Answer:
Step-by-step explanation:
Hello!
X: Cholesterol level of a woman aged 30-39. (mg/dl)
This variable has an approximately normal distribution with mean μ= 190.14 mg/dl
1. You need to find the corresponding Z-value that corresponds to the top 9.3% of the distribution, i.e. is the value of the standard normal distribution that has above it 0.093 of the distribution and below it is 0.907, symbolically:
P(Z≥z₀)= 0.093
-*or*-
P(Z≤z₀)= 0.907
Since the Z-table shows accumulative probabilities P(Z<Z₁₋α) I'll work with the second expression:
P(Z≤z₀)= 0.907
Now all you have to do is look for the given probability in the body of the table and reach the margins to obtain the corresponding Z value. The first column gives you the integer and first decimal value and the first row gives you the second decimal value:
z₀= 1.323
2.
Using the Z value from 1., the mean Cholesterol level (μ= 190.14 mg/dl) and the Medical guideline that indicates that 9.3% of the women have levels above 240 mg/dl you can clear the standard deviation of the distribution from the Z-formula:
Z= (X- μ)/δ ~N(0;1)
Z= (X- μ)/δ
Z*δ= X- μ
δ=(X- μ)/Z
δ=(240-190.14)/1.323
δ= 37.687 ≅ 37.7 mg/dl
I hope it helps!
Apply the properties of angles to solve for the missing angles. Angle y is what degrees. Angle x is what degrees.
Answer:
y= 65 degrees. and x=30 degrees
Step-by-step explanation:
i haven't done this for a while so i may or may not be correct i tried to help though :)
Answer: y = 52°. x = 64°
Step-by-step explanation:
y = 26 + 26 = 52 degrees ( exterior angle of a triangle is the sum of the two opposite interior angle of the triangle)
x = 180-(52+64)(sum of angles in a triangle is 180°)
x = 180-116
x = 64degrees ( base angles of an isosceles triangle are equal)
Identify the correct statements for the given functions of set {a, b, c, d} to itself.
1.
f(a) = b, f(b) = a, f(c) = c, f(d) = d is a one-to-one function as each element is an image of exactly one element.
2.
f(a) = b, f(b) = a, f(c) = c, f(d) = d is not a one-to-one function as a is mapped to b and b is mapped to a.
3.
f(a) = b, f(b) = b, f(c) = d, f(d) = c is a one-to-one function as each element is mapped to only one element.
4.
f(a) = b, f(b) = b, f(c) = d, f(d) = c is not a one-to-one function as b is an image of more than one elements.
5.
f(a) = d, f(b) = b, f(c) = c, f(d) = d is a one-to-one function as each element is mapped to only one element.
6.
f(a) = d, f(b) = b, f(c) = c, f(d) = d is not a one-to-one function as d is an image of more than one element.
Answer:
The correct answers are
option(1), option (4),option (5)
Step-by-step explanation:
One to one: Every image has exactly one unique pre-image in domain.
(1)
f(a) = b, f(b)=a, f(c)=c, f(d)=d
b has a pre-image in domain i.e a
a has a pre-image in domain i.e b
c has a pre-image in domain i.e c
d has a pre-image in domain i.e d
Here all elements have a unique pre- image in domain.
Therefore it is one to one.
(2)
f(a) = b, f(b)=a, f(c)=c, f(d)=d
b has a pre-image in domain i.e a
a has a pre-image in domain i.e b
c has a pre-image in domain i.e c
d has a pre-image in domain i.e d
Here all elements have a unique pre- image in domain.
Therefore it is one to one.
(3)
f(a) = b, f(b)=b, f(c)=d, f(d)=c
b has a pre-image in domain i.e a
b has a pre-image in domain i.e b
d has a pre-image in domain i.e c
c has a pre-image in domain i.e d
b has two pre image.
Here all elements have not a unique pre- image in domain.
Therefore it is not a one to one mapping.
(4)
f(a) = b, f(b)=b, f(c)=d, f(d)=c
b has a pre-image in domain i.e a
b has a pre-image in domain i.e b
d has a pre-image in domain i.e c
c has a pre-image in domain i.e d
b has two pre image.
Here all elements have not a unique pre- image in domain.
Therefore it is not a one to one mapping.
(5)
f(a) = d, f(b)=b, f(c)=d, f(d)=c
d has a pre-image in domain i.e a
b has a pre-image in domain i.e b
d has a pre-image in domain i.e c
c has a pre-image in domain i.e d
Here all elements have a unique pre- image in domain.
Therefore it is a one to one mapping.
(6)
f(a)=d, f(b)=b,f(c)=c,f(d)=d
d has a pre-image in domain i.e a
b has a pre-image in domain i.e b
c has a pre-image in domain i.e c
d has a pre-image in domain i.e d
Here all elements have a unique pre- image in domain.
Therefore it is a one to one mapping.
f(a) = b, f(b) = a, f(c) = c, f(d) = d and f(a) = d, f(b) = b, f(c) = c, f(d) = d are one-to-one function and f(a) = b, f(b) = b, f(c) = d, f(d) = c is not one-to-one function. Then the correct statements are 1, 4, and 5.
What is a function?The function is an expression, rule, or law that defines the relationship between one variable to another variable. Functions are ubiquitous in mathematics and are essential for formulating physical relationships.
One-to-one - every image has exactly one unique pre-image in the domain.
1. f(a) = b, f(b) = a, f(c) = c, f(d) = d
b has a pre image in a domian that is a
a has a pre image in a domian that is b
c has a pre image in a domian that is c
d has a pre image in a domian that is d
Here all elements have a unique pre-image in a domain.
Therefore it is one-to-one.
2. f(a) = b, f(b) = a, f(c) = c, f(d) = d
b has a pre image in a domian that is a
a has a pre image in a domian that is b
c has a pre image in a domian that is c
d has a pre image in a domian that is d
Here all elements have a unique pre-image in a domain.
Therefore it is one-to-one.
3. f(a) = b, f(b) = b, f(c) = d, f(d) = c
b has a pre image in a domian that is a
b has a pre image in a domian that is b
d has a pre image in a domian that is c
c has a pre image in a domian that is d
Here all elements do have not a unique pre-image in a domain.
Therefore it is not one-to-one.
4. f(a) = b, f(b) = b, f(c) = d, f(d) = c
b has a pre image in a domian that is a
b has a pre image in a domian that is b
d has a pre image in a domian that is c
c has a pre image in a domian that is d
Here all elements do have not a unique pre-image in a domain.
Therefore it is not one-to-one.
5. f(a) = d, f(b) = b, f(c) = c, f(d) = d
d has a pre image in a domian that is a
b has a pre image in a domian that is b
c has a pre image in a domian that is c
d has a pre image in a domian that is d
Here all elements have a unique pre-image in a domain.
Therefore it is one-to-one.
6. f(a) = d, f(b) = b, f(c) = c, f(d) = d
d has a pre image in a domian that is a
b has a pre image in a domian that is b
c has a pre image in a domian that is c
d has a pre image in a domian that is d
Here all elements have a unique pre-image in a domain.
Therefore it is one-to-one.
More about the function link is given below.
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Children inherit many traits from their parents. Some inherited traits are not desirable. For example, if both parents are carriers, there is a 25% chance any child will have sickle-cell disease. If we assume no identical twins in a family, and both individuals in a couple are carriers, what is the probability none of their four children has the disease
The probability that none of their four children has the disease = 0.316
Step-by-step explanation:
Step 1 :
The probability that the child will have the specified disease if both the parents are carriers = 25% = 0.25
So the probability that the child does not have the specified disease = 1 - 0.25 = 0.75
Step 2 :
There are four children. The probability that each child has the disease is independent of the condition of the other children
The probability of more than one independent events happening together can be obtained by multiplying probability of individual events
So, the probability that none of their 4 children has the sickle cell disease = [tex](0.75)^{4}[/tex] = 0.316
Step 3 :
Answer :
The probability no children has the specified cell disease = 0.316
Which of the following is equivalent to 60 Superscript one-half? StartFraction 60 Over 2 EndFraction StartRoot 60 EndRoot StartFraction 1 Over 60 squared EndFraction StartFraction 1 Over StartRoot 60 EndRoot EndFraction
Answer:
[tex](60)^{\frac{1}{2}} = \sqrt{60}[/tex]
Step-by-step explanation:
We are given the following expression:
[tex](60)^{\frac{1}{2}}[/tex]
We are given the following options:
[tex]A.~\dfrac{60}{2}\\\\B.~\sqrt{60}\\\\C.~(\dfrac{1}{60})^{2}\\\\D.~\dfrac{1}{\sqrt{60}}[/tex]
Exponent Properties:
[tex]x^{-a} = (\dfrac{1}{x})^{a}\\\\(x^m)^n = x^{mn}\\\\\dfrac{x^m}{x^n} = x^{m-n}[/tex]
Thus, the correct answer is
[tex](60)^{\frac{1}{2}} = \sqrt{60}[/tex]
Answer:
B or [tex]\sqrt{60}[/tex]
Step-by-step explanation:
A box contains five slips of paper, marked $1, $1, $10, $25, and $25. The winner of a contest selects two slips of paper at random and then gets the larger of the dollar amounts on the two slips. Define a random variable w by w = amount awarded. Determine the probability distribution of w. (Hint: Think of the slips as numbered 1, 2, 3, 4, and 5, so that an outcome of the experiment consists of two of these numbers.)
Answer:
w = $1 or $10 or $25
Probability distribution of w = ($1,$1 $1,$1 $1,$10 $1,$25 $1,$25 $1,$1 $1,$1 $1,$10 $1,$25 $1,$25 $10,$1 $10,$1 $10,$10 $10,$25 $10,$25 $25,$1 $25,$1 $25,$10 $25,$25 $25,$25 $25,$1 $25,$1 $25,$10 $25,$25 $25,$25)
Step-by-step explanation:
Since the winner gets the larger amount of the two slips picked, random variable 'w' which is the amount awarded could be $1 or $10 or $25.
The probability distribution of 'w' is the values that the statistic takes on. Which could be: $1,$1 $1,$1 $1,$10 $1,$25 $1,$25 $1,$1 $1,$1 $1,$10 $1,$25 $1,$25 $10,$1 $10,$1 $10,$10 $10,$25 $10,$25 $25,$1 $25,$1 $25,$10 $25,$25 $25,$25 $25,$1 $25,$1 $25,$10 $25,$25 $25,$25
A college senior who took the Graduate Record Examination exam scored 530 on the Verbal Reasoning section and 600 on the Quantitative Reasoning section. The mean score for Verbal Reasoning section was 460 with a standard deviation of 105, and the mean score for the Quantitative Reasoning was 429 with a standard deviation of 148. Suppose that both distributions are nearly normal. Round calculated answers to 4 decimal places unless directed otherwise.
Answer:
Step-by-step explanation:
Given that a college senior who took the Graduate Record Examination exam scored 530 on the Verbal Reasoning section and 600 on the Quantitative Reasoning section.
Quantitative reasons scores are N (429, 148)
Hence Z score for the college senior = [tex]\frac{530-429}{148} \\=0.6824[/tex]
The mean score for Verbal Reasoning section was 460 with a standard deviation of 105
Verbal reasoning score was N(460, 105)
The score of college senior = 530
Z score for verbal reasoning =[tex]\frac{530-460}{105} \\=0.6667[/tex]
comparing we can say he scored more on quantiative reasoning.
Thus comparison was possible only by converting to Z scores.
Suppose the null hypothesis, H0, is: a sporting goods store claims that at least 70% of its customers do not shop at any other sporting goods stores. What is the Type I error in this scenario? a. The sporting goods store thinks that less than 70% of its customers do not shop at any other sporting goods stores when, in fact, less than 70% of its customers do not shop at any other sporting goods stores. b. The sporting goods store thinks that at least 70% of its customers do not shop at any other sporting goods stores when, in fact, at least 70% of its customers do not shop at any other sporting goods stores. c. The sporting goods store thinks that less than 70% of its customers do not shop at any other sporting goods stores when, in fact, at least 70% of its customers do not shop at any other sporting goods stores. d. The sporting goods store thinks that at least 70% of its customers do not shop at any other sporting goods stores when, in fact, less than 70% of its customers do not shop at any other sporting goods stores.
Answer:
Null hypothesis: [tex]p \geq 0.7[/tex]
Alternative hypothesis: [tex]p<0.7[/tex]
A type of error I for this case would be reject the null hypothesis that the population proportion is greater or equal than 0.7 when actually is not true.
So the correct option for this case would be:
c. The sporting goods store thinks that less than 70% of its customers do not shop at any other sporting goods stores when, in fact, at least 70% of its customers do not shop at any other sporting goods stores.
Step-by-step explanation:
Previous concepts
A hypothesis is defined as "a speculation or theory based on insufficient evidence that lends itself to further testing and experimentation. With further testing, a hypothesis can usually be proven true or false".
The null hypothesis is defined as "a hypothesis that says there is no statistical significance between the two variables in the hypothesis. It is the hypothesis that the researcher is trying to disprove".
The alternative hypothesis is "just the inverse, or opposite, of the null hypothesis. It is the hypothesis that researcher is trying to prove".
Type I error, also known as a “false positive” is the error of rejecting a null hypothesis when it is actually true. Can be interpreted as the error of no reject an alternative hypothesis when the results can be attributed not to the reality.
Type II error, also known as a "false negative" is the error of not rejecting a null hypothesis when the alternative hypothesis is the true. Can be interpreted as the error of failing to accept an alternative hypothesis when we don't have enough statistical power.
Solution to the problem
On this case we want to test if the sporting goods store claims that at least 70^ of its customers, so the system of hypothesis would be:
Null hypothesis: [tex]p \geq 0.7[/tex]
Alternative hypothesis: [tex]p<0.7[/tex]
A type of error I for this case would be reject the null hypothesis that the population proportion is greater or equal than 0.7 when actually is not true.
So the correct option for this case would be:
c. The sporting goods store thinks that less than 70% of its customers do not shop at any other sporting goods stores when, in fact, at least 70% of its customers do not shop at any other sporting goods stores.
Seventy independent messages are sent from an electronic transmission center.Messages are processed sequentially, one after another. Transmission time of each message is Exponential with parameter λ = 5 min−1. Find the probability that all 70 messages are transmitted in less than 12 minutes. Use the Central Limit Theorem.
Answer:
The probability that all 70 messages are transmitted in less than 12 minutes is 0.117.
Step-by-step explanation:
Let X = the transmission time of each message.
The random variable X follows an Exponential distribution with parameter λ = 5 minutes.
The expected value of X is:
[tex]E(X)=\frac{1}{\lambda}=\frac{1}{5}=0.20[/tex]
The variance of X is:
[tex]V(X)=\frac{1}{\lambda^{2}}=\frac{1}{5^{2}}=0.04[/tex]
Now define a random variable T as:
T = X₁ + X₂ + ... + X₇₀
According to a Central limit theorem if a large sample (n > 30) is selected from a population with mean μ and variance σ² then the sum of random variables X follows a Normal distribution with mean, [tex]\mu_{s} = n\mu[/tex] and variance, [tex]\sigma^{2}_{s}=n\sigma^{2}[/tex].
Compute the probability of T < 12 as follows:
[tex]P(T<12)=P(\frac{T-\mu_{T}}{\sqrt{\sigma_{T}^{2}}}<\frac{12-(70\times0.20)}{\sqrt{70\times 0.04}})\\=P(Z<-1.19)\\\=1-P(Z<1.19)\\=1-0.883\\=0.117[/tex]
*Use a z-table for the probability.
Thus, the probability that all 70 messages are transmitted in less than 12 minutes is 0.117.
Following are the solution to the given question:
Let X signify the letter's transmission time [tex]i^{th}[/tex] , and [tex]i =1,2,3,...,70.[/tex] this is assumed that the [tex]X_i \sim Exp (\lambda =5)[/tex].
[tex]\to Mean =\frac{1}{\lambda} =\frac{1}{5}=0.2\\\\ \to Variance =\frac{1}{\lambda^2} =\frac{1}{5^2} =\frac{1}{25}=0.04\\\\[/tex]
Let [tex]T = X_1 + X_2 +...+ X_{70}[/tex], be the total transmission time. According to the Central Limit Theorem, T has a Normal distribution of mean, Variance.mean [tex]= E(T) = 70 \times 0.2=14[/tex]
Variance [tex]=V(T) =70 \times 0.04 = 2.8[/tex]
Therefore
[tex]\to P(T<12) = P (Z < \frac{12-14}{\sqrt{2.8}})[/tex]
[tex]= P(Z<-1.195) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (by sy memetry) \\\\=0.5-P(0<Z<1.19)\\\\=0.5-0.3830 \\\\=0.1170\\\\[/tex]
Therefore, the final answer is "0.1170".
Learn more about the Central Limit Theorem:
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(1 point) Find the general solution to the homogeneous differential equation. ????2y????????2−20????y????????+136y=0 Use c1 and c2 in your answer to denote arbitrary constants, and enter them as c1 and c2. y(????)= ?
Answer:
Question is not clear please post question clearly lots of question marks.
Your differential equation is not displayed well. It though looks like this:
2d²y/dx² - 20dy/dx + 136y = 0
If this is not the differential equation, the method of solving this would still be used in solving the correct one.
We first write an auxiliary equation to the differential equation.
The auxiliary equation is:
2m² - 20m + 136 = 0
Dividing by 2, we have
m² - 10m + 68 = 0
Next, we solve the auxiliary equation to obtain the values of m.
Solving using the quadratic formula
m = [-b ± √(b² - 4ac)]/2a
Where a = 1, b = -10, and c = 68
m = [10 ± √(100 - 272)]/2
= 5 ± (1/2)√(-172)
= 5 ± (1/2)i√172
= 5 ± 6.6i
For solutions of the form a ± ib, the complimentary solution is
y = e^(ax)[C1cosbx + C2sinbx]
Therefore, the complimentary solution is
y = e^(5x)[C1cos(6.6x) + C2sin(6.6x)]
In a new card game, you start with a well-shuffled full deck and draw 3 cards without replacement. If you draw 3 hearts, you win $50. If you draw 3 black cards, you win $25. For any other draws, you win nothing.
(a) Find the expected winnings for a single game.
(b) Find the standard deviation of the winnings.
(c) If the game costs $5 to play, what would be the expected value of the net profit (or loss)?
(Hint: profit = winnings - cost; X - 5)
(d) If the game costs $5 to play, what would be the standard deviation of the net profit (or loss)?
(e) If the game costs $5 to play, should you play this game?
There are [tex]\binom{52}3=\frac{52!}{3!(52-3)!}[/tex] (or "52 choose 3") ways of drawing any 3 cards from the deck.
There are 13 hearts in the deck, and 26 cards with a black suit. So there are [tex]\binom{13}3[/tex] and [tex]\binom{26}3[/tex] ways of drawing 3 hearts or 3 black cards, respectively. Then the probability of drawing 3 hearts is
[tex]P(\text{3 hearts})=\dfrac{\binom{13}3}{\binom{52}3}=\dfrac{11}{850}[/tex]
and the probability of drawing 3 black cards is
[tex]P(\text{3 black})=\dfrac{\binom{26}3}{\binom{52}3}=\dfrac2{17}[/tex]
All other combinations can be drawn with probability [tex]1-\frac{11}{850}-\frac2{17}=\frac{739}{850}[/tex].
Let [tex]W[/tex] be the random variable for one's potential winnings from playing the game. Then
[tex]P(W=w)=\begin{cases}\frac{11}{850}&\text{for }w=\$50\\\frac2{17}&\text{for }w=\$25\\\frac{739}{850}&\text{otherwise}\end{cases}[/tex]
a. For a single game, one can expect to win
[tex]E[W]=\displaystyle\sum_ww\,P(W=w)=\frac{\$50\cdot11}{850}+\frac{\$20\cdot2}{17}+\frac{\$0\cdot739}{850}=\$3[/tex]
b. For a single game, one's winnings have a variance of
[tex]V[W]=E[(W-E[W])^2]=E[W^2]-E[W]^2[/tex]
where
[tex]E[W^2]=\displaystyle\sum_ww^2\,P(W=w)=\frac{\$50^2\cdot11}{850}+\frac{\$20^2\cdot2}{17}+\frac{\$0^2\cdot739}{850}=\$^2\frac{1350}{17}\approx\$^279.41[/tex]
so that [tex]V[W]=\$^2\frac{1197}{17}\approx\$^270.41[/tex]. (No, that's not a typo, variance is measured in squared units.) Standard deviation is equal to the square root of the variance, so it is approximately $8.39.
c. With a $5 buy-in, the expected value of the game would be
[tex]E[W-\$5]=E[W]-\$5=-\$2[/tex]
i.e. a player can expect to lose $2 by playing the game (on average).
d. With the $5 cost, the variance of the winnings is the same, since the variance of a constant is 0:
[tex]V[W-\$5]=V[W][/tex]
so the standard deviation is the same, roughly $8.39.
e. You shouldn't play this game because of the negative expected winnings. The odds are not in your favor.
The expected winning for a single game defined is : $3.59
The standard deviation of winning is : $10.11
Expected winning if game costs $5 to play is :
- $0.79
The standard deviation of winning if game costs $5 to play is : $11.33
The game should not be played with a game play fee of -$5 as the expected winning value is negative.
Recall : selection is done without replacement :
Number of hearts in a deck = 13
Probability of drawing 3 hearts :
P(drawing 3 Hearts)
First draw × second draw × third draw
13/52 × 12/51 × 11/50 = 1716/132600 = 858/66300
Probability of selecting 3 black cards :
Number of black cards in a deck = 26
P(drawing 3 black cards) :
First draw × second draw × third draw
26/52 × 25/51 × 24/50 = 15600 / 132600 = 7800/66300
Probability of making any other draw :
P(3 hearts) + P(3 blacks) + P(any other draw) = 1
858/66300 + 7800/66300 + P(any other draw) = 1
P(any other draw) = 57642/66300
For a single game :
X _______ $50 ________ $25 _______ $0
P(X)_ 858/66300__ 7800/66300_57642/66300
E(X) = Σ[ X × p(X)]
E(X) =Σ[(50 × 858/66300)+(25 × 7800/66300)+0]
E(X) = $3.588
The standard deviation = √Var(X)
Var(X) = Σ[ X² × p(X)] - E(X)
Σ[ X²×p(X)] = Σ[(50² × 858/66300)+(25² × 7800/66300)+0] = 105.88235
Var(X) = 105.88235 - 3.588 = 102.29435
Standard deviation of winning = √102.29435 = $10.114
If the game cost $5 to play :
Net amount won if :
3 hearts are drawn = $50 - $5 = $45
3 blacks are drawn = $25 - $5 = $20
Any other combination are drawn= $0 - $5 = -$5
The distribution becomes :
X _______ $45 ________ $20 _______ -$5
P(X)_ 858/66300__ 7800/66300_57642/66300
E(X) = Σ[ X × p(X)]
E(X) =Σ[(50 × 858/66300)+(25 × 7800/66300)+ (-5 × 57642/66300)]
E(X) = - $0.7588
Standard deviation of winning :
The standard deviation = √Var(X)
Var(X) = Σ[ X² × p(X)] - E(X)
Σ[ X²×p(X)] = Σ[(50² × 858/66300)+(25² × 7800/66300)+ (-5² × 57642/66300)] = 127.61764
Var(X) = 127.61764 - (-0.7588) = 128.37644
Standard deviation of winning :
Std(X) = √Var(X) = √128.37644 = $11.330
With a game cost of - $5 ; the expected winning for a single game gives a negative value, therefore you should not play the game.
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The daily sales S (in thousands of dollars) that are attributed to an advertising campaign are given by S = 11 + 3 t + 3 − 18 (t + 3)2 where t is the number of weeks the campaign runs. (a) Find the rate of change of sales at any time t.
Answer:
The parking rate at the hospital a parking garage are 50 cents for the 1st hour and 25 cents for each additional hour if guan part in the hospital parking garage for 8 hours how much will the total of charge for parking be
Assume lim f(x)-6 and lim g(x)-9. Compute the following limit and state the limit laws used to justify the computation.
x→3 x →3
Lim 3√f(x).g(x)+10
x →3
Using the limit laws of addition and multiplication, the limit of the given equation is solved by combining the calculated limits of the function and the constant, giving an answer of approximately 13.78.
Explanation:To solve this problem, we can use the limit laws of addition and multiplication. The limit laws state that the limit of the sum of two functions is equal to the sum of their individual limits, and the limit of the product of two functions is equal to the product of their individual limits.
So, based on these laws, we first separate the function into two parts: the limit of 3√(f(x).g(x)) as x approaches 3 and the limit of 10 as x approaches 3.
Given lim f(x)=>6 and lim g(x)=>9 when x approaches 3, we multiply these values together to get a product of 54. The cubed root of 54 is approximately 3.78.
The constant 10 has a limit of 10, as constants maintain their value irrespective of the limit.
Adding these values together, 3.78 + 10, gives us a limit of approximately 13.78.
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Montoya (2007) asked 56 men and 82 women to rate 21 different body parts on a scale of 1 (no opinion) to 5 (very desirable). They found that men and women rated the eyes similarly, with an average rating of about 3.77 ± 1.23 (M ± SD). Assuming these data are normally distributed, answer the following questions. (Round your answers to two decimal places.) (a) What percentage of participants rated the eyes at least a 5 (very desirable)? % (b) What percentage rated the eyes at most a 1 (no opinion)? %
Answer:
a. 15.9%
b. 1.2%
Step-by-step explanation:
using the normal distribution we have the following expression
[tex]P(X\geq a)=P(\frac{X-u}{\alpha } \geq \frac{5-M}{SD}) \\[/tex]
Where the first expression in the right hand side is the z-scores, M is the mean of value 3.77 and SD is the standard deviation of value 1.23.
if we simplify and substitute values, we arrive at
[tex]P(X\geq a)=P(\frac{X-u}{\alpha } \geq \frac{a-M}{SD}) \\P(X\geq 5)=P(Z \geq \frac{5-3.77}{1.23})\\P(X\geq 5)=P(Z \geq 1)\\P(X\geq 5)=1- P(Z < 5)\\P(X\geq 5)=1-0.841\\P(X\geq 5)=0.159[/tex]
in percentage, we arrive at 15.9%
b. for the percentage rated the eyes most a 1
[tex]P(X\leq a)=P(\frac{X-u}{\alpha } \leq \frac{a-M}{SD})\\for a=1\\P(X\leq 1)=P(Z \leq \frac{1-3.77}{1.23})\\P(X\leq 1)=P(Z \leq -2.25})\\P(X\leq 1)=0.012\\[/tex]
In percentage we have 1.2%
he purpose of an x-bar chart is to determine whether there has been a: change in the AOQ. change in the number of defects in a sample. change in the percent defective in a sample. change in the dispersion of the process output. change in the central tendency of the process output.
Answer:
E. Change in the central tendency of the process output.
Step-by-step explanation:
The x-bar chart is a control chart for the central tendency of the process output. Therefore it's purpose is to determine whether there has been a change in the central tendency of the process output.
Answer:
The change in the central tendency of the process output.
Step-by-step explanation:
Lets examine each of the listed options
The change in the percent defective in a sample is associated with the change in average outgoing quality (AOQ).
The change in the dispersion of the process output is associated with R-chart since the function of R-chart is to detect changes in the dispersion.
The change in the number of defects in a sample is associated with C-chart since the function of a C-chart is to show the number of flaws per unit in a sample.
The change in the central tendency of the process output is associated with X-bar chart since the function of X-bar chart is to check whether the values are within the appropriate limits (central tendency) of the process output or not.
Therefore, the purpose of an X-bar chart is to determine whether there has been a change in the central tendency of the process output.
5. (20 pts) Consider the array 25, 14, 63, 29, 63, 47, 12, 21. Apply the Split procedure in Quicksort, as described in class, to this array using the first element as the pivot.
Answer:
Step-by-step explanation: see attachment
The accounting records of Blossom Company show the following data. Beginning inventory 3,090 units at $5 Purchases 7,930 units at $8 Sales 9,390 units at $10 Determine cost of goods sold during the period under a periodic inventory system using the FIFO method. (Round answer to 0 decimal places, e.g. 1,250.)
Answer:
The cost of goods sold is 65,850
Step-by-step explanation:
FIFO Perpetual chart is attached.
FIFO Perpetual chart shows purchases, sales and balance of the period.
The cost of goods sold is:
3,090 units x $5=$15,450
6,300 units x $8=$50,400
Total=65,850
1. In order to get more female customers, a new clothing store offers free gourmet coffee and pastry to its customers. The average daily revenue over the past five-week period has been $1,080 with a standard deviation of $260. Use this sample information to construct a 95% confidence interval for the average daily revenue. The store manager believes that the coffee and pastry strategy would lead to an average daily revenue of $1,200. Is the manager correct based on the 95% confidence interval?
Answer:
No, the manager is not correct based on the 95% confidence interval.
Step-by-step explanation:
We are given that the average daily revenue over the past five-week period has been $1,080 with a standard deviation of $260, i.e.; X bar = $1080 and s = $260 and sample size, n = 35 .
The Pivotal quantity for 95% confidence interval is given by;
[tex]\frac{Xbar - \mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]
where, X bar = sample mean = $1080
s = sample standard deviation = $260
n = sample size = 35 {five-week}
So, 95% confidence interval for average daily revenue, [tex]\mu[/tex] is given by;
P(-2.032 < [tex]t_3_4[/tex] < 2.032) = 0.95
P(-2.032 < [tex]\frac{Xbar - \mu}{\frac{s}{\sqrt{n} } }[/tex] < 2.032) = 0.95
P(-2.032 * [tex]{\frac{s}{\sqrt{n} }[/tex] < [tex]{Xbar - \mu}[/tex] < 2.032 * [tex]{\frac{s}{\sqrt{n} }[/tex] ) = 0.95
P(X bar - 2.032 * [tex]{\frac{s}{\sqrt{n} }[/tex] < [tex]\mu[/tex] < X bar + 2.032 * [tex]{\frac{s}{\sqrt{n} }[/tex] ) = 0.95
95% confidence interval for [tex]\mu[/tex] = [ X bar - 2.032 * [tex]{\frac{s}{\sqrt{n} }[/tex] , X bar + 2.032 * [tex]{\frac{s}{\sqrt{n} }[/tex] ]
= [ 1080 - 2.032 * [tex]{\frac{260}{\sqrt{35} }[/tex] , 1080 + 2.032 * [tex]{\frac{260}{\sqrt{35} }[/tex] ]
= [ 990.70 , 1169.30 ]
No, the manager is not correct based on the fact that the coffee and pastry strategy would lead to an average daily revenue of $1,200 because the calculate 95% confidence interval does not include value of $1200.
Therefore, the store manager believe is not correct.
The 95% confidence interval for the store's average daily revenue is calculated to be approximately ($993.97, $1166.03). Since $1200 is outside this interval, the manager's belief that the coffee and pastry strategy will lead to an average daily revenue of $1200 is not backed by this confidence level.
Explanation:In the field of statistics, a confidence interval (CI) is a type of interval estimate that is used to indicate the reliability of an estimate. The method for calculating a 95% confidence interval for the average daily revenue involves the sample mean, the standard deviation, and the z-score associated with a 95% confidence level, which is approximately 1.96. Let's use the provided data to calculate:
Calculate the standard error by dividing the standard deviation by the square root of the sample size. Here, the standard deviation is $260, and the sample size is 5 weeks * 7 days/week = 35 days. So, the standard error is $260 / sqrt(35) = $43.89.Multiply the standard error by the z-score to get the margin of error. So, $43.89 * 1.96 = $86.03.Calculate the lower and upper bounds of the 95% confidence interval by subtracting and adding the margin of error from/to the sample mean. So, ($1080 - $86.03, $1080 + $86.03) = ($993.97, $1166.03).The range of this 95% confidence interval is from $993.97 to $1166.03. This means we are 95% confident that the true average daily revenue lies within this interval. Since $1200 lies outside this interval, the manager's belief is not supported by this confidence interval.
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A builder of houses needs to order some supplies that have a waiting time Y for delivery, with a continuous uniform distribution over the interval from 1 to 4 days. Because she can get by without them for 2 days, the cost of the delay is fixed at $400 for any waiting time up to 2 days. After 2 days, however, the cost of the delay is $400 plus $50 per day (prorated) for each additional day. That is, if the waiting time is 3.5 days, the cost of the delay is $400 $50(1.5)
Find the expected value of the builder’s cost due to waiting for supplies.
Answer:
The Expected cosy of the builder is $433.3
Step-by-step explanation:
$400 is the fixed cost due to delay.
Given Y ~ U(1,4).
Calculating the Variable Cost, V
V = $0 if Y≤ 2
V = 50(Y-2) if Y > 2
This can be summarised to
V = 50 max(0,Y)
Cost = 400 + 50 max(0, Y-2)
Expected Value is then calculated by;
Waiting day =2
Additional day = at least 1
Total = 3
E(max,{0, Y - 2}) = integral of Max {0, y - 2} * ⅓ Lower bound = 1; Upper bound = 4, (4,1)
Reducing the integration to lowest term
E(max,{0, Y - 2}) = integral of (y - 2) * ⅓ dy Lower bound = 2; Upper bound = 4 (4,2)
E(max,{0, Y - 2}) = integral of (y) * ⅓ dy Lower bound = 0; Upper bound = 2 (2,0)
Integrating, we have
y²/6 (2,0)
= (2²-0²)/6
= 4/6 = ⅔
Cost = 400 + 50 max(0, Y-2)
Cost = 400 + 50 * ⅔
Cost = 400 + 33.3
Cost = 433.3
Answer:
the expected value of the builder’s cost due to waiting for supplies is $433.3
Step-by-step explanation:
Due to the integration symbol and also for ease of understanding, i have attached the explanation as an attachment.
In an article regarding interracial dating and marriage recently appeared in a newspaper. Of 1719 randomly selected adults, 311 identified themselves as Latinos, 322 identified themselves as blacks, 251 identified themselves as Asians, and 775 identified themselves as whites. Among Asians, 79% would welcome a white person into their families, 71% would welcome a Latino, and 66% would welcome a black person.
NOTE: If you are using a Student's t-distribution, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, though.)
Construct the 95% confidence intervals for the three Asian responses.
1. Welcome a white person ( , )
2. Welcome a Latino ( , )
3. Welcome a Black person ( , )
Answer:
Step-by-step explanation:
Hello!
The parameter of interest in this exercise is the population proportion of Asians that would welcome a person of other races in their family. Using the race of the welcomed one as categorizer we can define 3 variables:
X₁: Number of Asians that would welcome a white person into their families.
X₂: Number of Asians that would welcome a Latino person into their families.
X₃: Number of Asians that would welcome a black person into their families.
Now since we are working with the population that identifies as "Asians" the sample size will be: n= 251
Since the sample size is large enough (n≥30) you can apply the Central Limit Theorem and approximate the variable distribution to normal.
[tex]Z_{1-\alpha /2}= Z_{0.975}= 1.965[/tex]
1. 95% CI for Asians that would welcome a white person.
If 79% would welcome a white person, then the expected value is:
E(X)= n*p= 251*0.79= 198.29
And the Standard deviation is:
V(X)= n*p*(1-p)= 251*0.79*0.21=41.6409
√V(X)= 6.45
You can construct the interval as:
E(X)±Z₁₋α/₂*√V(X)
198.29±1.965*6.45
[185.62;210.96]
With a 95% confidence level, you'd expect that the interval [185.62; 210.96] contains the number of Asian people that would welcome a White person in their family.
2. 95% CI for Asians that would welcome a Latino person.
If 71% would welcome a Latino person, then the expected value is:
E(X)= n*p= 251*0.71= 178.21
And the Standard deviation is:
V(X)= n*p*(1-p)= 251*0.71*0.29= 51.6809
√V(X)= 7.19
You can construct the interval as:
E(X)±Z₁₋α/₂*√V(X)
178.21±1.965*7.19
[164.08; 192.34]
With a 95% confidence level, you'd expect that the interval [164.08; 192.34] contains the number of Asian people that would welcome a Latino person in their family.
3. 95% CI for Asians that would welcome a Black person.
If 66% would welcome a Black person, then the expected value is:
E(X)= n*p= 251*0.66= 165.66
And the Standard deviation is:
V(X)= n*p*(1-p)= 251*0.66*0.34= 56.3244
√V(X)= 7.50
You can construct the interval as:
E(X)±Z₁₋α/₂*√V(X)
165.66±1.965*7.50
[150.92; 180.40]
With a 95% confidence level, you'd expect that the interval [150.92; 180.40] contains the number of Asian people that would welcome a Black person in their family.
I hope it helps!
A certain college graduate borrows 7864 dollars to buy a car. The lender charges interest at an annual rate of 13%. Assuming that interest is compounded continuously and that the borrower makes payments continuously at a constant annual rate k dollars per year, determine the payment rate that is required to pay off the loan in 3 years. Also determine how much interest is paid during the 3-year period.
Answer:
Therefore rate of payment = $ 3145.72
Therefore the rate of interest = =$1573.17
Step-by-step explanation:
Consider A represent the balance at time t.
A(0)=$ 7864.
r=13 % =0.13
Rate payment = $k
The balance rate increases by interest (product of interest rate and current balance) and payment rate.
[tex]\frac{dB}{dt} = rB-k[/tex]
[tex]\Rightarrow \frac{dB}{dt} - rB=-k[/tex].......(1)
To solve the equation ,we have to find out the integrating factor.
Here p(t)= the coefficient of B =-r
The integrating factor [tex]=e^{\int p(t) dt[/tex]
[tex]=e^{\int (-r)dt[/tex]
[tex]=e^{-rt}[/tex]
Multiplying the integrating factor the both sides of equation (1)
[tex]e^{-rt}\frac{dB}{dt} -e^{-rt}rB=-ke^{-rt}[/tex]
[tex]\Rightarrow e^{-rt}dB - e^{-rt}rBdt=-ke^{-rt}dt[/tex]
Integrating both sides
[tex]\Rightarrow \int e^{-rt}dB -\int e^{-rt}rBdt=\int-ke^{-rt}dt[/tex]
[tex]\Rightarrow e^{-rt}B=\frac{-ke^{-rt}}{-r} +C[/tex] [ where C arbitrary constant]
[tex]\Rightarrow B(t)=\frac{k}{r} +Ce^{rt}[/tex]
Initial condition B=7864 when t =0
[tex]\therefore 7864= \frac{k}{r} - Ce^0[/tex]
[tex]\Rightarrow C= \frac{k}{r} -7864[/tex]
Then the general solution is
[tex]B(t)=\frac{k}{r}-( \frac{k}{r}-7864)e^{rt}[/tex]
To determine the payment rate, we have to put the value of B(3), r and t in the general solution.
Here B(3)=0, r=0.13 and t=3
[tex]B(3)=0=\frac{k}{0.13}-( \frac{k}{0.13}-7864)e^{0.13\times 3}[/tex]
[tex]\Rightarrow- 0.48\frac{k}{0.13} +11614.98=0[/tex]
⇒k≈3145.72
Therefore rate of payment = $ 3145.72
Therefore the rate of interest = ${(3145.72×3)-7864}
=$1573.17
The payment rate that is required to pay off the loan in 3 years is approximately 2949.51 dollars/year. The interest paid during the 3-year period is about 984.53 dollars.
Explanation:To solve this problem, the formula for continuously compounded interest should be used which is A = P * e^(rt), where A is the value of the investment at a future time t, P is the principal amount, r is the annual interest rate, and t is the time the money is invested or borrowed for.
First, set up the equation: 7864 = k * (1/(0.13)) * (e^(0.13 * 3) - 1), then solve for k: k = 7864 * (0.13) / (e^(0.13 * 3) - 1) ≈ 2949.51 dollars/year.
Using the formula A = P * e^(rt), the total amount paid is A = 2949.51 * 3 = 8848.53 dollars. The interest paid during the 3-year period is calculated by subtracting the loan amount from the total amount paid, that is 8848.53 - 7864 = 984.53 dollars.
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Solve using normalcdf
Answer:
0.209
Step-by-step explanation:
Find the sample mean and standard deviation.
μ = 1050
s = 218 / √50 = 30.8
Find the z-score.
z = (1075 − 1050) / 30.8
z = 0.81
Find the probability.
P(Z > 0.81) = 1 − 0.7910 = 0.2090
Is there more wood in a 60-foot-high tree trunk with a radius of 2.2 feet or in a 50-foot-high tree trunk with a radius of 2.6 feet? Assume that the trees can be regarded as right circular cylinders.
Answer:
V₂=1061.85 ft³
Step-by-step explanation:
To determine where more wood is found, just find the volume of each log and see which one has the largest volume
knowing that the volume of a straight cylinder is
V=πR²h, where R=radius and h=height
[tex]V_{1}=\pi (2.2)^{2}60= 912.31ft^{3}\\V_{2}=\pi (2.6)^{2}50= 1061.85ft^{3}[/tex]
As we can observe
V₂>V₁
solution to 2/3x = -5/2 + 2
Answer:
x=-0.75
Step-by-step explanation:
Combine Like terms
2/3x= -5/2+2
2/3x= -2.5+2
2/3x= -0.5
Multiply both sides by 3
2x= -1.5
Divide both sides by 2
x= -0.75
Answer:
x = -4/3
Step-by-step explanation:
The LCM for the fractions is 6x.
2/3x = -5/2 + 2
Multiply through by 6x
(6x) 2/3x = (6x) -5/2 + (6x)2
4 = -15x + 12x
4 = -3x
x = -4/3
Enjoy Maths!
Suppose that grade point averages of undergraduate students at one university have a bell-shaped distribution with a mean of 2.52 and a standard deviation of 0.38. Using the empirical rule, what percentage of the students have grade point averages that are between 1.76 and 3.28
Using the empirical rule for a bell-shaped distribution, which states that nearly all data for a normal distribution falls within three standard deviations of the mean, approximately 95% of undergraduate students at this university have grade point averages between 1.76 and 3.28.
Explanation:The student is asking for the percentage of GPA between 1.76 and 3.28 using the Empirical Rule which applies to a Bell-shaped distribution or normal distribution. The empirical rule states that for a normal distribution, nearly all of the data will fall within three standard deviations of the mean.
The mean in this case is 2.52 and the standard deviation is 0.38. Therefore the values 1.76 and 3.28 fall within the mean minus two standard deviations and mean plus two standard deviations respectively. Therefore, by the empirical rule, these values represent approximately 95% of the data.
In other words, according to the empirical rule, approximately 95% of undergraduate students at this university have grade point averages between 1.76 and 3.28.
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An average light bulb manufactured at The Lightbulb Company lasts and average of 300 days, with a standard deviation of 50 days. Suppose the lifespan of a light bulb from this company is normally distributed. (a) What is the probability that a light bulb from this company lasts less than 210 days? More than 330 days?(b) What is the probability that a light bulb from this company lasts between 280 and 380 days?(c) How would you characterize the lifespan of the light bulbs whose lifespans are among the shortest 2% of all bulbs made by this company?(d) If a pack of 6 light bulbs from this company are purchased, what is the probability that exactly 4 of them last more than 330 days?
Answer:
a). 0.03593, 0.27425
(bl 0.0703
(c) the lifespan of such bulb is less than 210 days
(d) 0.0298.
Step-by-step explanation:
Given that U = 300, sd= 50
Z = X-U/sd
(a) We find the probability
Pr(X<210) = Pr(Z<(210-300)/50)
= Pr(Z<-1.8)
= 0.03593
Pr(X >330) = Pr(Z >(330-300)/50))
= Pr(Z> 0.6)
= 1- PR(Z<0.6) = 1 - 0.72575
Pr(X>330)= 0.27425
(b) Pr(280< X< 330) = Pr( 280 < Z< 380)
= Pr([280-300/50] < Z < [380-300/50])
= Pr(0.4 < Z < 1.6)
= Pr(Z< 1.6) - Pr(Z < 0.4)
=0.72575 - 0.65542
= 0.07033
= 0.0703
(c) the lifespan of such bulbs is less than 210 days
(d) given n = 6, x = 4 ,
Pr(X>330) = 0.27425 = p
q = 1-p = 0.72575
Pr(X=4) = 6C4 × (0.27425)⁴ ×(0.72575)²
Pr(X=4) = 10 × 0.005657 × 0.52671
Pr(X= 4) = 0.029795
Pr(X= 4) = 0.0298
In a certain small dorm, the dorm council will consist of 8 students, ofwhom 3 must be women and 5 must be men, because there are a total of 15 women and 25 men currently living there. What is the total possible number of complete councils that could be selected
Answer:
The total possible number of complete councils that could be selected is 24,174,150.
Step-by-step explanation:
The order of the men and the women is not important. So we use the combinations formula to solve this question.
Combinations formula:
[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]
What is the total possible number of complete councils that could be selected
3 women from a set of 15
5 men from a set of 25
[tex]T = C_{15,3}*C_{25,5} = \frac{15!}{3!12!}*\frac{25!}{5!20!} = 455*53130 = 24174150[/tex]
The total possible number of complete councils that could be selected is 24,174,150.
The numerical course grades in a statistics course can be approximated by a normal model with a mean of 70 and a standard deviation of 10. The professor must convert the numerical grades to letter grades. She decides that she wants 10% A's, 30% B's, 40% C's, 15% D's, and 5% F's. a. What is the cutoff for an A grade?
Answer:
The cutoff for an A grade is 82.8.
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
[tex]\mu = 70, \sigma = 10[/tex]
a. What is the cutoff for an A grade?
The top 10% of the class get an A grade. So the cutoff is the value of X when Z has a pvalue of 1-0.1 = 0.9. So it is X when Z = 1.28
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]1.28 = \frac{X - 70}{10}[/tex]
[tex]X - 70 = 1.28*10[/tex]
[tex]X = 82.8[/tex]
The cutoff for an A grade is 82.8.
Jeremy and Brenda drove their cars in
opposite directions. When they stopped
after some time, they were already
126 miles apart. If Jeremy drove twice
as far as Brenda, how many miles did
Jeremy drive?
Answer: Jeremy drove 84 miles.
Step-by-step explanation:
Let x represent the number of miles that Brenda drove.
If Jeremy drove twice
as far as Brenda, it means that the distance covered by Jeremy would be 2x miles
When they stopped after some time, they were already
126 miles apart. This means that the total distance covered by both of them is 126 miles. Therefore,
x + 2x = 126
3x = 126
x = 126/3
x = 42 miles
The number of miles that Jeremy drove is
42 × 2 = 84 miles
A car is being driven at a rate of 40 ft/sec when the brakes are applied. The car decelerates at a constant rate of 10 ft/sec2. Calculate how far the car travels in the time it takes to stop. Round your answer to one decimal place.
Answer:
80 feet
Step-by-step explanation:
Given:
Initial speed of the car ([tex]v_0[/tex]) = 40 ft/sec
Deceleration of the car ([tex]\frac{dv}{dt}[/tex]) = -10 ft/sec²
Final speed of the car ([tex]v_x[/tex]) = 0 ft/sec
Let the distance traveled by the car be 'x' at any time 't'. Let 'v' be the velocity at any time 't'.
Now, deceleration means rate of decrease of velocity.
So, [tex]\frac{dv}{dt}=-10\ ft/sec^2[/tex]
Negative sign means the velocity is decreasing with time.
Now, [tex]\frac{dv}{dt}=\frac{dv}{dx}(\frac{dx}{dt})[/tex] using chain rule of differentiation. Therefore,
[tex]\frac{dv}{dx}\cdot\frac{dx}{dt}= -10\\\\But\ \frac{dx}{dt}=v.\ So,\\\\v\frac{dv}{dx}=-10\\\\vdv=-10dx[/tex]
Integrating both sides under the limit 40 to 0 for 'v' and 0 to 'x' for 'x'. This gives,
[tex]\int\limits^0_{40} {v} \, dv=\int\limits^x_0 {-10} \, dx\\\\\left [ \frac{v^2}{2} \right ]_{40}^{0}=-10x\\\\-10x=\frac{0}{2}-\frac{1600}{2}\\\\10x=800\\\\x=\frac{800}{10}=80\ ft[/tex]
Therefore, the car travels a distance of 80 feet before stopping.
Consider the following data on x = rainfall volume (m3) and y = runoff volume (m3) for a particular location. x 8 12 14 19 23 30 40 45 55 67 72 79 96 112 127 y 4 10 13 14 15 25 27 44 38 46 53 74 82 99 105 Use the general software output to decide whether there is a useful linear relationship between rainfall and runoff.
There is a useful linear relationship between rainfall and runoff and the relationship is y = 0.85x - 2.65
Deciding whether there is a useful linear relationship between rainfall and runoff
From the question, we have the following parameters that can be used in our computation:
x 8 12 14 19 23 30 40 45 55 67 72 79 96 112 127
y 4 10 13 14 15 25 27 44 38 46 53 74 82 99 105
Using a graphing tool, we have the following summary
Sum of X = 799Sum of Y = 649Mean X = 53.2667Mean Y = 43.2667Sum of squares (SSX) = 20086.9333Sum of products (SP) = 17313.9333The regression equation can be represented as
y = bx + a
Where
b = SP/SSX = 17313.93/20086.93 = 0.86
a = MY - bMX = 43.27 - (0.86*53.27) = -2.65
So, we have
y = 0.85x - 2.65
Hence, there is a useful linear relationship between rainfall and runoff