P(A) is 0.59. Thus, P(A | B) ≠ P(A).
So, therefore the 2 events are not independent.
A weighted average of the value of a random variable, where the probability function provides weights, is known as _____. a. a probability function b. a random variable c. the expected value d. None of the answers is correct.
Answer:
The answer is c. The expected value.
Step-by-step explanation:
Final answer:
The weighted average of the value of a random variable calculated with probability weights is known as the expected value. The expected value indicates the long-term average outcome for a random variable that has been experimented on multiple times. It is calculated using the sum of the products of the variable values and their corresponding probabilities.
Explanation:
The weighted average of the value of a random variable, where the probability function provides weights, is known as the expected value. The expected value is a key concept in probability and statistics, often referred to in equations as E(X) and symbolized as μ (mu), where X is a random variable. It represents the long-term average outcome of a random variable after many repetitions of an experiment. For a discrete random variable with a probability distribution function P(x), the expected value can be calculated using the formula: μ = Σ XP(x).
Random variables are typically defined in the context of a probability distribution, which can take on various forms like normal, uniform, or exponential distributions, depending on the nature of the experiment. When constructing a confidence interval for a population mean, the confidence level indicates the degree of certainty in the interval estimate. The error bound, which is part of the confidence interval calculation, would typically decrease if the confidence level is lowered because a lower confidence level means that we are willing to accept a greater chance that the interval does not contain the population mean, leading to a narrower interval.
Find the lateral (side) surface area of the cone generated by revolving the line segment y equals one fourth x , 0 less than or equals x less than or equals 7, about the x-axis. The lateral surface area of the cone generated by revolving the line segment y equals one fourth x , 0 less than or equals x less than or equals 7, about the x-axis is nothing. (Type an exact answer, using pi as needed.)
Answer:
lateral surface area of the cone =49π
Step-by-step explanation:
y= x/4
dx/dy = 4
Given x range as from 0 ≤ x ≤ 7 ⇒ x ranges between (a,b)
a= x/4 = 0/4 = 0
b = x/4= 7/4
lateral surface area of the cone
[tex]\int\limits^b_a {2\pi x\sqrt{1 + (\frac{dx}{dy})^2 } } \, dx \\\\= \int\limits^{\frac{7}{4}}_0 {2\pi (4y)\sqrt{1 + (4)^2 } } \, dx \\\\= 32\pi \int\limits^{\frac{7}{4}}_0 { y } \, dx \\\\= 32\pi [\frac{y^2}{2}]|^\frac{7}{4}_0[/tex]
=32π * 49/(16 *2)
=49π
Final answer:
To find the lateral surface area of the cone with the given line segment, use the formula [tex]L = \(\ r l\)[/tex], where r is the base radius, calculated from the given equation, and l is the slant height determined by the Pythagorean theorem.
Explanation:
The student has asked for help in finding the lateral surface area of a cone generated by revolving a line segment around the x-axis.
The specific line segment is given by the equation [tex]y = \frac{1}{4} x[/tex], for 0 \<= x \<= 7. To find this surface area, we can use the formulas for the surface area of a cone, which is \rl\, where \ is the radius of the base of the cone and \ is the slant height.
In this case, when x=7, y = [tex]y = \frac{1}{4} x[/tex] = 7/4, which gives us the radius of the base of the cone. The slant height can be calculated using the Pythagorean theorem, since we have a right triangle with the slant height as the hypotenuse, and the radius and height of the cone as the other two sides. Thus, [tex]l = {7^2+(7/4)^2}[/tex]
Finally, the formula for the lateral surface area of the cone is L = rl, which, after substituting the given values, results in the exact answer needed. The calculation will result in the following:
[tex]L = \frac{7}{4}\times\sqrt{7^2+(7/4)^2}[/tex]