Mathematics College

There are two machines available for cutting corks intended for use in bottles. The first produces corks with diameters that are normally distributed with mean 3 cm and standard deviation 0.1 cm. The second machine produces corks with diameters that have a normal distribution with mean 3.04 cm and standard deviation 0.02 cm. Acceptable corks have diameters between 2.9 cm and 3.1 cm. Which machine is more likely to produce an acceptable cork? What should the acceptable range for cork diameters be (from 3 − d cm to 3 + d cm) to be 90% certain for the first machine to produce an acceptable cork?

Answers

Answer 1

Answer:

a) The second machine is more likely to produce an acceptable cork.

b) Acceptable range for cork diameters produced by the first machine with a 90% confidence = (2.8355, 3.1645)

Step-by-step explanation:

This is a normal distribution problem

For the first machine,

Mean = μ = 3 cm

Standard deviation = σ = 0.1 cm

And we want to find which percentage of the population falls between 2.9 cm and 3.1 cm.

P(2.9 ≤ x ≤ 3.1) = P(x ≤ 3.1) - P(x ≤ 2.9)

We first standardize this measurements.

The standardized score for any value is the value minus the mean then divided by the standard deviation.

For 2.9 cm

z = (x - μ)/σ = (2.9 - 3.0)/0.1 = - 1.00

For 3.1 cm

z = (x - μ)/σ = (3.1 - 3.0)/0.1 = 1.00

P(x ≤ 3.1) = P(z ≤ 1.00) = 0.841

P(x ≤ 2.9) = P(z ≤ -1.00) = 0.159

P(2.9 ≤ x ≤ 3.1) = P(-1.00 ≤ z ≤ 1.00) = P(z ≤ 1.00) - P(z ≤ -1.00) = 0.841 - 0.159 = 0.682 = 68.2%

This means that 68.2% of the diameter of corks produced by the first machine lies between 2.9 cm and 3.1 cm.

For the second machine,

Mean = μ = 3.04 cm

Standard deviation = σ = 0.02 cm

And we want to find which percentage of the population falls between 2.9 cm and 3.1 cm.

P(2.9 ≤ x ≤ 3.1) = P(x ≤ 3.1) - P(x ≤ 2.9)

We standardize this measurements.

The standardized score for any value is the value minus the mean then divided by the standard deviation.

For 2.9 cm

z = (x - μ)/σ = (2.9 - 3.04)/0.02 = - 7.00

For 3.1 cm

z = (x - μ)/σ = (3.1 - 3.0)/0.02 = 3.00

P(x ≤ 3.1) = P(z ≤ 3.00) = 0.999

P(x ≤ 2.9) = P(z ≤ -7.00) = 0.0

P(2.9 ≤ x ≤ 3.1) = P(-7.00 ≤ z ≤ 3.00) = P(z ≤ 3.00) - P(z ≤ -7.00) = 0.999 - 0.0 = 0.999 = 99.9%

This means that 99.9% of the diameter of corks produced by the second machine lies between 2.9 cm and 3.1 cm.

Hence, we can conclude that the second machine is more likely to produce an acceptable cork.

b) Margin of error = (z-multiplier) × (standard deviation of the population)

For 90% confidence interval, z-multiplier = 1.645 (from literature and the z-tables)

Standard deviation for first machine = 0.1

Margin of error, d = 1.645 × 0.1 = 0.1645.

The acceptable range = (mean ± margin of error)

Mean = 3

Margin of error = 0.1645

Lower limit of the acceptable range = 3 - d = 3 - 0.1645 = 2.8355

Upper limit of the acceptable range = 3 + d = 3 + 0.1645 = 3.1645

Acceptable range = (2.8355, 3.1645)

Answer 2

Final answer:

To determine which machine is more likely to produce acceptable corks, we examine their distribution characteristics. Machine 2 may be more reliable due to its tighter control despite a slightly higher mean. To find a 90% certain acceptable range for Machine 1, we calculate using its standard deviation and the z-score for the 90th percentile.

Explanation:

The question involves comparing two machines based on their ability to produce corks within a specified acceptable diameter range using normal distribution properties, and calculating the range for diameters to ensure a 90% certainty of producing acceptable corks for the first machine.

Comparing the Two Machines

For the first machine with a mean diameter of 3 cm and a standard deviation of 0.1 cm, and the second machine with a mean diameter of 3.04 cm and a standard deviation of 0.02 cm, the question is which machine is more likely to produce corks within the acceptable range of 2.9 cm to 3.1 cm.

Machine 1 produces corks closer to the center of the acceptable range but with a wider spread (higher standard deviation), while Machine 2 produces corks that are skewed slightly larger but with a much tighter spread around their mean (lower standard deviation). To determine which machine is more likely to produce acceptable corks, we would need to calculate the z-scores for the acceptance limits for both machines and compare the probabilities. However, intuitively, Machine 2 might be seen as more reliable due to its tighter control (lower standard deviation), assuming its mean is not too far out of the acceptable range.

Finding the Acceptable Range for 90% Certainty

To ensure 90% certainty that a cork produced by Machine 1 falls within an acceptable diameter range, we need to determine d in the range of 3 − d cm to 3 + d cm. This involves finding the z-score that corresponds to the 5th and 95th percentiles due to the symmetric nature of normal distribution, then solving for d using the properties of normal distribution and the given standard deviation of 0.1 cm.

The z-score corresponding to the 5th and 95th percentiles (for a 90% certainty) typically falls around ±1.645. Using the formula for z-score, which is (X − μ) / σ, and solving for d, we can find the acceptable range of diameters for the first machine to produce an acceptable cork with 90% certainty.

Related Questions

In remodeling a kitchen, a builder decides to place a splash-guard behind the sink consisting of eight 6-inch-square ceramic tiles decorated with different botanical herbs. The tiles will be installed in a custom-made wooden panel. The tile supplier has 12 different herb designs to choose from, and the builder selects 8 of these 12 at random. Suppose the order in which the tiles are arranged on the splash-guard does not matter. Fill in the blanks. (Give your answers to four decimal places.)

Two of the 12 herb tiles contain a blue tint that matches the kitchen color scheme. The probability that these 2 tiles will be included in the splash-guard is __.

The family actually grows 5 of the 12 herbs in a backyard garden. The probability that all 5 of these will be included on the splash-guard is _

Answers

Answer:

(a) 0.4242

(b) 0.0707

Step-by-step explanation:

The total number of ways of selecting 8 herbs from 12 is

[tex]\binom{12}{8}=495[/tex]

(a) If 2 herbs are selected, then there are 8 - 2 = 6 herbs to be selected from 12 - 8 = 10. The number of ways of the selection is then

[tex]\binom{10}{6}= 210[/tex]

Note that this is the number of ways that both are included. We would have multiplied by 2! if any of them were to be included.

The probability = [tex]\dfrac{210}{495}=0.4242[/tex]

(b) If 5 herbs are selected, then there are 8 - 5 = 3 herbs to be selected from 12 - 5 = 7. The number of ways of the selection is then

[tex]\binom{7}{3}= 35[/tex]

This is the number of ways that both are included. We would have multiplied by 5! if any of them were to be included. In that case, our probability will exceed 1; this implies that certainly, at least, one of them is included.

The probability = [tex]\dfrac{35}{495}=0.0707[/tex]

The probability that the 2 tiles with a blue tint will be included in the splash-guard is 0.0455. The probability that all 5 of the herbs grown in the backyard garden will be included on the splash-guard is 0.0058.

The probability that the 2 tiles with a blue tint will be included in the splash-guard is 0.0455.

The probability that all 5 of the herbs grown in the backyard garden will be included on the splash-guard is 0.0058.

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A branch of a certain bank has six ATMs. Let X represent the number of machines in use at a particular time of day. The cdf of X is as follows: F(x) = 0 x < 0 0.08 0 ≤ x < 1 0.20 1 ≤ x < 2 0.40 2 ≤ x < 3 0.62 3 ≤ x < 4 0.83 4 ≤ x < 5 0.99 5 ≤ x < 6 1 6 ≤ x

Answers

Final answer:

The question pertains to calculating probabilities from a given discrete Cumulative Distribution Function (CDF) for the number of ATMs in use. By utilizing the step function provided in the CDF, probabilities for specific intervals or exact values can be determined.

Explanation:

The student's question involves understanding the Cumulative Distribution Function (CDF) for a discrete random variable, specifically the number of ATMs in use at a particular time.

The CDF is used to find the probability that the random variable X, representing the number of ATMs used, is less than or equal to a certain value x.

In this scenario, we have a step function showing the probabilities associated with different intervals of x. The function increases as x increases, which is typical of a CDF, as it accumulates the probabilities for all values up to x.

To calculate probabilities for specific intervals from the given CDF, one would subtract the CDF value just before the interval from the CDF value at the end of the interval. For example, to find the probability that exactly four ATMs are in use (P(X=4)), one would calculate P(X<5) - P(X<4), which is 0.83 - 0.62 = 0.21.

A 16-inch candle is lit and burns at a constant rate of 0.8 inches per hour. Let t represent the number of hours since the candle was lit, and suppose f is a function such that f(t) represents the remaining length of the candle (in inches) t hours after it was lit. a. Write a function formula for f. f(t)- 16-0.8t Previewb. What is the domain of f relative to this context? c. What is the range of f relative to this context?

Answers

Answer:

See the explanation.

Step-by-step explanation:

The initial length of the candle is 16 inch. It also given that, it burns with a constant rate of 0.8 inch per hour.

After one hour since the candle was lit, the length of the candle will be (16 - 0.8) = 15.2 inch.

After two hour since the candle was lit, the length of the candle will be (15.2 - 0.8) = 14.4 inch. The length of the candle after two hours can also be represented by {16 - 2(0.8)}.

Hence, the length of the candle after t hours when it was lit can be represented by the function, [tex]f(t) = 16 - 0.8t[/tex]. [tex]f(t) = 0[/tex] at t = 20.

The domain of the function is 0 to 20.

The range is 0 to 16.

Final answer:

The function formula for the given context is f(t) = 16 - 0.8*t. The domain of this function is from 0 to 20 hours, and the range is from 0 to 16 inches.

Explanation:

The function you want to define for the scenario presented represents the remaining length, in inches, of the candle after being lit for a certain number of hours (t). Given that the initial length of the candle is 16 inches and burns at the constant rate of 0.8 inches per hour, the function will follow a linear format: f(t) = 16 - 0.8*t.

The domain of this function, referring to the permissible values for t in this context, would be any value greater than or equal to 0 and less than or equal to 20 (since it would take 20 hours for the candle to completely burn down). Therefore, the domain of this function is [0, 20].

The range of this function will be the possible lengths the candle could have, depending on the time it has burned. As the candle was 16 inches long initially and reduces to 0 after burning for 20 hours, the range lies between 0 and 16 (inclusive). Thus, the range of this function is [0, 16].

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(1 point) Suppose we will flip a fair coin 100 times. (a) What does 35 heads correspond to on the standard scale? (enter exact answer) (b) What does z=2.4 on the standard scale correspond to on the number of heads scale? (enter exact answer)

Answers

Answer:

(a) 35 heads correspond to -3 on the standard scale.

(b) z = 2.4 corresponds to 62 heads on the number of heads scale.

Step-by-step explanation:

(a) If we flip a fair coin once, probability of getting head = 0.5

If we flip a fair coin 100 times, mean number of heads = 100(0.5) = 50

If there are N draws with a P probability of success, the standard deviation (SD) is given as:

[tex]SD = \sqrt{(N)(P)(1 - P)}[/tex]

Here, the probability of getting a head (P) is 0.5 while the number of draws (N) is 100. So,

[tex]SD = \sqrt{(100)(0.5)(1-0.5)}[/tex]

SD = 5

The standard scale value is: (35 - 50) / 5 = -3

Hence, 35 heads correspond to -3 on the standard scale.

(b) The standard scale value is 2.4 and we need to find the number of heads.

(X - 50) / 5 = 2.4

X - 50 = 12

X = 62

Hence, z = 2.4 on the standard scale corresponds to 62 on the number of heads scale.

The 35 heads corresponds to -3 on the standard scale, and the number of heads is 62

The head on the standard scale

The given parameter is:

n = 100

In a flip of a coin, the probability of a head is:

p = 1/2

So, the mean of the distribution is:

[tex]\bar x = np[/tex]

[tex]\bar x = 100 * 1/2[/tex]

[tex]\bar x = 50[/tex]

The standard deviation is:

[tex]\sigma = \sqrt{np(1 - p)}[/tex]

So, we have:

[tex]\sigma = \sqrt{100 * 1/2 * (1 - 1/2)}[/tex]

[tex]\sigma = 5[/tex]

The corresponding value on the standard scale is then calculated as:

[tex]z= \frac{x - \mu}{\sigma}[/tex]

For 35 heads, we have:

[tex]z= \frac{35 - 50}{5}[/tex]

[tex]z = -3[/tex]

Hence, the 35 heads corresponds to -3 on the standard scale

(b) The number of heads

We have:

[tex]z= \frac{x - \mu}{\sigma}[/tex]

When z = 2.4, the equation becomes

[tex]2.4= \frac{x - 50}{5}[/tex]

Multiply both sides by 5

[tex]12= x - 50[/tex]

Add 50 to both sides

[tex]x = 62[/tex]

Hence, the number of heads is 62

Answers

The rate of change of the distance between the two cars at that instant is 100344 meters per second.

Here, we have,

To find the rate of change of the distance between the two cars at a certain instant, we can use the concept of relative velocity.

The relative velocity between the two cars is the difference between their velocities.

Since the cars are moving towards each other, the relative velocity will be the sum of their individual velocities.

Let's denote the distance between the two cars at the certain instant as

d (in meters). The rate of change of this distance d with respect to time (t) is the derivative of d with respect to t (i.e., d/dt).

At the given instant, the first car's velocity (V₁) is 101010 meters per second, and the second car's velocity (V₂) is 666 meters per second.

The distance traveled by the first car (d₁) at the given instant is 444 meters, and the distance traveled by the second car (d₂) is 333 meters.

The distance between the two cars at the given instant (d) is the difference between the distances traveled by the first and second cars:

d=d₁−d₂

Now, let's find the rate of change of d with respect to time (t):

d/dt = d₁/dt - d₂/dt

Since the velocities of the cars are constants (not changing with time), the derivatives of d₁ and d₂ with respect to time are simply their velocities:

d/dt =V₁−V₂

Now, substitute the given values for V₁ and V₂:

d/dt=101010meters per second−666meters per second

d/dt =100344meters per second

Therefore, the rate of change of the distance between the two cars at that instant is 100344 meters per second.

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Final answer:

The rate of change of the distance between the cars at the given instant is 1210 m/s. We used the Pythagorean theorem to express the distance between the two cars at any time 't' and then differentiated this distance with respect to 't' to get the relative speed of the cars.

Explanation:

To calculate the rate of change of the distance between the two cars at a certain instance, we can use the Pythagorean theorem. Consider the two cars as points moving along perpendicular paths. At any time 't', car 1 is '1010t' m away from the intersection and car 2 is '666t' m away from the intersection. The distance 'd' between them is √[(1010t)² + (666t)²] which simplifies to d = 1210t.

The rate of change of the distance with respect to time, i.e., the derivative of 'd' with respect to 't' (dd/dt) would be the relative speed of the two cars at any instant. This could be found by differentiating 'd' with respect to 't'. Thus, at the moment when car 1 is 444 meters away from the crossing and car 2 is 333 meters away from the crossing, the rate of change of the distance between the cars (the relative velocity) at that instant would be d/dt = 1210 m/s.

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[6 points] Prove Bayes’ Theorem. Briefly explain why it is useful for machine learning problems, i.e., by converting posterior probability to likelihood and prior probability.

Answers

Answer: Mathematically Bayes’ theorem is defined as

P(A\B)=P(B\A) ×P(A)

P(B)

Bayes theorem is defined as where A and B are events, P(A|B) is the conditional probability that event A occurs given that event B has already occurred (P(B|A) has the same meaning but with the roles of A and B reversed) and P(A) and P(B) are the marginal probabilities of event A and event B occurring respectively.

Step-by-step explanation: for example, picking a card from a pack of traditional playing cards. There are 52 cards in the pack, 26 of them are red and 26 are black. What is the probability of the card being a 4 given that we know the card is red?

To convert this into the math symbols that we see above we can say that event A is the event that the card picked is a 4 and event B is the card being red. Hence, P(A|B) in the equation above is P(4|red) in our example, and this is what we want to calculate. We previously worked out that this probability is equal to 1/13 (there 26 red cards and 2 of those are 4's) but let’s calculate this using Bayes’ theorem.

We need to find the probabilities for the terms on the right-hand side. They are:

P(B|A) = P(red|4) = 1/2

P(A) = P(4) = 4/52 = 1/13

P(B) = P(red) = 1/2

When we substitute these numbers into the equation for Bayes’ theorem above we get 1/13, which is the answer that we were expecting.

Final answer:

Bayes’ theorem, featuring components of likelihood and prior probability, is fundamental to probability theory and statistics. It's a tool that allows us to update our previous assumptions based on new data, making it essential in machine learning.

Explanation:

Bayes’ theorem is a fundamental theorem in probability theory and statistics, named after Thomas Bayes. It formalizes the process of updating probabilities based on new data. The theorem can be derived from the definition of conditional probability, P(A|B), which is the probability of event A given that event B has occurred.

Bayes' theorem is generally represented by the formula:

P(A|B) = (P(B|A) * P(A)) / P(B)

Bayes’ theorem has two main components:

Likelihood: P(B|A) shows the likelihood of the data under a particular hypothesis.Prior Probability: P(A) represents our prior belief in the hypothesis before seeing any data.

In the realm of Machine Learning, Bayes' theorem allows us to update our previous assumptions (priors) with data to get updated, more accurate results (posterior). The prior and posterior are probability distributions over the same event space, but Bayesian inference allows us to adjust our initial assumptions in light of new evidence, making it a dynamic and flexible method.

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A company is interested in evaluating its current inspection procedure on large shipments of identical items. The procedure is to take a sample of 5 items and pass the shipment if no more than 1 item is found to be defective. It is known that items are defective at a 10% rate overall. a. What is the probability that the inspection procedure will pass the shipment? b. What is the expected number of defectives in this process of inspecting 5 items? c. If items are defective at a 20% rate overall, what is the probability that you will find 4 defectives in a sample of 5?

Answers

Answer:

(a) The probability that the inspection procedure will pass the shipment is 0.9185.

(b) The expected number of defectives in this process of inspecting 5 items is 0.50.

Step-by-step explanation:

Let X = number of defective items.

The probability of selecting a defective item is, p = 0.10.

A sample of n = 5 items is selected at random.

The random variable X follows a Binomial distribution with parameters n = 5 and p = 0.10.

The probability mass function of X is:

[tex]P(X=x)={5\choose x}0.10^{x}(1-0.10)^{5-x};\ x=0,1,2,3....[/tex]

It is provided that the shipment will pass if there are no more than 1 defective items in the selected 5 units.

(a)

Compute the probability that the inspection procedure will pass the shipment as follows:

P (X ≤ 1) = P (X = 0) + P (X = 1)

[tex]={5\choose 0}0.10^{0}(1-0.10)^{5-0}+{5\choose 1}0.10^{1}(1-0.10)^{5-1}\\=(1\times 1\times 0.59049) + (5\times 0.10\times 0.6561)\\=0.59049+0.32805\\=0.91854\\\approx0.9185[/tex]

Thus, the probability that the inspection procedure will pass the shipment is 0.9185.

(b)

The expected value of a Binomial distribution is:

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

Compute the expected number of defectives in this process of inspecting 5 items as follows:

[tex]E(X)=5\times0.10=0.50[/tex]

Thus, the expected number of defectives in this process of inspecting 5 items is 0.50.

(c)

The probability of finding a defective is now changed to 0.20.

Compute the probability that there will be 4 defectives in a sample of 5 as follows;

[tex]P(X=4)={5\choose 4}0.20^{4}(1-0.20)^{5-4}\\=5\times0.0016\times0.20\\=0.0016[/tex]

Thus, the probability that there will be 4 defectives in a sample of 5 is 0.0016.

Answer:

(a) Probability that the inspection procedure will pass the shipment = 0.91854

(b) E(X) = [tex]5 \times 0.1[/tex] = 0.5

(c) Probability of 4 defectives in a sample of 5 is 0.0064

Step-by-step explanation:

We are given that a company is interested in evaluating its current inspection procedure on large shipments of identical items. The procedure is to take a sample of 5 items and pass the shipment if no more than 1 item is found to be defective. It is known that items are defective at a 10% rate overall.

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 = 5 items

r = number of success

p = probability of success which in our question is % of items that are

defective, i.e., 10%

LET X = Number of defective items

Also, it is given that a sample of 5 items is taken,

So, it means X ~ [tex]Binom(n=5,p=0.10)[/tex]

(a) Probability that the inspection procedure will pass the shipment is given by the fact that if no more than 1 item is found to be defective, then only shipment is passed, i.e.;

P(X [tex]\leq[/tex] 1) = P(X = 0) + P(X = 1)

= [tex]\binom{5}{0}0.1^{0} (1-0.1)^{5-0} + \binom{5}{1}0.1^{1} (1-0.1)^{5-1}[/tex]

= [tex]1 \times 1 \times 0.9^{5} +5 \times 0.1 \times 0.9^{4}[/tex]

= 0.59049 + 0.32805 = 0.91854

(b) The expected number of defectives in this process of inspecting 5 items is given by = E(X) or Mean of X

So, mean of binomial distribution is, E(X) = [tex]n \times p[/tex]

So, E(X) = [tex]5 \times 0.1[/tex] = 0.5

Therefore, the expected number of defectives in this process of inspecting 5 items is 1 (after rounding to nearest integer).

(c) Now, the probability of success which is % of items that are defective is changed to 20% rate overall, i.e., p = 0.20 now.

So, probability that you will find 4 defectives in a sample of 5 = P(X = 4)

P(X = 4) = [tex]\binom{5}{4}0.2^{4} (1-0.2)^{5-4}[/tex]

= [tex]5 \times 0.2^{4} \times 0.8^{1}[/tex] = 0.0064

According to Scarborough Research, more than 85% of working adults commute by car. Of all U.S. cities, Washington, D.C., and New York City have the longest commute times. A sample of 25 commuters in the Washington, D.C., area yielded the sample mean commute time of 27.97 minutes and sample standard deviation of 10.04 minutes. Construct and interpret a 99% confidence interval for the mean commute time of all commuters in Washington D.C. area.

Answers

Answer:

The 99% confidence interval for the mean commute time of all commuters in Washington D.C. area is (22.35, 33.59).

Step-by-step explanation:

The (1 - α) % confidence interval for population mean (μ) is:

[tex]CI=\bar x\pm z_{\alpha /2}\frac{\sigma}{\sqrt{n}}[/tex]

Here the population standard deviation (σ) is not provided. So the confidence interval would be computed using the t-distribution.

The (1 - α) % confidence interval for population mean (μ) using the t-distribution is:

[tex]CI=\bar x\pm t_{\alpha /2,(n-1)}\frac{s}{\sqrt{n}}[/tex]

Given:

[tex]\bar x=27.97\\s=10.04\\n=25\\t_{\alpha /2, (n-1)}=t_{0.01/2, (25-1)}=t_{0.005, 24}=2.797[/tex]

*Use the t-table for the critical value.

Compute the 99% confidence interval as follows:

[tex]CI=27.97\pm 2.797\times\frac{10.04}{\sqrt{25}}\\=27.97\pm5.616\\=(22.354, 33.586)\\\approx(22.35, 33.59)[/tex]

Thus, the 99% confidence interval for the mean commute time of all commuters in Washington D.C. area is (22.35, 33.59).

A large sample of tires from cabs driven within a city have an average tire tread depth of 0.25cm at the end of the winter. If the city’s average winter tire tread depth is 2.2cm, with a standard deviation of 0.33cm at the end of the same winter, what is the probability cab tires depths would be shallower than 0.25cm.

Answers

Answer:

0% probability cab tires depths would be shallower than 0.25cm.

Step-by-step explanation:

Problems of normally distributed samples can be 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 = 2.2, \sigma = 0.33[/tex]

What is the probability cab tires depths would be shallower than 0.25cm.

This probability is the pvalue of Z when X = 0.25. So

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

[tex]Z = \frac{0.25 - 2.2}{0.33}[/tex]

[tex]Z = -5.9[/tex]

[tex]Z = -5.9[/tex] has a pvalue of 0.

0% probability cab tires depths would be shallower than 0.25cm.

Final answer:

The Z-score for a tire tread depth of 0.25 cm when compared to the city's average tread depth is -5.91, which suggests an extremely low probability, nearly zero, of finding a tire tread depth shallower than 0.25 cm.

Explanation:

To determine the probability that the tire tread depth for cab tires is shallower than 0.25 cm, we will use the standard normal distribution and the process of finding a Z-score. We are given the population mean (μ) to be 2.2 cm and the population standard deviation (σ) to be 0.33 cm.

The Z-score is calculated as:

Z = (X - μ) / σ

Where:

X is the value we are comparing to the mean (0.25 cm in this case)μ is the mean tire tread depth (2.2 cm)σ is the standard deviation of the tire tread depth (0.33 cm)

So, the Z-score for 0.25 cm is:

Z = (0.25 - 2.2) / 0.33 = -5.91

A Z-score of -5.91 is significantly beyond the typical range for a standard normal distribution table, which indicates that the probability of having a tire tread depth of less than 0.25 cm is extremely low, essentially approaching zero.

You are collecting a sample of 60 data points from a population that you know to follow an exponential distribution. It is not appropriate to use the results of the Central Limit Theorem (CLT) to calculate the confidence intervals for the mean.

Answers

Answer:

For the exponential distribution:

[tex] \mu = \frac{1}{\lambda}[/tex]

[tex] \sigma^2 = \frac{1}{\lambda^2}[/tex]

We know that the exponential distribution is skewed but the sample mean for this case using a sample size of 60 would be approximately normal, so then we can conclude that if we have a sample size like this one and an exponential distribution we can approximate the sample mean to the noemal distribution and indeed use the Central Limit theorem.

[tex] \bar X \sim N(\mu_{\bar X} , \frac{\sigma}{\sqrt{n}})[/tex]

[tex] \mu_{\bar X} = \bar X[/tex]

[tex]\sigma_{\bar X}= \frac{\sigma}{\sqrt{n}}[/tex]

Step-by-step explanation:

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

For this case we have a large sample size n =60 >30

The exponential distribution is the probability distribution that describes the time between events in a Poisson process.

For the exponential distribution:

[tex] \mu = \frac{1}{\lambda}[/tex]

[tex] \sigma^2 = \frac{1}{\lambda^2}[/tex]

[tex] \bar X \sim N(\mu_{\bar X} , \frac{\sigma}{\sqrt{n}})[/tex]

[tex] \mu_{\bar X} = \bar X[/tex]

[tex]\sigma_{\bar X}= \frac{\sigma}{\sqrt{n}}[/tex]

Let's use the results of the 2012 presidential election as our x0. Looking up the popular vote totals, we find that our initial distribution vector should be (0.5106, 0.4720, 0.0075, 0.0099)T. Enter the matrix P and this vector x0 in MATLAB:

Answers

Answer:

The code is as given below to be copied in a new matlab script m file. The screenshots are attached.

Step-by-step explanation:

As the question is not complete, the complete question is attached herewith.

The code for the problem is as follows:

%Defining the given matrices:

%P is the matrix showing the percentage of changes in voterbase

P = [ 0.8100 0.0800 0.1600 0.1000;

0.0900 0.8400 0.0500 0.0800;

0.0600 0.0400 0.7400 0.0400;

0.0400 0.0400 0.0500 0.7800];

%x0 is the vector representing the current voterbase

x0 = [0.5106; 0.4720; 0.0075; 0.0099];

%In MATLAB, the power(exponent) operator is defined by ^

%After 3 elections..

x3 = P^3 * x0;

disp("The voterbase after 3 elections is:");

disp(x3);

%After 6 elections..

x3 = P^6 * x0;

disp("The voterbase after 6 elections is:");

disp(x3);

%After 10 elections..

x10 = P^10 * x0;

disp("The voterbase after 10 elections is:");

disp(x10);

%After 30 elections..

x30 = P^30 * x0;

disp("The voterbase after 30 elections is:");

disp(x30);

%After 60 elections..

x60 = P^60 * x0;

disp("The voterbase after 60 elections is:");

disp(x60);

%After 100 elections..

x100 = P^100 * x0;

disp("The voterbase after 100 elections is:");

disp(x100);

The output is as well as the code in the matlab is as attached.

Answer:

The voter-base after 3 elections is:

0.392565, 0.400734, 0.109855, 0.096846

The voter-base after 6 elections is:

0.36168, 0.36294, 0.14176, 0.13362

The voter-base after 10 elections is:

0.35405, 0.34074, 0.15342, 0.15178

Step-by-step explanation:

This question is incomplete. I will proceed to give the complete question. Then I will add a screenshot of my code solution to this question. After which I will give the expected outputs.

P = [ 0.8100 0.0800 0.1600 0.1000;

0.0900 0.8400 0.0500 0.0800;

0.0600 0.0400 0.7400 0.0400;

0.0400 0.0400 0.0500 0.7800];

x0 = [0.5106; 0.4720; 0.0075; 0.0099];

According to our model, what should the party distribution vector be after three, six and ten elections?

Please find the code solution in the images attached to this question.

The voter-base after 3 elections is therefore:

0.392565, 0.400734, 0.109855, 0.096846

The voter-base after 6 elections is therefore:

0.36168, 0.36294, 0.14176, 0.13362

The voter-base after 10 elections is therefore:

0.35405, 0.34074, 0.15342, 0.15178

Set up but do not solve for the appropriate particular solution yp for the differential equation y′′+4y=5xcos(2x) using the Method of Undetermined Coefficients (primes indicate derivatives with respect to x). In your answer, give undetermined coefficients as A, B, etc.

Answers

Answer:

So, solution of the differential equation is

[tex]y(t)=-\frac{5x^2}{4}\cot 2x\cdot \cos 2x+c_1e^{-2it}+c_2e^{2it}\\[/tex]

Step-by-step explanation:

We have the given differential equation: y′′+4y=5xcos(2x)

We use the Method of Undetermined Coefficients.

We first solve the homogeneous differential equation y′′+4y=0.

[tex]y''+4y=0\\\\r^2+4=0\\\\r=\pm2i\\\\[/tex]

It is a homogeneous solution:

[tex]y_h(t)=c_1e^{-2i t}+c_2e^{2i t}[/tex]

Now, we finding a particular solution.

[tex]y_p(t)=A5x\cos 2x\\\\y'_p(t)=A5\cos 2x-A10x\sin 2x\\\\y''_p(t)=-A20\sin 2x-A20x\cos 2x\\\\\\\implies y''+4y=5x\cos 2x\\\\-A20\sin 2x-A20x\cos 2x+4\cdot A5x\cos 2x=5x\cos 2x\\\\-A20\sin 2x=5x\cos 2x\\\\A=-\frac{x}{4} \cot 2x\\[/tex]

we get

[tex]y_p(t)=A5\cos 2x\\\\y_p(t)=-\frac{5x^2}{4}\cot 2x\cdot \cos 2x\\\\\\y(t)=y_p(t)+y_h(t)\\\\y(t)=-\frac{5x^2}{4}\cot 2x\cdot \cos 2x+c_1e^{-2it}+c_2e^{2it}\\[/tex]

So, solution of the differential equation is

[tex]y(t)=-\frac{5x^2}{4}\cot 2x\cdot \cos 2x+c_1e^{-2it}+c_2e^{2it}\\[/tex]

Final answer:

To set up the particular solution for the differential equation using the Method of Undetermined Coefficients, we propose a function similar to the nonhomogeneous term with unknown coefficients. The form is a combination of polynomial and trigonometric terms, reflecting the equation's right-hand side.

Explanation:

To set up the particular solution yp for the differential equation y″+4y=5xcos(2x) using the Method of Undetermined Coefficients, first identify the form of the nonhomogeneous term 5xcos(2x). Since it's a product of a polynomial and a trigonometric function, we anticipate a particular solution that mirrors this form, but with unknown coefficients to solve for.

The lowest degree polynomial in the nonhomogeneous term is x, so we begin with Ax and will also need to account for the derivative of cosine, which is sine. Therefore, our yp will be of the form:

yp = x(Acos(2x) + Bsin(2x)) + Cx²cos(2x) + Dx²sin(2x).

Here, A, B, C, and D are the undetermined coefficients that will be found by differentiating this assumed form and substituting back into the original differential equation.

A player of a video game is confronted with a series of opponents and has an 80% probability of defeating each one. Success with any opponent is independent of previous encounters. Until defeated, the player continues to contest opponents. a. What is the probability mass function of the number of opponents contested in a game? b. What is the probability that a player defeats at least two opponents in a game? c. What is the expected number of opponents contested in a game? d. What is the probability that a player contests four or more opponents in a game? e. What is the expected number of game plays until a player contests four or more opponents?

Answers

Answer:

(a) The PMF of X is: [tex]P(X=k)=(1-0.20)^{k-1}0.20;\ k=0, 1, 2, 3....[/tex]

(b) The probability that a player defeats at least two opponents in a game is 0.64.

(d) The probability that a player contests four or more opponents in a game is 0.512.

(e) The expected number of game plays until a player contests four or more opponents is 2.

Step-by-step explanation:

Let X = number of games played.

It is provided that the player continues to contest opponents until defeated.

(a)

The random variable X follows a Geometric distribution.

The probability mass function of X is:

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

It is provided that the player has a probability of 0.80 to defeat each opponent. This implies that there is 0.20 probability that the player will be defeated by each opponent.

Then the PMF of X is:

[tex]P(X=k)=(1-0.20)^{k-1}0.20;\ k=0, 1, 2, 3....[/tex]

(b)

Compute the probability that a player defeats at least two opponents in a game as follows:

P (X ≥ 2) = 1 - P (X ≤ 2)

= 1 - P (X = 1) - P (X = 2)

[tex]=1-(1-0.20)^{1-1}0.20-(1-0.20)^{2-1}0.20\\=1-0.20-0.16\\=0.64[/tex]

Thus, the probability that a player defeats at least two opponents in a game is 0.64.

(c)

The expected value of a Geometric distribution is given by,

[tex]E(X)=\frac{1}{p}[/tex]

Compute the expected number of opponents contested in a game as follows:

[tex]E(X)=\frac{1}{p}=\frac{1}{0.20}=5[/tex]

Thus, the expected number of opponents contested in a game is 5.

(d)

Compute the probability that a player contests four or more opponents in a game as follows:

P (X ≥ 4) = 1 - P (X ≤ 3)

= 1 - P (X = 1) - P (X = 2) - P (X = 3)

[tex]=1-(1-0.20)^{1-1}0.20-(1-0.20)^{2-1}0.20-(1-0.20)^{3-1}0.20\\=1-0.20-0.16-0.128\\=0.512[/tex]

Thus, the probability that a player contests four or more opponents in a game is 0.512.

(e)

Compute the expected number of game plays until a player contests four or more opponents as follows:

[tex]E(X\geq 4)=\frac{1}{P(X\geq 4)}=\frac{1}{0.512}=1.953125\approx 2[/tex]

Thus, the expected number of game plays until a player contests four or more opponents is 2.

Let X denote the number of bars of service on your cell phone whenever you are at an intersection with the following probabilities: x 01 2 3 4 5 0.1 0.15 0.25 0.25 0.15 0.1 Determine the following: (a) F(x) (b) Mean and variance (c) P(X 2) (d) P(X 3.5)

Answers

Answer:

Step-by-step explanation:

Hello!

Given the variable

X: number of bars of service on your cell phone.

The posible values of this variable are {0, 1, 2, 3, 4, 5}

a. F(X) is the cummulative distribution function P(X≤x₀) you can calculate it by adding all the point probabilities for each value:

X: 0; 1; 2; 3; 4; 5

P(X): 0.15; 0.25; 0.25; 0.15; 0.1; 0.1

F(X): 0.15; 0.4; 0.65; 0.8; 0.9; 1

The maximum cumulative probability of any F(X) is 1, knowing this, you can calculate the probability of X=5 as: 1 - F(4)= 1 - 0.9= 0.1

The mean of the sample is:

E(X)= ∑Xi*Pi= (0*0.15)+(1*0.25)+(2*0.25)+(3*0.15)+(4*0.1)+(5*0.1)

E(X)= 2.1

V(X)= E(X²)-(E(X))²

V(X)= ∑Xi²*Pi- (∑Xi*Pi)²= 6.7- (2.1)²= 2.29

∑Xi²*Pi= (0²*0.15)+(1²*0.25)+(2²*0.25)+(3²*0.15)+(4²*0.1)+(5²*0.1)= 6.7

P(X<2)

You can read this expression as the probability of having less than 2 bars of service. This means you can have either one or zero service bars, you can rewrite it as:

P(X<2)= P(X≤1)= F(1)= 0.4

P(X≤3.5)

This expression means "the probability of having at most (or less or equal to) 3.5 service bars", this variable doesn't have the value 3.5 in its definition range so you have to look for the accumulated probability until the lesser whole number. This expression includes the probabilities of X=0, X=1, X=2, and X=3, you can express it as the accumulated probability until 3, F(3):

P(X≤3.5)= P(X≤3)= F(3)= 0.80

I hope it helps!

A password is required to be 12 to 16 characters in length. Characters can be digits (0-9), upper or lower-case letters (A-Z, a-z) or special characters. There are 10 permitted special characters. There is an additional rule that not all characters can be letters (i.e. there has to be at least one digit or one special character.) How many permitted passwords are there? Give your answer in un-evaluated/un-simpli ed form and explain it fully.

Answers

Answer:

Step-by-step explanation:

You can find your answer in attached document.

Final answer:

The number of passwords is the sum from i=12 to 16 of 66^i minus the sum from i=12 to 16 of 52^i, representing all combinations of characters from length 12 to 16, excluding all-letter combinations.

Explanation:

The subject of your question pertains to combinatorics, specifically about counting password possibilities. Given that we have 66 different characters (10 digits, 52 letters lower and upper-case and 10 special characters), we have 66^12 to 66^16 possibilities for the length. However, we need to exclude the passwords composed strictly of letters. There are 52 letter characters, so we have 52^12 to 52^16 such passwords. Hence, the number of permitted passwords, in un-evaluated/un-simplified form, is the sum from i=12 to 16 of 66^i minus the sum from i=12 to 16 of 52^i

Learn more about Combinatorics here:

https://brainly.com/question/31293479

#SPJ11

Write an equation (-2, 10), (10, -14)

Answers

[tex]\bf (\stackrel{x_1}{-2}~,~\stackrel{y_1}{10})\qquad (\stackrel{x_2}{10}~,~\stackrel{y_2}{-14}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{-14}-\stackrel{y1}{10}}}{\underset{run} {\underset{x_2}{10}-\underset{x_1}{(-2)}}}\implies \cfrac{-24}{10+2}\implies \cfrac{-24}{12}\implies -2[/tex]

[tex]\bf \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{10}=\stackrel{m}{-2}[x-\stackrel{x_1}{(-2)}]\implies y-10=-2(x+2) \\\\\\ y-10=-2x-4\implies y=-2x+6[/tex]

Of all airline flight requests received by a certain discount ticket broker, 90% are for domestic travel (D) and 10% are for international flights (I). Let x be the number of requests among the next three requests received that are for domestic flights. Assuming independence of successive requests, determine the probability distribution of x. (Hint: One possible outcome is DID, with the probability (0.9)(0.1)(0.9)

Answers

Answer:

Step-by-step explanation:

Given that of all airline flight requests received by a certain discount ticket broker, 90% are for domestic travel (D) and 10% are for international flights (I).

Let x be the number of requests among the next three requests received that are for domestic flights.

X can take values as 0,1, 2 or 3.

Since independence of successive requests is assumed we find that X having two outcomes and independence

X is binomial with n =3 and p= 0.90 (constant for each trial)

PDF of X would be

[tex]P(X=x) = 3Cx (0.9)^r (0.1)^{3-r} , r=0,1,2,3[/tex]

Thus pdf of X is

X 0 1 2 3

p 0.001 0.027 0.243 0.729

Final answer:

The probability distribution of x, where x is the number of domestic flight requests among the next three requests, follows a binomial distribution with n=3 and p=0.9. It can be calculated using the formula for binomial probabilities, considering the value of x can be 0, 1, 2, or 3. We get P = 0.081.

Explanation:

The question involves calculating the probability distribution of the random variable x, which represents the number of domestic flight requests among the next three requests received by a discount ticket broker. Since each request is independent and only two outcomes are possible (domestic or international), a binomial distribution can be used to determine the probabilities for x. For instance, the probability for one particular sequence of requests such as DID (domestic, international, domestic) is calculated as (0.9)(0.1)(0.9), which is the product of the probabilities for each request in that sequence. To find the probability distribution of x, we consider all possible sequences of three requests and their associated probabilities, which are determined by multiplying the individual probabilities for each request in the sequence following the rules of binomial distribution.

The random variable x can take on the following values: 0, 1, 2, or 3, representing the number of domestic flight requests out of three. We calculate probabilities for each of these values using the binomial formula:

[tex]P(x=0) = (0.1)^3\\P(x=1) = 3 * (0.9)(0.1)^2\\P(x=2) = 3 * (0.9)^2(0.1)\\P(x=3) = (0.9)^3[/tex]

P = 0.081.

A soccer ball manufacturer wants to estimate the mean circumference of soccer balls within 0.1 in. Determine the minimum sample size required to construct a 95% confidence interval for the population mean. Assume the population standard deviation is 0.25 in.

Answers

Answer:

[tex]n=(\frac{z_{\alpha/2} \sigma}{ME})^2[/tex] (2)

The critical value for 95% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.025,0,1)", and we got [tex]z_{\alpha/2}=1.96[/tex], replacing into formula (2) we got:

[tex]n=(\frac{1.96(0.25)}{0.1})^2 =24.01 [/tex]

So the answer for this case would be n=25 rounded up to the nearest integer

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

[tex]\bar X[/tex] represent the sample mean for the sample

[tex]\mu[/tex] population mean (variable of interest)

[tex]\sigma=0.25[/tex] represent the population standard deviation

n represent the sample size (variable of interest)

Solution to the problem

The confidence interval for the mean is given by the following formula:

[tex]\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex]

The margin of error is given by this formula:

[tex] ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex] (1)

And on this case we have that ME =0.1 and we are interested in order to find the value of n, if we solve n from equation (1) we got:

[tex]n=(\frac{z_{\alpha/2} \sigma}{ME})^2[/tex] (2)

[tex]n=(\frac{1.96(0.25)}{0.1})^2 =24.01 [/tex]

So the answer for this case would be n=25 rounded up to the nearest integer

Final answer:

To estimate the mean circumference of soccer balls within 0.1 in at a 95% confidence level, with a population standard deviation of 0.25 in, the minimum sample size required is 24.

Explanation:

The question involves estimating the minimum sample size required for a 95% confidence interval with a given precision and known population standard deviation. In this scenario, the soccer ball manufacturer wants to ensure that the mean circumference of soccer balls is estimated within 0.1 in, given a population standard deviation of 0.25 in. To find the minimum sample size, we use the formula: n = (Z*σ/E)², where Z is the Z-score corresponding to a 95% confidence level (~1.96), σ is the population standard deviation (0.25 in), and E is the margin of error (0.1 in). Plugging in the values gives us: n = (1.96*0.25/0.1)² = 23.04. Since we can't have a fraction of a sample, we round up to get a minimum sample size of 24.

Which matrix multiplication is possible?

Answers

When calculating A×B, the number of columns of A = number of rows of B.

Thus, only last case is possible.

5.10 Classify each random variable as discrete or continuous. a. The number of visitors to the Museum of Science in Boston on a randomly selected day. b. The camber-angle adjustment necessary for a front-end alignment. c. The total number of pixels in a photograph produced by a digital camera. d. The number of days until a rose begins to wilt after purchase from a flower shop. e. The running time for the latest James Bond movie. f. The blood alcohol level of th

Answers

Answer:

(a) Discrete

(b) Discrete

(d) Discrete

(e) Discrete

(f) Continuous

Step-by-step explanation:

A discrete random variable considers a finite set of values. Whereas a continuous random viable assumes infinite set of values.

One cannot count the values of a continuous random variable.

(a)

X = The number of visitors to the Museum of Science in Boston on a randomly selected day.

The number of visitors can be counted and on a selected day a finite number of people will visit the museum.