Answer:
a) For this case the intercept of 2.52 represent a common effect of measure for any student without taking in count the other variables analyzed, and we know that if HSGPA=0, ACT= 0 and skip =0 we got [tex] colGPA=2.52[/tex]
b) This value represent the effect into the ACT scores in the GPA, we know that:
[tex]\hat \beta_{ACT} = 0.015[/tex]
So then for every unit increase in the ACT score we expect and increase of 0.015 in the GPA or the predicted variable
c) If we are interested in analyze if we have a significant relationship between the dependent and the independent variable we can use the following system of hypothesis:
Null Hypothesis: [tex]\beta_i = 0[/tex]
Alternative hypothesis: [tex]\beta_i \neq 0[/tex]
Or in other wouds we want to check if an specific slope is significant.
The significance level assumed for this case is [tex]\alpha=0.05[/tex]
Th degrees of freedom for a linear regression is given by [tex]df=n-p-1 = 45-3-1 = 41[/tex], where p =3 the number of variables used to estimate the dependent variable.
In order to test the hypothesis the statistic is given by:
[tex]t=\frac{\hat \beta_i}{SE_{\beta_i}}[/tex]
And replacing we got:
[tex] t = \frac{-0.5}{0.0001}=-5000[/tex]
And for this case we see that if we find the p value for this case we will get a value very near to 0, so then we can conclude that this coefficient would be significant for the regression model .
Step-by-step explanation:
For this case we have the following multiple regression model calculated:
colGPA =2.52+0.38*HSGPA+0.015*ACT-0.5*skip
Part a
(a) Interpret the intercept in this model.
For this case the intercept of 2.52 represent a common effect of measure for any student without taking in count the other variables analyzed, and we know that if HSGPA=0, ACT= 0 and skip =0 we got [tex] colGPA=2.52[/tex]
(b) Interpret [tex]\hat \beta_{ACT}[/tex] from this model.
This value represent the effect into the ACT scores in the GPA, we know that:
[tex]\hat \beta_{ACT} = 0.015[/tex]
So then for every unit increase in the ACT score we expect and increase of 0.015 in the GPA or the predicted variable
(c) What is the predicted college GPA for someone who scored a 25 on the ACT, had a 3.2 high school GPA and missed 4 lectures. Show your work.
For this case we can use the regression model and we got:
[tex] colGPA =2.52 +0.38*3.2 +0.015*25 - 0.5*4 = 26.751[/tex]
(d) Is the estimate of skipping class statistically significant? How do you know? Is the estimate of skipping class economically significant? How do you know? (Hint: Suppose there are 45 lectures in a typical semester long class).
If we are interested in analyze if we have a significant relationship between the dependent and the independent variable we can use the following system of hypothesis:
Null Hypothesis: [tex]\beta_i = 0[/tex]
Alternative hypothesis: [tex]\beta_i \neq 0[/tex]
Or in other wouds we want to check if an specific slope is significant.
The significance level assumed for this case is [tex]\alpha=0.05[/tex]
Th degrees of freedom for a linear regression is given by [tex]df=n-p-1 = 45-3-1 = 41[/tex], where p =3 the number of variables used to estimate the dependent variable.
In order to test the hypothesis the statistic is given by:
[tex]t=\frac{\hat \beta_i}{SE_{\beta_i}}[/tex]
And replacing we got:
[tex] t = \frac{-0.5}{0.0001}=-5000[/tex]
And for this case we see that if we find the p value for this case we will get a value very near to 0, so then we can conclude that this coefficient would be significant for the regression model .
Conditional distribution. Give (in percents) the conditional distribution of income level among single men. Should your percents add to 100% (up to roundoff error)? Explain your reasoning.
Answer:
note:
solution is attached due to error in mathematical equation. please find the attachment
The conditional distribution of income level among single men should add up to 100%, since when calculating a conditional distribution we treat the subgroup as the total population. Thus, income levels among single men are all the possible outcomes for this group and should collectively correspond to 100% of cases.
Explanation:The conditional distribution in this scenario is the distribution of income level among single men, being key to understanding he relative frequency of various income levels within this specific group. Since we are looking at a subset of the total population (in this case, single men), the percentages should indeed add up to 100%. This is because when we calculate the conditional distribution, we treat this subgroup as if it were the total population. Therefore, the total should be 100%, reflecting the entire subgroup, not compared to the entire population. An example could be: If 10% of single men are in low income, 30% are in middle income, and 60% are in high income, these percentages represent the income level distribution among single men, and add up to 100%.
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Let p1 be the proportion of successes in the first population and let p2 be the proportion of successes in the second population. Suppose that you are testing the hypotheses: H0:p1−p2=0Ha:p1−p2≠0 Futhermore suppose that z∗=1.73, find and input the p-value for this test. Round your answer to 4 decimal places.
Answer:
0.0836
Step-by-step explanation:
Given that p1 be the proportion of successes in the first population and let p2 be the proportion of successes in the second population
[tex]H_0:p_1-p_2=0\\H_a:p_1-p_2\neq 0[/tex]
is the hypotheses created
This is two tailed
Test statistic Z = 1.73
corresponding p value
= 0.08363
=0.0836 (after rounding off to 4 decimals)
The p-value for this test is 0.0802.
The p-value for this two-tailed test can be calculated by doubling the probability of observing a test statistic at least as extreme as the observed z-score of 1.73, under the assumption that the null hypothesis is true.
Given that the z-score is 1.73, we need to find the area under the standard normal distribution curve that lies beyond this z-score. This area represents the probability of observing a value at least as extreme as 1.73 standard deviations from the mean under the null hypothesis.
Using a standard normal distribution table or a calculator, we can find the probability corresponding to a z-score of 1.73. Let's denote this probability as P(Z > 1.73).
P(Z > 1.73) = 1 - P(Z < 1.73)
Using a standard normal distribution table or calculator, we find that P(Z < 1.73) is approximately 0.9599. Therefore:
P(Z > 1.73) = 1 - 0.9599 = 0.0401
Since this is a two-tailed test, we need to consider the probability of observing a z-score at least as extreme in either direction (both tails of the distribution). Thus, we multiply the single-tail probability by 2:
p-value = 2 * P(Z > 1.73) = 2 * 0.0401 = 0.0802
Rounded to four decimal places, the p-value is 0.0802.
Therefore, the p-value for this test is 0.0802.
The answer is: 0.0802.
Mike brought a new car and financed $25,000 to make the purchase. He financed the car for 60 months with an APR of 6.5%. Determine each of the following:
A. Mike's monthly payment
B. Total cost if Mike's car
C. Total interest Mike pays over the life of the loan
Final answer:
Mike's monthly car payment is approximately $489.99. The total cost of the car including interest over 60 months is $29,399.40, and the total interest Mike pays over the loan's lifetime is $4,399.40.
Explanation:
To determine Mike's monthly car payment, total cost of the car, and total interest paid over the life of the loan, we need to use the formula for the monthly payment on an installment loan, which can be found using the formula:
P = (Pv * r) / (1 - (1 + r)-n)
Where:
P is the monthly paymentPv is the present value or the amount of the loan, which is $25,000r is the monthly interest rate (annual rate divided by 12), which is 0.065/12n is the total number of payments (months), which is 60Lets calculate the monthly payment (P):
P = ($25,000 * 0.065/12) / (1 - (1+0.065/12)-60)
= $489.99 approximately
Mike's total cost of the car is the monthly payment multiplied by the number of payments:
$489.99 * 60 = $29,399.40
The total interest Mike pays is the total cost minus the loan amount:
$29,399.40 - $25,000 = $4,399.40
Thus, Mike's monthly payment is approximately $489.99, the total cost of his car will be $29,399.40, and the total interest paid over the life of the loan is $4,399.40.
If mAC 93°, and mBD= 39°, what is m∠BED?
54°
32°
108°
27°
Answer: LAST OPTION.
Step-by-step explanation:
For this exercise it is important to remember that, by definition, the angle formed by two secants intersecting outside of a circle, can be found with the following formula:
[tex]Angle\ formed\ by\ two\ secants=\frac{Difference\ of\ intercepted\ arcs}{2}[/tex]
According to the information provided in the exercise, you know that:
[tex]mAC= 93\°\\\\mBD= 39\°[/tex]
Therefore, you can find the difference of the intercepted arcs:
[tex]Difference\ of\ intercepted\ arcs=93\°-39\°\\\\Difference\ of\ intercepted\ arcs=54\°[/tex]
The final step is to substitute the difference of the intercepted arcs calculated above, into the formula, in order to find the measure of the angle BED.
You get that this is:
[tex]Angle\ formed\ by\ two\ secants=\frac{54\°}{2}\\\\Angle\ formed\ by\ two\ secants=27\°[/tex]
Answer:
The answer should be 27
Step-by-step explanation:
A company has learned that the relationship between its advertising and sales shows diminishing marginal returns. That is, as it saturates consumers with ads, the benefits of increased advertising diminish. The company should expect to find linear association between its advertising and sales.
Answer:
A company has learned that the relationship between its advertising and sales shows diminishing marginal returns. That is, as it saturates consumers with ads, the benefits of increased advertising diminish. The company should expect to find a linear association between its advertising and sales - This statement is false
Step-by-step explanation:
According to the scenario given for the company, it was said that the marginal return diminished after a saturation point, therefore, the company should rather expect a non-linear pattern and not a linear pattern.
Therefore, the statement expressed in the question is false.
A 15-inch candle is lit and burns at a constant rate of 1.1 inches per hour. Let t represent the number of hours since the candle was lit, and suppose f is a function such that f ( t ) represents the remaining length of the candle (in inches) t hours after it was lit. Write a function formula for f . f ( t )
Answer: f(t) = 15 - 1.1t
Step-by-step explanation:
Let t represent the number of hours since the candle was lit.
A 15-inch candle is lit and burns at a constant rate of 1.1 inches per hour. This means that in t hours, the candle that would have burnt is 1.1t
The length of the candle that would be left after t hours is expressed as
15 - 1.1t
suppose f is a function such that f(t) represents the remaining length of the candle (in inches) t hours after it was lit, then
f(t) = 15 - 1.1t
help, ill mark brainliest
Answer:
Marcus rents the car for 5 days
Step-by-step explanation:
For 5 days:
Plan A would equal $150 (30x5)
Plan B would equal $140 (125+15)
Plan b is cheaper only when Marcus rents the car for 5 days.
HELP PLEASE 100 POINTS! Review the diagram below. Apply the properties of angles to solve for the missing angles. Angle x is ____ degrees. And please show how you got it.
Step-by-step explanation:
<y+60+58=180 (angle sum property)
<y+118=180
<y=180-118=62
<ACB+<ACD=180(degree of a line )
58+x=180
x=122
(btw u cant get a 100 points by answering one question)
Suppose that the problem is modified to specify that the chairs produced should accompany the tables so that of one-third of the tables produced each must have 6 accompanying chairs, of another one-third of the tables each must have 4 accompanying chairs, and of the other one-third of the tables each must have 2 accompanying chairs. How would this constraint be written
Answer:
C = 4T
Step-by-step explanation:
The question is not completed, the complete question and the solution is attached in the file below
The table in the shape of a circle has a diameter of 6 feet. How much fabric is needed to make a table cloth if it hangs 1 foot off the table all the way around?jdjdkndjcjjfnnfm
Answer: area of fabric needed is 50.24 ft²
Step-by-step explanation:
The table in the shape of a circle has a diameter of 6 feet. This means that the diameter of the fabric that would just fit the table is 6 feet. Therefore, the diameter of the fabric needed to make a table cloth if it hangs 1 foot off the table would be 6 + 1 + 1 = 8 feet
The formula for determining the area of a circle is expressed as
Area = πr²
Where
r represents radius of the circle.
π is a constant whose value is 3.14
Radius = diameter/2. Therefore
r = 8/2 = 4
Area of fabric = 3.14 × 4²
= 50.24 ft²
Final answer:
To make a tablecloth for a circular table with a diameter of 6 feet, and an overhang of 1 foot, the student would need approximately 50.27 square feet of fabric.
Explanation:
The student is asking about finding the amount of fabric needed to create a tablecloth for a circular table with specific dimensions. Given that the table has a diameter of 6 feet, and the tablecloth needs to hang 1 foot off the table all the way around, we need to calculate the diameter of the fabric required.
To solve this, we need to add the overhang to the diameter of the table, considering that the overhang occurs on both sides:
Diameter of the table: 6 feetOverhang on one side: 1 footOverhang on the other side: 1 footTotal diameter needed: 6 feet + 1 foot + 1 foot = 8 feetNow, to find the area of fabric needed, we apply the formula for the area of a circle which is π × radius². First, we find the radius by halving the diameter:
Radius of the fabric: 8 feet / 2 = 4 feetThen, we calculate the area:
Area of fabric: π × (4 feet)² = π × 16 feet²Finally, we can approximate π as 3.1416 to get an approximate area of:
Area of fabric ≈ 3.1416 × 16 feet² ≈ 50.2656 square feetTherefore, the student would need approximately 50.27 square feet of fabric to make the tablecloth.
Dave’s Automatic Door, referred to in Exercise 29, installs automatic garage door openers. Based on a sample, following are the times, in minutes, required to install 10 door openers: 28, 32, 24, 46, 44, 40, 54, 38, 32, and 42.
The question is not complete and the full question says;
Calculate the (a) range, (b) arithmetic mean, (c) mean deviation, and (d) interpret the values. Dave’s Automatic Door installs automatic garage door openers. The following list indicates the number of minutes needed to install a sample of 10 door openers: 28, 32, 24, 46, 44, 40, 54, 38, 32, and 42.
Answer:
A) Range = 30 minutes
B) Mean = 38
C) Mean Deviation = 7.2
D) This is well written in the explanation.
Step-by-step explanation:
A) In statistics, Range = Largest value - Smallest value. From the question, the highest time is 54 minutes while the smallest time is 24 minutes.
Thus; Range = 54 - 24 = 30 minutes
B) In statistics,
Mean = Σx/n
Where n is the number of times occurring and Σx is the sum of all the times occurring
Thus,
Σx = 28 + 32 + 24 + 46 + 44 + 40 + 54 + 38 + 32 + 42 = 380
n = 10
Thus, Mean(x') = 380/10 = 38
C) Mean deviation is given as;
M.D = [Σ(x-x')]/n
Thus, Σ(x-x') = (28-38) + (32-38) + (24-38) + (46-38) + (44-38) + (40-38) + (54-38) + (38-38) + (32-38) + (42-38) = 72
So, M.D = 72/10 = 7.2
D) The range of the times is 30 minutes.
The average time required to open one door is 38 minutes.
The number of minutes the time deviates on average from the mean of 38 minutes is 7.2 minutes
The average time to install an automatic garage door opener, based on the provided data, is approximately 38 minutes, though individual times can vary.
Explanation:The subject of this question is Mathematics, specifically statistics. The question provides a series of data points representing the time, in minutes, that it took to install 10 automatic garage door openers. To find the average installation time, you would add up all of the times and then divide by the number of data points, which in this case is 10.
The data points are: 28, 32, 24, 46, 44, 40, 54, 38, 32, and 42. So, average installation time = (28+32+24+46+44+40+54+38+32+42) / 10 = 38 minutes.
This suggests that on average, it takes about 38 minutes to install an automatic garage door opener. But remember, this is an average. Individual installation times can vary greatly, as seen in the range of times provided.
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Multiply.
(x - 6)(x - 4)
Answer:
[tex]x^{2} -10x+24[/tex]
Step-by-step explanation:
A student is asked to find the derivative of y = x sin2 (x) with respect to variable x, given x, y > 0. They provide the following answer: dy dx = sin2 (x) · x sin2 (x)−1 · cos2 (x) Is the student correct? If the student is correct, then explain how they used derivative rules correctly to find this derivative. If the student is incorrect, then give the correct answer and provide an explanation that you would use to correct the student’s thinking.
Answer:
[tex]\frac{dy}{dx} = 2x ( 2cos2x) +sin2x[/tex]
[tex]\frac{dy}{dx} = 4x( cos2x)+sin2x[/tex]
Step-by-step explanation:
Given y = x sin2x .....(1)
Applying UV formula [tex]\frac{d(UV)}{dx} = u \frac{dv}{dx} + v\frac{du}{dx}[/tex]
Differentiating with respective to 'x' we get
[tex]\frac{dy}{dx} = x ( 2cos2x)\frac{d(2x)}{dx} +sin2x (1)[/tex]
[tex]\frac{dy}{dx} = x ( 2cos2x)(2) +sin2x (1)[/tex]
Final answer:-
[tex]\frac{dy}{dx} = 4x( cos2x)+sin2x[/tex]
Suppose that a die is rolled twice. What are the possible values that the following random variables can take on: a. the maximum value to appear in the two rolls; b. the minimum value to appear in the two rolls; c. the sum of the two rolls; d. the value of the first roll minus the value of the second roll
Answer:
a. A = {1, 2, 3, 4, 5, 6}
b. B = {1, 2, 3, 4, 5, 6}
c. C = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}
d. D = {-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5}
Step-by-step explanation:
a. the maximum value to appear in the two rolls
Since only the maximum value is computed, the variable can assume any integer from 1 to 6:
A = {1, 2, 3, 4, 5, 6}
b. the minimum value to appear in the two rolls;
Since only the minimum value is computed, the variable can assume any integer from 1 to 6:
B = {1, 2, 3, 4, 5, 6}
c. the sum of the two rolls;
The minimum value would be from rolling two ones (sum is 2) and the maximum value would be from rolling two sixes (sum is 12). Every integer in-between is possible:
C = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}
d. the value of the first roll minus the value of the second roll
The minimum value would be from rolling a one and a six (result is -5) and the maximum value would be from rolling a six and a one (result is 5). Every integer in-between is possible:
D = {-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5}
Final answer:
Explanation of possible values for maximum, minimum, sum, and difference in rolling two dice.
Explanation:
a. The possible values for the maximum value:
1, if the same number appears twice on both rolls
2 to 6, for individual numbers in different scenarios
b. The possible values for the minimum value:
1 to 5, as the minimum value will exclude the maximum value
c. The possible values for the sum:
2 to 12, representing all possible sums from rolling two dice
d. The possible values for the difference between two rolls:
-5 to 5, as the difference can range from -5 to 5
Before every flight, the pilot must verify that the total weight of the load is less than the maximum allowable load for the aircraft. The aircraft can carry 41 passengers, and a flight has fuel and baggage that allows for a total passenger load of 6 comma 683 lb. The pilot sees that the plane is full and all passengers are men. The aircraft will be overloaded if the mean weight of the passengers is greater than StartFraction 6 comma 683 l b Over 41 EndFraction equals 163 lb. What is the probability that the aircraft is overloaded? Should the pilot take any action to correct for an overloaded aircraft? Assume that weights of men are normally distributed with a mean of 180.5 lb and a standard deviation of 38.2.
Answer:
The probability that the plane is oveloaded is P=0.9983.
The pilot should take out the baggage and send it in another plain or have less passengers in the plain to not overload.
Step-by-step explanation:
The aircraft will be overloaded if the mean weight of the passengers is greater than 163 lb.
If the plane is full, we have 41 men in the plane. This is our sample size.
The weights of men are normally distributed with a mean of 180.5 lb and a standard deviation of 38.2.
So the mean of the sample is 180.5 lb (equal to the population mean).
The standard deviation is:
[tex]\sigma=\frac{\sigma}{\sqrt{N}} =\frac{38.2}{\sqrt{41}}=\frac{38.2}{6.4} =5.97[/tex]
Then, we can calculate the z value for x=163 lb.
[tex]z=\frac{x-\mu}{\sigma}=\frac{163-180.5}{5.97}=\frac{-17.5}{5.97}= -2.93[/tex]
The probability that the mean weight of the men in the airplane is below 163 lb is P=0.0017
[tex]P(\bar X<163)=P(z<-2.93)=0.00169[/tex]
Then the probability that the plane is oveloaded is P=0.9983:
[tex]P(overloaded)=1-P(X<163)=1-0.0017=0.9983[/tex]
The pilot should take out the baggage or have less passengers in the plain to not overload.
Determine if the described set is a subspace. The subset of Rn (n even) consisting of vectors of the form v = v1 vn , such that v1 − v2 + v3 − v4 + v5 − − vn = 0. The set is a subspace. The set is not a subspace.
Answer:
The set is a subspace
Step-by-step explanation:
We need to check 3 things: whether the 0 vector is in the set, whether the sum of 2 elements of the set is an element of the set and whether the product of an element of the set for a real scalar is an element of the set.
0 is in the setYes: the 0 vector (0, 0, ..., 0) satysfies the set property: 0-0+0-0........-0 = 0.
Given 2 elements v = (v1, ..., vn), w = (w1, ..., wn), is the sum v+2 = (v1+w1, v2+w2, ..., vn+wn) an element of the set?Yes: Note that (v1+w1)-(v2+w2)+(v3+w3)- ..... - (vn+wn) = v1-v1+v3 - ... - vn + w1 - w2 + w3 - ... - wn = 0+0 = 0.
Given an element of the set v = (v1, ... ,vn), and a real number r, is rv = (rv1, ..., rvn) an element of the set?Yes: By taking r as common factor, we have rv1 - rv2 + rv3 - ... - rvn = r * (v1-v2+v3 - ... - vn) = r*0 = 0.
Thus, the described set is effectively a subspace.
The described set is a subspace of Rn (n even). It satisfies all three conditions of a subspace: containing the zero vector, closed under addition, and closed under scalar multiplication.
Explanation:The set described is a subspace of ℝn (where n is even).
To determine if the set is a subspace, we need to check if it satisfies three conditions:
It contains the zero vector: The zero vector satisfies v1 - v2 + v3 - v4 + v5 - ... - vn = 0, so it is in the set.
Since the set satisfies all three conditions, it is a subspace of ℝn (where n is even).
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The mean investment that employees put into their companies 401k per year is $10,000 with standard deviation of $500 Assuming the investment follow a normal distribution, determine the following a. what proportion of employees put between $9, 500 and $11,000 into the 401 k per year b. What proportion of employee put more than $11, 500 into the 401 k per year? c. What proportional of employees put less than $11,000 into the 401k per year?d. What proportional of employees put more than $9,000 into the 401k per year?e. What proportional of employees put between than $11,000 and $11, 500 into the 401k per year?f. How much would an employees need to put into his or her 401 K to be in the upper 10% of investors?
Answer:
a) 81.85% of employees put between $9, 500 and $11,000 into the 401 k per year
b) 0.13% of employee put more than $11, 500 into the 401 k per year
c) 97.72% of employees put less than $11,000 into the 401k per year.
d) 97.72% of employees put more than $9,000 into the 401k per year
e) 2.15% of employees put between than $11,000 and $11, 500 into the 401k per year
f) An employee would need to put $10,640 into his or her 401 K to be in the upper 10% of investors
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 = 10000, \sigma = 500[/tex]
a. what proportion of employees put between $9, 500 and $11,000 into the 401 k per year
This is the pvalue of Z when X = 11000 subtracted by the pvalue of Z when X = 9500. So
X = 11000
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{11000 - 10000}{500}[/tex]
[tex]Z = 2[/tex]
[tex]Z = 2[/tex] has a pvalue of 0.9772.
X = 9500
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{9500 - 10000}{500}[/tex]
[tex]Z = -1[/tex]
[tex]Z = -1[/tex] has a pvalue of 0.1587
0.9772 - 0.1587 = 0.8185
81.85% of employees put between $9, 500 and $11,000 into the 401 k per year
b. What proportion of employee put more than $11, 500 into the 401 k per year?
This is 1 subtracted by the pvalue of Z when X = 11500. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{11500 - 10000}{500}[/tex]
[tex]Z = 3[/tex]
[tex]Z = 3[/tex] has a pvalue of 0.9987
1 - 0.9987 = 0.0013
0.13% of employee put more than $11, 500 into the 401 k per year
c. What proportional of employees put less than $11,000 into the 401k per year?
This is the pvalue of Z when X = 11000. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{11000 - 10000}{500}[/tex]
[tex]Z = 2[/tex]
[tex]Z = 2[/tex] has a pvalue of 0.9772.
97.72% of employees put less than $11,000 into the 401k per year.
d. What proportional of employees put more than $9,000 into the 401k per year?
This is 1 subtracted by the pvalue of Z when X = 9000. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{9000 - 10000}{500}[/tex]
[tex]Z = -2[/tex]
[tex]Z = -2[/tex] has a pvalue of 0.0228.
1 - 0.0228 = 0.9772
97.72% of employees put more than $9,000 into the 401k per year
e. What proportional of employees put between than $11,000 and $11, 500 into the 401k per year?
This is the pvalue of Z when X = 11500 subtracted by the pvalue of Z when X = 11000. So
X = 11500
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{11500 - 10000}{500}[/tex]
[tex]Z = 3[/tex]
[tex]Z = 3[/tex] has a pvalue of 0.9987
X = 11000
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{11000 - 10000}{500}[/tex]
[tex]Z = 2[/tex]
[tex]Z = 2[/tex] has a pvalue of 0.9772.
0.9987 - 0.0972 = 0.0215
2.15% of employees put between than $11,000 and $11, 500 into the 401k per year
f. How much would an employees need to put into his or her 401 K to be in the upper 10% of investors?
This is the value of Z when X 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 - 10000}{500}[/tex]
[tex]X - 10000 = 500*1.28[/tex]
[tex]X = 10640[/tex]
An employee would need to put $10,640 into his or her 401 K to be in the upper 10% of investors
In a city, the distance between the library and the police station is 3 miles less than twice the distance between the police
station and the fire station. The distance between the library and the police station is 5 miles. How far apart are the police
station and the fire station?
0
O
O
O
1 mile
3 miles
4 miles
6 miles
Answer: the distance between the the police station and the fire station is 4 miles.
Step-by-step explanation:
Let x represent the distance between the library and the police station.
Let y represent the distance between the the police station and the fires station.
In the city, the distance between the library and the police station is 3 miles less than twice the distance between the police station and the fire station. This is expressed as
x = 2y - 3
The distance between the library and the police station is 5 miles. This means that
5 = 2y - 3
2y = 5 + 3 = 8
y = 8/2 = 4
Minute Maid states that a bottle of juice contains 473 mL. Consumer groups are interested in determining if the bottles contain less than the amount stated on the label. To test their claim, they sample 30 bottles. The sample mean was 472mL and the standard deviation is 0.2. What does mu represent here? Group of answer choices The average contents of all bottles of juice in the population, which is 472mL. The average contents of all bottles of juice in the sample, which is unknown. The average contents of all bottles of juice in the sample, which is 472mL. The average contents of all bottles of juice in the population, which is unknown.
Answer:
The average contents of all bottles of juice in the population
[tex]\mu = 473\text{ mL}[/tex]
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = 473 mL
Sample mean, [tex]\bar{x}[/tex] = 472 mL
Sample size, n = 30
Sample standard deviation, s = 0.2
First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu = 473\text{ mL}\\H_A: \mu < 473\text{ mL}[/tex]
Representation of [tex]\mu[/tex]
It is the population parameter for mean.Thus, it represents the average contents of all bottles of juice in the population, which is 473 mL.The average juice a bottle contain is the mean value of the juice.
[tex]\mathbf{\mu }[/tex] is the average content in all bottles of juice in the population, which is 472mL.
The given parameters are:
[tex]\mathbf{n = 30}[/tex] --- the sample size
[tex]\mathbf{\sigma = 0.2}[/tex] --- the standard deviation
[tex]\mathbf{\bar x = 472}[/tex] --- the sample mean
[tex]\mathbf{\mu = 473}[/tex] --- the population mean
The above highlights means that:
The parameter [tex]\mathbf{\mu }[/tex] represents the population mean
This means that:
[tex]\mathbf{\mu }[/tex] is the average content in all bottles of juice in the population.
From the question, the value is given as: 473
Hence, the true statement is:
[tex]\mathbf{\mu }[/tex] is the average content in all bottles of juice in the population, which is 473mL.
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In class we derived the MOM and MLE for an exponential distribution with parameter ????. Conduct a Bootstrap simulation to compare the estimation of λ with sample sizes of n = 10, n = 100, and n = 500. Choose true value λ = 0.2 and use B = 1000. Calculate and compare the mean and standard error for each set of simulations to each other as well as their theoretical values.
Answer:
Below is the R code for the bootstrapping in exponential distribution. The result is attached below.
####################################
rm(list=ls(all=TRUE))
set.seed(12345)
N=c(10,100,500)
Rate=0.2
B=1000
MN=SE=rep()
for(i in 1:length(N))
{
n=N[i]
X=rexp(n,rate=Rate)
EST=1/mean(X)
ESTh=rep()
for(j in 1:B)
{
Xh=rexp(n,rate=EST)
ESTh[j]=1/mean(Xh)
}
MN[i]=mean(ESTh)
SE[i]=sd(ESTh)
}
cbind(N,Rate,MN,SE)
Step-by-step explanation:
Use the properties of limits to help decide whether the limit exists. If the limit exists, find its value. ModifyingBelow lim With x right arrow infinity StartFraction 6 x cubed plus 5 x minus 7 Over 6 x Superscript 4 Baseline minus 4 x cubed minus 9 EndFraction
Answer:
The value of given limit problem is 0.
Step-by-step explanation:
The given limit problem is
[tex]lim_{x\rightarrow \infty}\dfrac{6x^3+5x-7}{6x^4-4x^3-9}[/tex]
We need to find the value of given limit problem.
Divide the numerator and denominator by the leading term of the denominator, i.e., [tex]x^4[/tex]
[tex]lim_{x\rightarrow \infty}\dfrac{\frac{6x^3+5x-7}{x^4}}{\frac{6x^4-4x^3-9}{x^4}}[/tex]
[tex]lim_{x\rightarrow \infty}\dfrac{\frac{6}{x}+\frac{5}{x^3}-\frac{7}{x^4}}{6-\frac{4}{x}-\frac{9}{x^4}}[/tex]
Apply limit.
[tex]\dfrac{\frac{6}{ \infty}+\frac{5}{ \infty}-\frac{7}{ \infty}}{6-\frac{4}{ \infty}-\frac{9}{ \infty}}[/tex]
We know that [tex]\frac{1}{\infty}=0[/tex].
[tex]\dfrac{0+0-0}{6-0-0}[/tex]
[tex]\dfrac{0}{6}[/tex]
[tex]0[/tex]
Hence, the value of given limit is 0.
The limit of the expression (6x^3 + 5x - 7)/(6x^4 - 4x^3 - 9) as x approaches infinity is 0.
Explanation:In mathematics, when we are asked to find the limit of an expression as x approaches infinity, we can use a technique known as the 'highest powers' method. This involves dividing all terms in the function by the highest power of x in the denominator. In this expression, the highest power of x in the denominator is x4. So, we'll divide all terms by x4.
The given expression is: (6x3+5x-7) / (6x4-4x3-9). Dividing all terms by x4, we get: (6/x+5/x3-7/x4) / (6-4/x-9/x4)
As x approaches infinity, all terms that have x in the denominator go to 0. So, our expression simplifies to (0/6) = 0. Hence, the limit of the given expression as x approaches infinity is 0.
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Suppose that MX=V is a linear system, for some matrix M and some vector V. Let the vector P be a particular solution to the system and the vector H a homogeneous solution to the system. Which of the following vectors must be a particular solution to the system? Select all that apply.
A. 2H-P
B. 2P+2H
C. 3H-P
D. H
E. 2P+H
F. P+3H
G. P
H. P+H
I. P-3H
J. 3P+H
K. 2H-P
L. P+2H
M. 3P+3H
N. P-2H
O. H-P
P. P-H
From what I understand, the general solution = particular+homogeneous.
I found that possible solutions could be P+H, a scalar * P, or a scalar * P + scalar*H. I tried these options and didn't get it right. Any help would be appreciated
Answer:
The Answer is "G. P"
Step-by-step explanation:
They ask for "Which of the following vectors must be a PARTICULAR solution to the system?" so the answer simply G. which is P
A relief fund is set up to collect donations for the families affected by recent storms. A random sample of 400 people shows that 28% of those 200 who were contacted by telephone actually made contributions compared to only 18% of the 200 who received first class mail requests. Which is the correct 95% confidence interval for the difference in the proportions of people who make donations if contacted by telephone or first class mail
Answer:
[tex](0.28-0.18) - 1.96 \sqrt{\frac{0.28(1-0.28)}{200} +\frac{0.18(1-0.18)}{200}}=0.0181[/tex]
[tex](0.28-0.18) - 1.96 \sqrt{\frac{0.28(1-0.28)}{200} +\frac{0.18(1-0.18)}{200}}=0.182[/tex]
And the 95% confidence interval would be given (0.0181;0.181).
We are confident at 95% that the difference between the two proportions is between [tex]0.0181 \leq p_B -p_A \leq 0.182[/tex]
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".
Solution to the problem
[tex]p_A[/tex] represent the real population proportion for telephone
[tex]\hat p_A =0.28[/tex] represent the estimated proportion for telephone
[tex]n_A=200[/tex] is the sample size required for telephone
[tex]p_B[/tex] represent the real population proportion for mail
[tex]\hat p_B =0.18[/tex] represent the estimated proportion for mail
[tex]n_B=200[/tex] is the sample size required for mail
[tex]z[/tex] represent the critical value for the margin of error
The population proportion have the following distribution
[tex]p \sim N(p,\sqrt{\frac{p(1-p)}{n}})[/tex]
The confidence interval for the difference of two proportions would be given by this formula
[tex](\hat p_A -\hat p_B) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}[/tex]
For the 95% confidence interval the value of [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2=0.025[/tex], with that value we can find the quantile required for the interval in the normal standard distribution.
[tex]z_{\alpha/2}=1.96[/tex]
And replacing into the confidence interval formula we got:
[tex](0.28-0.18) - 1.96 \sqrt{\frac{0.28(1-0.28)}{200} +\frac{0.18(1-0.18)}{200}}=0.0181[/tex]
[tex](0.28-0.18) - 1.96 \sqrt{\frac{0.28(1-0.28)}{200} +\frac{0.18(1-0.18)}{200}}=0.182[/tex]
And the 95% confidence interval would be given (0.0181;0.181).
We are confident at 95% that the difference between the two proportions is between [tex]0.0181 \leq p_B -p_A \leq 0.182[/tex]
The 95% confidence interval for the difference in the proportions of people who make donations if contacted by telephone or first class mail is between 1.14% and 18.86%.
Explanation:To calculate a 95% confidence interval for the difference in the proportions of people who make donations if contacted by telephone or first class mail, we use the formula for comparing two proportions.
We are given that 28% of the 200 contacted by telephone donated, which is a proportion of p1 = 0.28, and 18% of the 200 contacted by first class mail donated, which is a proportion of p2 = 0.18. Each group size is n1 = n2 = 200.
The formula for the standard error of the difference between two independent proportions is:
SE = sqrt(p1(1-p1)/n1 + p2(1-p2)/n2)
So, the standard error (SE) for our example is:
SE = sqrt(0.28(1-0.28)/200 + 0.18(1-0.18)/200) = sqrt(0.001248 + 0.000792) = sqrt(0.00204)
The Z-score for a 95% confidence interval is approximately 1.96. Therefore, the margin of error (MOE) is calculated as:
MOE = 1.96 * SE
We then compute the confidence interval as:
(p1 - p2) ± MOE = (0.28 - 0.18) ± 1.96 * sqrt(0.00204)
Calculate the MOE:
MOE = 1.96 * sqrt(0.00204) ≈ 1.96 * 0.0452 ≈ 0.0886
Then the 95% confidence interval is:
(0.10 ± 0.0886) which is (0.0114, 0.1886)
This means that we are 95% confident that the true difference in the proportion of people who would donate when contacted by telephone versus first class mail is between 1.14% and 18.86%.
tems produced by a manufacturing process are supposed to weigh 90 grams. However, there is variability in the items produced, and they do not all weigh exactly 90 grams. The distribution of weights can be approximated by a Normal distribution with a mean of 90 grams and a standard deviation of 1 gram. What percentage of the items will either weigh less than 87 grams or more than 93 grams? Group of answer choices
Based on the properties of the Normal Distribution and Standard Deviations, we find that about 0.3% of the items will weigh either less than 87 grams or more than 93 grams.
Explanation:To solve this question, we need to understand and apply concepts of the Normal Distribution and Standard Deviations. A normal distribution is a common type of statistical distribution representing various types of data, characterized by a bell-shaped curve symmetric about its mean.
In this case, the item weights are normally distributed with a mean (average) of 90 grams and a standard deviation of 1 gram. Standard deviation basically tells us how much the data varies around the mean value.
The question now asks us to find out what percentage of weights are either less than 87 grams or more than 93 grams. We know that within 1 standard deviation of the mean (between 89 and 91 grams in this case), we have about 68.2% of the data. Similarly, within 2 standard deviations of the mean (between 88 and 92 grams), we have about 95.4% of the data. And within 3 standard deviations (between 87 and 93 grams), we have about 99.7% of the data.
So, since 87 grams and 93 grams are 3 standard deviations away from the mean, we know that about 99.7% of the weights lie between these two values. This means that the other 0.3%, or 0.15% on either side, is either below 87 grams or above 93 grams. Hence, 0.3% of the items will either weigh less than 87 grams or more than 93 grams.
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Suppose that there are six prospective jurors, four men and two women, who might be impaneled to sit on the jury in a criminal case. Two jurors are randomly selected from these six to fill the two remaining jury seats.
a. List the simple events in the experiment
b. What is the probability that both impaneled jurors are women?
Exercise:
Jury Duty Three people are randomly selected from voter registration and driving records to report for jury duty. The gender of each person is noted by the county clerk.
a. Define the experiment.
b. List the simple events in S.
c. If each person is just as likely to be a man as a woman, what probability do you assign to each simple event?
d. What is the probability that only one of the three is a man?
e. What is the probability that all three are women?
Answer:
(1)
(a) Shown below.
(b) The probability that both impaneled jurors are women is 0.0667.
(2)
(a) Sampling 3 people and noting their gender.
(b) Shown below.
(c) The probability of each simple event is 0.125.
(d) The probability of selecting only one male is 0.375.
(e) The probability of selecting all 3 females is 0.125.
Step-by-step explanation:
(1)
Let the 4 men be denoted as: M₁, M₂, M₃ and M₄.
And the 2 women be denoted as: W₁ and W₂.
(a)
A jury of two is to be selected.
The simple events in this experiment are:
(M₁, M₂), (M₁, M₃), (M₁, M₄), (M₁, W₁), (M₁, W₂)
(M₂, M₃), (M₂, M₄), (M₂, W₁), (M₂, W₂)
(M₃, M₄), (M₃, W₁), (M₃, W₂)
(M₄, W₁), (M₄, W₂)
(W₁, W₂)
(b)
The total possible number of jury selections is, N = 15.
The possible combination such that both the jurors are woman is, n = 1.
Compute the probability of selecting two women jurors as follows:
[tex]P(2\ juror\ are\ women)=\frac{n}{N} =\frac{1}{15}=0.0667[/tex]
Thus, the probability that both impaneled jurors are women is 0.0667.
(2)
(a)
The experiment consists of sampling 3 people from the voter registration and driving records and noting the gender of each person.
(b)
The simple events are:
S = {(M, M, M), (M, M, F), (M, F, M), (F, M, M), (M, F, F), (F, F, M), (F, M, F), (F, F, F)}
Total number of simple events = 8.
(c)
If the probability of selecting a male is same as the probability of selecting a female, i.e. P (M) = P (F) = [tex]\frac{1}{2}[/tex] then,
The probability of each simple event is:
[tex]\frac{1}{2} \times\frac{1}{2} \times\frac{1}{2}=\frac{1}{8}=0.125[/tex]
(d)
The number of simple events with only 1 male is n = 3.
Compute the probability that only one of the three is a man as follows:
[tex]P(1\ male)=\frac{n}{N} =\frac{3}{8} =0.375[/tex]
Thus, the probability of selecting only one male is 0.375.
(e)
The number of simple events with all 3 females is n = 1.
Compute the probability that all 3 females are selected as follows:
[tex]P(3\ female)=\frac{n}{N} =\frac{1}{8} =0.125[/tex]
Thus, the probability of selecting all 3 females is 0.125.
There are 98 balls in a box. 25 of them are red, 19 of them are green, 30 of them are purple,and 24 of them are blue. Suppose Hao draws 22 balls from the box with replacement (hedraws the ball, records its color, and then puts it back into the box). Find the probabilitythat he draws 2 red balls, 5 green balls, 10 purple balls, and 5 blue balls.
Answer:
Since the balls are drawn with replacement, it means the individual probability remain constant,
I have solved this problem on paper (Figures Attached).
Thanks.
Answer:
Since this problem related to the replacement problem thus
Pr(2 red)=(25/98)(25/98)=0.0651
Pr(5 green)=(19/98)(19/98)(19/98)(19/98)(19/98)
Pr(5 green)=2.73*10^-4
Pr(10 purple)=((30/98))^10=7.22*10^-6
Pr(5 blue)=((24/98))^5=8.8*10^-4
The distribution of water fleas (Daphnia) in a given water pond is fairly random and the population density is fairly constant. The average number of water fleas caught by sweeping the water a single time with a standard net is 3.7 individuals. If tomorrow a net will be used once in the pond what is the probability of catching: a) 5 individuals?b) at least 2 individuals?
Answer:
(a) The probability of catching 5 individuals in the pond is 0.1429.
(b) The probability of catching at least 2 individuals in the pond is 0.8838.
Step-by-step explanation:
Let X = number of water fleas caught by sweeping the water a single time.
The random variable X follows a Poisson distribution with parameter λ = 3.7.
The probability mass function of the Poisson distribution is:
[tex]P(X=x)=\frac{e^{-3.7}3.7^{x}}{x!}[/tex]
(a)
Compute the value of P (X = 5) as follows:
[tex]P(X=5)=\frac{e^{-3.7}3.7^{5}}{5!}=\frac{17.1443}{120} =0.142869\approx0.1429[/tex]
Thus, the probability of catching 5 individuals in the pond is 0.1429.
(b)
Compute the value of P (X ≥ 2) as follows:
P (X ≥ 2) = 1 - P (X < 2)
= 1 - P (X = 0) - P (X = 1)
[tex]=1-\frac{e^{-3.7}3.7^{0}}{0!}-\frac{e^{-3.7}3.7^{1}}{1!}\\=1-0.0247-0.0915\\=0.8838[/tex]
Thus, the probability of catching at least 2 individuals in the pond is 0.8838.
In the final round of a TV game show, contestants have a chance to increase their current winnings of $1 million to $2 million. If they are wrong, their prize is decreased to $500,000. A contestant thinks his guess will be right 50 percent of the time. Should he play
Answer:
The contestant should not play.Step-by-step explanation:
As per the given question, their current winning is $1 million.
The probability of the guessing to be true is 50% = [tex]\frac{50}{100} = \frac{1}{2}[/tex].
There is also a possibility of 50% to be wrong, which can reduced the winning amount to $500,000 that is the half of the current amount.
Hence, the contestant should not play.
A brewery's filling machine is adjusted to fill bottles with a mean of 32.7 oz. of ale and a variance of 0.003. Periodically, a bottle is checked and the amount of ale noted.
(a) Assuming the amount of fill is normally distributed, what is the probability that the next randomly checked bottle contains more than 32.73 oz? (Give your answer correct to four decimal places.)
(b) Let's say you buy 95 bottles of this ale for a party. How many bottles would you expect to find containing more than 32.73 oz. of ale? (Round your answer up to the nearest whole number.) bottles You may need to use the appropriate table in Appendix B to answer this question.
Answer:
(a) P(X>32.73) = 0.2912
(b) Out of 95 bottles, 28 would contain more than 32.73 oz of ale.
Step-by-step explanation:
(a) The amount of fill is normally distributed so we will calculate the z-score and then use it to find the probability using the normal distribution probability table.
Let the amount of fill be denoted by X. The z-score can be computed using the formula:
z = (X - μ)/σ
where μ = mean value of fill
σ = standard deviation of value of fill = √Variance
P(X>32.73) = 1 - P(X<32.73)
= 1 - P((X-μ)/σ < (32.73 - 32.7)/√0.003)
= 1 - P(z<0.55)
Using the normal distribution table in Appendix B, we can see the probability at z=0.55 is 0.7088. So,
P(X>32.73) = 1 - 0.7088
P(X>32.73) = 0.2912
(b) We are buying 95 bottles and we need to calculate how many of them contain more than 32.73 oz. For that, we will multiply the total number of bottles by the probability of finding more than 32.73 oz which we have calculated in (a).
95 * 0.2912 = 27.664
Rounding off to a whole number we get 28 bottles.
Out of 95 bottles, 28 would contain more than 32.73 oz of ale.
The number and frequency of Atlantic hurricanes annually from 1940 through 2007 is shown here:
a. Find the probabilities of 0-8 hurricanes each season using these data
b. Assuming a Poisson distribution and using the mean number of hurricanes per season from the empirical data, compute the probabilities of experiencing 0-8 hurricanes in a season.
Compare these to your answer to part (a).
How good does a Poisson distribution model this phenomenon?
Number Frequency
0 5
1 16
2 19
3 13
4 3
5 5
6 4
7 2
8 1
Total 68
Answer:
The probability table is shown below.
A Poisson distribution can be used to approximate the model of the number of hurricanes each season.
Step-by-step explanation:
(a)
The formula to compute the probability of an event E is:
[tex]P(E)=\frac{Favorable\ no.\ of\ frequencies}{Total\ NO.\ of\ frequencies}[/tex]
Use this formula to compute the probabilities of 0 - 8 hurricanes each season.
The table for the probabilities is shown below.
(b)
Compute the mean number of hurricanes per season as follows:
[tex]E(X)=\frac{\sum x f_{x}}{\sum f_{x}}=\frac{176}{68}= 2.5882\approx2.59[/tex]
If the variable X follows a Poisson distribution with parameter λ = 7.56 then the probability function is:
[tex]P(X=x)=\frac{e^{-2.59}(2.59)^{x}}{x!} ;\ x=0, 1, 2,...[/tex]
Compute the probability of X = 0 as follows:
[tex]P(X=0)=\frac{e^{-2.59}(2.59)^{0}}{0!} =\frac{0.075\times1}{1}=0.075[/tex]
Compute the probability of X = 1 as follows:
[tex]\neq P(X=1)=\frac{e^{-2.59}(2.59)^{1}}{1!} =\frac{0.075\times7.56}{1}=0.1943[/tex]
Compute the probabilities for the rest of the values of X in the similar way.
The probabilities are shown in the table.
On comparing the two probability tables, it can be seen that the Poisson distribution can be used to approximate the distribution of the number of hurricanes each season. This is because for every value of X the Poisson probability is approximately equal to the empirical probability.
To address this question, probabilities from the given data are calculated first, after which Poisson distribution is applied to compute the probabilities using the mean number of hurricanes per season. A comparison of both results offers an insight into the accuracy of the Poisson distribution in modeling this phenomenon.
Explanation:To answer this question, we first have to calculate probabilities based on the empirical data and then compare them to probabilities computed under the assumption of a Poisson distribution.
First, we count total seasons from 1940 through 2007, which is 68. Using these frequencies, we can calculate the probability of having 0-8 hurricanes each season as follows: the number of seasons with a certain number of hurricanes divided by the total number of seasons.
For the Poisson distribution, we first need to calculate the average (mean) number of hurricanes per season, which is the sum of the product of the number of hurricanes and its frequency divided by the total number of seasons. After finding the mean, we can compute the probability of experiencing 0-8 hurricanes using the Poisson formula: e^(-mean) * (mean^n) / n!. After that, we can compare these probabilities with the ones derived from the empirical data.
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