Eli, a pretty smart ape in the Salt Lake City zoo, has correctly picked the winner of the last 7 Super Bowls. Read about Eli's prowess at selecting winners here. Question. If Eli is guessing when he chooses a Super Bowl winner (that is, he mentally flips a fair coin to decide which team he chooses as the winner), what is the probability that Eli chooses 7 winners in a row

Answers

Answer 1

Answer:

0.0078125

Step-by-step explanation:

If the probability of flipping a fair coin (mentally) correctly for 1 row is 0.5, then the probability of flipping it correctly for 2 rows is

0.5 * 0.5 = 0.25

for 3 rows: 0.5*0.5*0.5 = 0.125

...

For 7 rows:

[tex]0.5*0.5*0.5*0.5*0.5*0.5*0.5 = 0.5^7 = 0.0078125[/tex]


Related Questions

Apply the properties of angles to solve for the missing angles. Angle y is what degrees. Angle x is what degrees.

Answers

Answer:

y= 65 degrees. and x=30 degrees

Step-by-step explanation:

i haven't done this for a while so i may or may not be correct i tried to help though :)

Answer: y = 52°. x = 64°

Step-by-step explanation:

y = 26 + 26 = 52 degrees ( exterior angle of a triangle is the sum of the two opposite interior angle of the triangle)

x = 180-(52+64)(sum of angles in a triangle is 180°)

x = 180-116

x = 64degrees ( base angles of an isosceles triangle are equal)

Below are the jersey numbers of 11 players randomly selected from a football team. Find the​ range, variance, and standard deviation for the given sample data. What do the results tell​ us? 59 19 30 75 42 49 57 81 11 87 91

Answers

Answer:

Range = The difference between the highest and lowest which is

The mean is the sum of all values divided by the number of values

= 59+19+30+75+42+49+57+81+11+87+91 divided by n (which is 11)

601÷11

= 54.6

Variance = sum of squared deviations from the mean divided by n-1

=679.65

Standard Deviation = This is the square root of variance which gives us;

26.07

Jeremy and Brenda drove their cars in
opposite directions. When they stopped
after some time, they were already
126 miles apart. If Jeremy drove twice
as far as Brenda, how many miles did
Jeremy drive?

Answers

Answer: Jeremy drove 84 miles.

Step-by-step explanation:

Let x represent the number of miles that Brenda drove.

If Jeremy drove twice

as far as Brenda, it means that the distance covered by Jeremy would be 2x miles

When they stopped after some time, they were already

126 miles apart. This means that the total distance covered by both of them is 126 miles. Therefore,

x + 2x = 126

3x = 126

x = 126/3

x = 42 miles

The number of miles that Jeremy drove is

42 × 2 = 84 miles

One method used to distinguish between Granitic (G) and Basaltic (B) rocks is to examine a portion of the infrared spectrum of the sun’s energy reflected from the rock surface. Let R1, R2, and R3 denote measures of spectrum intensities at three different wave lengths; typically, for granite R1 < R2 < R3, whereas for basalt R3 < R1 < R2. When measurements are made remotely (using aircraft), various orderings of the Ri′s may arise whether the rock is basalt of granite. Flights over regions of know composition have yielded the following information: Reading R_1 P(basalt|R1 < R2 < R3). If measurements yielded R1 < R2 < R3, would you classify the rock as granite? b. If measurements yielded R1 < R3 < R2, how would you classify the rock? Answer the same question forR3

Answers

Answer:  

Step-by-step explanation:  

so from the question i have here, i will be giving a step by step analysis of the question.  

(a). Here we are showing a relationship, i.e.  

P(Granite ║ R₁ ∠ R₂ ∠ R₃) ˃ P (Basalt ║  R₁ ∠ R₂ ∠ R₃)  

from the LHS;  

P(Granite ║  R₁ ∠ R₂ ∠ R₃) = P(Granite ║  R₁ ∠ R₂ ∠ R₃) / P(R₁ ∠ R₂ ∠ R₃)  

= P(Granite)P(R₁ ∠ R₂ ∠ R₃ ║  Granite) / [ P(Granite  R₁ ∠ R₂ ∠ R₃) + P(Basaltic  R₁ ∠ R₂ ∠ R₃) ]  

= 0.25 × 0.6 / [(0.25×0.6)+(0.75×0.1)] = 0.667  

from the RHS;  

P (Basalt ║  R₁ ∠ R₂ ∠ R₃) = P(Basalt  R₁ ∠ R₂ ∠ R₃) / P(R₁ ∠ R₂ ∠ R₃)  

= P(Basalt)P(R₁ ∠ R₂ ∠ R₃ ║  Basalt) / [ P(Basalt  R₁ ∠ R₂ ∠ R₃) + P(Granite  R₁ ∠ R₂ ∠ R₃) ]  

= 0.75 × 0.1 / [(0.25 × 0.6)+(0.75 × 0.1)] = 0.333  

Therefore from this we can infer that;  

P(Granite ║  R₁ ∠ R₂ ∠ R₃) ˃ P (Basalt ║  R₁ ∠ R₂ ∠ R₃)  

(b). here we are asked to classify the rocks considering the measurement yield.  

Measurement yielded R1 < R3 < R2  

P(Granite ║  R₁ ∠ R₃ ∠ R₂) = P(Granite  R₁ ∠ R₃ ∠ R₂) / P(R₁ ∠ R₃ ∠ R₂)  

= P(Granite)P(R₁ ∠ R₃ ∠ R₂ ║  Granite) / [ P(Granite  R₁ ∠ R₃ ∠ R₂) + P(Basaltic  R₁ ∠ R₃ ∠ R₂) ]  

= 0.25 × 0.25 / [(0.25 × 0.25)+(0.75 × 0.2)] = 0.294  

also for RHS;  

P (Basalt ║  R₁ ∠ R₃ ∠ R₂) = P(Basalt ║  ∠ R₃ ∠ R₂) / P(R₁ ∠ R₃ ∠ R₂)  

= P(Basalt)P(R₁ ∠ R₃ ∠ R₂ ║  Basalt) / [ P(Basalt  R₁ ∠ R₃∠ R₂) + P(Granite  R₁ ∠ R₃ ∠ R₂) ]  

= 0.75 × 0.2 / [(0.25 × 0.25)+(0.75 × 0.2)] = 0.706  

from this we can infer that;

P(Granite ║  R₁ ∠ R₃ ∠ R₂) P (Basalt ║  R₁ ∠ R₃ ∠ R₂)  

Also considering measurements yielded R₃ ∠ R₁ ∠ R₂  

P(Granite ║  R₃ ∠ R₁ ∠ R₂) = P(Granite  R₃ ∠ R₁ ∠ R₂) / P(R₃ ∠ R₁ ∠ R₂)  

= P(Granite)P(R₃ ∠ R₁ ∠ R₂ ║  Granite) / [ P(Granite  R₃ ∠ R₁ ∠ R₂) + P(Basaltic  R₃ ∠ R₁ ∠ R₂) ]  

= 0.25 × 0.15 / [(0.25×0.15)+(0.75×0.7)] = 0.067

from the RHS;

P (Basalt ║  R₃ ∠ R₁ ∠ R₂) = P(Basalt  R₃ ∠ R₁ ∠ R₂) / P(R₃ ∠ R₁ ∠ R₂)  

= P(Basalt)P(R₃ ∠ R₁ ∠ R₂ ║  Basalt) / [ P(Basalt  R₃ ∠ R₁ ∠ R₂) + P(Granite  R₃ ∠ R₁ ∠ R₂) ]  

1 - 0.067 = 0.933  

from this we can infer that;

P(Granite ║  R₃ ∠ R₁ ∠ R₂) ∠ P (Basalt ║  R₃ ∠ R₁ ∠ R₂)  

cheers i hope this helps  

Final answer:

Measurements yielding R1 < R2 < R3 suggest a granitic composition, while the pattern R1 < R3 < R2 leans granitic, and R3 < R1 < R2 lean towards basaltic. The rock type can be further supported by the appearance of the rock, with granitic rocks being coarse and light-colored due to feldspar and quartz, while basaltic rocks are fine-grained and dark with ferromagnesian minerals. If granite contains basaltic inclusions, the granite is younger.

Explanation:

In analyzing rock compositions from infrared spectrum measurements, if measurements yield R1 < R2 < R3, based on typical patterns, the rock would more likely be classified as granitic because, for granite, the intensity measurements usually increase in that order. Conversely, when measurements produce R1 < R3 < R2, it does not fit neatly into the categories provided for basalt or granite, but it leans closer to a granitic composition given that R3 is not the least among the three, which is a characteristic of basaltic rocks. Similarly, if R3 presents a lower intensity compared to R1 and R2, and you have to classify without further information, it might suggest a basaltic nature following the typical pattern for basalt (R3 < R1 < R2), although it is important to note that real-world applications may require additional contextual information to accurately determine the rock type.

When studying rock samples, you should also consider differences in mineral size (coarse-textured like granite versus fine-textured like basalt), mineral content (dark, mafic minerals vs light, felsic minerals), and elemental content of the igneous rock types. For instance, granitic rock contains feldspar and quartz, which is reflected in its coarse and light-colored appearance. In contrast, basaltic rock, which is fine-grained and dark-colored, includes ferromagnesian minerals and feldspars. Additionally, if a granitic rock has inclusions of basalt (xenoliths), this implies that the granitic rock would typically be younger than the basalt since the inclusions must have been present already for the granite to incorporate them during its formation process.

Suppose that grade point averages of undergraduate students at one university have a bell-shaped distribution with a mean of 2.52 and a standard deviation of 0.38. Using the empirical rule, what percentage of the students have grade point averages that are between 1.76 and 3.28

Answers

Final answer:

Using the empirical rule for a bell-shaped distribution, which states that nearly all data for a normal distribution falls within three standard deviations of the mean, approximately 95% of undergraduate students at this university have grade point averages between 1.76 and 3.28.

Explanation:

The student is asking for the percentage of GPA between 1.76 and 3.28 using the Empirical Rule which applies to a Bell-shaped distribution or normal distribution. The empirical rule states that for a normal distribution, nearly all of the data will fall within three standard deviations of the mean.

The mean in this case is 2.52 and the standard deviation is 0.38. Therefore the values 1.76 and 3.28 fall within the mean minus two standard deviations and mean plus two standard deviations respectively. Therefore, by the empirical rule, these values represent approximately 95% of the data.

In other words, according to the empirical rule, approximately 95% of undergraduate students at this university have grade point averages between 1.76 and 3.28.

Learn more about Empirical Rule here:

https://brainly.com/question/35669892

#SPJ3

A college senior who took the Graduate Record Examination exam scored 530 on the Verbal Reasoning section and 600 on the Quantitative Reasoning section. The mean score for Verbal Reasoning section was 460 with a standard deviation of 105, and the mean score for the Quantitative Reasoning was 429 with a standard deviation of 148. Suppose that both distributions are nearly normal. Round calculated answers to 4 decimal places unless directed otherwise.

Answers

Answer:

Step-by-step explanation:

Given that a college senior who took the Graduate Record Examination exam scored 530 on the Verbal Reasoning section and 600 on the Quantitative Reasoning section.

Quantitative reasons scores are N (429, 148)

Hence Z score for the college senior = [tex]\frac{530-429}{148} \\=0.6824[/tex]

The mean score for Verbal Reasoning section was 460 with a standard deviation of 105

Verbal reasoning score was N(460, 105)

The score of college senior = 530

Z score for verbal reasoning =[tex]\frac{530-460}{105} \\=0.6667[/tex]

comparing we can say he scored more on quantiative reasoning.

Thus comparison was possible only by converting to Z scores.

Children inherit many traits from their parents. Some inherited traits are not desirable. For example, if both parents are carriers, there is a 25% chance any child will have sickle-cell disease. If we assume no identical twins in a family, and both individuals in a couple are carriers, what is the probability none of their four children has the disease

Answers

The probability that none of their four children has the disease = 0.316

Step-by-step explanation:

Step 1 :

The probability that the child will have the specified disease if both the parents are carriers = 25% = 0.25

So the probability that the child does not have the specified disease = 1 - 0.25 = 0.75

Step 2 :

There are four children. The probability that each child has the disease is independent of the condition of the other children

The probability of more than one independent events happening together can be obtained by multiplying  probability of individual events

So, the probability that none of their 4 children has the sickle cell disease  = [tex](0.75)^{4}[/tex] = 0.316

Step 3 :

Answer :

The probability no children has the specified cell disease = 0.316

An inverted pyramid is being filled with water at a constant rate of 45 cubic centimeters per second. The pyramid, at the top, has the shape of a square with sides of length 6 cm, and the height is 12 cm. Find the rate at which the water level is rising when the water level is 5 cm.

Answers

Answer:

The volume of a pyramid is 1/3 Sh, where S is the area of the base and h is the height.  Since the area of base (Square) is , S = [tex]s^{2}[/tex], where s is the side of the base.  So the volume is

V = 1/3 [tex]s^{2}[/tex]h.

Further solution is on paper (Pictures attached)

In a certain small dorm, the dorm council will consist of 8 students, ofwhom 3 must be women and 5 must be men, because there are a total of 15 women and 25 men currently living there. What is the total possible number of complete councils that could be selected

Answers

Answer:

The total possible number of complete councils that could be selected is 24,174,150.

Step-by-step explanation:

The order of the men and the women is not important. So we use the combinations formula to solve this question.

Combinations formula:

[tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

What is the total possible number of complete councils that could be selected

3 women from a set of 15

5 men from a set of 25

[tex]T = C_{15,3}*C_{25,5} = \frac{15!}{3!12!}*\frac{25!}{5!20!} = 455*53130 = 24174150[/tex]

The total possible number of complete councils that could be selected is 24,174,150.

The given family of functions is the general solution of the differential equation on the indicated interval. Find a member of the family that is a solution of the initial-value problem. y = c1 + c2 cos(x) + c3 sin(x), (−[infinity], [infinity]); y''' + y' = 0, y(π) = 0, y'(π) = 9, y''(π) = −1

Answers

To solve the initial-value problem for the differential equation y''' + y' = 0 with the general solution y = c1 + c2 cos(x) + c3 sin(x), we differentiate y to find y' and y'', apply the initial conditions at x = π, and solve for c1, c2, and c3. The specific solution is y = 1 + cos(x) - 9 sin(x).

We're given the general solution of the third-order differential equation y''' + y' = 0 as y = c1 + c2 cos(x) + c3 sin(x), and we need to find specific constants c1, c2, and c3 that satisfy the initial conditions y(π) = 0, y'(π) = 9, and y''(π) = -1.

First, differentiate y = c1 + c2 cos(x) + c3 sin(x) to find y' and y''.

Substitute x = π into the resulting equations to apply the initial conditions.

Solve the system of equations to find the values of c1, c2, and c3.

Let's carry out these steps:

Differentiation gives us:

y' = -c2 sin(x) + c3 cos(x)

y'' = -c2 cos(x) - c3 sin(x)

Applying initial conditions at x = π:

y(π) = c1 - c2 = 0

y'(π) = -c3 = 9

y''(π) = -c2 = -1

Solve the system:

c1 = c2

c2 = 1

c3 = -9

Therefore, the specific solution to the given initial-value problem is y = 1 + cos(x) - 9 sin(x).

Which of the following is equivalent to 60 Superscript one-half? StartFraction 60 Over 2 EndFraction StartRoot 60 EndRoot StartFraction 1 Over 60 squared EndFraction StartFraction 1 Over StartRoot 60 EndRoot EndFraction

Answers

Answer:

[tex](60)^{\frac{1}{2}} = \sqrt{60}[/tex]

Step-by-step explanation:

We are given the following expression:

[tex](60)^{\frac{1}{2}}[/tex]

We are given the following options:

[tex]A.~\dfrac{60}{2}\\\\B.~\sqrt{60}\\\\C.~(\dfrac{1}{60})^{2}\\\\D.~\dfrac{1}{\sqrt{60}}[/tex]

Exponent Properties:

  [tex]x^{-a} = (\dfrac{1}{x})^{a}\\\\(x^m)^n = x^{mn}\\\\\dfrac{x^m}{x^n} = x^{m-n}[/tex]

Thus, the correct answer is

[tex](60)^{\frac{1}{2}} = \sqrt{60}[/tex]

Answer:

B or [tex]\sqrt{60}[/tex]

Step-by-step explanation:

Military radar and missile detection systems are designed to warn a country of an enemy attack. A reliability question is whether a detection system will be able to identify an attack and issue a warning. Assume that a particular detection system has a 0.80 probability of detecting a missile attack. Use the binomial probability distribution to answer the following questions. (a) What is the probability that a single detection system will detect an attack

Answers

Answer:

0.8 is the probability that a single detection system will detect a missile attack.

Step-by-step explanation:

We are given the following information:

We treat detection a missile attack as a success.

P(detecting a missile attack) = 80% = 0.8

Then the number of missile attack follows a binomial distribution, where

[tex]P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}[/tex]

where n is the total number of observations, x is the number of success, p is the probability of success.

Now, we are given n = 1

We have to evaluate:

[tex]P(x = 1)\\= \binom{1}{1}(0.8)^1(1-0.8)^0\\= 0.8[/tex]

0.8 is the probability that a single detection system will detect a missile attack.

Answer:

(a) Probability that a single detection system will detect an attack is 0.80

Step-by-step explanation:

We are given that a reliability question is whether a detection system will be able to identify an attack and issue a warning. Assuming that a particular detection system has a 0.80 probability of detecting a missile attack.

The above situation can be represented through Binomial distribution;

[tex]P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....[/tex]

where, n = number of trials(samples) taken = 1 detection system

            r = number of success

            p = probability of success which in our question is probability of

                   detecting a missile attack, i.e., 80%

LET X = a particular detection system

Also, it is given that a single detection system is taken,

So, it means X ~ [tex]Binom(n=1,p= 0.80)[/tex]

Now, Probability that a single detection system will detect an attack is given by = P(X = 1)  

  P(X = 1) = [tex]\binom{1}{1}0.8^{1} (1-0.8)^{1-1}[/tex]

               = [tex]1 \times 0.8 \times 1[/tex] = 0.80 .

In an article regarding interracial dating and marriage recently appeared in a newspaper. Of 1719 randomly selected adults, 311 identified themselves as Latinos, 322 identified themselves as blacks, 251 identified themselves as Asians, and 775 identified themselves as whites. Among Asians, 79% would welcome a white person into their families, 71% would welcome a Latino, and 66% would welcome a black person.
NOTE: If you are using a Student's t-distribution, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, though.)
Construct the 95% confidence intervals for the three Asian responses.
1. Welcome a white person ( , )
2. Welcome a Latino ( , )
3. Welcome a Black person ( , )

Answers

Answer:

Step-by-step explanation:

Hello!

The parameter of interest in this exercise is the population proportion of Asians that would welcome a person of other races in their family. Using the race of the welcomed one as categorizer we can define 3 variables:

X₁: Number of Asians that would welcome a white person into their families.

X₂: Number of Asians that would welcome a Latino person into their families.

X₃: Number of Asians that would welcome a black person into their families.

Now since we are working with the population that identifies as "Asians" the sample size will be: n= 251

Since the sample size is large enough (n≥30) you can apply the Central Limit Theorem and approximate the variable distribution to normal.

[tex]Z_{1-\alpha /2}= Z_{0.975}= 1.965[/tex]

1. 95% CI for Asians that would welcome a white person.

If 79% would welcome a white person, then the expected value is:

E(X)= n*p= 251*0.79= 198.29

And the Standard deviation is:

V(X)= n*p*(1-p)= 251*0.79*0.21=41.6409

√V(X)= 6.45

You can construct the interval as:

E(X)±Z₁₋α/₂*√V(X)

198.29±1.965*6.45

[185.62;210.96]

With a 95% confidence level, you'd expect that the interval [185.62; 210.96] contains the number of Asian people that would welcome a White person in their family.

2. 95% CI for Asians that would welcome a Latino person.

If 71% would welcome a Latino person, then the expected value is:

E(X)= n*p= 251*0.71= 178.21

And the Standard deviation is:

V(X)= n*p*(1-p)= 251*0.71*0.29= 51.6809

√V(X)= 7.19

You can construct the interval as:

E(X)±Z₁₋α/₂*√V(X)

178.21±1.965*7.19

[164.08; 192.34]

With a 95% confidence level, you'd expect that the interval [164.08; 192.34] contains the number of Asian people that would welcome a Latino person in their family.

3. 95% CI for Asians that would welcome a Black person.

If 66% would welcome a Black person, then the expected value is:

E(X)= n*p= 251*0.66= 165.66

And the Standard deviation is:

V(X)= n*p*(1-p)= 251*0.66*0.34= 56.3244

√V(X)= 7.50

You can construct the interval as:

E(X)±Z₁₋α/₂*√V(X)

165.66±1.965*7.50

[150.92; 180.40]

With a 95% confidence level, you'd expect that the interval [150.92; 180.40] contains the number of Asian people that would welcome a Black person in their family.

I hope it helps!

A researcher would like to evaluate the claim that large doses of vitamin C can help prevent the common cold. One group of participants is given a large dose of the vitamin (500 mg per day), and a second group is given a placebo (sugar pill). The researcher records the number of colds each individual experiences during the 3-month winter season. a. Identify the dependent variable for this study.b. Is the dependent variable discrete or continuous?c. What scale of measurement (nominal, ordinal, or interval/ratio) is used to measure the dependent variable?d. What is the independent variable?e. What research method is being used (experimental or correlational?

Answers

Answer:

(a) the dependent variable here are the Participants.

(b) the dependent variable is discrete.

(c) The scale measurement of measurement used is interval/ratio to measure the dependent variable.

(d) the independent variable is Vitamin C

(e) The research method being used is experimental.

Step-by-step explanation:

The dependent variable (sometimes known as the responding variable) is what is being studied and measured in the experiment.

Examples of continuous dependent variable may include costs, profits and sales.But some dependent variables are discrete – that is, they take on a relatively small number of integer values.

Interval scale and ratio scale are the two variable measurement scales where they define the attributes of the variables quantitatively.A ratio scale is a measurement scale which has more or less all the properties of an interval scale. Ratio data on this scale has measurable intervals.

Experimental research is a study that strictly adheres to a scientific research design. It includes a hypothesis, a variable that can be manipulated by the researcher, and variables that can be measured, calculated and compared.

Answer: 1. The dependent variable is the common cold.

2. It is a discrete variable.

3. The interval/ratio scale.

4. The independent value is the large doses of Vitamin C administered.

5. The research method used is Experimental.

Step-by-step explanation:

1. The dependent variable is the factor that we are trying to understand. In this case, it is the common cold and how it is affected by large doses of vitamins.

2. A discrete value is computed by counting. So the dependent variable - the common cold was computed by counting the number of occurrences.

3. The Interval/ratio scale is used because both the order of measurement and the differences between them are observed.

4. The Independent variable is the control in the experiment which can be compared to changes in the dependent variable. The large doses of vitamin C affects the common cold.

5. Correlational research observes patterns in variables that occur naturally while Experimental research introduces a change and monitors its effect. The research is Experimental because a change in the form of the placebo is introduced and observed.

2. A canoe requires 8 hours of fabrication. A row boat requires 5 hours of fabrication. The fabrication department has at most 110 hours to labor each week. Write the equation and solve the problems.

Answers

Answer:

Let x is the number of the cannon and y is the number of the row boat.Thus the equation will become

8x+5y=110

Answer: 8C+5R≤110

Step-by-step explanation: Let hours of fabrication of canoes = C

Hours of fabrication of rowboat = R

If a canoe requires 8 hours of fabrication. And a row boat requires 5 hours of fabrication. Then

8C + 5R

The fabrication department has at most 110 hours to labor each week. 

Therefore

8C+5R≤110

At most means less than or equal to.

The numerical course grades in a statistics course can be approximated by a normal model with a mean of 70 and a standard deviation of 10. The professor must convert the numerical grades to letter grades. She decides that she wants 10% A's, 30% B's, 40% C's, 15% D's, and 5% F's. a. What is the cutoff for an A grade?

Answers

Answer:

The cutoff for an A grade is 82.8.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

[tex]\mu = 70, \sigma = 10[/tex]

a. What is the cutoff for an A grade?

The top 10% of the class get an A grade. So the cutoff is the value of X when Z has a pvalue of 1-0.1 = 0.9. So it is X when Z = 1.28

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]1.28 = \frac{X - 70}{10}[/tex]

[tex]X - 70 = 1.28*10[/tex]

[tex]X = 82.8[/tex]

The cutoff for an A grade is 82.8.

solution to 2/3x = -5/2 + 2

Answers

Answer:

x=-0.75

Step-by-step explanation:

Combine Like terms

2/3x= -5/2+2

2/3x= -2.5+2

2/3x= -0.5

Multiply both sides by 3

2x= -1.5

Divide both sides by 2

x= -0.75

Answer:

x = -4/3

Step-by-step explanation:

The LCM for the fractions is 6x.

2/3x = -5/2 + 2

Multiply through by 6x

(6x) 2/3x = (6x) -5/2 + (6x)2

4 = -15x + 12x

4 = -3x

x = -4/3

Enjoy Maths!

1. In order to get more female customers, a new clothing store offers free gourmet coffee and pastry to its customers. The average daily revenue over the past five-week period has been $1,080 with a standard deviation of $260. Use this sample information to construct a 95% confidence interval for the average daily revenue. The store manager believes that the coffee and pastry strategy would lead to an average daily revenue of $1,200. Is the manager correct based on the 95% confidence interval?

Answers

Answer:

No, the manager is not correct based on the 95% confidence interval.

Step-by-step explanation:

We are given that the average daily revenue over the past five-week period has been $1,080 with a standard deviation of $260, i.e.; X bar = $1080 and s = $260 and sample size, n = 35 .

The Pivotal quantity for 95% confidence interval is given by;

                [tex]\frac{Xbar - \mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]

where, X bar = sample mean = $1080

                s  = sample standard deviation = $260

                 n = sample size = 35 {five-week}

So, 95% confidence interval for average daily revenue, [tex]\mu[/tex] is given by;

P(-2.032 < [tex]t_3_4[/tex] < 2.032) = 0.95

P(-2.032 < [tex]\frac{Xbar - \mu}{\frac{s}{\sqrt{n} } }[/tex] < 2.032) = 0.95

P(-2.032 * [tex]{\frac{s}{\sqrt{n} }[/tex] < [tex]{Xbar - \mu}[/tex] < 2.032 * [tex]{\frac{s}{\sqrt{n} }[/tex] ) = 0.95

P(X bar - 2.032 * [tex]{\frac{s}{\sqrt{n} }[/tex] < [tex]\mu[/tex] < X bar + 2.032 * [tex]{\frac{s}{\sqrt{n} }[/tex] ) = 0.95

95% confidence interval for [tex]\mu[/tex] = [ X bar - 2.032 * [tex]{\frac{s}{\sqrt{n} }[/tex] , X bar + 2.032 * [tex]{\frac{s}{\sqrt{n} }[/tex] ]

                                            = [ 1080 - 2.032 * [tex]{\frac{260}{\sqrt{35} }[/tex] , 1080 + 2.032 * [tex]{\frac{260}{\sqrt{35} }[/tex] ]

                                             = [ 990.70 , 1169.30 ]

No, the manager is not correct based on the fact that the coffee and pastry strategy would lead to an average daily revenue of $1,200 because the calculate 95% confidence interval does not include value of $1200.

Therefore, the store manager believe is not correct.

Final answer:

The 95% confidence interval for the store's average daily revenue is calculated to be approximately ($993.97, $1166.03). Since $1200 is outside this interval, the manager's belief that the coffee and pastry strategy will lead to an average daily revenue of $1200 is not backed by this confidence level.

Explanation:

In the field of statistics, a confidence interval (CI) is a type of interval estimate that is used to indicate the reliability of an estimate. The method for calculating a 95% confidence interval for the average daily revenue involves the sample mean, the standard deviation, and the z-score associated with a 95% confidence level, which is approximately 1.96. Let's use the provided data to calculate:

Calculate the standard error by dividing the standard deviation by the square root of the sample size. Here, the standard deviation is $260, and the sample size is 5 weeks * 7 days/week = 35 days. So, the standard error is $260 / sqrt(35) = $43.89.Multiply the standard error by the z-score to get the margin of error. So, $43.89 * 1.96 = $86.03.Calculate the lower and upper bounds of the 95% confidence interval by subtracting and adding the margin of error from/to the sample mean. So, ($1080 - $86.03, $1080 + $86.03) = ($993.97, $1166.03).

The range of this 95% confidence interval is from $993.97 to $1166.03. This means we are 95% confident that the true average daily revenue lies within this interval. Since $1200 lies outside this interval, the manager's belief is not supported by this confidence interval.

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This exercise illustrates that poor quality can affect schedules and costs. A manufacturing process has 100 customer orders to fill. Each order requires one component part that is purchased froma supplier. However, typically, 2% of the components are identifiedas defective, and the components can be assumed to beindependent.a)If the manufacturer stocks 100 components, what is theprobability that the 100 orders can be filled without reorderingcomponents?b) If the manufacturer stocks 102 components, what is theprobability that the 100 orders can be filled without reorderingcomponents?c) If the manufacturer stocks 105 components, what is theprobability that the 100 orders can be filled without reorderingcomponents?

Answers

Answer:

(a) 0.1326

(b) 0.2732

(c) 0.0410

Step-by-step explanation:

Let X = number of defective components.

The probability of X is, P (X) = p = 0.02.

The random variable X follows a Binomial distribution with parameters n and p. The probability mass function of a Binomial distribution is:

[tex]P(X=x)={n\choose x}p^{x}(1-p)^{n-x};\ x=0,1,2,3,...[/tex]

(a)

Compute the probability that the 100 orders can be filled without reordering components as follows:

n = 100

[tex]P(X=0)={100\choose 0}0.02^{0}(1-0.02)^{100-0}=1\times1\times0.13262=0.1326[/tex]

Thus, the probability that the 100 orders can be filled without reordering components is 0.1326.

(b)

Compute the probability that out of 102 orders 2 orders needs reordering as follows:

n = 102

[tex]P(X=2)={102\choose 2}0.02^{2}(1-0.02)^{102-2}=5151\times0.0004\times0.13262=0.2732[/tex]

Thus, the probability that out of 102 orders 2 orders needs reordering is 0.2732.

(c)

Compute the probability that out of 105 orders 2 orders needs reordering as follows:

n = 105

[tex]P(X=5)={105\choose 5}0.02^{5}(1-0.02)^{105-5}=96560646\times0.0000000032\times0.13262=0.0410[/tex]

Thus, the probability that out of 105 orders 5 orders needs reordering is 0.0410.

5. (20 pts) Consider the array 25, 14, 63, 29, 63, 47, 12, 21. Apply the Split procedure in Quicksort, as described in class, to this array using the first element as the pivot.

Answers

Answer:

Step-by-step explanation: see attachment

(1 point) Find the general solution to the homogeneous differential equation. ????2y????????2−20????y????????+136y=0 Use c1 and c2 in your answer to denote arbitrary constants, and enter them as c1 and c2. y(????)= ?

Answers

Answer:

Question is not clear please post question clearly lots of question marks.

Your differential equation is not displayed well. It though looks like this:

2d²y/dx² - 20dy/dx + 136y = 0

If this is not the differential equation, the method of solving this would still be used in solving the correct one.

We first write an auxiliary equation to the differential equation.

The auxiliary equation is:

2m² - 20m + 136 = 0

Dividing by 2, we have

m² - 10m + 68 = 0

Next, we solve the auxiliary equation to obtain the values of m.

Solving using the quadratic formula

m = [-b ± √(b² - 4ac)]/2a

Where a = 1, b = -10, and c = 68

m = [10 ± √(100 - 272)]/2

= 5 ± (1/2)√(-172)

= 5 ± (1/2)i√172

= 5 ± 6.6i

For solutions of the form a ± ib, the complimentary solution is

y = e^(ax)[C1cosbx + C2sinbx]

Therefore, the complimentary solution is

y = e^(5x)[C1cos(6.6x) + C2sin(6.6x)]

he purpose of an x-bar chart is to determine whether there has been a: change in the AOQ. change in the number of defects in a sample. change in the percent defective in a sample. change in the dispersion of the process output. change in the central tendency of the process output.

Answers

Answer:

E. Change in the central tendency of the process output.

Step-by-step explanation:

The x-bar chart is a control chart for the central tendency of the process output. Therefore it's purpose is to determine whether there has been a change in the central tendency of the process output.

Answer:

The change in the central tendency of the process output.

Step-by-step explanation:

Lets examine each of the listed options

The change in the percent defective in a sample is associated with the change in average outgoing quality (AOQ).

The change in the dispersion of the process output is associated with R-chart since the function of R-chart is to detect changes in the dispersion.

The change in the number of defects in a sample is associated with C-chart since the function of a C-chart is to show the number of flaws per unit in a sample.

The change in the central tendency of the process output is associated with X-bar chart since the function of X-bar chart is to check whether the values are within the appropriate limits (central tendency) of the process output or not.

Therefore, the purpose of an X-bar chart is to determine whether there has been a change in the central tendency of the process output.

Define a random variable x = number of cups of coffee consumed on an average day. Let x = 4 represent four or more cups. Round your answers to four decimal places.

Answers

Answer:

E (X) = 6.4

Step-by-step explanation:

SOLUTION:

A random variable x = number of cups of coffee consumed on an average day.

∴Let x = 4 represent four or more cups. Round your answers to four decimal places.

X          Probability (X)

0             0.1  

1              0.15

2             0.3

3             0.75

4             0. 25

5             0.21

∴ E (X) = Ux(Mean)

0x.0.1 + 1 x.15 + 2 x 0.3 + 3 x 0.75 + 4 x 0.25 + 5 x 0.21 =  6.4

The least-squares regression model y =−3.4+5.2x and correlation coefficient r=−0.66 were calculated for a set of bivariate data with variables x and y . What is closest to the proportion of the variation in y that cannot be explained by the explanatory variable?

Answers

Answer:

[tex]r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^2 -(\sum x)^2][n\sum y^2 -(\sum y)^2]}}[/tex]  

For this case the value of r = -0.66

Now we can calculate the determination coeffcient:

[tex] r^2 = (-0.66)^2 = 0.4356[/tex]

And then we can conclude that 43.56% of the variation in y can be explained by the explanatory variable

And then 100-43.56 = 56.44 % of the variation in y that cannot be explained by the explanatory variable

Step-by-step explanation:

For this case we need to calculate the slope with the following formula:

[tex]m=\frac{S_{xy}}{S_{xx}}[/tex]

Where:

[tex]S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}[/tex]

[tex]S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}[/tex]

[tex]\bar x= \frac{\sum x_i}{n}[/tex]

[tex]\bar y= \frac{\sum y_i}{n}[/tex]

And we can find the intercept using this:

[tex]b=\bar y -m \bar x[/tex]

And the model obtained for this case is:

[tex] y = -3.4 +5.2 x[/tex]

The correlation coefficient is a "statistical measure that calculates the strength of the relationship between the relative movements of two variables". It's denoted by r and its always between -1 and 1.

And in order to calculate the correlation coefficient we can use this formula:  

[tex]r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^2 -(\sum x)^2][n\sum y^2 -(\sum y)^2]}}[/tex]  

For this case the value of r = -0.66

Now we can calculate the determination coeffcient:

[tex] r^2 = (-0.66)^2 = 0.4356[/tex]

And then we can conclude that 43.56% of the variation in y can be explained by the explanatory variable

And then 100-43.56 = 56.44 % of the variation in y that cannot be explained by the explanatory variable

The Coefficient of determination gives the proportion of variation which can be explained by the regression line.

The proportion of variation that cannot be explained by the given regression line is 56.44%

From the question, the correlation Coefficient, R = - 0.66

The Coefficient of determination, R² can be calculated from the R value given. R² = - 0.66² = 0.4356

This means that :

Percentage of variation that can be explained = (0.4356) × 100% = 43.56%

The percentage of variation that cannot be explained = 100% - 43.56% = 56.44%

Therefore, proportion of the variation in y that cannot be explained is 56.44%

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A subway has good service 70% of the time and runs less frequently 30% of the time because of signal problems. When there are signal problems, the amount of time in minutes that you have to wait at the platform is described by the pdf probability density function with signal problems = pT|SP(t) = .1e^(−.1t). But when there is good service, the amount of time you have to wait at the platform is probability density function with good service = pT|Good(t) = .3e^(−.3t) You arrive at the subway platform and you do not know if the train has signal problems or running with good service, so there is a 30% chance the train is having signal problems. (a) After 1 minute of waiting on the platform, you decide to re-calculate the probability that the train is having signal problems based on the fact that your wait will be at least 1 minute long. What is that new probability? (b) After 5 minutes of waiting, still no train. You re-calculate again. What is the new probability? (c) After 10 minutes of waiting, still no train. You re-calculate again. What is the new probability?

Answers

Answer:

Part a: The probability is 0.3436

Part b: The probability is 0.5381

Part c: The probability is 0.7600

Step-by-step explanation:

As per the given data

[tex]p_{T | SP} = e^{0.1}\\p_{T | Good} = e^{0.3}\\[/tex]

(a)

Probability that train is delayed by more than 1 minute = P(T > 1) = P(SP) * P(T > 1 | SP) + P(Good) * P(T > 1 | Good)

[tex]= 0.3 e^{-.1 \times 1 }+ 0.7 * e^{-.3 \times 1}\\ = 0.3 e^{-.1} + 0.7 e^{-.3}[/tex]

Probability that after 1 minute of waiting, probability that train has signal problems = P(SP | T > 1)

= P(T > 1 | SP) * P(SP) / P(T > 1) (By Bayes theorem)

[tex]= \dfrac{0.3 e^{-.1 }}{ 0.3 e^{-.1 }+ 0.7 e^{-.3 }}\\\\= \dfrac{0.2714512}{0.790024}\\\\= 0.3435987[/tex]

(b)

Probability that train is delayed by more than 5 minutes = P(T > 5) = P(SP) * P(T > 5 | SP) + P(Good) * P(T > 5 | Good)

[tex]= 0.3 e^{-.1 \times 5 }+ 0.7 * e^{-.3 \times 5}\\ = 0.3 e^{-.5} + 0.7 e^{-1.5}[/tex]

Probability that after 5 minute of waiting, probability that train has signal problems = P(SP | T > 5)

= P(T > 5 | SP) * P(SP) / P(T > 5) (By Bayes theorem)

[tex]= \dfrac{0.3 e^{-.5 }}{ 0.3 e^{-.5 }+ 0.7 e^{-1.5 }}\\\\= \dfrac{0.1819592}{0.3381503}\\\\= 0.5381015[/tex]

(c)

Probability that train is delayed by more than 10 minutes = P(T > 10) = P(SP) * P(T > 10 | SP) + P(Good) * P(T > 10 | Good)

[tex]= 0.3 e^{-.1 \times 10 }+ 0.7 * e^{-.3 \times 10}\\ = 0.3 e^{-1.0} + 0.7 e^{-3.0}[/tex]

Probability that after 5 minute of waiting, probability that train has signal problems = P(SP | T > 10)

= P(T > 10 | SP) * P(SP) / P(T > 10) (By Bayes theorem)

[tex]= \dfrac{0.3 e^{-1.0 }}{ 0.3 e^{-1.0 }+ 0.7 e^{-3.0 }}\\\\= \dfrac{0.1103638}{0.1452148}\\\\= 0.7600038[/tex]

(a) The probabilities recalculated based on the waiting time are: 21.4% after 1 minute,

(b) 54.0% after 5 minutes, and

(c) 75.9% after 10 minutes.

These calculations use Bayes' theorem and the exponential distribution functions provided.

The question pertains to conditional probability and involves exponential distributions.

Let’s solve this step-by-step:

Part (a)

Given:
pT|SP(t) = [tex]0.1e^(-0.1t)[/tex]
pT|Good(t) = [tex]0.3e^(-0.3t)[/tex]
P(Signal Problems) = [tex]0.30[/tex]
P(Good Service) = [tex]0.70[/tex]

We need to find P(Signal Problems | T ≥ 1). Using Bayes’ theorem:

P(T ≥ 1 | Signal Problems) = [tex]\int_{1}^{\infty} 0.1 \, dx[/tex] [tex]e^(-0.1t)[/tex] [tex]dt =[/tex] [tex]e^(-0.1)[/tex]P(T ≥ 1 | Good Service) = [tex]\int_{1}^{\infty} 0.3 \, dx[/tex] [tex]e^(-0.3t)[/tex] [tex]dt =[/tex] [tex]e^(-0.3)[/tex]

Using the law of total probability:

P(T ≥ 1) = P(T ≥ 1 | Signal Problems)P(Signal Problems) + P(T ≥ 1 | Good Service)P(Good Service)
= [tex]e^(-0.1)[/tex] [tex]* 0.30 +[/tex] [tex]e^(-0.3)[/tex] [tex]* 0.70[/tex]

Now applying Bayes’ theorem:
P(Signal Problems | T ≥ 1) = [tex][[/tex][tex]e^(-0.1)[/tex] [tex]* 0.30] / [[/tex][tex]e^(-0.1)[/tex] [tex]* 0.30 +[/tex] [tex]e^(-0.3)[/tex] [tex]* 0.70][/tex]
[tex]\approx [0.9048 * 0.30] / [0.9048 * 0.30 + 0.7408 * 0.70][/tex]
[tex]\approx 0.214[/tex]

Part (b)

Similarly for T ≥ 5, the probabilities become:

P(T ≥ 5 | Signal Problems) = [tex]e^(-0.5)[/tex]P(T ≥ 5 | Good Service) = [tex]e^(-1.5)[/tex]

Using the law of total probability:
P(T ≥ 5) = [tex]e^(-0.5)[/tex] [tex]* 0.30 +[/tex] [tex]e^(-1.5)[/tex][tex]* 0.70[/tex]

Applying Bayes’ theorem:
P(Signal Problems | T ≥ 5) = [tex][[/tex][tex]e^(-0.5)[/tex] [tex]* 0.30] / [[/tex][tex]e^(-0.5)[/tex] [tex]* 0.30 +[/tex] [tex]e^(-1.5)[/tex] [tex]* 0.70][/tex]
≈ [tex][0.6065 * 0.30] / [0.6065 * 0.30 + 0.2231 * 0.70][/tex]
≈ [tex]0.540[/tex]

Part (c)

For T ≥ 10:

P(T ≥ 10 | Signal Problems) = [tex]e^{-1}[/tex]P(T ≥ 10 | Good Service) = [tex]e^{-3}[/tex]

Using the law of total probability:
P(T ≥ 10) = [tex]e^{-1} \cdot 0.30 + e^{-3} \cdot 0.70[/tex]

Applying Bayes’ theorem:
P(Signal Problems | T ≥ 10) = [tex]\frac{e^{-1} \cdot 0.30}{e^{-1} \cdot 0.30 + e^{-3} \cdot 0.70}[/tex]
≈ [tex][0.3679 * 0.30] / [0.3679 * 0.30 + 0.0498 * 0.70][/tex]
≈ [tex]0.759[/tex]

Is there more wood in a 60​-foot-high tree trunk with a radius of 2.2 feet or in a 50​-foot-high tree trunk with a radius of 2.6 ​feet? Assume that the trees can be regarded as right circular cylinders.

Answers

Answer:

V₂=1061.85 ft³

Step-by-step explanation:

To determine where more wood is found, just find the volume of each log and see which one has the largest volume

knowing that the volume of a straight cylinder is

V=πR²h, where R=radius and h=height

[tex]V_{1}=\pi (2.2)^{2}60= 912.31ft^{3}\\V_{2}=\pi (2.6)^{2}50= 1061.85ft^{3}[/tex]

As we can observe

V₂>V₁

Different species can interact in interesting ways. One type of grass produces the toxin ergovaline at levels about 1.0 part per million in order to keep grazing animals away. However, a recent study27 has found that the saliva from a moose counteracts these toxins and makes the grass more appetizing (for the moose). Scientists estimate that, after treatment with moose drool, mean level of the toxin ergovaline (in ppm) on the grass is 0.183. The standard error for this estimate is 0.016.

a.Give notation for the quantity being estimated, and define any parameters used.

b.Give notation for the quantity that gives the best estimate, and give its value.

c.Give a 95% confidence interval for the quantity being estimated. Interpret the interval in context.

Answers

Answer:

a) [tex]\mu[/tex]

b) [tex]\bar{x}[/tex]

c) (0.152, 0.214)

Step-by-step explanation:

We are given the following in the question:

Sample mean = 0.183 ppm

Standard error = 0.016

a) quantity being estimated

We have to estimate the population mean.

Notation for population mean:

[tex]\mu[/tex]

b) Best estimate for population mean is the sample mean

Notation for sample mean:

[tex]\bar{x}[/tex]

The point estimate for population mean is

[tex]\mu = \bar{x} = 0.183[/tex]

c) 95% confidence interval

[tex]\bar{x}\pm z_{critical}(\text{Standard Error})[/tex]

[tex]z_{critical}\text{ at}~\alpha_{0.05} = 1.96[/tex]

Putting values, we get,

[tex]0.183 \pm 1.96(0.016)\\=0.183\pm 0.03136\\=(0.15164, 0.21436)\\\approx (0.152, 0.214)[/tex]

Thus, the 95% confidence interval is:

(0.152, 0.214)

Interpretation:

After treatment with moose drool, we are 95% certain or 95% confident that the interval (0.152, 0.214) contains the true mean of the population that is the mean level of the toxin ergovaline on the grass.

Final answer:

The quantity being estimated is the mean level of the toxin ergovaline on the grass after treatment with moose drool. The best estimate for this quantity is 0.183 ppm. The 95% confidence interval for this estimate is (0.15164, 0.21436) ppm.

Explanation:

a. The quantity being estimated is the mean level of the toxin ergovaline on the grass after treatment with moose drool. The parameter used is the standard error, which measures the variability of the estimate.

b. The notation for the quantity that gives the best estimate is the mean level of the toxin ergovaline after treatment with moose drool, denoted as μ.

c. To calculate a 95% confidence interval, we need to determine the margin of error. Since the standard error is given as 0.016, the margin of error is 1.96 times the standard error, which is approximately 0.03136. The 95% confidence interval is then calculated by subtracting and adding the margin of error from the mean level of the toxin ergovaline after treatment with moose drool, resulting in an interval of (0.15164, 0.21436). This means that we are 95% confident that the true mean level of the toxin ergovaline after treatment with moose drool is between 0.15164 and 0.21436 ppm.

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A car is being driven at a rate of 40 ft/sec when the brakes are applied. The car decelerates at a constant rate of 10 ft/sec2. Calculate how far the car travels in the time it takes to stop. Round your answer to one decimal place.

Answers

Answer:

80 feet

Step-by-step explanation:

Given:

Initial speed of the car ([tex]v_0[/tex]) = 40 ft/sec

Deceleration of the car ([tex]\frac{dv}{dt}[/tex]) = -10 ft/sec²

Final speed of the car ([tex]v_x[/tex]) = 0 ft/sec

Let the distance traveled by the car be 'x' at any time 't'. Let 'v' be the velocity at any time 't'.

Now, deceleration means rate of decrease of velocity.

So, [tex]\frac{dv}{dt}=-10\ ft/sec^2[/tex]

Negative sign means the velocity is decreasing with time.

Now, [tex]\frac{dv}{dt}=\frac{dv}{dx}(\frac{dx}{dt})[/tex] using chain rule of differentiation. Therefore,

[tex]\frac{dv}{dx}\cdot\frac{dx}{dt}= -10\\\\But\ \frac{dx}{dt}=v.\ So,\\\\v\frac{dv}{dx}=-10\\\\vdv=-10dx[/tex]

Integrating both sides under the limit 40 to 0 for 'v' and 0 to 'x' for 'x'. This gives,

[tex]\int\limits^0_{40} {v} \, dv=\int\limits^x_0 {-10} \, dx\\\\\left [ \frac{v^2}{2} \right ]_{40}^{0}=-10x\\\\-10x=\frac{0}{2}-\frac{1600}{2}\\\\10x=800\\\\x=\frac{800}{10}=80\ ft[/tex]

Therefore, the car travels a distance of 80 feet before stopping.

In a new card game, you start with a well-shuffled full deck and draw 3 cards without replacement. If you draw 3 hearts, you win $50. If you draw 3 black cards, you win $25. For any other draws, you win nothing.

(a) Find the expected winnings for a single game.

(b) Find the standard deviation of the winnings.

(c) If the game costs $5 to play, what would be the expected value of the net profit (or loss)?

(Hint: profit = winnings - cost; X - 5)

(d) If the game costs $5 to play, what would be the standard deviation of the net profit (or loss)?

(e) If the game costs $5 to play, should you play this game?

Answers

There are [tex]\binom{52}3=\frac{52!}{3!(52-3)!}[/tex] (or "52 choose 3") ways of drawing any 3 cards from the deck.

There are 13 hearts in the deck, and 26 cards with a black suit. So there are [tex]\binom{13}3[/tex] and [tex]\binom{26}3[/tex] ways of drawing 3 hearts or 3 black cards, respectively. Then the probability of drawing 3 hearts is

[tex]P(\text{3 hearts})=\dfrac{\binom{13}3}{\binom{52}3}=\dfrac{11}{850}[/tex]

and the probability of drawing 3 black cards is

[tex]P(\text{3 black})=\dfrac{\binom{26}3}{\binom{52}3}=\dfrac2{17}[/tex]

All other combinations can be drawn with probability [tex]1-\frac{11}{850}-\frac2{17}=\frac{739}{850}[/tex].

Let [tex]W[/tex] be the random variable for one's potential winnings from playing the game. Then

[tex]P(W=w)=\begin{cases}\frac{11}{850}&\text{for }w=\$50\\\frac2{17}&\text{for }w=\$25\\\frac{739}{850}&\text{otherwise}\end{cases}[/tex]

a. For a single game, one can expect to win

[tex]E[W]=\displaystyle\sum_ww\,P(W=w)=\frac{\$50\cdot11}{850}+\frac{\$20\cdot2}{17}+\frac{\$0\cdot739}{850}=\$3[/tex]

b. For a single game, one's winnings have a variance of

[tex]V[W]=E[(W-E[W])^2]=E[W^2]-E[W]^2[/tex]

where

[tex]E[W^2]=\displaystyle\sum_ww^2\,P(W=w)=\frac{\$50^2\cdot11}{850}+\frac{\$20^2\cdot2}{17}+\frac{\$0^2\cdot739}{850}=\$^2\frac{1350}{17}\approx\$^279.41[/tex]

so that [tex]V[W]=\$^2\frac{1197}{17}\approx\$^270.41[/tex]. (No, that's not a typo, variance is measured in squared units.) Standard deviation is equal to the square root of the variance, so it is approximately $8.39.

c. With a $5 buy-in, the expected value of the game would be

[tex]E[W-\$5]=E[W]-\$5=-\$2[/tex]

i.e. a player can expect to lose $2 by playing the game (on average).

d. With the $5 cost, the variance of the winnings is the same, since the variance of a constant is 0:

[tex]V[W-\$5]=V[W][/tex]

so the standard deviation is the same, roughly $8.39.

e. You shouldn't play this game because of the negative expected winnings. The odds are not in your favor.

The expected winning for a single game defined is : $3.59

The standard deviation of winning is : $10.11

Expected winning if game costs $5 to play is :

- $0.79

The standard deviation of winning if game costs $5 to play is : $11.33

The game should not be played with a game play fee of -$5 as the expected winning value is negative.

Recall : selection is done without replacement :

Number of hearts in a deck = 13

Probability of drawing 3 hearts :

P(drawing 3 Hearts)

First draw × second draw × third draw

13/52 × 12/51 × 11/50 = 1716/132600 = 858/66300

Probability of selecting 3 black cards :

Number of black cards in a deck = 26

P(drawing 3 black cards) :

First draw × second draw × third draw

26/52 × 25/51 × 24/50 = 15600 / 132600 = 7800/66300

Probability of making any other draw :

P(3 hearts) + P(3 blacks) + P(any other draw) = 1

858/66300 + 7800/66300 + P(any other draw) = 1

P(any other draw) = 57642/66300

For a single game :

X _______ $50 ________ $25 _______ $0

P(X)_ 858/66300__ 7800/66300_57642/66300

E(X) = Σ[ X × p(X)]

E(X) =Σ[(50 × 858/66300)+(25 × 7800/66300)+0]

E(X) = $3.588

The standard deviation = √Var(X)

Var(X) = Σ[ X² × p(X)] - E(X)

Σ[ X²×p(X)] = Σ[(50² × 858/66300)+(25² × 7800/66300)+0] = 105.88235

Var(X) = 105.88235 - 3.588 = 102.29435

Standard deviation of winning = √102.29435 = $10.114

If the game cost $5 to play :

Net amount won if :

3 hearts are drawn = $50 - $5 = $45

3 blacks are drawn = $25 - $5 = $20

Any other combination are drawn= $0 - $5 = -$5

The distribution becomes :

X _______ $45 ________ $20 _______ -$5

P(X)_ 858/66300__ 7800/66300_57642/66300

E(X) = Σ[ X × p(X)]

E(X) =Σ[(50 × 858/66300)+(25 × 7800/66300)+ (-5 × 57642/66300)]

E(X) = - $0.7588

Standard deviation of winning :

The standard deviation = √Var(X)

Var(X) = Σ[ X² × p(X)] - E(X)

Σ[ X²×p(X)] = Σ[(50² × 858/66300)+(25² × 7800/66300)+ (-5² × 57642/66300)] = 127.61764

Var(X) = 127.61764 - (-0.7588) = 128.37644

Standard deviation of winning :

Std(X) = √Var(X) = √128.37644 = $11.330

With a game cost of - $5 ; the expected winning for a single game gives a negative value, therefore you should not play the game.

Learn more on expected value : https://brainly.com/question/22097128

Montoya (2007) asked 56 men and 82 women to rate 21 different body parts on a scale of 1 (no opinion) to 5 (very desirable). They found that men and women rated the eyes similarly, with an average rating of about 3.77 ± 1.23 (M ± SD). Assuming these data are normally distributed, answer the following questions. (Round your answers to two decimal places.) (a) What percentage of participants rated the eyes at least a 5 (very desirable)? % (b) What percentage rated the eyes at most a 1 (no opinion)? %

Answers

Answer:

a. 15.9%

b. 1.2%

Step-by-step explanation:

using the normal distribution we have the following expression

[tex]P(X\geq a)=P(\frac{X-u}{\alpha } \geq \frac{5-M}{SD}) \\[/tex]

Where  the first expression in the right hand side is the z-scores, M is the mean of value 3.77 and SD is the standard deviation of value 1.23.

if we simplify and substitute values, we arrive at

[tex]P(X\geq a)=P(\frac{X-u}{\alpha } \geq \frac{a-M}{SD}) \\P(X\geq 5)=P(Z \geq \frac{5-3.77}{1.23})\\P(X\geq 5)=P(Z \geq 1)\\P(X\geq 5)=1- P(Z < 5)\\P(X\geq 5)=1-0.841\\P(X\geq 5)=0.159[/tex]

in percentage, we arrive at 15.9%

b. for the percentage rated the eyes most a 1

[tex]P(X\leq a)=P(\frac{X-u}{\alpha } \leq \frac{a-M}{SD})\\for a=1\\P(X\leq 1)=P(Z \leq \frac{1-3.77}{1.23})\\P(X\leq 1)=P(Z \leq -2.25})\\P(X\leq 1)=0.012\\[/tex]

In percentage we have 1.2%

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