Answer:
a) [tex]P(X<2.52)=P(Z<\frac{2.52-2.55}{0.035})=P(Z<-0.857)=0.196[/tex]
b) [tex]P(\bar X <2.52) = P(Z<-1.714)=0.043[/tex]
c) [tex]P(\bar X <2.52) = P(Z<-4.286)=0.000[/tex]
d) For part a we are just finding the probability that an individual bottle would have a value of 2.52 liters or less. So we can't compare the result of part a with the results for parts b and c.
If we see part b and c are similar but the difference it's on the sample size for part b we just have a sample size 4 and for part c we have a sample size of 25. The differences are because we have a higher standard error for part b compared to part c.
Step-by-step explanation:
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". The letter [tex]\phi(b)[/tex] is used to denote the cumulative area for a b quantile on the normal standard distribution, or in other words: [tex]\phi(b)=P(z<b)[/tex]
Let X the random variable that represent the amount of water in a bottle of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(2.55,0.035)[/tex]
a. What is the probability that an individual bottle contains less than 2.52 liters?
We are interested on this probability
[tex]P(X<2.52)[/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<2.52)=P(\frac{X-\mu}{\sigma}<\frac{2.52-\mu}{\sigma})[/tex]
And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.
[tex]P(X<2.52)=P(Z<\frac{2.52-2.55}{0.035})=P(Z<-0.857)=0.196[/tex]
b. If a sample of 4 bottles is selected, what is the probability that the sample mean amount contained is less than 2.52 liters? (Round to three decimal places as noeded)
And let [tex]\bar X[/tex] represent the sample mean, the distribution for the sample mean is given by:
[tex]\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}})[/tex]
On this case [tex]\bar X \sim N(2.55,\frac{0.035}{\sqrt{4}})[/tex]
The z score on this case is given by this formula:
[tex]z=\frac{\bar x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
And if we replace the values that we have we got:
[tex]z=\frac{2.52-2.55}{\frac{0.035}{\sqrt{4}}}=-1.714[/tex]
For this case we can use a table or excel to find the probability required:
[tex]P(\bar X <2.52) = P(Z<-1.714)=0.043[/tex]
c. If a sample of 25 bottles is selected, what is the probability that the sample mean amount contained is less than 2.52 liters? (Round to three decimal places as needed.)
The z score on this case is given by this formula:
[tex]z=\frac{\bar x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
And if we replace the values that we have we got:
[tex]z=\frac{2.52-2.55}{\frac{0.035}{\sqrt{25}}}=-4.286[/tex]
For this case we can use a table or excel to find the probability required:
[tex]P(\bar X <2.52) = P(Z<-4.286)=0.0000091[/tex]
d. Explain the difference in the results of (a) and (c)
For part a we are just finding the probability that an individual bottle would have a value of 2.52 liters or less. So we can't compare the result of part a with the results for parts b and c.
If we see part b and c are similar but the difference it's on the sample size for part b we just have a sample size 4 and for part c we have a sample size of 25. The differences are because we have a higher standard error for part b compared to part c.
The problem involves using z-scores, normal distribution, central limit theorem and law of large numbers for probability calculations related to the amount of water in sampled bottles. The increase in sample size from 1 to 25 should show a higher probability for the sample's mean to be close to the population mean.
Explanation:In this problem, we will apply the concept of the normal distribution and the central limit theorem to calculate probabilities and compare results. First, we will calculate the z-scores for (a), (b), and (c). The z-score formula is Z = (X - μ) / σ. In (a), X is 2.52 liters, μ (mean) is 2.55 liters, and σ (standard deviation) is 0.035 liters.
In (b) and (c), the σ of a sample mean is σ/√n, where n is the sample size (4 in (b) and 25 in (c)).
Next, we look up the z-score in the z-score table (or use a normal distribution calculator) to get the probabilities.
To answer part (d), probabilistically, as sample size increases, the sample mean tends to get closer to the population mean according to the law of large numbers. Hence, the probabilities in (c) should be higher than (a).
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One hundred eight Americans were surveyed to determine the number of hours they spend watching television each month. It was revealed that they watched an average of 151 hours each month with a standard deviation of 32 hours. Assume that the underlying population distribution is normal.
Construct a 99% confidence interval for the population mean hours spent watching television per month.
Fill in the blank: Round to two decimal places. ( , )
Answer: (143.07, 158.93)
Step-by-step explanation:
The formula to find the confidence interval is given by :-
[tex]\overline{x}\pm z^*\dfrac{\sigma}{\sqrt{n}}[/tex]
where n= sample size
[tex]\overline{x}[/tex] = Sample mean
z* = critical z-value (two tailed).
[tex]\sigma[/tex] = Population standard deviation
We assume that the underlying population distribution is normal.
As per given , we have
n= 108
[tex]\overline{x}=151[/tex]
[tex]\sigma=32[/tex]
Critical value for 99% confidence level = 2.576 (By using z-table)
Then , the 99% confidence interval for the population mean hours spent watching television per month :-
[tex]151\pm (2.576)\dfrac{32}{\sqrt{108}}[/tex]
[tex]151\pm (2.576)\dfrac{32}{10.3923048454}[/tex]
[tex]151\pm (2.576)(3.07920143568)[/tex]
[tex]151\pm (7.93202289831)\approx151\pm7.93\\\\=(151-7.93,\ 151+7.93)\\\\=(143.07,\ 158.93 )[/tex]
Hence, the required 99% confidence interval for the population mean hours spent watching television per month. = (143.07, 158.93)
The 99% confidence interval for the average number of hours all Americans spend watching television per month, based on the given sample, is (143.76, 158.23). This is computed using the confidence interval formula with the given sample mean, standard deviation, and the z-score for a 99% confidence interval.
Explanation:The question involves the concept of the confidence interval in statistics. Here we are given the sample size (n=108), the sample mean ([tex]\overline{X}[/tex] = 151), and the sample standard deviation (s=32). We are required to compute the 99% confidence interval.
To calculate a confidence interval, we apply this formula: [tex]\overline{X}[/tex] ± (z-value * (s/√n)) Where '[tex]\overline{X}[/tex]' is the sample mean, 'z-value' is the Z-score (which for a 99% confidence interval is 2.58), 's' is the standard deviation and 'n' is the sample size.
Substitute the given values into the formula: 151 ± (2.58 * (32/√108))
This results in: (143.76, 158.23)
So, we can say with 99% confidence that the average number of hours all Americans spend watching television per month is between 143.76 hours and 158.23 hours.
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A lab technician is tested for her consistency by making multiple measurements of the cholesterol level in one blood sample. The target precision is a standard deviation of 1.2 mg/dL or less. If 12 measurements are taken and the standard deviation is 1.8 mg/dL, is there enough evidence to support the claim that her standard deviation is greater than the target, at = .01? (Show the answers to all 5 steps of the hypothesis test.)
Step-by-step explanation:
Given precision is a standard deviation of s=1.8, n=12, target precision is a standard deviation of σ=1.2
The test hypothesis is
H_o:σ <=1.2
Ha:σ > 1.2
The test statistic is
chi square = [tex]\frac{(n-1)s^2}{\sigma^2}[/tex]
=[tex]\frac{(12-1)1.8^2}{1.2^2}[/tex]
=24.75
Given a=0.01, the critical value is chi square(with a=0.01, d_f=n-1=11)= 3.05 (check chi square table)
Since 24.75 > 3.05, we reject H_o.
So, we can conclude that her standard deviation is greater than the target.
We know that narrower confidence intervals give us a more precise estimate of the true population proportion. Which of the following could we do to produce higher precision in our estimates of the population proportion?
A. We can select a lower confidence level and increase the sample size.
B. We can select a higher confidence level and decrease the sample size.
C. We can select a higher confidence level and increase the sample size.
D. We can select a lower confidence level and decrease the sample size.
Answer:
A. We can select a lower confidence level and increase the sample size.
Step-by-step explanation:
The length of a confidence interval is:
Direct proportional to the confidence interval. This means that the higher the confidence level, the higher the length of the interval is.
Inverse proportional to the size of the sample.This means that the higher the size of the sample, the lower, or narrower, the length of the interval is.
Which of the following could we do to produce higher precision in our estimates of the population proportion?
We want a narrower interval. So the correct answer is:
A. We can select a lower confidence level and increase the sample size.
A random sample of 500 registered voters in Phoenix is asked if they favor the use of oxygenated fuels year-round to reduce air pollution. If more than 314 voters respond positively, we will conclude that at least 60% of the voters favor the use of these fuels. Round your answers to four decimal places (e.g. 98.7654).
a) Find the probability of type I error if exactly 60% of the voters favor the use of these fuelsb) What is the Type II error probability (Beta) β if 75% of the voters favor this action?
Answer:
a) 0.0853
b) 0.0000
Step-by-step explanation:
Parameters given stated that;
H₀ : p = 0.6
H₁ : p = 0.6, this explains the acceptance region as;
p° ≤ [tex]\frac{315}{500}[/tex]=0.63 and the region region as p°>0.63 (where p° is known as the sample proportion)
a).
the probability of type I error if exactly 60% is calculated as :
∝ = P (Reject H₀ | H₀ is true)
= P (p°>0.63 | p=0.6)
where p° is represented as pI in the subsequent calculated steps below
= P [tex][\frac{p°-p}{\sqrt{\frac{p(1-p)}{n}}} >\frac{0.63-p}{\sqrt{\frac{p(1-p)}{n}}} |p=0.6][/tex]
= P [tex][\frac{p°-0.6}{\sqrt{\frac{0.6(1-0.6)}{500}}} >\frac{0.63-0.6}{\sqrt{\frac{0.6(1-0.6)}{500}}} ][/tex]
= P [tex][Z>\frac{0.63-0.6}{\sqrt{\frac{0.6(1-0.6)}{500} } } ][/tex]
= P [Z > 1.37]
= 1 - P [Z ≤ 1.37]
= 1 - Ф (1.37)
= 1 - 0.914657 ( from Cumulative Standard Normal Distribution Table)
≅ 0.0853
b)
The probability of Type II error β is stated as:
β = P (Accept H₀ | H₁ is true)
= P [p° ≤ 0.63 | p = 0.75]
where p° is represented as pI in the subsequent calculated steps below
= P [tex][\frac{p°-p} \sqrt{\frac{p(1-p)}{n} } }\leq \frac{0.63-p}{\sqrt{\frac{p(1-p)}{n} } } | p=0.75][/tex]
= P [tex][\frac{p°-0.6} \sqrt{\frac{0.75(1-0.75)}{500} } }\leq \frac{0.63-0.75}{\sqrt{\frac{0.75(1-0.75)}{500} } } ][/tex]
= P[tex][Z\leq\frac{0.63-0.75}{\sqrt{\frac{0.75(1-0.75)}{500} } } ][/tex]
= P [Z ≤ -6.20]
= Ф (-6.20)
≅ 0.0000 (from Cumulative Standard Normal Distribution Table).
Final answer:
The probability of Type I error can be calculated using the formula P(Type I error) = P(Z > Zα), and the probability of Type II error (Beta) can be calculated using the formula Beta = P(Z < Zβ) + P(Z > Z1-β).
Explanation:
a) The probability of Type I error can be calculated using the formula:
P(Type I error) = P(Z > Zα)
where Zα is the standard score corresponding to the desired level of significance.
b) The probability of Type II error (Beta) can be calculated using the formula:
Beta = P(Z < Zβ) + P(Z > Z1-β)
where Zβ is the standard score corresponding to the desired power level and Z1-β is the standard score corresponding to the complement of the desired power level.
Suppose a realtor wants to determine the current percentage of customers who have a family of five or more. How many customers should the realtor survey in order to be 98% confident that the estimated (sample) proportion is within 2 percentage points of the true population proportion of customers who have a family of five or more?
Answer:
n=3394
Step-by-step explanation:
Previous concepts
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The population proportion have the following distribution
[tex]p \sim N(p,\sqrt{\frac{p(1-p)}{n}})[/tex]
Solution to the problem
The confidence interval for the mean is given by the following formula:
[tex]\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex]
In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 98% of confidence, our significance level would be given by [tex]\alpha=1-0.98=0.02[/tex] and [tex]\alpha/2 =0.01[/tex]. And the critical value would be given by:
[tex]z_{\alpha/2}=-2.33, z_{1-\alpha/2}=2.33[/tex]
The margin of error for the proportion interval is given by this formula:
[tex] ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex] (a)
And on this case we have that [tex]ME =\pm 0.02[/tex] and we are interested in order to find the value of n, if we solve n from equation (a) we got:
[tex]n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}[/tex] (b)
We can use as estimation of [tex]\hat p=0.5[/tex] since we don't have any other info provided. And replacing into equation (b) the values from part a we got:
[tex]n=\frac{0.5(1-0.5)}{(\frac{0.02}{2.33})^2}=3393.06[/tex]
And rounded up we have that n=3394
find y from the picture plz
It's a six sided polygon. For any polygon the external angles add to 360 degrees. The internal angles shown are the supplements of the external angles. We have
(180 - θ₁) + (180 - θ₂) + ... + (180 - θ₆) = 360
6(180) - 360 = θ₁ + θ₂ + θ₃ + θ₄ + θ₅ + θ₆
720 = θ₁ + θ₂ + θ₃ + θ₄ + θ₅ + θ₆
The six angles add up to 720 degrees, and five of them add to
126+101+135+147+96=605
So y = 720 - 605 = 115
The degree sign is external to y so not part of the answer:
Answer: 115
Answer:y = 115 degrees
Step-by-step explanation:
The given polygon has 6 sides. It is a hexagon. The sum of the interior angles of a polygon is
180(n - 2)
Where
n is the number of sides that the polygon has. This means that n = 6
Therefore, the sum of the interior angles would be
180(6 - 2) = 720 degrees
Therefore,
126 + 101 + 135 + 147 + 96 + y = 720
605 + y = 720
Subtracting 605 from both sides of the equation, it becomes
y = 720 - 605 = 115 degrees
A solid lies between planes perpendicular to the x-axis at x=0 and x=8. The cross-sections perpendicular to the axis on the interval 0
Answer:
The volume of the solid is 256 cubic units.
Step-by-step explanation:
Given:
The solid lies between planes [tex]x=0\ and\ x=8[/tex]
The cross section of the solid is a square with diagonal length equal to the distance between the parabolas [tex]y=-2\sqrt{x}\ and\ y=2\sqrt{x}[/tex].
The distance between the parabolas is given as:
[tex]D=2\sqrt x-(-2\sqrt x)\\\\D=2\sqrt x+2\sqrt x\\\\D=4\sqrt x[/tex]
Now, we know that, area of a square with diagonal 'D' is given as:
[tex]A=\frac{D^2}{2}[/tex]
Plug in [tex]D=4\sqrt x[/tex]. This gives,
[tex]A=\frac{(4\sqrt x)^2}{2}\\\\A=\frac{16x}{2}\\\\A=8x[/tex]
Now, volume of the solid is equal to the product of area of cross section and length [tex]dx[/tex]. So, we integrate it over the length from [tex]x=0\ to\ x=8[/tex]. This gives,
[tex]V=\int\limits^8_0 {A} \, dx\\\\V=\int\limits^8_0 {(8x)} \, dx\\\\V=8\int\limits^8_0 {(x)} \, dx\\\\V=8(\frac{x^2}{2})_{0}^{8}\\\\V=4[8^2-0]\\\\V=4\times 64\\\\V=256\ units^3[/tex]
Therefore, the volume of the solid is 256 cubic units.
This question is about volume calculation using calculus. The solid between two planes at x=0 and x=8 has cross-sections which, when described by a function of x A(x), the volume of the object can be computed via integration of A(x) dx from x=0 to x=8.
Explanation:The subject of this question falls under the field of Calculus, specifically, it's about Volume Calculation. The question describes a solid which is located between two planes at x=0 and x=8, perpendicular to the x-axis. Cross-sections perpendicular to the axis of this solid can be visualized like slices of the solid made along the x-axis.
If the area of these cross-sections can be represented by a function of x, A(x), then the volume of the entirety of the solid, V, can be calculated using the definite integral from x=0 to x=8 of A(x) dx. Essentially, this is summing up the volumes of the infinitesimal discs that make up the solid along the x-axis, from x=0 to x=8.
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A polling company conducts an annual poll of adults about political opinions. The survey asked a random sample of 419 adults whether they think things in the country are going in the right direction or in the wrong direction, 54% said that things were going in the wrong direction. How many people would need to ve surveyed for a 90% confidence interval to ensure the margin or error would be less than 3%?
Answer: 747
Step-by-step explanation:
When prior estimate of population proportion (p) is given , then the formula to find the sample size is given by :-
[tex]n=p(1-p)(\dfrac{z^*}{E})^2[/tex]
, where z* = Critical value and E = Margin of error.
As per given , we have
p= 0.54
E= 0.03
Critical value for 90% confidence : z* = 1.645
Then, the required sample size is given by :-
[tex]n=0.54(1-0.54)(\dfrac{1.645}{0.03})^2[/tex]
[tex]n=0.2484(54.8333333333)^2[/tex]
[tex]n=746.862899999\approx747[/tex]
Hence, the number of people would be needed = 747
An urn contains 17 red marbles and 18 blue marbles. 16 marbles are chosen. In how many ways can 6 red marbles be chosen?
Answer:
Total number of ways 6 red marbles can be chosen=541549008
Step-by-step explanation:
16 marbles are chosen in which 6 are red marbles and remaining marbles ,which are 10, are blue marbles.
In order to find in how many ways 6 red marbles can be chosen we will proceed as:
Out of 17 red marbles 6 are chosen and out of 18 blue marbles 10 are chosen.
Total number of ways 6 red marbles can be chosen= [tex]17_{C_6} * 18_{C_1_0}[/tex]
Total number of ways 6 red marbles can be chosen=[tex]\frac{17!}{6!*(17-6)!} * \frac{18!}{10!*(18-10)!}[/tex]
Total number of ways 6 red marbles can be chosen= 12376*43758
Total number of ways 6 red marbles can be chosen=541549008
Answer: N = 541,549,008
Therefore the number of ways to select 6 red marbles is 541,549,008
Step-by-step explanation:
Given;
Number of red marbles total = 17
Number of blue marbles total = 18
Number of red marbles to be selected = 6
Number of blue marbles to be selected = 16 - 6 = 10
To determine the number of ways 6 red marbles can be selected N.
N = number of ways 6 red marbles can be selected from 17 red marbles × number of ways 10 blue marbles can be selected from 18 blue marbles
N = 17C6 × 18C10
N = 17!/(6! × (17-6)!) × 18!/(10! × (18-10)!)
N = 17!/(6! × 11!) × 18!/(10! × 8!)
N = 541,549,008
Therefore the number of ways to select 6 red marbles is 541,549,008
[6.18] ([1] 7.52) Resistors to be used in a circuit have average resistance 200 ohms and standard deviation 10 ohms. Suppose 25 of these resistors are randomly selected to be used in a circuit. a) What is the probability that the average resistance for the 25 resistors is between 199 and 202 ohms? b) Find the probability that the total resistance does not exceed 5100 ohms.
Answer: a) 0.5328 b) 0.9772
Step-by-step explanation:
Given : Resistors to be used in a circuit have average resistance 200 ohms and standard deviation 10 ohms.
[tex]\mu=200[/tex] and [tex]\sigma=10[/tex]
We assume that the resistance in circuits are normally distributed.
a) Let x denotes the average resistance of the circuit.
Sample size : n= 25
Then, the probability that the average resistance for the 25 resistors is between 199 and 202 ohms :-
[tex]P(199<x<200)=P(\dfrac{199-200}{\dfrac{10}{\sqrt{25}}}<\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}<\dfrac{202-200}{\dfrac{10}{\sqrt{25}}})\\\\=P(-0.5<z<1)\ \ [\because z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}]\\\\=P(z<1)-P(z<-0.5)\\\\=P(z<1)-(1-P(z<0.5))\ \ [\because\ P(Z<-z)=1-P(Z<z)]\\\\=0.8413-(1-0.6915)\ \ [\text{By z-table}]\\\\=0.5328[/tex]
b) Total resistors = 25
Let Z be the total resistance of 25 resistors.
To find P(Z≤5100 ohms) , first we find the mean and variance for Z.
Mean= E(Y) = E(25 X)=25 E(X)=25(200)= 5000 ohm
[tex]Var(Y)=Var(25\ X)=25^2(\dfrac{\sigma^2}{n})=25^2\dfrac{(10)^2}{25}=2500[/tex]
The probability that the total resistance does not exceed 5100 ohms will be :
[tex]P(Y\leq5000)=P(\dfrac{Y-\mu}{\sqrt{Var(Y)}}<\dfrac{5100-5000}{\sqrt{2500}})\\\\=P(z\leq2)=0.9772\ \ [\text{By z-table}][/tex]
Hence, the probability that the total resistance does not exceed 5100 ohms = 0.9772
Solve the following equation by taking the square root 12 - 6n2 = -420 i need help
Answer:
[tex]n=\pm 6\sqrt{2}[/tex]
Step-by-step explanation:
It may work well to divide by 6, subtract 2, and multiply by -1 before you take the square root.
[tex]12-6n^2=-420\\2-n^2=-70 \qquad\text{divide by 6}\\-n^2=-72 \qquad\text{subtract 2}\\n^2=72 \qquad\text{multiply by -1}\\\\n=\pm\sqrt{36\cdot 2} \qquad\text{take the square root}\\\\n=\pm 6\sqrt{2} \qquad\text{simplify}[/tex]
In the following sequence, each number (except the first two) is the sum of the previous two numbers: 0, 1, 1, 2, 3, 5, 8, 13, .... This sequence is known as the Fibonacci sequence. We speak of the i'th element of the sequence (starting at 0)-- thus the 0th element is 0, the 1st element is 1, the 2nd element is 1, the 3rd element is 2 and so on. Given the positive integer n, associate the nth value of the fibonacci sequence with the variable result. For example, if n is associated with the value 8 then result would be associated with 21.
Final answer:
To find the nth Fibonacci number, dynamic programming stores previously calculated values in an array, which allows for efficient linear time computation by summing the two previous numbers to obtain the nth value.
Explanation:
The Fibonacci sequence is defined such that each number in the sequence is the sum of the two preceding ones, starting from 0 and 1. To calculate the nth Fibonacci number, denoted as Fib(n), we start by setting Fib(0) and Fib(1) equal to 0 and 1, respectively. For n ≥ 2, Fib(n) is defined recursively as Fib(n) = Fib(n - 1) + Fib(n - 2). A naive recursive algorithm could be inefficient due to repeated calculations. Using dynamic programming or memoization improves efficiency by storing intermediate results, thus avoiding unnecessary recalculations.
Computing Fibonacci Numbers Using Dynamic Programming
To compute the nth Fibonacci number using dynamic programming, we create an array or list to save previously computed Fibonacci numbers. The nth value, for instance Fib(8) = 21, is then easily found by summing up the n-1th and n-2th values from the array, which are already computed and stored. This approach leads to a time complexity that is linear, i.e., O(n), instead of exponential.
Evaluate the integral Integral from nothing to nothing ∫ StartFraction 3 Over t Superscript 4 EndFraction 3 t4 sine left parenthesis StartFraction 1 Over t cubed EndFraction minus 6 right parenthesis sin 1 t3 −6dt
Answer:
[tex]\cos (\frac{1}{t^3}-6)} + c[/tex]
Step-by-step explanation:
Given function:
[tex]\int {\frac{3}{t^4}\sin (\frac{1}{t^3}-6)} \, dt[/tex]
Now,
let [tex]\frac{1}{t^3}-6[/tex] be 'x'
Therefore,
[tex]d(\frac{1}{t^3}-6)[/tex] = dx
or
[tex]\frac{-3}{t^4}dt[/tex] = dx
on substituting the above values in the equation, we get
⇒ ∫ - sin (x) . dx
or
⇒ cos (x) + c [ ∵ ∫sin (x) . dx = - cos (x)]
Here,
c is the integral constant
on substituting the value of 'x' in the equation, we get
[tex]\cos (\frac{1}{t^3}-6)} + c[/tex]
As part of a research project on student debt at TWU, a researcher interviewed a sample of 35 students that were chosen at random concerning their monthly credit card balance. On average, these students had a balance of $2,573. The range of the data ran from a high of $22,315 to a low of $0. The median (Md) was $2,455 and the variance was $4,252. If a student selected at random had a credit card balance of $1,700; then he would have a Z Score of___________.
a. -13.4.
b. -9.73
c. 0.4
d. 9.73
Answer:
Option a is right
Step-by-step explanation:
Given that as part of a research project on student debt at TWU, a researcher interviewed a sample of 35 students that were chosen at random concerning their monthly credit card balance.
Sample average = 2573
Variance = 4252
Sample size = 35
STd deviation of X = [tex]\sqrt{4252} \\=65.21[/tex]
Score of student selected at random X=1700
Corresponding Z score = [tex]\frac{1700-2573}{65.201} \\=-13.38[/tex]
Rounding of we get Z score = -13.4
option a is right
Find the seventh term of an increasing geometric progression if the first term is equal to 9−4sqrt5 and each term (starting with the second) is equal to the difference of the term following it and the term preceding it.
Answer:
7th term = 1.
Step-by-step explanation:
Given that, first term of increasing geometric progression is 9-4√5.
each term (starting with the second) is equal to the difference of the term following it and the term preceding it.
let first term of geometric progression be a and the increasing ratio be r.
⇒ The geometric progression is a , ar , ar² , ar³, ....... so on.
Given, each term (starting with the second) is equal to the difference of the term following it and the term preceding it.
⇒ second term = (third term - first term)
⇒ ar = (ar² - a)
⇒ r = r² - 1
⇒ r² - r -1 =0
⇒ roots of this equation is r = [tex]\frac{1+\sqrt{5} }{2}[/tex] , [tex]\frac{1-\sqrt{5} }{2}[/tex]
(roots of ax²+bx+c are [tex]\frac{-b+\sqrt{b^{2} -4ac} }{2a}[/tex] and [tex]\frac{-b-\sqrt{b^{2} -4ac} }{2a}[/tex])
and it is given, increasing geometric progression
⇒ r > 0.
⇒ r = [tex]\frac{1+\sqrt{5} }{2}[/tex].
Now, nth term in geometric progression is arⁿ⁻¹.
⇒ 7th term = ar⁷⁻¹ = ar⁶.
= (9-4√5)([tex]\frac{1+\sqrt{5} }{2}[/tex])⁶
= (0.05572809)(17.94427191) = 1
⇒ 7th term = 1.
Automated manufacturing operations are quite precise but still vary, often with distributions that are close to Normal. The width in inches of slots cut by a milling machine follows approximately the N(0.8750, 0.0012) distribution. The specifications allowslot widths between 0.8725 and 0.8775 inch. What proportion of slots meet these specifications?
Answer:
96.2% of slots meet these specifications.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 0.8750
Standard Deviation, σ = 0.0012
We are given that the distribution of width in inches of slots is a bell shaped distribution that is a normal distribution.
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
P( widths between 0.8725 and 0.8775 inch)
[tex]P(0.8725 \leq x \leq 0.8775) = P(\displaystyle\frac{0.8725 - 0.8750}{0.0012} \leq z \leq \displaystyle\frac{0.8775-0.8750}{0.0012}) = P(-2.083 \leq z \leq 2.083)\\\\= P(z \leq 2.083) - P(z < -2.083)\\= 0.981 - 0.019 = 0.962 = 96.2\%[/tex]
[tex]P(0.8725 \leq x \leq 0.8775) = 96.2\%[/tex]
96.2% of slots meet these specifications.
The question asks for the proportion of slots meeting width specifications within a normal distribution with defined mean and standard deviation. We calculate the corresponding Z-scores for the lower and upper specification limits and then determine the probability of a slot falling within these limits.
Explanation:The problem involves finding the proportion of slots that meet the specified width requirements in a normal distribution. In this case, the slot widths follow a normal distribution with a mean (μ) of 0.8750 inches, and a standard deviation (σ) of 0.0012 inches. The specifications require that slot widths be between 0.8725 inches and 0.8775 inches.
To find the proportion of slots that meet these specifications, we calculate the Z-scores for both the lower specification limit of 0.8725 and the upper specification limit of 0.8775. The Z-score formula is given by Z = (X - μ) / σ, where X is the value for which we want to find the Z-score.
For the lower limit, we have:
Z(lower) = (0.8725 - 0.8750) / 0.0012 = -2.083…
For the upper limit, we have:
Z(upper) = (0.8775 - 0.8750) / 0.0012 = 2.083…
Next, we use the standard normal distribution to find the probability corresponding to these Z-scores. The area under the curve between these two Z-scores represents the proportion of slots that are within the specifications. This can be found using standard normal distribution tables or a calculator with statistical functions.
Suppose that your company has just developed a new screening test for a disease and you are in charge of testing its validity and feasibility.You decide to evaluate the test on 1000 individuals and compare the results of the new test to the gold standard.You know the prevalence of disease in your population is 30%.The screening test gave a positive result for 292 individuals.285 of these individuals actually had the disease on the basis of the gold standard determination.
Calculate the sensitivity of the new screening test.
95.0% 97.6% 99.0% 96.9%
Answer:
The sensitivity of the new screening test is 97.6%
Step-by-step explanation:
The sensitivy of a test, or true positive rate, is defined as the proportion of positive results that are correctly identified. It is complementary to the proportion of "false positives".
In this case the test gave 292 positive results. Of this 292 tests, 285 of these individuals actually had the disease.
So the sensititivity is equal to the ratio of the true positives (285) and the total positives (292)
[tex]Sensitivity=\frac{TP}{P}=\frac{285}{292}=0.976=97.6\%[/tex]
In order to determine whether or not there is a significant difference between the hourly wages of two companies, the following data have been accumulated.
Company 1 Company 2 n1 = 80 n2 = 60 x̄1 = $10.80 x̄2 = $10.00 σ1 = $2.00 σ2 = $1.50 Refer to Exhibit 10-13. The point estimate of the difference between the means (Company 1 – Company 2) is _____.
a. .8
b. –20
c. .50
d. 20
Answer:
a. .8
Step-by-step explanation:
The point estimate of the difference between the means of Company 1 and Company 2 can be calculated as:
point estimate = mu1 - mu2 where
mu1 is the sample mean hourly wage of Company 1mu2 is the sample mean hourly wage of Company 2Therefore point estimate = $10.80- $10 =$ .8
A researcher matched 30 participants on intelligence (hence 15 pairs of participants), and then compared differences in emotional responsiveness to two experimental stimuli between each pair. For this test, what are the critical values, assuming a two-tailed test at a 0.05 level of significance?
(A) ±2.042
(B) ±2.045
(C) ±2.131
(D) ±2.145
company manufactures and sells x cellphones per week. The weekly price-demand and cost equations are given below. p equals 500 minus 0.5 xp=500−0.5x and Upper C (x )equals 25 comma 000 plus 140 xC(x)=25,000+140x (A) What price should the company charge for the phones, and how many phones should be produced to maximize the weekly revenue
Answer:
The number of cellphones to be produced per week is 500.
The cost of each cell phone is $250.
The maximum revenue is $1,25,000
Step-by-step explanation:
We are given the following information in the question:
The weekly price-demand equation:
[tex]p(x)=500-0.5x[/tex]
The cost equation:
[tex]C(x) = 25000+140x[/tex]
The revenue equation can be written as:
[tex]R(x) = p(x)\times x\\= (500-0.5x)x\\= 500x - 0.5x^2[/tex]
To find the maximum value of revenue, we first differentiate the revenue function:
[tex]\displaystyle\frac{dR(x)}{dx} = \frac{d}{dx}(500x - 0.5x^2) = 500-x[/tex]
Equating the first derivative to zero,
[tex]\displaystyle\frac{dR(x)}{dx} = 0\\\\500-x = 0\\x = 500[/tex]
Again differentiating the revenue function:
[tex]\displaystyle\frac{dR^2(x)}{dx^2} = \frac{d}{dx}(500 - x) = -1[/tex]
At x = 500,
[tex]\displaystyle\frac{dR^2(x)}{dx^2} < 0[/tex]
Thus, by double derivative test, R(x) has the maximum value at x = 500.
So, the number of cellphones to be produced per week is 500, in order to maximize the revenue.
Price of phone:
[tex]p(500)=500-0.5(500) = 250[/tex]
The cost of each cell phone is $250.
Maximum Revenue =
[tex]R(500) = 500(500) - 0.5(500)^2 = 125000[/tex]
Thus, the maximum revenue is $1,25,000
Twenty years ago, entering male high school students of Central High could do an average of 24 pushups in 60 seconds. To see whether this remains true today, a random sample of 36 freshmen was chosen. Suppose their average was 22.5 with a sample standard deviation of 3.1,
(a) Test, using the p-value approach, whether the mean is still equal to 24 at the 5 percent level of significance.
(b) Calculate the power of the test if the true mean is 23.
Answer:
a) [tex]t=\frac{22.5-24}{\frac{3.1}{\sqrt{36}}}=-2.903[/tex]
[tex]p_v =2*P(t_{(35)}<-2.903)=0.0064[/tex]
If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, so we can't conclude that the true mean is still equal to 24 at 5% of significance.
b) Power =0.4626+0.000172=0.463
See explanation below.
Step-by-step explanation:
Part a
Data given and notation
[tex]\bar X=22.5[/tex] represent the sample mean
[tex]s=3.1[/tex] represent the sample standard deviation
[tex]n=36[/tex] sample size
[tex]\mu_o =24[/tex] represent the value that we want to test
[tex]\alpha=0.05[/tex] represent the significance level for the hypothesis test.
t would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value for the test (variable of interest)
State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the mean is still equal to 24, the system of hypothesis would be:
Null hypothesis:[tex]\mu = 24[/tex]
Alternative hypothesis:[tex]\mu \neq 24[/tex]
If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".
Calculate the statistic
We can replace in formula (1) the info given like this:
[tex]t=\frac{22.5-24}{\frac{3.1}{\sqrt{36}}}=-2.903[/tex]
P-value
The first step is calculate the degrees of freedom, on this case:
[tex]df=n-1=36-1=35[/tex]
Since is a bilateral test the p value would be:
[tex]p_v =2*P(t_{(35)}<-2.903)=0.0064[/tex]
Conclusion
If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, so we can't conclude that the true mean is still equal to 24 at 5% of significance.
Part b
The Power of a test is the probability of rejecting the null hypothesis when, in the reality, it is false.
For this case the power of the test would be:
P(reject null hypothesis| [tex]\mu=23[/tex])
If we see the null hypothesis we reject it when we have this:
The critical values from the t distribution with 35 degrees of freedom and at 5% of significance are -2.03 and 2.03. From the z score formula:
[tex]t=\frac{\bar x-\mu}{\frac{s}{\sqrt{n}}}[/tex]
If we solve for [tex]\bar x[/tex] we got:
[tex]\bar X= \mu \pm t \frac{s}{\sqrt{n}}[/tex]
Using the two critical values we have the critical values four our sampling distribution under the null hypothesis
[tex]\bar X= 24 -2.03 \frac{3.1}{\sqrt{36}}=22.951[/tex]
[tex]\bar X= 24 +2.03 \frac{3.1}{\sqrt{36}}=25.049[/tex]
So we reject the null hypothesis if [tex]\bar x<22.951[/tex] or [tex]\bar X >25.049[/tex]
So for our case:
P(reject null hypothesis| [tex]\mu=23[/tex]) can be founded like this:
[tex]P(\bar X <22.951|\mu=23)=P(t<\frac{22.951-23}{\frac{3.1}{\sqrt{36}}})=P(t_{35}<-0.0948)=0.4626[/tex]
[tex]P(\bar X >25.012|\mu=23)=P(t<\frac{25.049-23}{\frac{3.1}{\sqrt{36}}})=P(t_{35}>3.966)=0.000172[/tex]
And the power on this case would be the sum of the two last probabilities:
Power =0.4626+0.000172=0.463
Final answer:
To test whether the mean length of time students spend doing homework each week has increased, we can conduct a hypothesis test using the null hypothesis that the mean time is still 2.5 hours and the alternative hypothesis that the mean time has increased.
Explanation:
To test whether the mean length of time students spend doing homework each week has increased, we can conduct a hypothesis test. The null hypothesis, denoted as H0, would be that the mean time is still 2.5 hours. The alternative hypothesis, denoted as Ha, would be that the mean time has increased. In this case, the alternative hypothesis would be Ha: µ > 2.5, where µ represents the population mean.
To conduct the hypothesis test, we can use a t-distribution because the population standard deviation is not known. We can calculate the test statistic by using the formula: t = (x - µ) / (s/√n), where x is the sample mean, µ is the hypothesized mean, s is the sample standard deviation, and n is the sample size. Once we calculate the test statistic, we can compare it to the critical value from the t-distribution table or calculate the p-value to determine the level of significance.
Find the percent of the data that can be explained by the regression line and regression equation given that the correlation coefficient = -.72 (Give your answer as a percent rounded to the hundredth decimal place. Include the % sign)
Answer:
51.84%
Step-by-step explanation:
The percentage of data explained by regression line is assessed using R-square. Here, in the given scenario correlation coefficient r is given. We simply take square of correlation coefficient to get r-square. r-square=(-0.72)^2=0.5184=51.84%
We wish to obtain a 90% confidence interval for the standard deviation of a normally distributed random variable. To accomplish this we obtain a simple random sample of 16 elements from the population on which the random variable is defined. We obtain a sample mean value of 20 with a sample standard deviation of 12. Give the 90% confidence interval (to the nearest integer) for the standard deviation of the random variable. a) 83 to 307 b) 9 to 18 c) 91 to 270 d) 15 to 25 e) 20 to 34
Answer: d) 15 to 25
Step-by-step explanation:
Given : Sample size : n= 16
Degree of freedom = df =n-1 = 15
Sample mean : [tex]\overline{x}=20[/tex]
sample standard deviation : [tex]s= 12[/tex]
Significance level : [tex]\alpha= 1-0.90=0.10[/tex]
Since population standard deviation is unavailable , so the confidence interval for the population mean is given by:-
[tex]\overline{x}\pm t_{\alpha/2, df}\dfrac{s}{\sqrt{n}}[/tex]
Using t-distribution table , we have
Critical value = [tex]t_{\alpha/2, df}=t_{0.05 , 15}=1.7530[/tex]
90% confidence interval for the mean value will be :
[tex]20\pm (1.7530)\dfrac{12}{\sqrt{16}}[/tex]
[tex]20\pm (1.7530)\dfrac{12}{4}[/tex]
[tex]20\pm (1.7530)(3)[/tex]
[tex]20\pm (5.259)[/tex]
[tex](20-5.259,\ 20+5.259)[/tex]
[tex](14.741,\ 25.259)\approx(15,\ 25 )[/tex][Round to the nearest integer]
Hence, the 90% confidence interval (to the nearest integer) for the standard deviation of the random variable.= 15 to 25.
Final answer:
To obtain a 90% confidence interval for the standard deviation of a normally distributed random variable with a sample size of 16, sample mean of 20, and sample standard deviation of 12, use the chi-square distribution to calculate the lower and upper bounds. The 90% confidence interval is approximately 86 to 283.
Explanation:
To obtain a 90% confidence interval for the standard deviation of a normally distributed random variable, we can use the chi-square distribution. Given a simple random sample of 16 elements with a sample mean of 20 and a sample standard deviation of 12, we can calculate the lower and upper bounds of the confidence interval.
Step 1: Calculate the chi-square values for the lower and upper bounds using the following formulas:
Lower bound: (n-1)s² / X², where n is the sample size, s is the sample standard deviation, and X² is the chi-square value for a 90% confidence level with (n-1) degrees of freedom.
Upper bound: (n-1)s² / X², where n is the sample size, s is the sample standard deviation, and X² is the chi-square value for a 10% significance level with (n-1) degrees of freedom.
Substituting the values into the formulas, we get:
Lower bound: (15)(144) / 24.996 = 86.437
Upper bound: (15)(144) / 7.633 = 283.368
Rounding to the nearest integer, the 90% confidence interval for the standard deviation of the random variable is approximately 86 to 283.
A least squares regression line was found. Using technology, it was determined that the total sum of squares (SST) was 46.8 and the sum of squares of regression (SSR) was 14.55. Use these values to calculate the percent of the variability in y that can be explained by variability in the regression model. Round your answer to the nearest integer.
Answer: 31%
Step-by-step explanation:
Formula : Percent of the variability = [tex]R^2\times100=\dfrac{SSR}{SST}\times100[/tex]
, where [tex]R^2[/tex] = Coefficient of Determination.
SSR = sum of squares of regression
SST = total sum of squares
[tex]R^2[/tex] is the proportion of the variation of Y that can be attributed to the variation of x.
As per given , we have
SSR = 14.55
SST= 46.8
Then, the percent of the variability in y that can be explained by variability in the regression model =[tex]\dfrac{14.55}{46.8}\times100=31.0897435897\%\approx31\%[/tex]
Hence, the percent of the variability in y that can be explained by variability in the regression model = 31%
Answer: 14.55/46.8= .3109
.3109x100=31.09
Step-by-step explanation:
A sample of 12 radon detectors of a certain type was selected, and each was exposed to 100 pCI/L of radon. The resulting readings were as follows: 105.6, 90.9, 91.2, 96.9, 96.5, 91.3, 100.1, 105.0, 99.6, 107.7, 103.3, 92.4 Does this data suggest that the population mean reading under these conditions differs from 100? Set up an appropriate hypothesis test to answer this question.
Answer:
Null hypothesis:[tex]\mu = 100[/tex]
Alternative hypothesis:[tex]\mu \neq 100[/tex]
[tex]t=\frac{98.375-100}{\frac{6.109}{\sqrt{12}}}=-0.921[/tex]
[tex]p_v =2*P(t_{11}<-0.921)=0.377[/tex]
Step-by-step explanation:
1) Data given and notation
Data: 105.6, 90.9, 91.2, 96.9, 96.5, 91.3, 100.1, 105.0, 99.6, 107.7, 103.3, 92.4
We can calculate the sample mean and deviation for this data with the following formulas:
[tex]\bar X =\frac{\sum_{i=1}^n X_i}{n}[/tex]
[tex]s=\sqrt{\frac{\sum_{i=1}^n (X_i- \bar X)^2}{n-1}}[/tex]
The results obtained are:
[tex]\bar X=98.375[/tex] represent the sample mean
[tex]s=0.6.109[/tex] represent the sample standard deviation
[tex]n=12[/tex] sample size
[tex]\mu_o =100[/tex] represent the value that we want to test
[tex]\alpha[/tex] represent the significance level for the hypothesis test.
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value for the test (variable of interest)
2) State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the mean is equal to 100pCL/L, the system of hypothesis are :
Null hypothesis:[tex]\mu = 100[/tex]
Alternative hypothesis:[tex]\mu \neq 100[/tex]
Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".
3) Calculate the statistic
We can replace in formula (1) the info given like this:
[tex]t=\frac{98.375-100}{\frac{6.109}{\sqrt{12}}}=-0.921[/tex]
4) P-value
First we need to find the degrees of freedom for the statistic given by:
[tex]df=n-1=12-1=11[/tex]
Since is a two sided test the p value would given by:
[tex]p_v =2*P(t_{11}<-0.921)=0.377[/tex]
5) Conclusion
If we compare the p value and the significance level assumed [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true mean is not significant different from 100 at 5% of significance.
The question asked is about setting a hypothesis test to see if the mean reading from radon detectors differs from 100 pCI/L. The null and alternative hypotheses should be formed, calculations should be performed and the p-value compared with the significance level to determine if the null hypothesis should be retained or rejected.
Explanation:The subject of this question is setting up a hypothesis test for determining if the mean reading from the radon detectors differs from 100 pCI/L. The first step is to set up the null and alternate hypothesis. The null hypothesis (H0) would be that the population mean reading is 100 pCI/L (µ = 100), whereas the alternative hypothesis (H1) would propose that the mean reading differs from 100 pCI/L (µ ≠ 100).
Next, we would calculate the sample mean and sample standard deviation. Using these calculations, you can perform a t-test to compare the sample mean to the proposed population mean of 100 pCI/L. The decision of rejecting or not rejecting the null hypothesis relies on the comparison of the p-value obtained from the test statistic with the significance level.
Without the actual calculations, it is not possible to conclude whether the data suggests that the population mean reading under these conditions differs from 100 pCI/L or not. However, if the p-value is less than the significance level (commonly 0.05), we would reject the null hypothesis and conclude that the data provides enough evidence to suggest that the population mean differs from 100.
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A bank with branches located in a commercial district of a city and in a residential district has the business
objective of developing an improved process for serving customers during the noon-to-1 P.M. lunch
period. Management decides to first study the waiting time in the current process. The waiting time is
defined as the time that elapses from when the customer enters the line until he or she reaches the teller
window. Data are collected from a random sample of 15 customers at each branch.
The following is the data sample of the wait times, in minutes, from the commercial district branch.
4.14 5.66 3.04 5.34 4.82 2.69 3.32 3.41
4.42 6.01 0.15 5.11 6.59 6.43 3.72
The following is the data sample of the wait times, in minutes, from the residential district branch.
9.99 5.89 8.06 5.91 8.64 3.77 8.21 8.52
10.46 6.87 5.53 4.23 6.25 9.88 5.59
Determine the test statistic.
Answer:
test statistic is 4.27
Step-by-step explanation:
[tex]H_{0}[/tex] : mean waiting time in a residential district branch is the same as a commercial district branch
[tex]H_{a}[/tex] : mean waiting time in a residential district branch is more than a commercial district branch
commercial district branch:
mean waiting time: [tex]\frac{4.14+5.66+3.04+5.34+4.82+2.69+3.32+3.41+4.42+6.01+0.15+5.11+6.59+6.43+3.72}{15} =4.32[/tex]
standard deviation:
mean squared differences from the mean = 1.63
residential district branch.
mean waiting time: [tex]\frac{9.99+5.89+8.06+5.91+8.64+3.77+8.21+8.52+10.46+6.87+5.53+4.23+6.25+9.88+5.59}{15} =7.19[/tex]
standard deviation:
mean squared differences from the mean = 2.03
test statstic can be calculated using the formula:
[tex]z=\frac{X-Y}{\sqrt{\frac{s(x)^2}{N(x)}+\frac{s(y)^2}{N(y)}}}[/tex] where
X is the mean mean waiting time for residential district branch. (7.19)Y is the mean mean waiting time for commercial district branch. (4.32)s(x) is the sample standard deviation for residential district branch (2.03)s(y) is the sample standard deviation for commercial district branch.(1.93)N(x) is the sample size for residential district branch (15)N(y) is the sample size for commercial district branch.(15)[tex]z=\frac{7.19-4.32}{\sqrt{\frac{2.03^2}{15}+\frac{1.63^2}{15}}}[/tex] ≈4.27
A linear enzyme is formed by four alpha and two beta protein subunits. How manydifferent arrangements are there?
Answer:
15
Step-by-step explanation:
We are given that
Number of alpha protein subunits=4
Number of beta protein subunits=2
Total number of protein sub-units=2+4=6
We have to find the number of different arrangements are there.
When r identical letters and y identical letters and total object are n then arrangements are
[tex]\frac{n!}{r!x!}[/tex]
n=6,r=2,x=4
By using the formula
Then, we get
Number of different arrangements =[tex]\frac{6!}{2!4!}[/tex]
Number of different arrangements=[tex]\frac{6\times 5\times 4!}{2\times 1\times 4!}[/tex]
Number of different arrangements=15
Hence, different arrangements are there= 15
The life span of a species of fruit fly have a bell-shaped distribution, with a mean of 33 days and a standard deviation of 4 days.
What percentage corresponds to the life span of fruit flies that are between 29 days and 37 days?
Answer:
68.26% corresponds to the life span of fruit flies that are between 29 days and 37 days
Step-by-step explanation:
Given that the life span of a species of fruit fly have a bell-shaped distribution, with a mean of 33 days and a standard deviation of 4 days.
Let X be the life span
X is N(33,4)
we can convert X into Z standard normal variate by
[tex]z=\frac{x-33}{4}[/tex]
To find the percentage corresponds to the life span of fruit flies that are between 29 days and 37 days
For this let us find probability for x lying between 29 and 37 days
[tex]29\leq x\leq 37\\=-1\leq z\leq 1[/tex]
Probability = P(|z|<1) = 0.6826
Convert to percent
68.26% corresponds to the life span of fruit flies that are between 29 days and 37 days
To find the percentage that corresponds to the life span of fruit flies between 29 days and 37 days, we need to calculate the z-scores for both values and then use the standard normal distribution table to find the area under the curve between those z-scores. The percentage is approximately 68%.
Explanation:To find the percentage that corresponds to the life span of fruit flies between 29 days and 37 days, we need to calculate the z-scores for both values and then use the standard normal distribution table to find the area under the curve between those z-scores.
To calculate the z-scores, we use the formula: z = (x - μ) / σ, where x is the value, μ is the mean, and σ is the standard deviation.
For 29 days: z = (29 - 33) / 4 = -1
For 37 days: z = (37 - 33) / 4 = 1
Looking at the standard normal distribution table or using a calculator, we find that the area between -1 and 1 is approximately 68%. Therefore, the percentage that corresponds to the life span of fruit flies between 29 days and 37 days is 68%.
The price to earnings ratio (P/E) is an important tool in financial work. A random sample of 14 large U.S. banks (J. P. Morgan, Bank of America, and others) gave the following P/E ratios†.24 16 22 14 12 13 17 22 15 19 23 13 11 18
The sample mean is x ≈ 17.1. Generally speaking, a low P/E ratio indicates a "value" or bargain stock.
Suppose a recent copy of a magazine indicated that the P/E ratio of a certain stock index is μ = 18.
Let x be a random variable representing the P/E ratio of all large U.S. bank stocks.
We assume that x has a normal distribution and σ = 5.1.
Do these data indicate that the P/E ratio of all U.S. bank stocks is less than 18? Use α = 0.01.(a) What is the level of significance?(b) What is the value of the sample test statistic? (Round your answer to two decimal places.)(c) Find (or estimate) the P-value. (Round your answer to four decimal places.)
Answer:
a) [tex]\alpha=0.01[/tex] is the significance level given
b) [tex]z=\frac{17.1-18}{\frac{5.1}{\sqrt{14}}}=-0.6603[/tex]
c) Since is a one side left tailed test the p value would be:
[tex]p_v =P(Z<-0.6603)=0.2545[/tex]
Step-by-step explanation:
Data given and notation
[tex]\bar X=17.1[/tex] represent the mean P/E ratio for the sample
[tex]\sigma=5.1[/tex] represent the sample standard deviation for the population
[tex]n=14[/tex] sample size
[tex]\mu_o =18[/tex] represent the value that we want to test
[tex]\alpha=0.01[/tex] represent the significance level for the hypothesis test.
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value for the test (variable of interest)
State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the mean for the P/E ratio is less than 18, the system of hypothesis would be:
Null hypothesis:[tex]\mu \geq 18[/tex]
Alternative hypothesis:[tex]\mu < 18[/tex]
If we analyze the size for the sample is < 30 but we know the population deviation so is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:
[tex]z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}[/tex] (1)
z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".
(a) What is the level of significance?
[tex]\alpha=0.01[/tex] is the significance level given
(b) What is the value of the sample test statistic?
We can replace in formula (1) the info given like this:
[tex]z=\frac{17.1-18}{\frac{5.1}{\sqrt{14}}}=-0.6603[/tex]
(c) Find (or estimate) the P-value. (Round your answer to four decimal places.)
Since is a one side left tailed test the p value would be:
[tex]p_v =P(Z<-0.6603)=0.2545[/tex]
Conclusion
If we compare the p value and the significance level given [tex]\alpha=0.01[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL reject the null hypothesis, so we can conclude that the true mean for the P/E ratio is not significantly less than 18.
The weight of people on a college campus are normally distributed with mean 185 pounds and standard deviation 20 pounds. What's the probability that a person weighs more than 200 pounds? (round your answer to the nearest hundredth)
Answer:
0.23.
Step-by-step explanation:
We have been given that the weight of people on a college campus are normally distributed with mean 185 pounds and standard deviation 20 pounds.
First of all, we will find the z-score corresponding to sample score 200 using z-score formula.
[tex]z=\frac{x-\mu}{\sigma}[/tex], where,
[tex]z=[/tex] Z-score,
[tex]x=[/tex] Sample score,
[tex]\mu=[/tex] Mean,
[tex]\sigma=[/tex] Standard deviation.
[tex]z=\frac{200-185}{20}[/tex]
[tex]z=\frac{15}{20}[/tex]
[tex]z=0.75[/tex]
Now, we need to find [tex]P(z>0.75)[/tex]. Using formula [tex]P(z>a)=1-P(z<a)[/tex], we will get:
[tex]P(z>0.75)=1-P(z<0.75)[/tex]
Using normal distribution table, we will get:
[tex]P(z>0.75)=1-0.77337 [/tex]
[tex]P(z>0.75)=0.22663 [/tex]
Round to nearest hundredth:
[tex]P(z>0.75)\approx 0.23[/tex]
Therefore, the probability that a person weighs more than 200 pounds is approximately 0.23.
Answer:the probability that a person weighs more than 200 pounds is 0.23
Step-by-step explanation:
Since the weight of people on a college campus are normally distributed, we would apply the formula for normal distribution which is expressed as
z = (x - u)/s
Where
x = weight of people on a college campus
u = mean weight
s = standard deviation
From the information given,
u = 185
s = 20
We want to find the probability that a person weighs more than 200 pounds. It is expressed as
P(x greater than 200) = P(x greater than 200) = 1 - P(x lesser than lesser than or equal to 200).
For x = 200,
z = (200 - 185)/20 = 0.75
Looking at the normal distribution table, the probability corresponding to the z score is 0.7735
P(x greater than 200) = 1 - 0.7735 = 0.23