Answer: The second option, x^4 - 8x^2 - 16.
Step-by-step explanation:
Polynomials in standard form start with the highest degree, from greatest to least exponent. After all terms with exponents are in order, alphabetical variables are next. In this case there's only x. Last are constant terms, which are by itself, with no variable next to it/an exponent to the right of it.
x^4 - 8x^2 - 16 is in standard form because it follows the criteria above. 4 is the highest degree since it's the highest exponent in the polynomial expression, which is why it starts off with x^4. Other terms with lesser exponents are next. In this case, it's 8x^2 with the less exponent of 2. Finally, it ends with your constant term, -16.
The standard form of the polynomial is,
⇒ x⁴ - 8x² - 16
Given that,
All the polynomials are,
⇒ 5 - 2x
⇒ x⁴ - 8x² - 16
⇒ 5x³ + 4x⁴ - 3x + 1
Since we know that,
In standard form, a polynomial is arranged in descending order of the exponents of its terms.
This means that the term with the highest degree is listed first, followed by the terms with lower degrees.
Hence, Based on this definition, the polynomial in standard form among the options is,
⇒ x⁴ - 8x² - 16
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It is known that IQ scores form a normal distribution with a mean of 100 and a standard deviation of 15. If a researcher obtains a sample of 16 students’ IQ scores from a statistics class at UT. What is the shape of this sampling distribution?
Answer:
[tex]X \sim N(100,15)[/tex]
Where [tex]\mu=100[/tex] and [tex]\sigma=15[/tex]
We select a sample of n=16 and we are interested on the distribution of [tex]\bar X[/tex], since the distribution for X is normal then we can conclude that the distribution for [tex] \bar X [/tex] is also normal and given by:
[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]
Because by definition:
[tex] \bar X = \frac{\sum_{i=1}^n X_i}{n}[/tex]
[tex] E(\bar X) = \mu[/tex]
[tex] Var(\bar X) = \frac{\sigma^2}{n}[/tex]
And for this case we have this:
[tex] \mu_{\bar X}= \mu = 100[/tex]
[tex] \sigma_{\bar X} = \frac{15}{\sqrt{16}}= 3.75[/tex]
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 IQ scores of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(100,15)[/tex]
Where [tex]\mu=100[/tex] and [tex]\sigma=15[/tex]
We select a sample of n=16 and we are interested on the distribution of [tex]\bar X[/tex], since the distribution for X is normal then we can conclude that the distribution for [tex] \bar X [/tex] is also normal and given by:
[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]
Because by definition:
[tex] \bar X = \frac{\sum_{i=1}^n X_i}{n}[/tex]
[tex] E(\bar X) = \mu[/tex]
[tex] Var(\bar X) = \frac{\sigma^2}{n}[/tex]
And for this case we have this:
[tex] \mu_{\bar X}= \mu = 100[/tex]
[tex] \sigma_{\bar X} = \frac{15}{\sqrt{16}}= 3.75[/tex]
Professor Jennings claims that only 35% of the students at Flora College work while attending school. Dean Renata thinks that the professor has underestimated the number of students with part-time or full-time jobs. A random sample of 83 students shows that 38 have jobs.
Do the data indicate that more than 35% of the students have jobs? Use a 5% level of significance.
What is the value of the sample test statistic? (Round your answer to two decimal places.)
Answer:
[tex]z=\frac{0.458 -0.35}{\sqrt{\frac{0.35(1-0.35)}{83}}}=2.06[/tex]
[tex]p_v =P(z>2.063)=0.020[/tex]
So the p value obtained was a low value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of students with jobs is significantly higher than 0.35.
Step-by-step explanation:
Data given and notation
n=83 represent the random sample taken
X=38 represent the students with jobs
[tex]\hat p=\frac{38}{83}=0.458[/tex] estimated proportion of students with jobs
[tex]p_o=0.35[/tex] is the value that we want to test
[tex]\alpha=0.05[/tex] represent the significance level
Confidence=95% or 0.95
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 higher than 0.35:
Null hypothesis:[tex]p \leq 0.35[/tex]
Alternative hypothesis:[tex]p > 0.35[/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{0.458 -0.35}{\sqrt{\frac{0.35(1-0.35)}{83}}}=2.06[/tex]
Statistical decision
It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.
The significance level provided [tex]\alpha=0.05[/tex]. The next step would be calculate the p value for this test.
Since is a right tailed test the p value would be:
[tex]p_v =P(z>2.06)=0.020[/tex]
So the p value obtained was a low value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of students with jobs is significantly higher than 0.35.
Computer chips often contain surface imperfections. For a certain type of computer chip, 9% contain no imperfections, 22% contain 1 imperfection, 26% contain 2 imperfections, 20% contain 3 imperfections, 12% contain 4 imperfections, and the remaining 11% contain 5 imperfections. Let X represent the number of imperfections in a randomly chosen chip. Is X discrete or continuous
X is a discrete random variable.
A discrete random variable can only take on specific, distinct values with gaps in between.
In this case, X represents the number of imperfections in a computer chip, and it can only take on integer values: 0, 1, 2, 3, 4, or 5.
These values are countable and separate, indicating that X is a discrete random variable.
In contrast, a continuous random variable would have an infinite number of possible values within a range, and you would typically use intervals or real numbers to describe it.
For example, if we were measuring the weight of computer chips, it could be a continuous random variable because it could take on any value within a range, including fractions or decimals.
However, in this scenario, we are dealing with a countable and finite set of values for the number of imperfections, making X a discrete random variable.
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Researchers wanted to compare the effectiveness of a water softener used with a filtering process with a water softener used without filtering, Ninety locations were randomly divided into two groups of equal size. Group A locations used a water softener and the filtering process, while group B used only the water softener. At the end of three months, a water sample was tested at each location for its level of softness. (Water softness was measured on a scale of 1 to 5, with 5 being the softest water.) The results were as follows. x1-2.1 s1-0.7 x2-1.7 82 0.4 State the null and alternate hypothesis. Graph and shade the critical region. Find the critical value, the point estimate for the difference in population means given by these samples, and it's test statistic. Label these values and areas on your graph above. Find and explain the meaning of the P-value. Shade a graph showing the area equal to the p-value. Clearly state your initial and final conclusion
Answer:
Step-by-step explanation:
Hello!
The researcher's objective is to compare the effectiveness of a water softener when used with a filtering process against its effectiveness when used without filtering.
To do so 90 locations were randomly divided into two equal groups.
Group A locations used the water softener with filtering.
Group B locations used the water softener without filtering.
At the end of three months, a water sample was taken of each location and its level of softness was registered (Scale 1 to 5, 5 represents the softest water)
X₁: Softness of water of a location from group A
n₁= 45 locations
X[bar]₁= 2.1
S₁= 0.7
X₂: Softness of water of a location from group B
n₂= 45 locations
X[bar]₂= 1.7
S₂= 0.4
To compare the effectiveness of the softener with and without a filtering process the parameter of interest is the difference between both population means:
Parameter: μ₁ - μ₂
The point estimation of the difference between the population means is the difference of the sample means: X[bar]₁ - X[bar]₂= 2.1-1.7= 0.4
Since the objective is to test if there is any difference with or without the filtering process, the hypothesis test to make is two-tailed:
H₀: μ₁ - μ₂ = 0
H₁: μ₁ - μ₂ ≠ 0
α: 0.05
Since there is no information about the distribution of both variables, you have to apply the central limit theorem and approximate the distribution of X[bar]₁ and X[bar]₂ to normal. Once both samples mean distribution is approximate to normal you can use the statistic:
[tex]Z= \frac{(X[bar]₁ - X[bar]₂)-(Mu_1-Mu_2)}{\sqrt{\frac{S_1^2}{n_1} +\frac{S_2^2}{n_2} } }[/tex]
[tex]Z_{H_0}= \frac{(2.1-1.7)-0}{\sqrt{\frac{0.49}{45} +\frac{0.16}{45} } } = 3.3282[/tex]
As said before, this test is two-tailed, so you will have two critical values:
Critical value 1: [tex]Z_{\alpha /2}= Z_{0.025}= -1.95[/tex]
Critical value 2: [tex]Z_{1-\alpha /2}= Z_{0.975}= 1.965[/tex]
The p-value of this test is also two tailed, you can calculate it as:
P(Z≥3.33) + P(Z≤3.33)= (1 - P(Z≤3.33))+P(Z≤-3.33)= (1-0.999566)+0.000434= 0.000868
p-value: 0.000868
This value means that 0.0868% of the sample size 45 taken from this population will provide natural evidence that there is no difference between the population means of the effectiveness of the water softener used with and without a filtering process.
A little reminder, the p-value is defined as the probability corresponding to the calculated statistic if possible under the null hypothesis (i.e. the probability of obtaining a value as extreme as the value of the statistic under the null hypothesis).
Both using the critical value method and the p-value method the decision is to reject the null hypothesis. This means that with a 5% level of significance there is a difference between the true average of the effectiveness of the water softener used with a filtering process and the true average effectiveness of the water softener used without a filtering process.
I hope it helps!
Dr. Sabbaghi is taking two flights today. The flight time for the first flight is Normally distributed with a mean of 90 minutes and a standard deviation of 3 minutes. The flight time for the second flight is Normally distributed with a mean of 110 minutes and a standard deviation of 4 minutes. His total flight time today has what type of distribution with what mean and standard deviation?
Answer:
[tex] E(Z) = E(X+Y) = E(X) +E(Y) = \mu_X +\mu_Y = 90+110=200[/tex]
[tex] Var(Z) = Var(X+Y) = Var(X) +Var(Y) +2 Cov(X,Y)[/tex]
Since X and Y are independent then [tex] Cov(X,Y) =0[/tex] and we have this:
[tex] Var(Z)= \sigma^2_X +\sigma^2_Y = 3^2 +4^2 = 9+16 =25[/tex]
And the deviation would be given by:
[tex] Sd(Z) = \sqrt{25}= 5[/tex]
And then the distribution for the total time would be given by:
[tex] Z= X+Y \sim N( \mu_Z= 200, \sigma_Z= 5)[/tex]
Step-by-step explanation:
For this case we can assume that X represent the flight time for the first filght and we know that:
[tex] X \sim N (\mu_X= 90. \sigma_x =3)[/tex]
And let Y the random variable that represent the time for the second filght and we know this:
[tex] Y \sim N(\mu_Y = 110, \sigma_Y =4)[/tex]
And we can define the random variable Z= X+Y as the total time for the two flights.
We can asume that X and Y are independent so then we have this:
[tex] E(Z) = E(X+Y) = E(X) +E(Y) = \mu_X +\mu_Y = 90+110=200[/tex]
[tex] Var(Z) = Var(X+Y) = Var(X) +Var(Y) +2 Cov(X,Y)[/tex]
Since X and Y are independent then [tex] Cov(X,Y) =0[/tex] and we have this:
[tex] Var(Z)= \sigma^2_X + \sigma^2_Y = 3^2 +4^2 = 9+16 =25[/tex]
And the deviation would be given by:
[tex] Sd(Z) = \sqrt{25}= 5[/tex]
And then the distribution for the total time would be given by:
[tex] Z= X+Y \sim N( \mu_Z= 200, \sigma_Z= 5)[/tex]
To prepare for surgery, Anne mixes an anesthetic solution using two different concentrations: 40 mL of 25% solution and 60 mL of 40% solution.
What is the concentration of the mixed solution?
Answer:
34%
Step-by-step explanation:
Amount of anesthetic in the 25% solution + amount of anesthetic in the 40% solution = amount of anesthetic in the mixed solution
0.25 (40) + 0.40 (60) = x (40 + 60)
10 + 24 = 100x
x = 0.34
The concentration of the mixed solution is 34%.
A study of Machiavellian traits in lawyers was performed. Machiavellian describes negative character traits such as manipulation, cunning, duplicity, deception, and bad faith. A Mach rating score was determined for each in a sample of lawyers. The lawyers were then classified as having high comma moderate comma or low Mach rating scores. The researcher investigated the impact of both Mach score classification and gender on the average income of a lawyer. For this experiment, identify Bold a. the experimental unit, Bold b. the response variable, Bold c. the factors, Bold d. the levels of each factor, and Bold e. the treatments.
a. The experimental unit:
The experimental units would be the lawyers that participate in this experiment. The experimental units are the subjects upon which the experiment is performed.
b. The response variable:
The response variable would be income. This is the variable that measures the response or outcome of the study.
c. The factors:
The factors are the variables whose levels are manipulated by the researcher. In this case, these would be the Mach score, classification and gender.
d. The levels of each factor:
The levels would include the levels of the Mach score (high, moderate, low) and the levels of gender (male, female).
e. The treatments:
The treatments are all the possible combinations of one level of each factor. Therefore, these are: High and male, high and female, moderate and men, moderate and female, low and male, low and female.
Two hikers come to a ravine and want to know how wide it is. They set up two similar triangles as shown in the diagram. How far is it across the ravine?
Answer:
d = 80 ft
Step-by-step explanation:
Start with angle BCA, tan(BCA) = 30/15 so BCA = 63.43 degrees.
BCA = ECD since they are opposite angles.
tan ECD = d/40
40 tan(63.4degrees) = d
d = 80 ft
Alternatively, you can use similar triangles since the angles are the same, BCA = ECD, CED = CAB, and CBA = EDC. In that case use proportions to get 30/15 = d/40 so 2 = d/40 and d = 80.
1. Write an equivalent expression for 27x+18
2. Write the inequality this number line represents
3.erin is going to paint a wall in her house she needs to find the area of the wall so she knows how much paint to purchase what is the area of her wall
4.walt received a package that is 12 1/3 inches long 6 3/4 inches high and 8 1/2 inches wide what is the surface area of the package
Answer:
1. 27x+18 = x+x+x+x+x......+x + 18
You sum "x" 27 times.
2. [tex](36,\infty)[/tex]
3. [tex]285/2 = 142.4[/tex]
4. 2*(12 1/3 )*(8 1/2) + 2*(12 1/3 )*(6 3/4)+2*(6 3/4 )*(8 1/2)
Step-by-step explanation:
1. Remember that multiplication is a simplification of the sum, so, when you say for example, 4*3, that actually means 3+3+3+3, similarly, when you say, 27x, that means x+x+x...+x 27 times.
2. From the image you can see that
The 36 is NOT taken, and then you go all the way to infinity, therefore we say [tex](36,\infty)[/tex]. Suppose that 36 was taken, then we would say [tex][36,\infty)[/tex].
3. From the attached photo
you can see that we can compute first the area of the rectangle with length = 15 and height = 7, and also note that at the top a triangle with base 15 and height 5 is formed, so the area of the whole figure would be the area of the rectangle at the bottom plus the area of the triangle on top. That would be 7*15+(15*5)/2 = 285/2
4. Remember that in general the formula for surface area would be
[tex]2lw +2lh+2wh[/tex]
Where l = length , w = wide, h = height. In this case l = 12 1/3 , w = 8 1/2 and h = 6 3/4
Please help!! It’s due @ midnight
How much should Marc deposit weekly into an account at 8% compounded weekly in order to have
$4500 available for a round trip plane ticket, hotel, and spending money for a trip to Sweden in 2 years?
Please give step by step!
Answer:
ooh i just learned this, not 100% sure but the amount he should have to deposit is $3835.12
if the weekly one means depositing money every week then it would be $36.88 I think.
Step-by-step explanation:
p=?
r=.08
n=52
t=2
4500 = P(1+.08/52)^(52 x 2)
divide both sides by (1+.08/52)^(52 x 2)
and you are left with $3835.12
if i take into account that Marc is depositing the money every week the i would divide it by 104 (that is 52 x 2 because 52 weeks in a year and it says 2 years) you would be left with $36.88.
Hope I was any help.
Answer: Marc should deposit $39.87 weekly.
Step-by-step explanation:
We would apply the formula for determining future value involving deposits at constant intervals. It is expressed as
S = R[{(1 + r)^n - 1)}/r][1 + r]
Where
S represents the future value of the investment.
R represents the regular payments made(could be weekly, monthly)
r = represents interest rate/number of interval payments.
n represents the total number of payments made.
From the information given,
S = $4500
Assuming there are 52 weeks in a year, then
r = 0.08/52 = 0.0015
n = 52 × 2 = 104
Therefore,
4500 = R[{(1 + 0.0015)^104 - 1)}/0.0015][1 + 0.0015]
4500 = R[{(1.0015)^104 - 1)}/0.0015][1.0015]
4500 = R[{(1.169 - 1)}/0.0015][1.0015]
4500 = R[{(0.169)}/0.0015][1.0015]
4500 = R[112.67][1.0015]
4500 = 112.839R
R = 4500/112.839
R = 39.87
According to the manufacturer of the candy Skittles, 20% of the candy produced are red. If we take a random sample of 100 bags of Skittles, what is the probability that the proportion in our sample of red candies will be less than 16%?
Answer:
0.15651
Step-by-step explanation:
This can be approximated using a Poisson distribution formula.
The Poisson distribution formula is given by
P(X = x) = (e^-λ)(λˣ)/x!
P(X ≤ x) = Σ (e^-λ)(λˣ)/x! (Summation From 0 to x)
where λ = mean of distribution = 20 red bags of skittles (20% of 100 bags of skittles means 20 red bags of skittles)
x = variable whose probability is required = less than 16 red bags of skittles
P(X < x) = Σ (e^-λ)(λˣ)/x! (Summation From 0 to (x-1))
P(X < 16) = Σ (e^-λ)(λˣ)/x! (Summation From x=0 to x=15)
P(X < 16) = P(X=0) + P(X=1) + P(X=2) +......+ P(X=15)
Solving this,
P(X < 16) = 0.15651
The question asks for the probability that the proportion of red Skittles is less than 16% from a sample of 100 bags. This involves the application of sampling distribution of a sample proportion, wherein the distribution should follow a normal distribution if certain conditions are met. The question can be solved by calculating a z-score and finding the corresponding probability from a standard normal distribution table.
Explanation:This question involves the concept of sampling distribution of a sample proportion. Here, under certain conditions, the sampling distribution of p' (the sample proportion) tends to follow a normal distribution. The mean (expected value) of the distribution is equal to the population proportion (p), and the standard deviation (standard error) of the distribution is sqrt [ p(1 - p) / n ], where n is the size of the sample.
Given that the population proportion (p) = 0.20 and n = 100. We are asked to find the probability that the sample proportion (p') is less than 0.16 (16%). So, we can represent this situation as: P( p' < 0.16 ).
To find this probability we need to standardize our value of interest (0.16), resulting in a z-score. The z-score = ( p' - p ) / sqrt [ p(1 - p) / n ]. Plugging our values in, you can calculate the z-score, and apply it into a standard normal distribution table or use a calculator that can calculate probabilities using normal distribution to find the probability.
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An investment earns 13% the first year, earns 20% the second year, and loses 15% the third year. The total compound return over the 3 years was ______.
Answer:
The total compound return over the 3 years is 15.26%
Step-by-step explanation:
Let the initial investment sum be assumed to be X
The total return after each year can be calculated as follows:
After First year: X + (13% of X) = 1.13X
After Second year: 1.13X + (20% of 1.13X) = 1.13X + 0.226X = 1.356X
After Third year: 1.356X - (15% of 1.356X) = 1.356X - 0.2034X = 1.1526X
It is apparent from here that after the third year, the investment has increased the initial X, by 0.1526X, which is 15.26%.
The total compound return over the 3 years is 15.26%
After scoring a touchdown, a football team may elect to attempt a two-point conversion, by running or passing the ball into the end zone. If successful, the team scores two points. For a certain football team, the probability that this play is successful is 0.80.1. Let X = 1 if successful, X = 0 if not. Find the mean and the variance of X. Round the answers to two decimal places.The mean of X is = .The variance of X is =2. Let Y be the number of points scored. Find the mean and variance of Y. Round the answers to two decimal places.The mean of Y is =The variance of Y is =
Answer:
Step-by-step explanation:
given that after scoring a touchdown, a football team may elect to attempt a two-point conversion, by running or passing the ball into the end zone. If successful, the team scores two points.
X=1 if successful and
X=0 if not
pdf of X is
X 1 0
p 0.8 0.2
E(x) = [tex]1(0.8)+0(0.2)\\=0.8[/tex]
[tex]E(x^2) = 1^2(0.8) = 0.8[/tex]
Var(x) = 0.8-0.8*0.8
= 0.16
Now let us consider Y.
Y is the no of points scored.
Y 2 0
p 0.8 0.2
E(Y) = [tex]2(0.8)+0(0.2)\\=1,.6[/tex]
[tex]2^2(0.8)+0(0.2)\\= 3.2[/tex]
The mean and variance for X and Y are calculated using their definitions in probability theory. For X, the mean is 0.80 and variance is 0.16. For Y, the mean is 1.60 and variance is 0.64.
Explanation:The random variable X is a Bernoulli random variable because it has only two outcomes: success (X=1) and failure (X=0). The mean and variance for a Bernoulli random variable can be found using the formulas: Mean (E[X]) = p and Variance (Var[X]) = p(1-p).
For X:
The mean E[X] = p = 0.80. So, X = 1 with probability 0.80.
The variance Var[X] = p(1-p) = 0.80(1-0.80) = 0.16
The random variable Y represents the number of points scored which can either be 0 or 2. We let p be the probability of scoring 2 points i.e., p = 0.80. Hence, if the two-point conversion is successful, we will score 2 points, otherwise, we score 0.
For Y:
The mean E[Y] = 0*(1-p) + 2*p = 0 + 2*0.80 = 1.60
The variance Var[Y] = (0-E[Y])^2*(1-p) + (2-E[Y])^2*p = (0-1.60)^2*(1-0.80) + (2-1.60)^2*0.80 = 0.64
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1) A home improvement store sold wind chimes for w dollars each. A customer signed up for a free membership card and received a 5% discount off the price. Sales tax of 6% was applied after the discount. Write an algebraic expression to represent the final price of the wind chime.
Answer:
Step-by-step explanation:
The original price of the wind chimes at the home improvement store is $w.
A customer signed up for a free membership card and received a 5% discount off the price. The value of the discount is
5/100 × w = 0.05w
The discounted price would be
w - 0.05w = 0.95w
Sales tax of 6% was applied after the discount. The amount of sales tax applied would be
6/100 × 0.95w = 0.057w
The algebraic expression to represent the final price of the wind chime is
0.95w + 0.057w
= 1.007w
The probability that a person in the United States has type B+ blood is 13%.
Four unrelated people in the United States are selected at random.
Complete parts (a) through(d).
(a) Find the probability that all four have type B+ blood.The probability that all four have type B+ blood is?
(Round to six decimal places as needed.)
(b) Find the probability that none of the four have type B+ blood.The probability that none of the four have type B+ blood is?
(Round to three decimal places as needed.)
(c) Find the probability that at least one of the four has type B+ blood.The probability that at least one of the four has type B+ blood is?
(Round to three decimal places as needed.)
(d) Which of the events can be considered unusual? Explain. Select all that apply.
A.None of these events are unusualNone of these events are unusual.
B.The event in part (a) is unusual because its probability is less than or equal to 0.05.
C.The event in part (b) is unusual because its probability is less than or equal to 0.05.
D.The event in part (c) is unusual because its probability is less than or equal to 0.05.
a) Probability that all four have type B+ blood = 0.00031213
b) Probability that none of the four have type B+ blood = 0.57289761
c) Probability that at least one of the four has type B+ blood = 0.42710239
d) B. The event in part (a) is unusual because its probability is less than or equal to 0.05.
To solve these probability problems, we'll use the binomial probability formula:
[tex]P(X=k) = (n, k) \times p^k \times (1-p)^{(n-k)[/tex]
Where:
P(X=k) is the probability of having exactly k successes in n trials.
(n choose k) is the number of ways to choose k successes from n trials (n! / (k! (n-k)!), where n! is the factorial of n).
p is the probability of success (having type B+ blood in this case).
q = 1 - p is the probability of failure (not having type B+ blood).
n is the number of trials.
Given:
p = 0.13 (probability of having type B+ blood)
q = 1 - p = 0.87 (probability of not having type B+ blood)
n = 4 (number of trials)
Let's solve each part step by step:
(a) Probability that all four have type B+ blood:
[tex]P(X=4) = (4, 4) \times 0.13^4 \times 0.87^{(4-4)[/tex]
[tex]P(X=4) = 1 \times 0.00031213 \times 1 \\\\= 0.00031213[/tex]
(b) Probability that none of the four have type B+ blood:
[tex]P(X=0) = (4, 0) \times 0.13^0 \times 0.87^4 \\\\P(X=0) = 1 \times 1 \times 0.57289761 \\\\= 0.57289761[/tex]
(c) Probability that at least one of the four has type B+ blood:
P(at least one) = 1 - P(none)
P(at least one) = 1 - 0.57289761
= 0.42710239
Now, let's determine which events are considered unusual. Generally, an event with a probability less than or equal to 0.05 is considered unusual.
Let's compare the probabilities:
Probability in part (a): 0.00031213 (less than 0.05)
Probability in part (b): 0.57289761 (greater than 0.05)
Probability in part (c): 0.42710239 (greater than 0.05)
Based on the comparison, the only event that can be considered unusual is the event in part (a) because its probability is less than 0.05. Therefore, the correct answers are:
B. The event in part (a) is unusual because its probability is less than or equal to 0.05.
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The calculations show that the probability of all four people having B+ blood is 0.000028561, and the likelihood of none of them having B+ blood is 0.569532. The chance of at least one of them having B+ blood is 0.430468. Thus, only the event in part (a) is considered unusual due to its low probability.
Explanation:This question is about using probability principles to figure out the likelihood of having certain blood types in a population.
(a) To find the probability that all four individuals have type B+ blood, we need to multiply the individual probabilities together. The probability that one person has B+ blood is given as 13% or 0.13. So, the probability that all four have B+ blood is 0.13*0.13*0.13*0.13 = 0.000028561.
(b) The probability that none of the four have type B+ blood is the complement of the probability that one person has B+ blood. This is 1 - 0.13 = 0.87. We now raise this to the power of four to find the probability that all four selected people do not have B+ blood: 0.87*0.87*0.87*0.87 = 0.569532.
(c) The probability that at least one has type B+ blood is the complement of the result in part b. We subtract our answer from part b from 1: 1 - 0.569532 = 0.430468.
(d) An event is considered unusual if its probability is less than or equal to 0.05. Here, the event in part (a) is unusual because its probability (0.000028561) is less than or equal to 0.05.
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The grade point averages for 10 randomly selected high school students are listed below. Assume the grade point averages are normally distributed. 2.0 3.2 1.8 2.9 0.9 4.0 3.3 2.9 3.6 0.8 Find a 98% confidence interval for the true mean.
Answer:
[tex]2.54-2.82\frac{1.110}{\sqrt{10}}=1.55[/tex]
[tex]2.54+2.82\frac{1.110}{\sqrt{10}}=3.53[/tex]
So on this case the 98% confidence interval would be given by (1.55;3.53)
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".
[tex]\bar X[/tex] represent the sample mean for the sample
[tex]\mu[/tex] population mean (variable of interest)
s represent the sample standard deviation
n represent the sample size
Solution to the problem
The confidence interval for the mean is given by the following formula:
[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)
In order to calculate the mean and the sample deviation we can use the following formulas:
[tex]\bar X= \sum_{i=1}^n \frac{x_i}{n}[/tex] (2)
[tex]s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}[/tex] (3)
The mean calculated for this case is [tex]\bar X=2.54[/tex]
The sample deviation calculated [tex]s=1.110[/tex]
In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:
[tex]df=n-1=10-1=9[/tex]
Since the Confidence is 0.98 or 98%, the value of [tex]\alpha=0.02[/tex] and [tex]\alpha/2 =0.01[/tex], and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-T.INV(0.01,9)".And we see that [tex]t_{\alpha/2}=2.82[/tex]
Now we have everything in order to replace into formula (1):
[tex]2.54-2.82\frac{1.110}{\sqrt{10}}=1.55[/tex]
[tex]2.54+2.82\frac{1.110}{\sqrt{10}}=3.53[/tex]
So on this case the 98% confidence interval would be given by (1.55;3.53)
The 98% confidence interval would be given by (1.55;3.53)
A range of values that, with a particular level of confidence, is likely to encompass a population value is called a confidence interval. A population mean is typically stated as a percentage that falls between an upper and lower interval.
The range of values in a confidence interval below and above the sample statistic is called the margin of error.
The 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".
show the sample mean for the given sample.
population mean (the relevant variable)
The sample standard deviation is denoted by s.
n stands for the number of samples.
Resolution of the issue
The following formula produces the mean's confidence interval:
[tex]\bar x[/tex] ± [tex]\frac{t_a}{2} \frac{s}{\sqrt{n} }[/tex] ____________(1)
In order to calculate the mean and the sample deviation we can use the following formulas:
[tex]\bar x =[/tex] [tex]\sum_{i =1}^n \frac{x_i}{n}[/tex]_______(2)
[tex]s = \sqrt{\frac{\sum^n_{i=1}(x_i-\bar x)}{n-1} }[/tex] ____________(3)
The mean calculated for this case is X = 2.54
The sample deviation calculated s = 1.110
t In order to calculate the critical value [tex]\frac{t_a}{2}[/tex] we need freedom, given by: to find first the degrees of
df = n-1=10-19
Since the Confidence is 0.98 or 98%, the value of a = 0.02 and a/2 = 0.01, and we can use excel, a calculator or a tabel to find the critical value.
[tex]\frac{t_a}{2}[/tex] = 2.82
Now we have everything in order to replace into formula (1):
2.54 - 2.82 [tex]\frac{1.110}{\sqrt{10} }[/tex] = 1.55
2.54 + 2.82 [tex]\frac{1.110}{\sqrt{10} }[/tex] = 3.53
Absorption rates are important considerations in the creation of a generic version of a brand-name drug. A pharmaceutical company wants to test if the absorption rate of a new generic drug (G) is the same as its brand-name counterpart (B).They run a small experiment to test H subscript 0 : space mu subscript G minus mu subscript B equals 0 against the alternative H subscript A : space mu subscript G minus mu subscript B not equal to 0 . Which of the following is a Type I error?a. Deciding that the absorption rates are different, when in fact they are not.b. The researcher cannot make a Type I error, since he has run an experiment.c. Deciding that the absorption rates are different, when in fact they are.d. Deciding that the absorption rates are the same, when in fact they are.e. Deciding that the absorption rates are the same, when in fact they are not.
Answer:
The type 1 error here is a. Deciding that the absorption rates are different, when in fact they are not.
Step-by-step explanation:
A type I error is the rejection of a true null hypothesis (also known as a "false positive" finding or conclusion).
More generally, a Type I error occurs when a significance test results in the rejection of a true null hypothesis. By one common convention, if the probability value is below 0.05, then the null hypothesis is rejected.
In inferential statistics, the null hypothesis is a general statement or default position that there is nothing significantly different happening, like there is no association among groups or variables, or that there is no relationship between two measured phenomena.
The guidance system of a ship is controlled by a computer that has 3 major modules. In order for the computer to function properly, all 3 modules must function. Two of the modules have reliabilities of 0.97 and the other has a reliability of 0.99.
a) What is the reliability of the computer?
b) A backup computer identical to the one being used will be installed to improve overall reliability. Assuming the new computer automatically functions if the main one falls, determine the resulting reliability.
c) If the backup computer must be activated by a switch in the event that the first computer fails, and the switch has a reliability of 0.98, what is the overall reliability of the system? (Both the switch and the backup computer must function in order for the backup to take over.)
Answer:
a) 0.931491
b) 0.995307
c) 0.994030
Step-by-step explanation:
a) Since all components must be working, the reliability of the computer is the product of the reliability of the three components:
[tex]R_1 = 0.97*0.97*0.99\\R_1=0.931491[/tex]
b) The resulting reliability is now the reliability of the first computer, added to the possibility of failure of the first computer multiplied by the reliability of the second computer:
[tex]R= R_1 +(1-R_1)*R_2\\R= 0.931491+(1-0.931491)*0.97*0.97*0.99\\R=0.995307[/tex]
c) If a switch with reliability of 0.98 must be activated to turn on the second computer, the switch's reliability must be taken into account as follows:
[tex]R= R_1 +(1-R_1)*R_2*R_S\\R= 0.931491+(1-0.931491)*0.97*0.97*0.99*0.98\\R=0.994030[/tex]
The reliability of the system is simply its probability of not failing
(a) The reliability of the computer
This is the product of the reliabilities of the three modules.
So, we have:
[tex]R = 0.97 \times 0.97 \times 0.99[/tex]
[tex]R = 0.931491[/tex]
Approximate
[tex]R = 0.9315[/tex]
Hence, the reliability of the computer is 0.9315
(b) The reliability when a backup is used
In (a), the reliability of the computer is 0.9315
When the computer fails, the reliabilities of the other two are 1 - 0.9315 and 1 - 0.9315.
So, the reliability when a backup is used is calculated using the following complement rule
[tex]R = 1 - [(1 - 0.9315) \times (1 - 0.9315)][/tex]
[tex]R = 0.99530775[/tex]
Approximate
[tex]R = 0.9953[/tex]
Hence, the reliability of the computer when a backup is 0.9953
(c) The overall reliability of the system
In (a), the reliability of the computer is 0.9315.
Also, the reliability of the switch is 0.98
So, the reliability of the backup is:
[tex]R = 0.9315 \times 0.98[/tex]
[tex]R = 0.9129[/tex]
So, the overall system has:
Main computer with reliability of 0.9315 Back up of the computer system with reliability of 0.9129
The reliability of the overall system is then calculated using the following complement rule
[tex]R = 1 - [(1 - 0.9315) \times (1 - 0.9129)][/tex]
[tex]R = 0.9940[/tex]
Hence, the reliability of the overall system is 0.9940
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In human resource management, performance of employees is measured as a numerical score which is assumed to be normally distributed. The mean score is 150 and the standard deviation 13. What is the probability that a randomly selected employee will have a score less than 120?
Answer:
[tex]P(X<120)=P(\frac{X-\mu}{\sigma}<\frac{120-\mu}{\sigma})=P(Z<\frac{120-150}{13})=P(z<-2.308)[/tex]
And we can find this probability using the normal standard table or excel and we got:
[tex]P(z<-2.308)=0.0105[/tex]
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 scores of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(150,13)[/tex]
Where [tex]\mu=150[/tex] and [tex]\sigma=13[/tex]
We are interested on this probability
[tex]P(X<120)[/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<120)=P(\frac{X-\mu}{\sigma}<\frac{120-\mu}{\sigma})=P(Z<\frac{120-150}{13})=P(z<-2.308)[/tex]
And we can find this probability using the normal standard table or excel and we got:
[tex]P(z<-2.308)=0.0105[/tex]
A store sells 15 2/3 pounds of carrots, 12 1/3 pounds of asparagus, and 3 1/3 of cabbage. How many pounds did the store sell altogether?
Answer: the store sold 31 1/3 pounds
Step-by-step explanation:
The store sold 15 2/3 pounds of carrots. Converting to improper fraction, it becomes 47/3 pounds.
The store sold 12 1/3 pounds of asparagus. Converting to improper fraction, it becomes 37/3 pounds.
The store sold 3 1/3 pounds of cabbage. Converting to improper fraction, it becomes 10/3 pounds.
Therefore, the total number of pounds that the store sold is
47/3 + 37/3 + 10/3 = 94/3 = 31 1/3 pounds
(a) Estimate a second point on the tangent line. (t, p) = Correct: Your answer is correct. (b) Calculate the rate of change of the function at the labeled point. (Round your answer to one decimal place.) 12.5 Incorrect: Your answer is incorrect. thousand employees per year (c) Calculate the percentage rate of change of the function at the labeled point. (Round your answer to three decimal places.) 0.833 Incorrect: Your answer is incorrect. % per year
Complete Question
The complete question is shown on the first uploaded image
Answer:
a
The Estimated second point is (t,p) =(5,9320)
b
The rate is 180 thousand employer per year
c
The percentage rate of change of the function is 1.939%
Step-by-step explanation:
Looking at the graph
First is to obtain the scale of the graph what i mean is what the distance between each line segment
Considering the y-axis each line segment is
[tex]\frac{9300-9200}{5} = \frac{100}{5} =20[/tex]
So this means that after 9300 the next line segment is 9320
Considering the x-axis each line segment is
[tex]\frac{4-2}{4} = \frac{2}{4} = 0.5[/tex]
What this means is that 2 line segment after 4 is 4 +2 ×(0.5) =5
So looking this two points (new_t,new_p) = (5 , 9320) = we see that they form a coordinate
B) The labeled point that we are to consider are
[tex](t_1,p_2) = (4.8, 9284) \ (t_2,p_2) = (5, 9320)[/tex]
The rate change
[tex]= \frac{p_2-p_1}{t_2-t_1}=\frac{9320-9284}{5-4.8} = \frac{36}{0.2} = 180[/tex]
So the rate is 180 thousand employer per year
C)
So to obtain the percentage rate of change of the function
Now
[tex]f(4.8) = 9284 \ Thousand[/tex]
[tex]f'(4.8) = 180 \ Thousand[/tex]
Note: This is so because differentiation is the same as slope of the graph
Hence the percentage rate of change
[tex]\frac{f'(4.8)}{f(4.8)} *\frac{100}{1} = \frac{180 \ 000}{9284 \ 000} * \frac{100}{1}[/tex]
= 1.939%
The question seems to be related to calculus, discussing concepts of tangents and rates of change for a function. To find another point on a tangent line or calculate the rate of change, we need the function and additional details. The respective formulas for these calculations are mentioned, but we can't provide a factual answer without more information.
Explanation:Given the question, it seems this is related to the field of calculus, specifically addressing concepts of tangents and rates of change of a function. However, to find another point on the tangent line or calculate the rate of change at a specific point, we need more context or the actual function.
Usually, for the first question, you can use the formula slope = (y2-y1) / (x2-x1), assuming (t, p) is a point on the tangent and you know the slope of the tangent. For the second question, the rate of change at a point is the derivative of the function at that point. For the third question, the percentage rate of change at a point is the derivative at that point divided by the function value at that point, multiplied by 100 to get percentage.
Without the function and some additional information, we cannot arrive at a factual answer. You might want to check the question again and include the necessary details.
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The level of nitrogen oxides (NOX) in a exhaust of cars of a particular model varies normally with mean 0.25 grams per miles and standard deviation 0.05 g/mi. government regulations call for NOX emissions no higher than 0.3 g/mi.
a. What is the probability that a single car of this model fails to meet the NOX requirement?
b. A company has 4 cars of this model in its fleet. What is the probability that the average NOX level of these cars are above 0.3 g/mi limit?
Answer:
a) 15.87% probability that a single car of this model fails to meet the NOX requirement.
b) 2.28% probability that the average NOX level of these cars are above 0.3 g/mi limit
Step-by-step explanation:
We use the normal probability distribution and the central limit theorem to solve this question.
Normal probability distribution:
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.
Central limit theorem:
The Central Limit Theorem estabilishes that, for a random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], a large sample size can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex]
In this problem, we have that:
[tex]\mu = 0.25, \sigma = 0.05[/tex]
a. What is the probability that a single car of this model fails to meet the NOX requirement?
Emissions higher than 0.3, which is 1 subtracted by the pvalue of Z when X = 0.3. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{0.3 - 0.25}{0.05}[/tex]
[tex]Z = 1[/tex]
[tex]Z = 1[/tex] has a pvalue of 0.8417.
1 - 0.8413 = 0.1587.
15.87% probability that a single car of this model fails to meet the NOX requirement.
b. A company has 4 cars of this model in its fleet. What is the probability that the average NOX level of these cars are above 0.3 g/mi limit?
Now we have [tex]n = 4, s = \frac{0.05}{\sqrt{4}} = 0.025[/tex]
The probability is 1 subtracted by the pvalue of Z when X = 0.3. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{0.3 - 0.25}{0.025}[/tex]
[tex]Z = 2[/tex]
[tex]Z = 2[/tex] has a pvalue of 0.9772
1 - 0.9772 = 0.0228
2.28% probability that the average NOX level of these cars are above 0.3 g/mi limit
The probability that a single car of this model fails to meet the NOX requirement is 15.87%.
What is z score?Z score is used to determine by how many standard deviations the raw score is above or below the mean. It is given by:
z = (raw score - mean) / standard deviation
Given; mean of 0.25 g and a standard deviation of 0.05 g/mi
a) For > 0.3:
z = (0.3 - 0.25)/0.05 = 1
P(z > 1) = 1 - P(z < 1) = 1 - 0.8413 = 0.1587
b) For > 0.3, sample size = 4
z = (0.3 - 0.25)/(0.05 ÷√4) = 2
P(z > 2) = 1 - P(z < 2) = 1 - 0.9772 = 0.0228
The probability that a single car of this model fails to meet the NOX requirement is 15.87%.
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Convert the measurement as indicated.
73 inches = ? ft ?in
Answer:
6.08333
Step-by-step explanation:
6ft 0.8inches
Answer:
6 ft 1 in
Step-by-step explanation:
Rod is a drummer who purchases his drumsticks online. When practicing with the newest pair, he notices they feel heavier than usual. When he weighs one of the sticks, he finds that it is 2.33 oz. The manufacturer's website states that the average weight of each stick is 1.75 oz with a standard deviation of 0.22 oz. Assume that the weight of the drumsticks is normally distributed. What is the probability of the stick's weight being 2.33 oz or greater? Give your answer as a percentage precise to at least two decimal places.
Answer:
The probability of the stick's weight being 2.33 oz or greater is 0.41%.
Step-by-step explanation:
Test statistic (z) = (weight - mean)/sd
weight of stick = 2.33 oz
mean = 1.75 oz
sd = 0.22 oz
z = (2.33 - 1.75)/0.22 = 2.64
The cumulative area of the test statistic is the probability that the weight is 2.33 oz or less. The cumulative area is 0.9959.
The probability the weight is 2.33 or greater = 1 - 0.9959 = 0.0041 = 0.41%
A study on educational aspirations of high school students (Crysdale, Int. J. Comp. Sociol., 16, 19-36, 1975) measured aspirations using the sale (some high school, high school graduate, some college, college graduate). For students whose family income was low, the counts in these categories were (9, 44, 13, 10); when the family income was middle, the counts were (11, 52, 23, 22); when the family income was high, the counts were (9, 41, 12, 27). a. Use SAS/R to test whether the aspirations and family income are independent, reporting both the X2 and G2 statistics. b. No matter your answer in part a, do the standardized residuals suggest any interesting patterns? c. Using SAS/R, conduct a more powerful test than those in part a
Answer:
I code an example question with answer.
Step-by-step explanation:
A study on educational aspirations of High School Students ( S. Crysdale, International Journal of Comparative Sociology, Vol 16, 1975, pp 19-36) measured aspirations using the scale (some high school, high school graduate, some college, college graduate). For students whose family income was low, the counts in these categories were (9, 44, 13, 10); when family income was middle, the counts were (11, 52, 23, 22); when family income was high, the counts were (9, 41, 12, 27)
A. Construct a suitable contingency table for the above data.
B. Find the conditional distribution on aspirations for those whose family income was high.
C. Conduct a Chi-square test of Independence between educational aspirations and income levels.
D. Explain what further analyses you could do that would be more informative than a chi-squared test.
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Chi-Squared Test for Homogeneity of Several Categorical Populations
Null Hypothesis: populations of people are homogeneous with respect to the four levels of education (low, med, high income groups educated same)
Alternative Hypothesis: populations not homogeneous.
Chi-Square Test: Some High School, Grad High School, Some College, Grad College
Expected counts are printed below observed counts
Chi-Square contributions are printed below expected counts
Chi-Square Test: Some High School, Grad High School, Some College, Grad College
Expected counts are printed below observed counts
Chi-Square contributions are printed below expected counts
Some Grad
High High Some Grad
School School College College Total
Low Income
9 44 13 10 76 [observed counts]
8.07 38.14 13.36 16.42 [expected counts]
0.106 0.901 0.010 2.513 [Chi-Square contr]
Medium Income
11 52 23 22 108
11.47 54.20 18.99 23.34
0.019 0.089 0.847 0.077
High Income
9 41 12 27 89
9.45 44.66 15.65 19.23
0.022 0.300 0.851 3.135
Total 29 137 48 59 273
Chi-Sq = 8.871, DF = 6, P-Value = 0.181
"P-Value" = 0.181 > 0.10 [90% confidence interval]; thus Null Hypothesis of homogeneity should be rejected (low, med, high income groups educated same). Alternative Hypothesis should be accepted (education dependent upon income level)
The task is a statistical analysis using SAS/R to test the independence of two variables, economic status and educational aspirations. This involves chi-square and likelihood ratio tests, investigating standardized residuals, and performing more powerful tests for comprehensive results.
Explanation:The question involves a statistical analysis task using SAS/R to determine the independence between two variables: economic status and educational aspirations. The Chi-square (X2) and likelihood ratio (G2) tests can be utilized to analyze the independence. Standardized residuals can help diagnose potential patterns and significance of divisions, while more powerful tests such as Fisher's exact test or Monte Carlo simulation could provide further insights.
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Determine the original set of data. 1 0 1 5 2 1 4 4 7 9 3 3 5 5 5 7 9 4 0 1 Legend: 1|0 represents 10The originat set of the data is?
The data set is S = {10, 11, 15, 21, 24, 24, 27, 29, 33, 35, 35, 37, 39, 40, 40}
A stem-and-leaf plot is a method to represent the data in tabular form.
The stem consist of the first digits of the data values arranged in ascending order.
The leaf consist of the remaining digits.
The data provided is:
Stem | Leaf
1 | 0 1 5
2 | 1 4 4 7 9
3 | 3 5 5 7 9
4 | 0 1
The original data is:
10, 11, 15, 21, 24, 24, 27, 29, 33, 35, 35, 37, 39, 40, 40
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The original set of data consists of number sequences as follows: 10 12 4 7 11 4 3 10 0, 10 4 14 11 13 2 4 6, 12 6 9 10, 5 13 4, 10 14 12 11, and 6 10 11 0 11 13 2.
The original set of data is:
10 12 4 7 11 4 3 10 010 4 14 11 13 2 4 612 6 9 105 13 410 14 12 116 10 11 0 11 13 2
Simplify.
25
−
40
A) 3
10
B) 10
10
C) 5 − 2
10
D) 2
10
− 5
Answer:
5 -2√10 = C
Step-by-step explanation:
√25 -√40
√25 = √5 x5 = √5² = 5
√40 = √5 x 8 = √5x 2x4 = √10x4 = √10x 2² = 2√10 (when the 2 comes back into the square root it become 2²)
√25 -√40 = 5 -2√10
Suppose Team A has a 0.75 probability to win their next game and Team B has a 0.85 probability to win their next game. Assume these events are independent. What is the probability that Team A wins and Team B loses
The probability that Team A wins and Team B loses is 0.112.
Given that,
Suppose Team A has a 0.75 probability to win their next game and Team B has a 0.85 probability to win their next game.
Assume these events are independent.
We have to determine,
What is the probability that Team A wins and Team B loses?
According to the question,
In an independent event and probability, the outcomes in an experiment are termed as events. Ideally, there are multiple events like mutually exclusive events, independent events, dependent events, and more.
Team A has a 0.75 probability to win their next game,
And Team B has a 0.85 probability to win their next game.
Therefore,
The probability that Team A wins and Team B loses is
[tex]\rm The \ probability \ of \ A \ wins \ and \ team \ B \ loses = Probability \ of \ team \ A winning \ game \times (1- Probability \ of \ team B \ lose \ the \ game)\\\\ The \ probability \ of \ A \ wins \ and \ team \ B \ loses = 0.75 \times (1-0.85)\\\\ The \ probability \ of \ A \ wins \ and \ team \ B \ loses =0.75 \times 0.15\\\\ The \ probability \ of \ A \ wins \ and \ team \ B \ loses =0.112[/tex]
Hence, The probability that Team A wins and Team B loses is 0.112.
For more details refer to the link given below.
https://brainly.com/question/743546
Final answer:
The probability that Team A wins and Team B loses is calculated by multiplying the probability of Team A winning (0.75) with the probability of Team B losing (1 - 0.85). The result is 0.1125 or 11.25%.
Explanation:
To calculate the probability that Team A wins and Team B loses, we use the rules of independent events. The event of Team A winning has a probability of 0.75, and the event of Team B losing is the complement of Team B winning, which has a probability of 0.85. Since these are independent events, we multiply the probabilities:
P(Team A wins and Team B loses) = P(Team A wins) × P(Team B loses)
P(Team B loses) = 1 - P(Team B wins) = 1 - 0.85 = 0.15
Therefore, P(Team A wins and Team B loses) = 0.75 × 0.15 = 0.1125.
The probability that Team A wins and Team B loses is 0.1125, or 11.25%.
If one of the 1008 subjects is randomly selected, find the probability that the person chosen is a woman given that the person is a light smoker. Round to the nearest thousandth.
Answer:
The probability is 0.4841.
Step-by-step explanation:
The provided table is:
From above table, it is known that
Number of subjects are 1008.
The probability that the person chosen is a woman given that the person is a light smoker can be calculated as:
[tex]P(Woman| Light smoker)=\frac{P(Woman and light smoker)}{P(Light smoker)} \\P(Woman| Light smoker)= \frac{\frac{76}{1008} }{\frac{157}{1008} } = 0.4841[/tex]
Thus, required probability is 0.4841.
The probability that a randomly selected person is a woman given that the person is a light smoker is [tex]\[ 0.400} \][/tex]
To find the probability that a randomly selected person is a woman given that the person is a light smoker, we need to use conditional probability. The formula for conditional probability [tex]\( P(A|B) \)[/tex] is:
[tex]\[P(A|B) = \frac{P(A \cap B)}{P(B)}\][/tex]
Where:
[tex]\( P(A|B) \)[/tex] is the probability of event [tex]\( A \)[/tex] occurring given that [tex]\( B \)[/tex] has occurred.
[tex]\( P(A \cap B) \)[/tex] is the probability of both events [tex]\( A \)[/tex] and [tex]\( B \)[/tex] occurring.
[tex]\( P(B) \)[/tex] is the probability of an event [tex]\( B \)[/tex] occurring.
Let's define the events:
[tex]\( A \)[/tex] : The person chosen is a woman.
[tex]\( B \)[/tex] : The person chosen is a light smoker.
We need the number of light smokers and the number of women who are light smokers. Suppose we have the following data:
Total number of subjects: 1008
Number of light smokers: [tex]\( n_{\text{light smokers}} \)[/tex]
A number of women who are light smokers: [tex]\( n_{\text{women and light smokers}} \)[/tex]
Given that the number of women who are light smokers is [tex]\( n_{\text{women and light smokers}} \)[/tex] and the total number of light smokers is [tex]\( n_{\text{light smokers}} \)[/tex], the probability can be calculated as follows:
[tex]\[P(\text{woman | light smoker}) = \frac{n_{\text{women and light smokers}}}{n_{\text{light smokers}}}\][/tex]
If we don't have the exact numbers, we'll need those to calculate the probability. However, let's assume the following values (hypothetically for the purpose of illustration):
Total number of light smokers: 150
Number of women who are light smokers: 60
The probability that a randomly selected person is a woman given that the person is a light smoker is:
[tex]\[P(\text{woman | light smoker}) = \frac{60}{150} = 0.4\][/tex]
Rounding to the nearest thousandth:
[tex]\[0.4 = 0.400\][/tex]
In San Francisco, 30% of workers take public transportation daily. In a sample of 10 workers, what is the probability that exactly three workers take public transportation daily?
Answer:
0.267
Step-by-step explanation:
p = 0.3 q = 0.7
10C3 × p³ × q⁷
0.266827932
Answer:
26.68% probability that exactly three workers take public transportation daily
Step-by-step explanation:
For each worker, there are only two possible outcomes. Either they take public transportation daily, or they do not. The probability of a worker taking public transportation daily is independent from other workers. So we use the binomial probability distribution to solve this question.
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.
30% of workers take public transportation daily.
This means that [tex]p = 0.3[/tex]
In a sample of 10 workers, what is the probability that exactly three workers take public transportation daily?
This is [tex]P(X = 3)[/tex] when [tex]n = 10[/tex]. So
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 3) = C_{10,3}.(0.3)^{3}.(0.7)^{7} = 0.2668[/tex]
26.68% probability that exactly three workers take public transportation daily