Final answer:
The probability of selecting the color of the first ball is 3/10. The probability of selecting the same color for the second ball depends on the remaining number of balls of that color divided by the total remaining number of balls.
Explanation:
The probability of selecting the color of the first ball is determined by the number of balls of that color divided by the total number of balls. In this case, there are 3 red balls out of 10, so the probability is 3/10.
After selecting the first ball without replacement, the probability of selecting the same color for the second ball depends on the remaining number of balls of that color divided by the total remaining number of balls. If the first ball was red, there are 2 red balls left out of 9, resulting in a probability of 2/9.
The probability of selecting a ball of the same color as the first ball drawn, without replacement, is as follows:
For red:[tex]\boxed{0.22}[/tex]
For white:[tex]\boxed{0.33}[/tex]
For blue:[tex]\boxed{0.22}[/tex]
Let's think step by step.To solve this problem, we need to calculate the probability of selecting a ball of the same color as the first ball drawn, without replacement. We will consider the two events separately: first, the probability of selecting a ball of the same color as the first ball when there are 10 balls in total, and second, the probability of selecting a ball of the same color as the first ball when there are only 9 balls left in the bucket (since the first ball is not replaced).
First Ball Selection:
When the first ball is picked, there are 10 balls in total, with 3 red, 4 white, and 3 blue balls. The probability of picking a ball of any specific color is the number of balls of that color divided by the total number of balls.
For example, if the first ball drawn is red, the probability of drawing a red ball is:
[tex]\[ P(\text{first red}) = \frac{\text{Number of red balls}}{\text{Total number of balls}} = \frac{3}{10} \][/tex]
Similarly, if the first ball is white, the probability is:
[tex]\[ P(\text{first white}) = \frac{\text{Number of white balls}}{\text{Total number of balls}} = \frac{4}{10} \][/tex]
And if the first ball is blue, the probability is:
[tex]\[ P(\text{first blue}) = \frac{\text{Number of blue balls}}{\text{Total number of balls}} = \frac{3}{10} \][/tex]
Second Ball Selection:
After the first ball is drawn and not replaced, there are 9 balls left in the bucket.
If the first ball drawn was red, there are now 2 red balls left out of 9 total balls. The probability of drawing another red ball is:
[tex]\[ P(\text{second red} | \text{first red}) = \frac{\text{Number of red balls left}}{\text{Total number of balls left}} = \frac{2}{9} \][/tex]
If the first ball was white, there are now 3 white balls left out of 9 total balls. The probability of drawing another white ball is:
[tex]\[ P(\text{second white} | \text{first white}) = \frac{\text{Number of white balls left}}{\text{Total number of balls left}} = \frac{3}{9} = \frac{1}{3} \][/tex]
If the first ball was blue, there are now 2 blue balls left out of 9 total balls. The probability of drawing another blue ball is:
[tex]\[ P(\text{second blue} | \text{first blue}) = \frac{\text{Number of blue balls left}}{\text{Total number of balls left}} = \frac{2}{9} \][/tex]
Now, let's calculate the probabilities for each color and round them to two decimal places:
For red:
[tex]\[ P(\text{second red} | \text{first red}) = \frac{2}{9} \approx 0.22 \][/tex]
For white:
[tex]\[ P(\text{second white} | \text{first white}) = \frac{1}{3} \approx 0.33 \][/tex]
For blue:
[tex]\[ P(\text{second blue} | \text{first blue}) = \frac{2}{9} \approx 0.22 \][/tex]
Final
The probability of selecting a ball of the same color as the first ball drawn, without replacement, is as follows:
For red:[tex]\boxed{0.22}[/tex]
For white:[tex]\boxed{0.33}[/tex]
For blue:[tex]\boxed{0.22}[/tex]
These probabilities are rounded to two decimal places.
The answer is: 0.22.
A group of students bakes 100 cookies to sell at the school bake sale. The students want to ensure that the price of each cookie offsets the cost of the ingredients. If all the cookies are sold for $0.10 each, the net result will be a loss of $4. If all the cookies are sold for $0.50 each. The students will make a $36 profit. First, write the linear function p(x) that represents the net profit from selling all the cookies, where x is the price of each cookie. Then, determine how much profit the students will make if they sell the coolies for $0.60 each. Explain. Tell how your answer is reasonable.
Answer:
(a) [tex]p=100x-14[/tex]
(b) [tex]p=\$46[/tex]
Step-by-step explanation:
Linear Modeling
It consists of finding an equation of a line that fits the conditions of a certain situation in real life. We'll use a linear model for the cookies of the students.
(a) We know that we have a total of n=100 cookies. If sold for $0.10 each, they lose $4. We have an initial condition (x,p) = (0.10,-4), where x is the price of each cookie and p(x) is the net profit from selling all the cookies. The second conditions are that when then the price is $0.50 each, there is a positive profit of $36, which is a second point (0.5,36). That is enough to build the linear function, that can be found by
[tex]\displaystyle p-p_1=\frac{p_2-p_1}{x_2-x_1}(x-x_1)[/tex]
[tex]\displaystyle p+4=\frac{36+4}{0.5-0.1}(x-0.1)[/tex]
Reducing
[tex]p=100x-14[/tex]
(b) If the students sell the cookies for x=0.60 each, the profit will be
[tex]p=100(0.6)-14=46[/tex]
[tex]p=\$46[/tex]
It's a reasonable answer because we have found that increasing the price, the profit will increase also. The model doesn't have any restriction for the price
A plane delivers two types of cargo between two destinations. Each crate of cargo I is 9 cubic feet in volume and 187 pounds in weight, and earns $30 in revenue. Each crate of cargo II is 9 cubic feet in volume and 374 pounds in weight, and earns $45 in revenue. The plane has available at most 540 cubic feet and 14,212 pounds for the crates. Finally, at least twice the number of crates of I as II must be shipped. Find the number of crates of each cargo to ship in order to maximize revenue. Find the maximum revenue. crates of cargo I crates of cargo II maximum revenue $
Answer:
So maximum when 46 of I grade and 16 of II grade are produced.
Max revenue = 2100
Step-by-step explanation:
Given that a plane delivers two types of cargo between two destinations
Crate I Crate II
Volume 9 9
Weight 187 374
Revenue 30 45
Let X be the no of crate I and y that of crate II
Then
[tex]9x+9y\leq 540\\187x+374y\leq 14212\\x\geq 2y[/tex]
Simplify these equations to get
[tex]x+y\leq 60\\x+2y\leq 76\\x\geq 2y[/tex]
Solving we get
[tex]y\leq 16\\x\leq 46 and x\geq 32\\32\leq x\leq 46[/tex]
REvenue = 30x+45y
The feasible region would have corner points as (60,0) or (32,16) or (46,16)
Revenue for (60,0) = 1800
(32,16) = 1680
(46,16)=2100
So maximum when 46 of I grade and 16 of II grade are produced.
Max revenue = 2100
To maximize revenue, we need to determine the number of crates of each cargo that should be shipped. The problem can be solved using linear programming techniques to find the optimal solution.
Explanation:To maximize revenue, we need to determine the number of crates of each cargo that should be shipped. Let's assume the number of crates of cargo I is x and the number of crates of cargo II is y.
Based on the given information, the constraints for the problem are:
Volume constraint: 9x + 9y ≤ 540Weight constraint: 187x + 374y ≤ 14,212Relationship constraint: x ≥ 2yTo find the maximum revenue, we need to maximize the objective function: Revenue = 30x + 45y.
The problem can be solved using linear programming techniques, such as graphical or simplex method, to find the optimal solution. However, since these methods require plotting and iterations, the detailed calculations are beyond the scope of this response. The optimal solution will provide the values of x and y, which can be used to determine the maximum revenue.
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Private colleges and universities rely on money contributed by individuals and corporations for their operating expenses. Much of this money is put into a fund called an endowment, and the college spends only the interest earned by the fund. A recent survey of 8 private colleges in the United States revealed the following endowments (in millions of dollars): 60.2, 47.0, 235.1, 490.0, 122.6, 177.5, 95.4, and 220.0. Summary statistics yield Upper X overbarequals180.975 and Sequals143.042. Calculate a 95% confidence interval for the mean endowment of all the private colleges in the United States assuming a normal distribution for the endowments.
Answer:
[tex]180.975 - 2.365\frac{143.042}{\sqrt{8}}=61.370[/tex]
[tex]180.975 + 2.365\frac{143.042}{\sqrt{8}}=300.580[/tex]
So on this case the 95% confidence interval would be given by (61.370;300.580)
Step-by-step explanation:
Previous concepts
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".
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".
Solution to the problem
[tex]\bar X=180.975[/tex] represent the sample mean
[tex]\mu[/tex] population mean (variable of interest)
[tex]s=143.042[/tex] represent the sample standard deviation
n=8 represent the sample size
The confidence interval on this case is given by:
[tex]\bar X \pm t_{\alpha/2} \frac{s}{\sqrt{n}} [/tex] (1)
We can find the degrees of freedom and we got:
[tex] df = n-1= 8-1=7[/tex]
The next step would be find the value of [tex]\t_{\alpha/2}[/tex], [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2=0.025[/tex]
Using the t table with df =7, excel or a calculator we see that:
[tex]t_{\alpha/2}=2.365[/tex]
Since we have all the values we can replace:
[tex]180.975 - 2.365\frac{143.042}{\sqrt{8}}=61.370[/tex]
[tex]180.975 + 2.365\frac{143.042}{\sqrt{8}}=300.580[/tex]
So on this case the 95% confidence interval would be given by (61.370;300.580)
A climber on Mount Everest is meters from the start of his trail and at elevation meters above sea level. At meters from the start, the elevation of the trail is meters above sea level. If for near , what is the approximate elevation another 8 meters along the trail
We are not given the value of
Mount Everest
Elevation
h'(x)
So we are going to based our parameters needed to solve this question o assumptions. The main thing is to understand the process in solving this question.
So here is it!.
A climber on Mount Everest is 8000 meters from the start of his trail and at elevation 10000 meters above sea level. At (x) meters from the start, the elevation of the trail is h(x) meters above sea level. If h' (x) = 0.5 for near , what is the approximate elevation another 8 meters along the trail
Answer:
10004 meters
Step-by-step explanation:
The rate of change of elevation, h' (x) = 0.5 near 8000 meters.
Thus; we can say that the elevation increases by 0.5 for each meter traveled at a given distance(x)
So, we need to determine the new elevation after 8 meters traveled from ( 8000 to 8008).Then the elevation change can now be written as:
= (0.5 × 8)
= 4 meters.
And also the new elevation will be:
10000 + 4 meters
= 10004 meters
Of the 50 states, 39 are currently under court order to alleviate overcrowding and poor conditions in one or more of their prisons. If a state is selected at random, find the probability that it is not currently under such a court order. Give your answer as a reduced fraction.
Answer:
11/50
Step-by-step explanation:
39 of 50 states are under court order, so 11 of 50 states are not under court order.
Final answer:
The probability that a randomly selected state is not under a court order to alleviate overcrowding and poor conditions in its prisons is 11/50.
Explanation:
To find the probability that a randomly selected state is not currently under court order to alleviate overcrowding and poor conditions in one or more of their prisons, we can use the complement rule in probability. This rule states that the probability of an event not happening is equal to one minus the probability of the event happening. Given that 39 states are under such a court order, the probability that a state is under court order is 39/50. Therefore, the probability that a state is not under court order is 1 - (39/50).
Calculating this, we get:
P(Not under court order) = 1 - (39/50) = (50/50) - (39/50) = 11/50
The probability that a randomly selected state is not currently under a court order to alleviate overcrowding and poor conditions in its prisons is 11/50, when expressed as a reduced fraction.
According to one theory of learning, the number of items, w(t), that a person can learn after t hours of instruction is given by: w(t) = 15 3 t2, 0 ≤ t ≤ 64 Find the rate of learning at the end of eight hours of instruction.
Answer:
The rate of study is 5 items per hour.
Step-by-step explanation:
Number of items a person can learn after t hours of instruction, w(t) is given by:
[tex]w(t)=15\sqrt[3]{t^{2}}[/tex]
We want to determine the rate of learning at any time t. The rate is the derivative of w(t) with respect to time.
[tex]\frac{dw(t)}{dt} =\frac{d}{dt} 15\sqrt[3]{t^{2}}[/tex]
[tex]\frac{dw(t)}{dt} =15\frac{d}{dt} {t^{2/3}}[/tex]
[tex]\frac{dw(t)}{dt} =15X\frac{2}{3} {t^{2/3-1}}[/tex]
[tex]\frac{dw(t)}{dt} =10 {t^{-\frac{1}{3} }}=\frac{10}{t^\frac{1}{3}}[/tex]
Therefore, the rate of learning at any time t
[tex]\frac{dw(t)}{dt} =\frac{10}{t^\frac{1}{3}}[/tex]
At the end of 8 hours, t=8
[tex]\frac{dw(t)}{dt} =\frac{10}{8^\frac{1}{3}}[/tex]
[tex]\frac{dw(t)}{dt} =\frac{10}{2}[/tex]=5
The rate of study is 5 items per hour.
For its first year of operations, Marcus Corporation reported pretax accounting income of $274,800. However, because of a temporary difference in the amount of $19,200 relating to depreciation, taxable income is only $255,600. The tax rate is 39%. What amount should Marcus report as its deferred income tax liability in its balance sheet at the end of that year
Answer:
Please find attached file for complete answer solution and explanation of same question
Step-by-step explanation:
Answer:
$374,484
Step-by-step explanation:
Amount payable as income tax = 0.39 X $255,600 = $99,684
The amount Marcus receives if he deferred income tax that year = $(274,800 + 99,684) = $374,484
A new article reported that college students who have part-time jobs work an average of 15 hour per week. The staff of a college from for newspaper thought that the average might be different from 15 hours per week for their college. Data were collected on the number of hours worked per week for a random sample of students at the college who have part-time jobs. The data were used to test the hypotheses H_o: mu = 15 H_a: mu notequalto 15. where mu is the mean number of hours worked per week for all students at the college with part-time jobs. The results of the test are shown in the table below. Assuming all conditions for inference were met, which of the following represents a 95 percent confidence interval for mu?
(A) 13.755 plusminus 0.244
(B) 13.755 plusminus 0.286
(C) 13.755 plusminus 0.707
(D) 13.755 plusminus 1.245
E) 13, 755 plusminus 1.456
Answer:
E (13.755 ± 1.456)
Step-by-step explanation:
with 95% confidence interval and df=25
we can find t score on the t table : t*=2.06
(because the table has already provided us standard error , we DON'T have to calculate it by using σsample=S/[tex]\sqrt{n}[/tex]) (standard error = standard deviation)
so the interval should be 13.755± 2.06(0.707)= 13.755± 1.456
The 95% confidence interval for [tex]\(\mu\)[/tex] is e. [tex]\(13.755 \pm 1.456\)[/tex], based on a sample mean of 13.755, standard error of 0.707, and 25 degrees of freedom.
To construct a 95% confidence interval for the mean [tex]\(\mu\)[/tex], we need to use the sample mean, the standard error, and the appropriate critical value for a t-distribution.
Given data:
- Sample mean [tex](\(\bar{x}\))[/tex]: 13.755
- Standard error (SE): 0.707
- Degrees of freedom (df): 25
For a 95% confidence interval with 25 degrees of freedom, we need the critical value [tex]\( t^* \).[/tex] We can find [tex]\( t^* \)[/tex] using a t-table or statistical software.
For 25 degrees of freedom, the critical value [tex]\( t^* \)[/tex] for a 95% confidence interval is approximately 2.064.
The formula for the confidence interval is:
[tex]\[\bar{x} \pm t^* \times \text{SE}\][/tex]
Substitute the values:
[tex]\[13.755 \pm 2.064 \times 0.707\][/tex]
Calculate the margin of error:
[tex]\[2.064 \times 0.707 \approx 1.459\][/tex]
Thus, the 95% confidence interval is:
[tex]\[13.755 \pm 1.459\][/tex]
So, the correct answer is:
e. [tex]\[\boxed{13.755 \pm 1.456}\][/tex]
Complete Question:
A developmental psychologist is i nterested in assessing t he "" emotional i ntelligence"" o f co llege s tudents. The e xperimental des ign ca lls f or a dministering a questionnaire that measures emotional intelligence to a sample of 100 undergraduate student volunteers who are enrolled in an introductory psychology course currently b eing t aught at her u niversity. A ssume t his is the only sample being used for this study and discuss the adequacy of the sample
Answer:
In terms of size it is adequate . Sample isn't randomly selected
Step-by-step explanation:
In order for the sampling ot be adequate the sample size must be large enough which in this case it is. The samples must also be randomly selected. Here, the sample includes students from one course only. In order for the study to represent whole population that is college students, students from other courses must also be included. Junior as well as senior students must also be part of this study.
Two balanced dice are rolled. Let X be the sum of the two dice. Obtain the probability distribution of X (i.e. what are the possible values for X and the probability for obtaining each value?). Check that the probabilities sum to one.
1) What is the probability for obtaining X \geq 8?
2) What is the average value of X?
Answer:
Step-by-step explanation:
Each die has faces numbered from 1 to 6. Hence, the sums range from 2 to 12.
There are 6 × 6 = 36 possible combinations of the dice.
The sums, their possible combinations and their probabiities are as below:
2 = 1+1 1/36
3 = 1+2 = 2+1 2/36= 1/18
4 = 1+3 = 2+2 = 3+1 3/36= 1/12
5 = 1+4 = 2+3 = 3+2 = 4+1 4/36= 1/9
6 = 1+5 = 2+4 = 3+3 = 4+2 = 5+1 5/36
7 = 1+6 = 2+5 = 3+4 = 4+3 = 5+2 = 6+1 6/36= 1/6
8 = 2+6 = 3+5 = 4+4 = 5+3 = 6+2 5/36
9 = 3+6 = 4+5 = 5+4 = 6+3 4/36= 1/9
10 = 4+6 = 5+5 = 6+4 3/96= 1/12
11 = 5+6 = 6+5 2/36= 1/18
12 = 6+6 1/36
Sum of probabilities = 1/36 + 1/18 + 1/12 + 1/9 + 5/36 + 1/6 + 5/36 + 1/9 + 1/12 + 1/18 + 1/36 = 1
1) [tex]P(X\ge 8) = 5/36+1/9+1/12+1/18+1/36=15/36[/tex]
2) The average value of X = 1/36*(2+12) + 1/18*(3+11) + 1/12*(4+10)+ 1/9(5+9)+ 5/36(6+8)+ 1/6(7) = 14*(5/6)+7/6 = 77/6
Department 1 of a two department production process shows: Units Beginning Work in Process 9900 Ending Work in Process 49000 Total units to be accounted for 180200 How many units were transferred out to Department 2?
Answer:
131200 Units
Step-by-step explanation:
Now, in the production process:
Units to be accounted for= Beginning Work in Process+Work Started
Units Accounted for=Ending Work in Process + Completed and Transferred out
Units to be accounted for = Units Accounted for
Beginning Work in Process =9900 Ending Work in Process = 49000 Total units to be accounted for= 180200
Since
Units to be accounted for = Units Accounted for
Then:
Units Accounted for=Ending Work in Process + Completed and Transferred out
180200=49000+Completed and Transferred out
Completed and Transferred out=180200-49000
=131200
131200 units were transferred to Department 2.
The number of units transferred out to Department 2 can be found by subtracting the Ending Work in Process from the Total units to be accounted for. In this case, that would be 180200 - 49000, which equals 131200 units.
Explanation:In the scenario described, the number of units transferred out to Department 2 can be calculated by deducting the Ending Work in Process from the Total units to be accounted for. This is because the Total units to be accounted for includes all units at the start (Beginning Work in Process), all units processed during the period, and all units still in process at the end (Ending Work in Process). So, the units transferred to Department 2 were processed by Department 1 but are not part of Department 1's ending work in process.
Hence, using the provided numbers:
Total units to be accounted for = 180200Ending Work in Process = 49000
Units transferred out to Department 2 = Total units to be accounted for - Ending Work in Process = 180200 - 49000 = 131200
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Starting from rest, a DVD steadily accelerates to 500 rpm in 1.0 s, rotates at this angular speed for 3.0 s, then steadily decelerates to a halt in 2.0 s. How many revolutions does it make
Answer:
37.5 revolutions
Step-by-step explanation:
The average rotation speed for the first second and for the last two seconds is:
[tex]V_1 = \frac{0+500}{2}\\ V_1 = 250\ rpm[/tex]
For the next 3.0 seconds, the rotation speed is V = 500 rpm.
The total number of revolutions, converting rpm to rps, is given by:
[tex]n=\frac{1*V_1+3*V+2*V_1}{60}\\n=\frac{1*250+3*500*2*250}{60}\\n=37.5\ revolutions[/tex]
The DVD makes 37.5 revolutions.
Suppose you work for Fender Guitar Company and you are responsible for testing the integrity of a new formulation of guitar strings. To perform your analysis, you randomly select 52 'high E' strings and put them into a machine that simulates string plucking thousands of times per minute. You record the number of plucks each string takes before failure and compile a dataset. You find that the average number of plucks is 5,314.4 with a standard deviation of 116.68. A 90% confidence interval for the average number of plucks to failure is (5,287.3, 5,341.5).
From the option listed below, what is the appropriate interpretation of this interval?
1) We are 90% confident that the average number of plucks to failure for all 'high E' strings tested is between 5,287.3 and 5,341.5
2) We cannot determine the proper interpretation of this interval.
3) We are 90% confident that the proportion of all 'high E' guitar strings fail with a rate between 5,287.3 and 5,341.5
4) We are certain that 90% of the average number of plucks to failure for all 'high E' strings will be between 5,287.3 and 5,341.5 5
5) We are 90% confident that the average number of plucks to failure for all 'high E' strings is between 5,287.3 and 5,341.5
Answer:
[tex]5314.4-1.675\frac{116.68}{\sqrt{52}}=5287.30[/tex]
[tex]5314.4+1.675\frac{116.68}{\sqrt{52}}=5341.5[/tex]
So on this case the 90% confidence interval would be given by (5287.3;5341.5)
And the best intrpretation is:
5) We are 90% confident that the average number of plucks to failure for all 'high E' strings is between 5,287.3 and 5,341.5
Step-by-step explanation:
Previous concepts
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
[tex]\bar X=5314.4[/tex] represent the sample mean
[tex]\mu[/tex] population mean (variable of interest)
[tex]s=116.68[/tex] represent the sample standard deviation
n=52 represent the sample size
90% confidence interval
The confidence interval for the mean is given by the following formula:
[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)
The degrees of freedom are given by:
[tex] df = n-1= 52-1=51[/tex]
Since the Confidence is 0.90 or 90%, the value of [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.05[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,51)".And we see that [tex]t_{\alpha/2}=1.675[/tex]
Now we have everything in order to replace into formula (1):
[tex]5314.4-1.675\frac{116.68}{\sqrt{52}}=5287.30[/tex]
[tex]5314.4+1.675\frac{116.68}{\sqrt{52}}=5341.5[/tex]
So on this case the 90% confidence interval would be given by (5287.3;5341.5)
And the best intrpretation is:
5) We are 90% confident that the average number of plucks to failure for all 'high E' strings is between 5,287.3 and 5,341.5
A penny is tossed. We observe whether it lands heads up or tails up. Suppose the penny is a fair coin; that is, the probability of heads is one-half and the probability of tails is one-half. What does this mean?A) if I flip the coin many, many times, the proportion of heads will be approximately 1/2, and this proportion will tend to
get closer and closer to 1/2 as the number of tosses increases.
B) regardless of the number of flips, half will be heads and half tails.
C) every occurrence of a head must be balanced by a tail in one of the next two or three tosses.
D) all of the above.
Answer:
So we conclude that the answer is under (B).
Step-by-step explanation:
We know that the penny is a fair coin; that is, the probability of heads is one-half and the probability of tails is one-half.
So when we throw a coin we have an equal chance of getting either a head or a tail.
So we conclude that the answer is under (B).
B) regardless of the number of flips, half will be heads and half tails.
Use seq() to generate a sequence 1, 3, ..., 27. (b) Use log() to generate a new sequence where each element is log-transformed from the sequence in (a). (c) Remove the second to fifth elements in the resulting sequence in (b). (d) Use length() to obtain the length of the resulting sequence in (c). (e) Sort the resulting sequence in (c) from high to low using sort().
Answer:
Step-by-step explanation:
sequence=seq(1,27, by=3)
new_seq=c()
for (i in sequence){
n_seq.append(log(i))
}
new_seq<- n_seq[-(2:5)]
seq_length=length(new_seq)
sorted_seq=sort(new_seq,decreasing=TRUE)
Some parts of California are particularly earthquake-prone. Suppose that in one metropolitan area, 31% of all homeowners are insured against earthquake damage. Four homeowners are to be selected at random. Let X denote the number among the four who have earthquake insurance.
(a) Find the probability distribution of X. [Hint: Let S denote a homeowner that has insurance and F one who does not. Then one possible outcome is SFSS, with probability (0.31)(0.69)(0.31)(0.31) and associated X value 3. There are 15 other outcomes.] (Round your answers to four decimal places.)
(b) What is the most likely value for X?
(c) What is the probability that at least two of the four selected have earthquake insurance? (Round your answer to four decimal places.)
Answer:
a.) 0.0822
b.) 1
c.) 0.3659
Step-by-step explanation:
Probability distribution formula is often denoted by :
P(X=r) = nCr × p^r × q^n-r
Where n = total number of samples
r = number of successful outcome of sample
p = probability of success
q = probability of failure.
If we take 4 samples,
3 of this 4 samples are successful
the success rate =S= 31% = 0.31
Failure rate = F= 0.69
a.) Then probability distribution of X becomes:
P(X=3) = 4C3 × 0.31³ × 0.69¹
P(X=3) = 0.0822 (4d.p),
b. Most likely value of X = expected value = np
= 4 × 0.31
= 1.24 ≈ 1
c.) probability that at least 2 out of the 4 have insurance = Probability that 2 have insurance) + probability that 3 have insurance + probability that 4 have insurance.
P(X=2) = 4C2 × 0.31² × 0.69² = 0.2745
P(X=3), as calculated earlier = 0.0822
P(X=4) = 4C4 × 0.31^4 × 0.69^0 = 0.0092
Total probability of having at least 2 out of those 4 insured = 0.2745 + 0.0822 + 0.0092 =0.3659.
A survey is conducted of students enrolled during both the Winter 2013 and Winter 2014 semesters. 12% of students got the flu during Winter 2013. 18% of students came got the flu during Winter 2014. 5% of students got the flu during both years. What is the probability that a randomly selected survey participant didnt?
Answer:
The probability that a randomly selected survey participant didn't got the flew = 0.75
Step-by-step explanation:
Students got the flu during Winter 2013 = 12 %
Students got the flu during Winter 2014 = 18 %
Students got the flu during both years = 5 %
Students got the flu at least in one year = 12 + 18 - 5 = 25 %
Students didn't got the flew = 100 - 25 = 75 %
⇒ The probability that a randomly selected survey participant didn't got the flew = [tex]\frac{75}{100}[/tex]
⇒ 0.75
This is the probability that a randomly selected survey participant didn't got the flew.
A recent study reported that 28% of the residents of a particular community lived in poverty. Suppose a random sample of 300 residents of this community is taken. We wish to determine the probability that 33% or more of our sample will be living in poverty. Complete parts (a) and (b) below.
Before doing any calculations, determine whether this probability is greater than 50% or less than 50%. Why? The answer should be less than 50%, because the resulting z-score will be negative and the sampling distribution is approximately Normal. The answer should be greater than 50%, because 0.24 is greater than the population proportion of 0.20 and because the sampling distribution is approximately Normal. The answer should be less than 50%, because 0.24 is greater than the population proportion of 0.20 and because the sampling distribution is approximately Normal. The answer should be greater than 50%, because the resulting z-score will be positive and the sampling distribution is approximately Normal. Calculate the probability that 24% or more of the sample will be living in poverty. Assume the sample is collected in such a way that the conditions for using the CLT are met. P (p ge 0.24) = (Round to three decimal places as needed.)
Answer:
a) The answer should be less than 50%, because 0.33 is greater than the population proportion of 0.28 and because the sampling distribution is approximately Normal.
b) P(x ≥ 0.33) = 2.68% = 0.027 to 3 d.p
Step-by-step explanation:
a) The required probability is P(x ≥ 0.33)
The population proportion of people living in poverty has already been given as 0.28, So, whatever the standard deviation of the distribution of sample means is, the sample proportion of 0.33 is more than the population proportion, So, it gives a z-score that is greater than 0. The probability from the z-score of the mean (0) to the end of distribution is exactly 50%; So, a Probability that only covers from a particular positive z-score to the end of the distribution will definitely be less than 50%.
So, because 33% is more than population proportion of 28%, and the sampling distribution is approximately normal, the probability of 33% or more of the sample living in poverty is less than 50%.
b) P(x ≥ 0.33)
The sample mean = population mean
μₓ = μ = 0.28
Standard deviation of the distribution of sample means = √[p(1-p)/n] (this is possible due to the central limit theorem for n greater than 30)
σₓ = √[(0.28×0.72)/300] = 0.0259
We then normalize 0.33
The standardized score for any value is the value minus the mean then divided by the standard deviation.
z = (x - μ)/σ = (0.33 - 0.28)/0.0259 = 1.93
To determine the probability of 33% or more of the sample size is living in poverty.
P(x ≥ 0.33) = P(z ≥ 1.93)
We'll use data from the normal probability table for these probabilities
P(x ≥ 0.33) = P(z ≥ 1.93) = 1 - P(z < 1.93)
= 1 - 0.9732 = 0.0268 = 2.68%
It is indeed way less than 50%.
Hope this Helps!!!
The answer to whether the probability is greater than 50% or less than 50% is; less than 50% because 0.33 is greater than the population proportion
What is the probability of the normal distribution?A) We want to determine the probability that 33% or more of our sample will be living in poverty and this is expressed as P(x ≥ 0.33)
Population proportion is; p = 0.28.
Formula for z-score is;
z = (x' - μ)/σ
Since our sample proportion is greater than our population proportion, it means the z-score will be greater than 0
Finally, due to the fact that 33% is more than population proportion of 28%, and the sampling distribution is approximately normal, the probability of 33% or more of the sample living in poverty will definitely be less than 50%.
b) We want to now calculate the above probability;
P(x ≥ 0.33)
The sample mean will be equal to population mean as;
μₓ = μ = 0.28
Standard deviation of the distribution of sample means is;
σₓ = √(p(1 - p)/n)
σₓ = √((0.28×0.72)/300)
σₓ = 0.0259
Thus, z-score is;
z = (x - μ)/σ
z = (0.33 - 0.28)/0.0259
z = 1.93
Thus, the of 33% or more of the sample size is living in poverty is;
P(x ≥ 0.33) = P(z ≥ 1.93) = 1 - P(z < 1.93)
P(x ≥ 0.33) = 1 - 0.9732
P(x ≥ 0.33) = 0.0268
P(x ≥ 0.33) = 0.027
It is indeed way less than 50%.
Read more about normal distribution at; https://brainly.com/question/4079902
The research question below describes a relationship between two quantitative variables. Which variable should be plotted on the horizontal X-axis? Is the sales price of a townhouse in San Francisco related to the number of square feet in the townhouse?
Answer:
number of square feet in the townhouse.
Step-by-step explanation:
When the relationship of two quantitative variables is assessed through plot then the scatter plot is made. Scatter plot shows relationship between two qualitative variables by showing dependent variable on y axis and independent variable on x axis.
When the relation between sale price of house and number of square feet in house is assessed then the sale price is dependent variable and square feet in house. Sale price is dependent variable because the price of house depends on the number of square feet of house. Thus, number of square feet in the town house will be plotted on the horizontal X-axis.
A market research firm conducts telephone surveys with a 44% historical response rate. What is the probability that in a new sample of 400 telephone numbers, at least 150 individuals will cooperate and respond to the questions?
Answer:
So, the probability is P=0.9953.
Step-by-step explanation:
We know that a market research firm conducts telephone surveys with a 44% historical response rate.
We get that:
[tex]p=44\%=0.44=\mu_{\hat{x}}\\\\n=400\\\\\hat{p}=\frac{150}{400}=0.375\\\\[/tex]
We calculate the standar deviation:
[tex]\sigma_{\hat{p}}=\sqrt{\frac{0.44(1-0.44)}{400}}\\\\\sigma_{\hat{p}}=0.025[/tex]
So, we get
[tex]z=\frac{\hat{p}-\mu_{\hat{p}}}{\sigma{\hat{p}}}\\\\z=\frac{0.375-0.44}{0.025}}\\\\z=-2.6[/tex]
We use a probability table to calculate it
[tex]P(\hat{p}>0.375)=P(z>-2.6)=1-P(z<-2.6)=1-0.0047=0.9953[/tex]
So, the probability is P=0.9953.
To find the probability of at least 150 of 400 individuals responding given a 44% response rate, calculate the mean and standard deviation and then find the z-score for 150 responses. The probability of getting at least 150 responses is the area to the right of the z-score in the standard normal distribution.
Explanation:The question asks for the probability that in a new sample of 400 telephone numbers, at least 150 individuals will respond, given a historical response rate of 44%. To determine this probability, we can approximate the binomial distribution to a normal distribution because the sample size is large (n=400).
First, we calculate the mean (μ) and the standard deviation (σ) for the number of responses. The mean is given by μ = np = 400 × 0.44 = 176. The standard deviation is σ = √(np(1-p)) = √(400 × 0.44 × 0.56) ≈ 9.92.
To calculate the probability of getting at least 150 responses, we would find the z-score for 150, which is z = (X - μ)/σ = (150 - 176)/9.92 ≈ -2.62. We then look up this z-score in a standard normal distribution table or use a calculator to determine the area to the right of this z-score, which represents the probability of getting more than 150 responses.
The question is related to market research and involves using statistical methods to calculate probabilities, which requires an understanding of binomial distributions and normal approximations.
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A Chinese restaurant offers 10 different lunch specials. Each weekday for one week, Fiona goes to the restaurant and selects a lunch special. How many different ways are there for her to select her lunches for the week
Answer:
10⁵
Step-by-step explanation:
10×10×10×10×10
= 10⁵ or 100,000
A random sample of 16 students selected from the student body of a large university had an average age of 25 years. We want to determine if the average age of all the students at the university is significantly different from 24. Assume the distribution of the population of ages is normal with a standard deviation of 2 years.At a .05 level of significance, it can be concluded that the mean age is _____. a. significantly less than 25 b. not significantly different from 24 c. significantly less than 24 d. significantly different from 24
Answer:
Option D) significantly different from 24
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = 24
Sample mean, [tex]\bar{x}[/tex] = 25
Sample size, n = 16
Alpha, α = 0.05
Population standard deviation, σ = 2
First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu = 24\text{ years}\\H_A: \mu \neq 24\text{ years}[/tex]
We use Two-tailed z test to perform this hypothesis.
Formula:
[tex]z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }[/tex]
Putting all the values, we have
[tex]z_{stat} = \displaystyle\frac{25 - 24}{\frac{2}{\sqrt{16}} } = 2[/tex]
Now, [tex]z_{critical} \text{ at 0.05 level of significance } = \pm 1.96[/tex]
Since,
The calculated z statistic does not lie in the acceptance region, we fail to accept the null hypothesis and reject it. We accept the alternate hypothesis.
Thus, it can be concluded that the mean age is
Option D) significantly different from 24
A food processor packages orange juice in small jars. The weights of the filled jars are approximately normally distributed with a mean of 10.5 ounces and a standard deviation of 0.3 ounce. Find the proportion of all jars packaged by this process that have weights that fall above 10.983 ounces.
Answer:
[tex]P(X>10.983)=P(\frac{X-\mu}{\sigma}>\frac{10.983-\mu}{\sigma})=P(Z>\frac{10.983-10.5}{0.3})=P(z>1.61)[/tex]
And we can find this probability using the complement rule and with excel or the normal standard table:
[tex]P(z>1.61)=1-P(z<1.61)=1-0.946=0.054[/tex]
Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Solution to the problem
Let X the random variable that represent the weights of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(10.5,0.3)[/tex]
Where [tex]\mu=10.5[/tex] and [tex]\sigma=0.3[/tex]
We are interested on this probability
[tex]P(X>10.983)[/tex]
And the best way to solve this problem is using the normal standard distribution and the z score given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
If we apply this formula to our probability we got this:
[tex]P(X>10.983)=P(\frac{X-\mu}{\sigma}>\frac{10.983-\mu}{\sigma})=P(Z>\frac{10.983-10.5}{0.3})=P(z>1.61)[/tex]
And we can find this probability using the complement rule and with excel or the normal standard table:
[tex]P(z>1.61)=1-P(z<1.61)=1-0.946=0.054[/tex]
The number of square feet per house are normally distributed with a population standard deviation of 137 square feet and an unknown population mean. A random sample of 19 houses is taken and results in a sample mean of 1350 square feet. Find the margin of error for a 80% confidence interval for the population mean. z0.10z0.10z0.05z0.05z0.025z0.025z0.01z0.01z0.005z0.005 1.2821.6451.9602.3262.576 You may use a calculator or the common z values above. Round the final answer to two decimal places.
Answer:
The MOE for 80% confidence interval for μ is 5.59.
Step-by-step explanation:
The random variable X is defined as the number of square feet per house.
The random variable X is Normally distributed with mean μ and standard deviation σ = 137.
The margin of error for a (1 - α) % confidence interval for population mean is:
[tex]MOE=z_{\alpha /2}\times\frac{\sigma}{\sqrt{n}}[/tex]
Given:
n = 19
σ = 137
[tex]z_{\alpha /2}=z_{0.20/2}=z_{0.10}=1.282[/tex]
Compute MOE for 80% confidence interval for μ as follows:
[tex]MOE=1.282\times\frac{137}{\sqrt{19}}=1.282\times4.36=5.58952\approx5.59[/tex]
Thus, the MOE for 80% confidence interval for μ is 5.59.
Which of the following meets the requirements of a stratified random sample? Multiple Choice A population contains 10 members under the age of 25 and 20 members over the age of 25. The sample will include six people who volunteer for the sample. A population contains 10 members under the age of 25 and 20 members over the age of 25. The sample will include six people chosen at random, without regard to age. A population contains 10 members under the age of 25 and 20 members over the age of 25. The sample will include six males chosen at random, without regard to age. A population contains 10 members under the age of 25 and 20 members over the age of 25. The sample will include two people chosen at random under the age of 25 and four people chosen at random over 25.
Answer:
Correct option is D.
Step-by-step explanation:
Random sampling implies the selection of of values or individuals in a random pattern.
A stratified random sampling is a sampling method where:
First divide the entire population into homogeneous subgroups, known as strata.Then take a random sample form each of the strata such that the sample size is proportional to the size of the strata.In this case the it is provided that the population consists of 30 members, 10 under the age of 25 and 20 over the age of 25.
So the stratas are:
10 members under the age of 25.20 members over the age of 25.Now to a random sample of size 2 is selected from strata 1 and a random sample of size 4 is selected from strata 2.
This forms a stratified random sample.
Correct option is:
A population contains 10 members under the age of 25 and 20 members over the age of 25. The sample will include two people chosen at random under the age of 25 and four people chosen at random over 25.
Final answer:
The correct choice for a stratified random sample is selecting two people randomly under the age of 25 and four people randomly over 25, as it ensures that the sample is proportional and representative of the two age groups within the population.
Explanation:
To determine which option meets the requirements of a stratified random sample, let's first define it. A stratified random sample divides the population into separate groups, known as strata, and a random sample is taken from each group. The key point here is that the sample from each stratum is proportional to the size of the stratum within the whole population, ensuring the sample is representative.
Given the multiple-choice options provided:
Sampling based on volunteers is a convenience sample, not stratified.Choosing people at random without regard to age is simple random sampling.Selecting only males is not stratified, as it does not account for another relevant characteristic, such as age.Selecting two people randomly under the age of 25 and four people randomly over 25 is the correct choice, as it ensures the sample is representative of the two age groups within the population.This type of sampling accounts for the different proportions of the sub-groups (under and over 25) in the population, which is a key aspect of stratified sampling.
You collect 100 samples from a large butterfly population. Fifty specimens are dark brown, 20 are speckled, and 30 are white. Coloration in this species of butterfly is controlled by one gene locus: BB individuals are brown, Bb are speckled, and bb are white. What are the allele frequencies for the coloration gene in this population
Answer:
Total Allele: 100
BB = 100 ; Bb = 40 ; bb = 60
[tex]\rho = 0.6\\\delta = 0.4[/tex]
Step-by-step explanation:
There are two allele in each gene. Since we have 100 samples of butterfly genes, the total number of allele are 100 x 2 = 200.
For each species:
Dark Brown (BB) → 50 x 2 = 100 allele
Speckled (Bb) → 20 x 2 = 40 allele
White (bb) → 30 x 2 = 60 allele
So, if we let [tex]\rho[/tex] be the frequency of the B allele and [tex]\delta[/tex] be the frequency of the b alleles, then:
[tex]\rho = \frac{ ((50 \times 2) + 20)}{200} \\\ \\\rho = \frac{100+20}{200} = \frac{120}{200}\\\\\\rho = 0.6[/tex]
[tex]\delta = \frac{((30 x 2) + 20)}{200}\\\\\delta = \frac{60+20}{200} = \frac{80}{200}\\\\\delta = 0.4[/tex]
We want to get the allele frequencies for the coloration gene in the population of butterflies, we will get:
The frequency for BB (brown) is 50%The frequency for Bb (speckled) is 20%The frequency for bb (white) is 30%How to get the frequencies?
Assuming that the sample is a good representation of the butterfly population, the frequencies are just given by the quotient between the number of each type of butterflies and the total number of butterflies in the sample, times 100%.
There are 100 butterflies, 50 are dark brown, 20 are speckled, and 30 are white.
The frequency for BB (brown) is:
[tex]F_{BB} = (50/100)*100\% = 50\% [/tex]
The frequency for Bb (speckled) is:
[tex]F_{Bb} = (20/100)*100\% = 20\%[/tex]
The frequency for bb (white) is:
[tex]F_{bb} = (30/100)*100\% = 30\%[/tex]
If you want to learn more about frequencies, you can read:
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If, based on a sample size of 950, a political candidate finds that 563 people would vote for him in a two-person race, what is the 99% confidence interval for his expected proportion of the vote? Would he be confident of winning based on this poll?
Answer:
The 99% confidence interval for his expected proportion of the vote is (0.5516, 0.6336). Since the interval for the proportion is above 50%, he would be confident of winning based on the pool.
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which
z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].
For this problem, we have that:
[tex]n = 950, p = \frac{563}{950} = 0.5926[/tex]
99% confidence level
So [tex]\alpha = 0.01[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.01}{2} = 0.995[/tex], so [tex]Z = 2.575[/tex].
The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.5926 - 2.575\sqrt{\frac{0.5926*0.4074}{950}} = 0.5516[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.5926 + 2.575\sqrt{\frac{0.5926*0.4074}{950}} = 0.6336[/tex]
The 99% confidence interval for his expected proportion of the vote is (0.5516, 0.6336). Since the interval for the proportion is above 50%, he would be confident of winning based on the pool.
Suppose a computer engineer is interested in determining the average weight of a motherboard manufactured by a certain company. A summary of a large sample provided to the engineer suggest a mean weight of 11.8 ounces and an estimated standard deviation, sigma = 0.75. How large a sample size is required if want a 99% confidence interval, with a tolerable interval width of 0.4? How large a sample would we need if were interested in a 95% confidence interval with a tolerable width of 0.5?
Answer:
We need a sample size of at least 23 for a 99% confidence interval, with a tolerable interval width of 0.4.
We need a sample size of at least 9 for a 95% confidence interval with a tolerable width of 0.5,
Step-by-step explanation:
How large a sample size is required if want a 99% confidence interval, with a tolerable interval width of 0.4?
We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:
[tex]\alpha = \frac{1-0.99}{2} = 0.005[/tex]
Now, we have to find z in the Ztable as such z has a pvalue of [tex]1-\alpha[/tex].
So it is z with a pvalue of [tex]1-0.005 = 0.995[/tex], so [tex]z = 2.575[/tex]
Now, find the margin of error(width) as such
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.
For this item, we have:
[tex]M = 0.4, \sigma = 0.75[/tex]. So
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
[tex]0.4 = 2.575*\frac{0.75}{\sqrt{n}}[/tex]
[tex]0.4\sqrt{n} = 1.93125[/tex]
[tex]\sqrt{n} = \frac{1.93125}{0.4}[/tex]
[tex]\sqrt{n} = 4.828125[/tex]
[tex]\sqrt{n}^{2} = (4.828125)^{2}[/tex]
[tex]n = 23[/tex]
We need a sample size of at least 23 for a 99% confidence interval, with a tolerable interval width of 0.4.
How large a sample would we need if were interested in a 95% confidence interval with a tolerable width of 0.5?
We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:
[tex]\alpha = \frac{1-0.95}{2} = 0.025[/tex]
Now, we have to find z in the Ztable as such z has a pvalue of [tex]1-\alpha[/tex].
So it is z with a pvalue of [tex]1-0.025 = 0.975[/tex], so [tex]z = 1.96[/tex]
Now, find the margin of error(width) as such
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.
For this item, we have:
[tex]M = 0.5, \sigma = 0.75[/tex]. So
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
[tex]0.5 = 1.96*\frac{0.75}{\sqrt{n}}[/tex]
[tex]0.5\sqrt{n} = 1.47[/tex]
[tex]\sqrt{n} = \frac{1.47}{0.5}[/tex]
[tex]\sqrt{n} = 2.94[/tex]
[tex]\sqrt{n}^{2} = (2.94)^{2}[/tex]
[tex]n \cong 9[/tex]
We need a sample size of at least 9 for a 95% confidence interval with a tolerable width of 0.5,
Suppose that we have collected a sample of 8 observations with values 12.6, 12.9, 13.4, 12.3, 13.6, 13.5, 12.6, and 13.1. What are the observed sample mean, observed sample variance, and observed sample standard deviation
Answer:
The sample mean is 13.
The sample variance is 0.2286.
The sample standard deviation is 0.4781.
Step-by-step explanation:
The sample is:
S = {12.6, 12.9, 13.4, 12.3, 13.6, 13.5, 12.6, 13.1}
The sample is of size n = 8.
The formula to compute the sample mean, sample variance and sample standard deviation are:
[tex]\bar x=\frac{1}{n} \sum x[/tex]
[tex]s^{2}=\frac{1}{n}\sum (x-\bar x)^{2} \\s=\sqrt{\frac{1}{n}\sum (x-\bar x)^{2} }\\[/tex]
Compute the sample mean as follows:
[tex]\bar x=\frac{1}{n} \sum x\\=\frac{1}{8}(12.6+ 12.9+ 13.4+ 12.3+ 13.6+ 13.5+ 12.6+ 13.1)\\=\frac{104}{8}\\ =13[/tex]
The sample mean is 13.
Compute the sample variance as follows:
[tex]s^{2}=\frac{1}{n-1}\sum (x-\bar x)^{2} \\=\frac{1}{8-1}[(12.6-13)^{2}+(12.9-13)^{2}+(13.4-13)^{2}+...+(13.1-13)^{2}] \\=\frac{1}{7}\times1.6\\=0.2286[/tex]
The sample variance is 0.2286.
Compute the sample standard deviation as follows:
[tex]s=\sqrt{s^{2}}\\=\sqrt{0.2286}\\=0.4781[/tex]
The sample standard deviation is 0.4781.
Answer:
The sample mean is 13.
The sample variance is 0.2286.
The sample standard deviation is 0.4781.
Step-by-step explanation:
Good luck
A super has a key ring with 12 keys. He has forgotten which one opens the apartment he needs to enter. What is the probability he is able to enter before reaching the 4th key?
Answer:
Therefore, the probability is P=1/4.
Step-by-step explanation:
We know that a super has a key ring with 12 keys. He has forgotten which one opens the apartment he needs to enter.
We conclude that each key has an equal probability of opening an apartment. Since there are 12 keys, it follows that the probability for each key is equal,
p = 1/12.
We calculate the probability he is able to enter before reaching the 4th key. So he will try three keys by then. We get:
[tex]P=p+p+p\\\\P=\frac{1}{12}+\frac{1}{12}+\frac{1}{12}\\\\P=\frac{3}{12}\\\\P=\frac{1}{4}[/tex]
Therefore, the probability is P=1/4.