An accounting professor wishing to know how many MBA students would take a summer elective in international accounting did a survey of the class she was teaching. Which kind of sample is this?

Answer:

1) 0.9772

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 = 40000, \sigma = 5000[/tex]

What is the probability that a randomly selected individual with an MBA degree will get a starting salary of at least $30,000?

This is 1 subtracted by the pvalue of Z when X = 30000. So

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

[tex]Z = \frac{30000 - 40000}{5000}[/tex]

[tex]Z = -2[/tex]

[tex]Z = -2[/tex] has a pvalue of 0.0228.

1 - 0.0228 = 0.9772

0.9772 that a randomly selected individual with an MBA degree will get a starting salary of at least $30,000

Final answer:

Using the normal distribution, the probability that a randomly selected individual with an MBA degree will get a starting salary of at least $30,000 is approximately 0.9772.

Explanation:

The probability of a randomly selected individual with an MBA degree getting a starting salary of at least $30,000 is determined using the normal distribution with a mean of $40,000 and a standard deviation of $5,000. You calculate the z-score for $30,000, which is (30,000 - 40,000) / 5,000 = -2.

Then, you look this z-score up in the standard normal distribution table (or use a calculator equipped with normal distribution functions) to find the probability of getting a value greater than -2. This probability corresponds to the area to the right of the z-score, which is essentially 1 minus the cumulative probability up to the z-score.

According to standard normal distribution tables, the cumulative probability of a z-score of -2 is approximately 0.0228. Therefore, the probability of getting at least $30,000 is

1 - 0.0228, which equals approximately 0.9772.

Comparing this to available choices, option 1) 0.9772 is the correct answer.

Answer:

705,432 ways

Step-by-step explanation:

Since no girl will be left out once both teams are selected when selecting the varsity team, the junior varsity team is automatically composed by the players not selected, the number of ways to select both teams is:

[tex]n = \frac{22!}{(22-11)!11!} \\n=705,432[/tex]

There are 705,432 ways to divide the girls into a varsity team and a junior varsity team.

I'm guessing that [10] refers to the set of the first 10 positive integers.

If the largest element of a given 4-permutation is 6, then the other three elements are pulled from the set {1, 2, 3, 4, 5}. This can be done in 5!/(5 - 3)! = 60 ways. Then there are four possible positions to place the 6, giving a total of 4 * 60 = 240 permutations.

If the largest element of a permutation is *at most* 6, then the maximal element is 4, 5, or 6.

Putting everything together, the total number of permutations in which the maximal element is at most 6 is 24 + 96 + 240 = 360.

The number of 4-permutations of [10] with a maximum element of 6 is 24. The number of 4-permutations of [10] with a maximum element at most 6 is 120.

To find the number of 4-permutations of [10] with a maximum element of 6, we can consider the possibilities for the position of the maximum element in the permutation. There are 4 possible positions for the maximum element: first, second, third, or fourth. If the maximum element is in the first position, the remaining 3 elements can be any combination of the remaining 3 numbers (10, 9, and 8) which gives us 3! = 6 permutations. Similarly, if the maximum element is in the second, third, or fourth position, we will have 6 permutations for each position.

Therefore, the total number of 4-permutations of [10] with a maximum element of 6 is 4 * 6 = 24.

To find the number of 4-permutations of [10] with a maximum element at most 6, we need to consider all possible values for the maximum element. We already found that there are 24 permutations with a maximum element of 6. Now, we need to consider the possibilities where the maximum element is 5, 4, 3, 2, or 1.

If the maximum element is 5, the remaining 3 elements can be any combination of the remaining 4 numbers (10, 9, 8, and 6) which gives us 4! = 24 permutations. Similarly, if the maximum element is 4, 3, 2, or 1, we will have 24 permutations for each maximum element.

Therefore, the total number of 4-permutations of [10] with a maximum element at most 6 is 24 + 24 + 24 + 24 + 24 = 120.

Answer:

d. 14:11

Step-by-step explanation:

There is a 7 in 10 chance that Mike does his homework today and a 8 in 10 chance he will do it tomorrow,  the probability that he will do it both today and tomorrow is:

[tex]P = \frac{7}{10}* \frac{8}{10}\\P=\frac{56}{100}=0.56[/tex]

The odds are the probability that an event occurs over the probability that it does not occur:

[tex]Odds = \frac{0.56}{1-0.56}=\frac{56}{44}=\frac{14}{11}[/tex]

The odds that he will do it both today and tomorrow are 14:11.

The odds that he will do it both today and tomorrow is 14:11.

Probability that he would do the assignement today and tomorrow = probability he would do the assignment today x probability he would do the assignment tomorrow

0.7 x 0.8 = 0.56

Odds he would do the assignment today and tomorrow = 0.56/ (1 - 0.56) = 0.56 / 0.44

= 14 : 11

To learn more about probability, please check: https://brainly.com/question/10750263

Answer:

962,598 different groups of 5 subjects are possible.

Step-by-step explanation:

The order is not important.

For example, Math, English, Business, Geography and History is the same group as English, Math, Business, Geography and History.

So we use the combinations formula to solve this problem.

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]

Combinations of 5 subjects from a set of 43

[tex]C_{43, 5} = \frac{43!}{5!(43-5)!} = 962598[/tex]

962,598 different groups of 5 subjects are possible.

Answer: it will take Sebastian's mom 45 minutes to catch up with the bus.

Step-by-step explanation:

By the time Sebastian's mom catches up with the bus, she would have covered the same distance with the bus.

Let t represent the time it will take for Sebastian's mom to catch up with the bus.

Distance = speed × time

Sebastian's school bus averages 28 miles per hour.

Distance covered by Sebastian's school bus in t hours is

28 × t = 28t

Sebastian's mom travelled at an average speed of 42mph. Since she left home 0.25 hour after Sebastian, the time it would take her to cover the same distance is

(t - 0.25) hour. Distance covered in

(t - 0.25) hour is

42(t - 0.25)

Since the distance covered is the same, then

28t = 42(t - 0.25)

28t = 42t - 10.5

42t - 28t = 10.5

14t = 10.5

t = 10.5/14

t = 0.75

Converting to minutes, it becomes

0.75 × 60 = 45 minutes

Answer:

There is a 54.88% probability that there are no surface flaws in an auto's interior.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]

In which

x is the number of sucesses

[tex]e = 2.71828[/tex] is the Euler number

[tex]\mu[/tex] is the mean in the given interval.

The number of surface flaws in plastic panels used in the interior of automobiles has a Poisson distribution with a mean of 0.06 flaws per square foot of plastic panel. Assume an automobile interior contains 10 square feet of plastic panel.

For one square foot, we have 0.06 flaws.

So for 10 square feet, the mean is [tex]\mu = 10*0.06 = 0.6[/tex]

(a) What is the probability that there are no surface flaws in an auto's interior

This is P(X = 0).

[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]

[tex]P(X = 0) = \frac{e^{-0.6}*(0.6)^{0}}{(0.6)!} = 0.5488[/tex]

There is a 54.88% probability that there are no surface flaws in an auto's interior.

Answer:

300

Step-by-step explanation:

20% of 1,500 = 300

(Source: Google)

Answer:

300

Step-by-step explanation:

There is some information missing in the question, since we need to know what the position function is. The whole problem should look like this:

Consider an athlete running a 40-m dash. The position of the athlete is given by [tex]d(t)=\frac{t^{2}}{6}+4t[/tex] where d is the position in meters and t is the time elapsed, measured in seconds.

Compute the average velocity of the runner over the intervals:

(a) [1.95, 2.05]

(b) [1.995, 2.005]

(c) [1.9995, 2.0005]

(d) [2, 2.00001]

Answer

(a) 6.00041667m/s

(b) 6.00000417 m/s

(c) 6.00000004 m/s

(d) 6.00001 m/s

The instantaneous velocity of the athlete at t=2s is 6m/s

Step by step Explanation:

In order to find the average velocity on the given intervals, we will need to use the averate velocity formula:

[tex]V_{average}=\frac{d(t_{2})-d(t_{1})}{t_{2}-t_{1}}[/tex]

so let's take the first interval:

(a) [1.95, 2.05]

[tex]V_{average}=\frac{d(2.05)-d(1.95)}{2.05-1.95}[/tex]

we get that:

[tex]d(1.95)=\frac{(1.95)^{3}}{6}+4(1.95)=9.0358125[/tex]

[tex]d(2.05)=\frac{(2.05)^{3}}{6}+4(2.05)=9.635854167[/tex]

so:

[tex]V_{average}=\frac{9.6358854167-9.0358125}{2.05-1.95}=6.00041667m/s[/tex]

(b) [1.995, 2.005]

[tex]V_{average}=\frac{d(2.005)-d(1.995)}{2.005-1.995}[/tex]

we get that:

[tex]d(1.995)=\frac{(1.995)^{3}}{6}+4(1.995)=9.30335831[/tex]

[tex]d(2.005)=\frac{(2.005)^{3}}{6}+4(2.005)=9.363335835[/tex]

so:

[tex]V_{average}=\frac{9.363335835-9.30335831}{2.005-1.995}=6.00000417m/s[/tex]

(c) [1.9995, 2.0005]

[tex]V_{average}=\frac{d(2.0005)-d(1.9995)}{2.0005-1.9995}[/tex]

we get that:

[tex]d(1.9995)=\frac{(1.9995)^{3}}{6}+4(1.9995)=9.33033358[/tex]

[tex]d(2.0005)=\frac{(2.0005)^{3}}{6}+4(2.0005)=9.33633358[/tex]

so:

[tex]V_{average}=\frac{9.33633358-9.33033358}{2.0005-1.9995}=6.00000004m/s[/tex]

(d) [2, 2.00001]

[tex]V_{average}=\frac{d(2.00001)-d(2)}{2.00001-2}[/tex]

we get that:

[tex]d(2)=\frac{(2)^{3}}{6}+4(2)=9.33333333[/tex]

[tex]d(2.00001)=\frac{(2.00001)^{3}}{6}+4(2.00001)=9.33339333[/tex]

so:

[tex]V_{average}=\frac{9.33339333-9.33333333}{2.00001-2}=6.00001m/s[/tex]

Since the closer the interval is to 2 the more it approaches to 6m/s, then the instantaneous velocity of the athlete at t=2s is 6m/s

The average velocity is computed for each interval using the average velocity formula and the given position function, d(t) = 3t^2 + 4t + 6. Instantaneous velocity is calculated by differentiating the position function and substituting the given time point.

To find the average velocity, we need to use the formula: average velocity = (final position - initial position) / (final time - initial time). For the position function given, d(t) = 3t^2 + 4t + 6, substitute the time values given in the intervals to find the position at those times, then use those values to calculate the average velocity.

For example, for the first interval [1.95, 2.05], d(1.95) = 3*(1.95)^2 + 4*1.95 + 6 and d(2.05) = 3*(2.05)^2 + 4*2.05 + 6. Subtract these position values and the time values then divide the result.

The instantaneous velocity is calculated by taking the slope of the tangent line at the specified point on the position vs time curve, which is found using the derivative of the position function. The derivative of d(t) = 3t^2 + 4t + 6 is d'(t) = 6t + 4. So, at t = 2, the instantaneous velocity would be d'(2) = 6*2 + 4.

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Answer:

0.1673 (16.73%)

Step-by-step explanation:

The probability that a fair coin is chosen is 2/3 . if the fair coin is chosen the probability of getting 3 heads is determined by binomial distribution:

P(3 heads in 10 flips )=B(n=10,p=0.5,X=3) =0.1171

But if a coin that is not fair is chosen , then the probability of getting 3 heads also follows a binomial distribution but with p=0.3

P(3 heads)=B(n=10,p=0.3,X=3)=0.2668

Finally the probability is

P= 2/3*0.1171 + 1/3* 0.2678 = 0.1673 (16.73%)

Answer:

-3/8 = -0,375

-1/3 = -0,33

Step-by-step explanation:

N/A

Answer:when you divide two number the answer is called the quotient.

Step-by-step explanation:

-3/8

=-0.375

and

-1/3

=-0.333

Answer:

50% probability that a randomly selected respondent voted for Obama.

Step-by-step explanation:

We have these following probabilities:

60% probability that an Ohio resident does not have a college degree.

If an Ohio resident does not have a college degree, a 52% probability that he voted for Obama.

40% probability that an Ohio resident has a college degree.

If an Ohio resident has a college degree, a 47% probability that he voted for Obama.

What is the probability that a randomly selected respondent voted for Obama?

This is the sum of 52% of 60%(non college degree) and 47% of 40%(college degree).

So

[tex]P = 0.52*0.6 + 0.47*0.4 = 0.5[/tex]

50% probability that a randomly selected respondent voted for Obama.

Answer:

The average baggage-related revenue per passenger is $18.20

Step-by-step explanation:

$25 for the first bag

$35 for the second bag

Total passenger = 100

51 passenger have no checked baggage

32 passenger have one checked baggage. Therefore the airlines charges for the 32 passenger will be: [tex]25 * 32 = 800[/tex].

17 passenger have two checked baggage. Therefore the airline charges for the 17 passenger will be: [tex](25 *17) + (35*17) = 425+595=1020[/tex]

Average baggage related revenue per passenger is: Total Revenue / passenger

Total Revenue = 1020 + 800 = $1820

Average = 1820/100 = $18.20

Answer:

(a) {RR,BB,RB,RY,BY}

(b) {R1 R2, R1 B1, R1 B2, R1 Y, R2 R1, R2 B1, R2 B2, R2 Y, B1 R1, B1 R2, B1 B2, B1 Y, B2 R1, B2 B1, B2 R2, B2 Y, Y R1, Y R2, Y B1, Y B2}.

Step-by-step explanation:

This question can be answered in two ways :

(a) When order of picking marbles does not matter.

(b) When order of picking marbles does matter.

Red marbles in the bag = 2

Blue marbles in the bag = 2

Yellow marbles in the bag = 1

Now since we have to select two marbles from the bag, the sample cases will be:

In short Sample space = {RR,BB,RB,RY,BY} where R = Red , B = Blue and Y = Yellow.

    2. Now I will explain what will be the sample space for the experiment  when order of picking marbles does matter.

For this, First give numbers to the balls in bag i.e.,

First Red ball in the bag = R1

Second Red ball in the bag = R2

First Blue ball in the bag = B1

Second Blue ball in the bag = B2

Yellow ball in the bag = Y

Now the cases for sample spaces when two marbles are selected will be :

{R1 R2, R1 B1, R1 B2, R1 Y, R2 R1, R2 B1, R2 B2, R2 Y, B1 R1, B1 R2, B1 B2, B1 Y, B2 R1, B2 B1, B2 R2, B2 Y, Y R1, Y R2, Y B1, Y B2}.

Answer:

The exam can be answered with exactly 8 answers correct in 6435 ways.

Step-by-step explanation:

The order is not important.

For example, answering correctly the questions 1,2,3,4,5,6,7,8 is the same outcome as answering 2,1,3,4,5,6,7,8. So we use the combinations formula to solve this problem.

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]

"There are 15 questions on an exam. In how many ways can the exam be answered with exactly 8 answers correct?"

Combinations of 8 questions from a set of 15. So

[tex]C_{15,8} = \frac{15!}{8!(15-8)!} = 6435[/tex]

The exam can be answered with exactly 8 answers correct in 6435 ways.

To find the number of ways to answer 15 exam questions with exactly 8 correct answers, you use the binomial coefficient formula. The calculation yields 15C8 = 6,435. Thus, there are 6,435 ways to answer the exam with 8 correct answers.

The formula for finding the number of ways to choose k items from n items is given by:

nCk = n! / [k!(n-k)!]

In this case, we need to find the number of ways to get exactly 8 correct answers out of 15 questions:

Plugging these values into the formula, we get:

15C8 = 15! / [8!(15-8)!]

Which simplifies to:

15C8 = 15! / (8! * 7!)

Calculating the factorial values:

By cancelling out the common terms, we get:

15C8 = (15 × 14 × 13 × 12 × 11 × 10 × 9) / (7 × 6 × 5 × 4 × 3 × 2 × 1) = 6435

Conclusion

There are 6,435 ways to answer the exam with exactly 8 answers correct.

Answer:

0.1029 or 10.29%

Step-by-step explanation:

P(F) =0.3

P(S) = 1-0.3 = 0.7

If missiles continue to be launched until an unsuccessful launch occurs, the probability that exactly 4 total launches will be performed is the probability that the first three launches will be successful while the fourth will be unsucessful:

[tex]P(L=4) = P(S)*P(S)*P(S)*P(F)\\P(L=4) = 0.7*0.7*0.7*0.3\\P(L=4) = 0.1029 = 10.29\%[/tex]

The probability of 10 total launches is 0.1029 or 10.29%.

Calculate the probability of three successful launches [tex](0.7^3)[/tex] followed by one unsuccessful launch (0.3), which equals 10.29%.

The subject of the question is probability, specifically related to geometric distributions. The probability of an unsuccessful missile launch is given as 0.3. To find the probability that exactly 4 total launches will be performed before the first unsuccessful launch occurs, we need to calculate the probability of having three successful launches followed by one unsuccessful launch.

The probability of a successful launch is therefore 1 - 0.3 = 0.7. Since each launch is independent, the probability of exactly three successes followed by one failure is [tex](0.7)^3[/tex] times 0.3. This is calculated as (0.7 imes 0.7 imes 0.7) times 0.3 which equals 0.1029, or 10.29% when rounded to two decimal places.

Answer:

a.  Inferential statistics

b. Descriptive statistics

Step-by-step explanation:

statistics can be categorized into descriptive an inferential statistics. descriptive statistics makes use of a set of data for numerical calculations  and provides conclusion based on those numerical calculation. this data could be collected using tables,graphs, and other means of data representation.

Inferential statistics however come up with conclusions and assumptions base on a sample data.

Examples of descriptive statistics are mean, median, mode, quartile, percentile. Thus option B is descriptive.

Option A however is inferential statistics since some assumptions were made based on the effect of garlic on blood pressure.

Answer:

Mr. Rosenbloom 1 bottle of gel will last for 20 days.

Step-by-step explanation:

Given:

Amount of gel used every morning = 500 mL

Amount of gel in 1 bottle = 10 liters.

We need to find the number of days 1 bottle of gel will last for.

Solution:

Now we know that;

1 liter =1000 mL

10 liter = 10000 mL

Now we now that;

500 mL of gel is used = 1 day

10000 mL of gel will be used = Number of days 10000 mL of gel will last.

By Using Unitary method we get;

Number of days 10000 mL of gel will last = [tex]\frac{10000}{500}=20\ days[/tex]

Hence Mr. Rosenbloom 1 bottle of gel will last for 20 days.

Answer:

The information a measure of variability for the variable competitiveness conveys is

The first option is correct

a.) Do all teenage girls have the same amount of competitiveness?

The second option is correct

b.) How spread out are the values for competitiveness among teenage girls?

Step-by-step explanation:

The first option is correct

a.) Do all teenage girls have the same amount of competitiveness?

The second option is correct

b.) How spread out are the values for competitiveness among teenage girls?

The third option is NOT correct. The given problem is a measure of dispersion not central tendency.

c.) What is the central tendency for the variable competitiveness among teenage girls?

The fourth option is also NOT correct. The problem given does not talk about a specific value of the competitiveness variable but rather it talks about the variability of the competitiveness variable

d.) What is the most common value for the variable competitiveness among teenage girls?

Final answer:

a. The probability that your bid will be accepted is 60%. b. The probability that your bid will be accepted is 10%. c. To maximize the probability of getting the property, you should bid $14,599. d. To maximize expected profit, bid $14,599 with an expected profit of $700.50.

Explanation:

a. To calculate the probability that your bid will be accepted, we need to find the probability that your bid is higher than the competitor's bid. Since the competitor's bid, x, is uniformly distributed between $10,400 and $14,600, the probability that the competitor's bid is less than $12,000 is given by:

P(x < 12,000) = (12,000 - 10,400) / (14,600 - 10,400) = 0.4

Therefore, the probability that your bid will be accepted is 1 - P(x < 12,000) = 1 - 0.4 = 0.6, or 60%.

b. Using the same method, the probability that your bid will be accepted when you bid $14,000 is:

P(x < 14,000) = (14,000 - 10,400) / (14,600 - 10,400) = 0.9

Therefore, the probability that your bid will be accepted is 1 - P(x < 14,000) = 1 - 0.9 = 0.1, or 10%.

c. To maximize the probability that you get the property, you should bid an amount that is slightly higher than the competitor's bid, but still below $14,600. This is because if your bid is equal to the competitor's bid, there is a 50% chance that the seller will choose your bid and a 50% chance that the seller will choose the competitor's bid. Therefore, to maximize your chances, you should bid $14,599.

d. If your objective is to maximize the expected profit, you should consider the probability of winning the property and the profit you will make if you win. Since you know someone is willing to pay you $16,000 for the property, the expected profit can be calculated as:

Expected Profit = Probability of Winning × (Selling Price - Bidding Amount)

According to part (c), if you bid $14,599, the probability of winning is 0.5. Therefore, the expected profit is:

Expected Profit = 0.5 × (16,000 - 14,599) = $700.50

a. Probability of bid acceptance when bidding $12,000 is approximately 0.38.

b. Probability of bid acceptance when bidding $14,000 is approximately 0.86.

c. To maximize chances, bid $12,500.

d. For maximum expected profit, bid $13,200, with an expected profit of approximately $3,000.

Let's break down the problem step by step:

a. Probability of bid acceptance when bidding $12,000:

Given that the competitor's bid ( x ) is uniformly distributed between $10,400 and $14,600, we need to find the probability that our bid is higher than $10,400 but less than $14,600.

[tex]\[ P(10400 < x < 12000) = \frac{12000 - 10400}{14600 - 10400} = \frac{1600}{4200} \][/tex]

b. Probability of bid acceptance when bidding $14,000:

[tex]\[ P(10400 < x < 14000) = \frac{14000 - 10400}{14600 - 10400} = \frac{3600}{4200} \][/tex]

c. To maximize the probability of winning, we need to find the midpoint of the range $10,400 to $14,600, which is $12,500. So, we should bid $12,500.

d. Expected profit when bidding $12,500:

If someone is willing to pay $16,000 for the property, and we win the bid, our profit would be $16,000 - $12,500 = $3,500. The probability of winning the bid when bidding $12,500 is the same as calculated in part c. So, the expected profit would be:

[tex]\[ E(Profit) = (Probability \ of \ winning) \times (Profit \ if \ won) \][/tex]

[tex]\[ E(Profit) = \frac{3600}{4200} \times 3500 \][/tex]

Now, let's calculate:

a. [tex]\[ P(10400 < x < 12000) = \frac{1600}{4200} \approx 0.38 \][/tex]

b.[tex]\[ P(10400 < x < 14000) = \frac{3600}{4200} \approx 0.86 \][/tex]

c. Midpoint: $12,500

d. [tex]\[ E(Profit) = \frac{3600}{4200} \times 3500 \approx 3000 \][/tex]

So, to maximize expected profit, the bid should be $13,200.

Answer:

a) [tex] \bar X =117.8[/tex]

[tex] Median= \frac{117+118}{2}=117.5[/tex]

The mode on this case is the most repeated value 128 with a frequency of 3

b) [tex] Range = Max -Min = 150-88=62[/tex]

[tex] s^2 = 225.334[/tex]

[tex] s= \sqrt{225.334}= 15.011[/tex]

c) [tex] y \pm s[/tex]

[tex] Lower = 117.8 -15.011=102.809[/tex]

[tex] Upper = 117.8 +15.011=132.831[/tex]

[tex] y \pm 2s[/tex]

[tex] Lower = 117.8 -2*15.011=87.797[/tex]

[tex] Upper = 117.8 +2*15.011=147.842[/tex]

[tex] y \pm 3s[/tex]

[tex] Lower = 117.8 -3*15.011=72.787[/tex]

[tex] Upper = 117.8 +3*15.011=162.85[/tex]

d) For this case we can calculate the position where we have accumulated 70% of the data below.

50*0.7 = 35

So on the position 35th from the dataset ordered we see that the value is 128 and this value would represent the 70th percentile on this case.

Step-by-step explanation:

For this case we consider the following data:

128,119,95,97,124,128,142,98,108,120,113,109,124,132,97,138,133,136,120,112,146,128,103,135,114,109,100,111,131,113,124,131,133,131,88,118,116,98,112,138,100,112,111,150,117,122,97,116,92,122

Part a

For this case we can calculate the mean with the following formula:

[tex] \bar X = \frac{\sum_{i=1}^{50} X_i}{50}[/tex]

And after replace we got [tex] \bar X =117.8[/tex]

In order to calculate the median first we order the dataset and we got:

88  92  95  97  97  97  98  98 100 100 103 108 109 109 111 111 112 112 112 113 113  114 116 116 117 118 119 120 120 122 122 124 124 124 128 128 128 131 131 131 132 133  133 135 136 138 138 142 146 150

The median would be the average between the position 25 and 26 from the data ordered and we got:

[tex] Median= \frac{117+118}{2}=117.5[/tex]

The mode on this case is the most repeated value 128 with a frequency of 3

Part b

the range is defined as the difference between the maximun and minimum so we got:

[tex] Range = Max -Min = 150-88=62[/tex]

The sample variance can be calculated with this formula:

[tex] s^2 = \frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}[/tex]

And after calculate we got: [tex] s^2 = 225.334[/tex]

And the deviation is just the square root of the variance and we got:

[tex] s= \sqrt{225.334}= 15.011[/tex]

Part c

For this case we can construct the interval with 1 , 2 and 3 deviation from the mean like this:

[tex] y \pm s[/tex]

[tex] Lower = 117.8 -15.011=102.809[/tex]

[tex] Upper = 117.8 +15.011=132.831[/tex]

[tex] y \pm 2s[/tex]

[tex] Lower = 117.8 -2*15.011=87.797[/tex]

[tex] Upper = 117.8 +2*15.011=147.842[/tex]

[tex] y \pm 3s[/tex]

[tex] Lower = 117.8 -3*15.011=72.787[/tex]

[tex] Upper = 117.8 +3*15.011=162.85[/tex]

Part d

For this case we can calculate the position where we have accumulated 70% of the data below.

50*0.7 = 35

So on the position 35th from the dataset ordered we see that the value is 128 and this value would represent the 70th percentile on this case.

Answer:

The amount at the end of 4 years is $2,252.79.

Step-by-step explanation:

The amount formula for the compound interest compounded quarterly is:

[tex]A=P[1+\frac{r}{4}]^{4t}[/tex]

Here,

A = Amount after t years

P = Principal amount

t = number of years

r = interest rate

Given:

P = $1,250, r = 0.15, t = 4 years.

The amount at the end of 4 years is:

[tex]A=P[1+\frac{r}{4}]^{4t}\\=1250\times[1+\frac{0.15}{4}]^{4\times4}\\=1250\times1.80223\\=2252.7875\\\approx2252.79[/tex]

Thus, the amount at the end of 4 years is $2,252.79.

Final answer:

To calculate the amount of money you will have at the end of four years with quarterly deposits and compounded interest, use the formula for compound interest:

. Substituting the given values, the result is approximately $1,776.40.

Explanation:

To calculate the amount of money you will have at the end of four years with quarterly deposits and compounded interest, you can use the formula for compound interest:

A = P(1 + r/n)^(nt)

Where:

A is the total amount of money after the specified time period

P is the principal amount (initial deposit)

r is the annual interest rate (15% in this case)

n is the number of times interest is compounded per year (4 in this case for quarterly compounding)

t is the specified time period in years (4 in this case)

Let's substitute the given values into the formula:

A ≈ 1250 * 1.82212

A ≈ $1,1776.4

Therefore, you will have approximately $1,776.40 at the end of four years.

Answer:

1300 mm  

Step-by-step explanation:

You want to convert metres to millimetres, so you multiply the metres by a conversion factor:

1.3 m × conversion factor = x mm

1 m = 1000 mm.

So,  the conversion factor is either (1 m/1000 mm) or (1000 mm/1 m).

We choose the latter, because it has the desired units on top.

The calculation becomes

[tex]\text{Length} = \text{1.3 m} \times \dfrac{\text{1000 mm}}{\text{1 m}} = \textbf{1300 mm}[/tex]

Answer:

a) [tex] E(X) =13.5*0.17 + 15.9*0.57 + 19.1*0.26 = 16.324[/tex]

[tex] E(X^2) =13.5^2*0.17 + 15.9^2*0.57 + 19.1^2*0.26 = 269.9348[/tex]

[tex] Var(X) = E(X^2) -[E(X)]^2 = 269.9348-(16.324)^2 = 3.462[/tex]

b) [tex] E(Y)= E(28X-8.5) = E(28X) - E(8.5) = 28 E(X) -8.5[/tex]

And replacing the result from part a we got:

[tex] E(Y) = 28*16.324 -8.5= 448.572[/tex]

c) [tex] Var(28X-8.5) = Var (28X)= 28^2 Var(X)= 784*3.462=2714.208[/tex]

d) [tex] E(H) = E(X -0.02 X^2) = E(X) -0.02 E(X^2) = 16.324-0.02(269.9348)= 10.925[/tex]

Step-by-step explanation:

For this case we have the following probability function given:

x 13.5 15.9 19.1

p(x) 0.17 0.57 0.26

The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.

The variance of a random variable X represent the spread of the possible values of the variable. The variance of X is written as Var(X).  

Part a

We can calculate the expected value with the following formula:

[tex] E(X) = \sum_{i=1}^n X_i P(X_i)[/tex]

And if we replace we got:

[tex] E(X) =13.5*0.17 + 15.9*0.57 + 19.1*0.26 = 16.324[/tex]

For the second moment we can use this definition:

[tex] E(X^2) = \sum_{i=1}^n X^2_i P(X_i)[/tex]

And if we replace we got:

[tex] E(X^2) =13.5^2*0.17 + 15.9^2*0.57 + 19.1^2*0.26 = 269.9348[/tex]

The variance is defined:

[tex] Var(X) = E(X^2) -[E(X)]^2 = 269.9348-(16.324)^2 = 3.462[/tex]

Part b

For this case we define this new random variable Y = 28 X -8.5. And we want to find the expected value, so we have this:

[tex] E(Y)= E(28X-8.5) = E(28X) - E(8.5) = 28 E(X) -8.5[/tex]

And replacing the result from part a we got:

[tex] E(Y) = 28*16.324 -8.5= 448.572[/tex]

Part c

For the variance we can use the following property:

[tex]Var(X+Y) = Var(X) + Var(Y) + 2 Cov(X,Y)[/tex]

And using this formula we have:

[tex] Var(28X -8.5) = Var(28X)+ Var(8.5)+ 2 Cov(28X,-8.5)[/tex]

The variance for a constant is 0 so then Var(8.5)=0 and Cov(28X, -8.5) = 0 since by properties if X is a random variable and a represent a constant [tex] Cov(X,a)=0[/tex], so then we just have this:

[tex] Var(28X-8.5) = Var (28X)[/tex]

Using the following property [tex] Var(aX)= a^2 Var(X)[/tex] we have:

[tex] Var(28X-8.5) = Var (28X)= 28^2 Var(X)= 784*3.462=2714.208[/tex]

Part d

For this case we define [tex] H = X -0.02 X^2[/tex]

And if we find the expected value we have this:

[tex] E(H) = E(X -0.02 X^2) = E(X) -0.02 E(X^2) = 16.324-0.02(269.9348)= 10.925[/tex]

To compute E(X), E(X²), and V(X), multiply storage space values by their corresponding probabilities. The expected price paid is found by substituting X into the price equation and computing E(28X - 8.5). To find the expected actual capacity, substitute X into the equation h(X) = X - 0.02X² and compute E(h(X)) by multiplying each storage space value by its corresponding probability and summing up the results.

To compute the expected value (E(X)), we multiply each storage space value by its corresponding probability and sum up the results. For E(X²), we square each storage space value, multiply it by its corresponding probability, and sum up the results. To find the variance (V(X)), we subtract the square of E(X) from E(X²). The expected price paid is found by substituting X into the price equation and computing E(28X - 8.5). The variance of the price is found by substituting X into the price equation, computing V(28X - 8.5), and rounding to the nearest whole number. To find the expected actual capacity, we substitute X into the equation h(X) = X - 0.02X² and compute E(h(X)) by multiplying each storage space value by its corresponding probability and summing up the results.

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Answer:

The system is INCONSISTENT.

Equation of two parallel lines differ only by a constant.

Step-by-step explanation:

The given equations are:

2x + y = -11

6x + 3y = 15

From the graph we see that these lines are parallel. Any two parallel lines never meet. Hence, they do not have a solution.

Hence, the system is called an Inconsistent system.

Also, note that it can be seen easily from the equations are parallel without the help of a graph.

The equation 6x + 3y = 3(2x + y)

The terms (except for the constant term) are proportional. That means they represent parallel lines.

Hence, the answer.

Answer:

1) 0.3

2) 0.64

3) 0.75

4) 0.33                              

Step-by-step explanation:

We are given the following probabilities. We have to probability of that event not happening.

That is we have to find the complement of the event,

Complement of event:

1. P(E) = 0.7

[tex]P(E') = 1 - P(E) = 1 - 0.7 =0.3[/tex]

2. P(E) = 0.36

[tex]P(E') = 1 - P(E) = 1 - 0.36 = 0.64[/tex]

3. P(E) = 1/4

[tex]P(E') = 1 = P(E) = 1 -\dfrac{1}{4} = \dfrac{3}{4} = 0.75[/tex]

4. P(E) = 2/3

[tex]P(E') = 1 = P(E) = 1 -\dfrac{2}{3} = \dfrac{1}{3} = 0.33[/tex]

Answer:

a) Binomial probability distribution. Only two outcomes possible for each machine, with independent probabilities.

b) [tex]P(X = 4) = 0.00045[/tex]

c) [tex]P(X \geq 4) = 0.00046[/tex]

d) [tex]E(X) = 0.5[/tex]

e) [tex]V(X) = 0.45[/tex]

Step-by-step explanation:

For each machine, there is only two possibilities. On a given day, either they will break down, or they will not. The probabilities for each machine breaking down are independent. So we use the binomial probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

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

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

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

And p is the probability of X happening.

The expected value of the binomial distribution is:

[tex]E(X) = np[/tex]

The variance of the binomial distribution is:

[tex]V(X) = np(1-p)[/tex]

In this problem we have that:

[tex]n = 5, p = 0.1[/tex]

a. What is the appropriate probability distribution for x? Explain how x satisfies the properties of the distribution.

Binomial probability distribution. Only two outcomes possible for each machine, with independent probabilities.

b. Compute the probability that 4 machines will break down.

This is [tex]P(X = 4)[/tex].

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

[tex]P(X = 4) = C_{5,4}.(0.1)^{4}.(0.9)^{1} = 0.00045[/tex]

c. Compute the probability that at least 4 machines will break down.

This is

[tex]P(X \geq 4) = P(X = 4) + P(X = 5)[/tex]

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

[tex]P(X = 4) = C_{5,4}.(0.1)^{4}.(0.9)^{1} = 0.00045[/tex]

[tex]P(X = 5) = C_{5,5}.(0.1)^{5}.(0.9)^{0} = 0.00001[/tex]

[tex]P(X \geq 4) = P(X = 4) + P(X = 5) = 0.00045 + 0.00001 = 0.00046[/tex]

d. What is expected number of machines that will break down in a day?

[tex]E(X) = np = 5*0.1 = 0.5[/tex]

e. What is the variance of the number of machines that will break down in a day?

[tex]V(X) = np(1-p) = 5*0.1*0.9 = 0.45[/tex]

The random variable x follows a binomial distribution with n=5 and p=0.1. The probability of exactly 4 machines breaking down is approximately 0.00045, while the probability of at least 4 machines breaking down is approximately 0.00046. The expected number of machines that will break down is 0.5 and the variance is 0.45.

The appropriate probability distribution for x is the binomial distribution. A binomial distribution has two outcomes (a machine breaks down, or it doesn't), a fixed number of trials (5 machines), and a constant probability of success (0.1, a machine breaks down). x satisfies all these properties.

To compute the probability that 4 machines will break down, we can use the binomial theorem: P(x=k) = C(n,k) * (p^k) * (1-p)^(n-k). Here, n=5, k=4, p=0.1. The calculation gives us approximately 0.00045 as the probability that exactly 4 machines will break down.

To compute the probability that at least 4 machines will break down, we calculate the sum of the probabilities that 4 and 5 machines will break down. It's approximately 0.00046.

The expected number of machines that will break down in a day is calculated by n*p, which gives us 0.5 machines.

The variance of the number of machines that will break down in a day is np(1-p), which gives us 0.45.

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Answer:

a. 68% of the workers will earn between $47300 and $69700.

b. 2.5% of workers will earn above $89000

c. Approximately 0

Step-by-step explanation:

The standard normal distribution curve in the attached graph is used to solve this question.

a. The value $47300 is a standard deviation below the mean i.e. 58500-11200=47300. While $69700 is a standard deviation above the mean. I.e. 58500+12000=69700.

Between the first deviation below and above the mean, you have 34%+34%=68% of the salary earners within this range. So we have 68%of staffs earning within this range

b. The second standard deviation above the mean is $80900. i.e. 58500+11200+11200=$80900

We have 50%+13.5%+2.5%= 97.5% earning below $80900. Therefore, 100-97.5= 2.5% of the workers earn above this amount.

c. From the Standard Deviation Rule, the probability is only about (1 -0 .997) / 2 = 0.0015 that a normal value would be more than 3 standard deviations away from its mean in one direction or the other. The probability is only 0.0002 that a normal variable would be more than 3.5 standard deviations above its mean. Any more standard deviations than that, and we generally say the probability is approximately zero.

To answer the question, we use z-scores and a z-table to find the percentages of workers who earn within certain ranges or above certain amounts.

To answer this question, we can use the concept of the standard normal distribution. First, we convert the given earnings into z-scores by subtracting the mean and dividing by the standard deviation. With these z-scores, we can then use a z-table to find the percentage of workers who earn within a certain range or above a certain amount.

a. To find the percentage of workers who earn between $47,300 and $69,700, we need to convert these values into z-scores and find the area between these two z-scores on the z-table.

b. To find the percentage of workers who earn more than $80,900, we need to convert this value into a z-score and find the area to the right of this z-score on the z-table.

c. To determine how likely it is that someone earns more than $100,000, we need to convert this value into a z-score and find the area to the right of this z-score on the z-table.

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Answer:

The two numbers are 37 and 26.

Step-by-step explanation:

This question can be solved by a simple system of equations.

Building the system:

I am going to say that our numbers are x and y.

Two numbers total 63

This means that [tex]x + y = 63[/tex]

difference of 11

This means that [tex]x - y = 11[/tex].

Solving the system

[tex]x + y = 63[/tex]

[tex]x - y = 11[/tex]

Using the addition method

x + x + y - y = 63 + 11

2x = 74

x = 37

[tex]x - y = 11[/tex]

[tex]y = x - 11[/tex]

[tex]y = 37 - 11 = 26[/tex]

The two numbers are 37 and 26.

Answer:

The total 2 numbers are 37 and 26

hope i helped

Answer:

1 spider, 3 cows, 5 birds.

Step-by-step explanation:

Cows (C): 4 legs, 2 eyes.

Birds (B): 2 legs, 2 eyes.

Spiders (S): 8 legs, 4 eyes.

The number of legs and eyes are given, respectively by:

[tex]30 = 4C+2B+8S\\20 = 2C+2B+4S[/tex]

Multiplying the second equation by -2 and adding it to the first one gives us the number of birds:

[tex]30 -40= 4C-4C+2B-4B+8S-8S\\-10 = -2B\\B=5[/tex]

Rewriting the original equations with B =5:

[tex]30 = 4C+2*5+8S\\20 = 2C+2*5+4S\\20 = 4C+8S\\10 = 2C+4S\\C=5-2S[/tex]

Since the number of cows cannot be negative, and both C and S must be odd numbers, the only possible value of S is 1:

[tex]C=5-2S\\S=1\\C=3[/tex]

There are 3 cows, 5 birds, and 1 spider in the field.

To solve this problem, we need to establish equations based on the given information

1. Define Variables

Let c be the number of cows,

b be the number of birds,

and s be the number of spiders

2. Create Equations

The total number of eyes is given to be 20, and knowing that spiders have 4 eyes, birds have 2 eyes, and cows have 2 eyes, we get:

2c + 2b + 4s = 20

Simplifying this, we get:

c + b + 2s = 10  (i)

The total number of legs is given to be 30, with cows having 4 legs, birds having 2 legs, and spiders having 8 legs, we get:

4c + 2b + 8s = 30

Simplifying this, we get:

2c + b + 4s = 15  (ii)

3. Solve Simultaneous Equations

We can subtract equation (i) from (ii):

(2c + b + 4s) - (c + b + 2s) = 15 - 10

c + 2s = 5  (iii)

Substituting (iii) into (i):

(5 - 2s) + b + 2s = 10

5 + b = 10

b = 5;

We know from (iii) that:

c + 2s = 5

Let’s check for odd values for s:

If s = 1, then c = 5 - 2(1) = 3 (valid since cows = 3)

Hence, the solution is:

c = 3, b = 5, s = 1

Checking:

Total eyes = 2(3) + 2(5) + 4(1) = 6 + 10 + 4 = 20 (correct)

Total legs = 4(3) + 2(5) + 8(1) = 12 + 10 + 8 = 30 (correct)

There are 3 cows, 5 birds, and 1 spider in the field.