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MATH 3020: Exam 2 Study Set 1

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For continuous random variable, P(X = x) = ______.

Probability density functions, denoted f(x), have the following properties: f(x) ≥ ______.

The probability density function, f(x) is equal to ______ for values of x that cannot occur for the random variable.

Which FOUR of the following are valid values for a probability? Select all that apply. -0.587 0 1 150 1.5 1/2 0.384

0, 1, 1/2, 0.384 A probability must be between 0 and 1. So we rule out the first one (less than 1), the fourth (way larger than 1), and fifth (greater than 1).

Probability density functions, denoted f(x), have the following properties: ∫_(−∞)^(∞) [f(x) dx] = ______.

Suppose the random variable X has pdf f(x) = (3x^2)/16 for −2 ≤ X ≤ 2. Choose the correct response for each of the following integrals. ∫_(−2)^(2) [(3x^2)/16 dx] must be equal to __________________________________.

The Cumulative Density Function for CRVs has the following properties: As x tends to infinity, F(x) tends to (8) [ Select ] ["cannot be determined", "0", "1"]

Consider any continuous random variable, X. Select the correct answers for each of the following statements. Consider any continuous random variable, X. Select the correct answers for each of the following statements. P(X ≥ x) = P(X > x) = __________.

1 - F(x)

Consider a random variable, X, which follows a normal distribution with a mean of 16 and a standard deviation 0.2. For each R command below, choose the correct answer describing what the R command accomplishes. qnorm(0.25, mean = 16, sd = 0.2)

1st Quartile (Q1) for X

Consider a random variable, X, which follows a normal distribution with a mean of 77 and a standard deviation of 3.8. Write a line of code in R to find the probability that X exceeds 81.5. Round your answer to 4 decimal places. Review the examples involving finding probabilities in Lesson 7. Pay particular attention to examples which involve calculating P(X>a), i.e. greater than a.

2 approaches here: round(pnorm(81.5, mean - 77, sd = 3.8, lower.tail=F),4) 1-round(pnorm(81.5, mean = 77, sd = 3.8),4) Note: We can either look at the right area, where we'll need to turn lower.tail. Or we can look at the left area, leaving lower.tails on, and use the complement rule.

3rd Quartile (Q3) for X

Suppose the fill amounts from a bottling process follow a normal distribution with a mean of 12.05 ounces and a standard deviation of 0.02 ounces. Based upon the empirical rule, what are the correct range of values that X lies between? 68%

68% means the data point is within 1 standard deviation of the mean. (12.05-0.02) < X < (12.05+0.02) 12.03 < X < 12.07

Consider a random variable, X, which follows a normal distribution with a mean of 78 and a standard deviation of 3.6. Based upon the empirical rule, what are the correct range of values that X lies between for a probability of 68%

68% means the data point is within 1 standard deviation of the mean. (78-3.6) < X < (78+3.6) 74.4 < X < 81.6

95% means the data point is within 2 standard deviations of the mean. (12.05-2*0.02) < X < (12.05+2*0.02) 12.01 < X < 12.09

95% means the data point is within 2 standard deviations of the mean. (78-2*3.6) < X < (78+2*3.6) 70.8 < X < 85.2

99.7% means the data point is within 2 standard deviations of the mean. (78-3*3.6) < X < (78+3*3.6) 67.2 < X < 88.8

99.7% means the data point is within 3 standard deviations of the mean. (12.05-3*0.02) < X < (12.05+3*0.02) 121.99 < X < 12.11

What is mutual exclusivity?

A set of objects are mutually exclusive if they cannot have the same label. For example, a smartphone can only be an iPhone, Google Pixel, Samsung, Huawei, etc. A car can only be a Tesla, Volkswagen, Acura, etc. You get the idea.

Suppose the fill amounts from a bottling process follow a normal distribution with a mean of 12.05 ounces and a standard deviation of 0.02 ounces. Question: Write a line of code in R to find P(X > 12.1) and round the result to 4 decimal places

Can be done in 2 ways: round(pnorm(12.1, mean = 12.05, sd = 0.02, lower.tail=F), 4) round(1 - pnorm(12.1, mean = 12.05, sd = 0.02),4) Note: We are looking at the right area of the normal distribution graph. We can either find this directly by turning lower.tails off, or by using the complement rule with the left area.

Circuit boards are produced and sampled. For each description below, select the correct type of random variable. Name of the operator on duty during production of a circuit board.

Categorical Variable, Not a Random Variable

For each description below, select the correct probability term which applies. Set of all elements that are not in an event, E.

Complement

Circuit boards are produced and sampled. For each description below, select the correct type of random variable. Weight of a circuit board in grams.

Continuous Random Variable

The __________ distribution function of a continuous random variable X is F(x) = P(X ≤ x).

Cumulative

The __________ distribution function of a discrete random variable X is F(x)=P(X ≤ x).

Cumulative

For a given continuous random variable, X: The notation F(x) denotes the _________________________________.

Cumulative Density Function (CDF)

Consider a discrete random variable, X. For each of the notations given below, select the correct term. The notation F(x) denotes the ________________________________.

Cumulative Distribution Function

Suppose the random variable X has pdf f(x) = (3x^2)/16 for −2 ≤ X ≤ 2. Choose the correct response for each of the following integrals. ∫_(−2)^(x) [(3u^2)/16 du] represents the ______.

Cumulative Distribution Function (CDF)

The distribution for both discrete and continuous random variables can be specified using the ______________________________________.

Cumulative Distribution Function (CDF)

Suppose the fill amounts from a bottling process follow a normal distribution with a mean of 12.05 ounces and a standard deviation of 0.02 ounces. Question: Write a line of code in R to find the value which represents the cutoff for the top 10% of all fill amounts and round your answer to 2 decimal places

Dealing with percentiles, so use the qnorm command. There are 2 valid approaches here: round(qnorm(0.9, mean = 12.05, sd = 0.02), 2) round(qnorm(0.1, mean = 12.05, sd = 0.02, lower.tail=F), 2) Note: the top 10% is the 90th percentile. So we can look at 90% of the data (left area - leave lower.tails on) OR look at the remaining 10% (right area - turn lower.tails off)

The Cumulative Density Function for CRVs has the following properties: Taking the derivative, F′(x) gives the probability _______ function

Density

Circuit boards are produced and sampled. For each description below, select the correct type of random variable. Number of solder joints on a circuit board.

Discrete Random Variable

A result of this fact is that the following probabilities are all __________. P(x_1 ≤ X ≤ x_2) P(x_1 ≤ X < x_2) P(x_1 < X ≤ x_2) P(x_1 < X < x_2)

Equivalent

Consider any continuous random variable, X. Select the correct answers for each of the following statements. P(a < X < b) = _______________.

F(b) - F(a)

Consider any continuous random variable, X. Select the correct answers for each of the following statements. P(X ≤ x) = P(X < x) = __________.

F(x)

True or False: Discrete random variables must have a finite number of possible values. Hint: Look carefully at the definition and Example: Surface Contamination of a Semiconductor Wafer. This is an important point about discrete random variables.

False

Consider the following PMF. You will need to find the missing probability before answering the question (or you will need to approach this question using the complement rule). Probability Mass Function x | 0 | 1 | 2 | 3 | 4 -----------------|--------|---------|--------|--------|-------- f(x) = P(X = x) | | 0.315 | 0.216 | 0.068 | 0.002 Question: Find F(3) for this random variable. Round your answer to 3 decimal places. Be sure you are using the values given in this question. Hint: The sum of all of the probabilities must equal 1.

Find the missing probability first. We'll need it to find F(3) f(0) = P(X = 0) = 1 - (0.315 + 0.216 + 0.068 + 0.002) = 0.399 F(3) = f(1) + f(2) + f(3) = 0.399 + 0.315 + 0.216 = 0.93 Note: f(3) IS included in the calculation, since we are adding up all the probabilities that are less than or equal to 3. So that is f(1), f(2), AND f(3).

Consider the following PMF. You will need to find the missing probability before answering the question (or you will need to approach tμ = E(X) = ∑^(i=1)^n [x_i * f(x_i)]his question using the complement rule). Probability Mass Function x | 0 | 1 | 2 | 3 | 4 -----------------|--------|----------|--------|--------|------- f(x) = P(X = x) | 0.08 | 0.19 | | 0.25 | 0.08 Question: Find the variance of X. Round your answer to 2 decimal places. Be sure you are using the values given in this question. You should be using the formulas: μ = E(X) = ∑_(i=1)^n [x_i * f(x_i)] σ^2 = V(x) = ∑_(i=1)^n [x_i^2 f(x_i)] − μ^2 = {∑_(i=1)^n [x_i^2 f(x_i)]} − μ^2 (note that the brackets are NOT necessary... only provided for absolute clarity!)

Find the missing probability: f(2) = P(X = 2) = 1 - (0.08 + 0.19 + 0.25 + 0.08) = 0.4 Next find the mean: μ = E(X) = ∑_(i=1)^n [x_i * f(x_i)] = 0*f(0) + 1*f(1) + 2*f(2) + 3*f(3) + 4*f(4) = 0(0.08) + 1(0.19) + 2(0.4) + 3(0.25) + 4(0.08) = 2.06 Finally, find the variance: σ^2 = V(x) = ∑_(i=1)^n [(x_i)^2 * f(x_i)] − μ^2 = {∑_(i=1)^n [(x_i)^2 * f(x_i)]} − μ^2 = [0^2 * f(0) + 1^2 * f(1) + 2^2 * f(2) + 3^2 * f(3) + 4^2 * f(4)] - μ^2 = [0^2 * (0.08) + 1^2 * (0.19) + 2^2 * (0.4) + 3^2 * (0.25) + 4^2 *(0.08)] - (2.06)^2 = 1.08

Consider the following PMF. You will need to find the missing probability before answering the question (or you will need to approach tμ = E(X) = ∑^(i=1)^n [x_i * f(x_i)]his question using the complement rule). Probability Mass Function x | 0 | 1 | 2 | 3 | 4 -----------------|--------|----------|--------|--------|------- f(x) = P(X = x) | 0.341 | 0.233 | | 0.176 | 0.229 Question: Find the standard deviation of X. Round your answer to 2 decimal places. Be sure you are using the values given in this question. You should be using the formulas: μ = E(X) = ∑_(i=1)^n [x_i * f(x_i)] σ^2 = V(x) = ∑_(i=1)^n [x_i^2 f(x_i)] − μ^2 = {∑_(i=1)^n [x_i^2 f(x_i)]} − μ^2 (note that the brackets are NOT necessary... only provided for absolute clarity!) σ = sqrt(σ^2)

Find the missing probability: f(2) = P(X = 2) = 1 - (0.341 + 0.233 + 0.176 + 0.229) = 0.021 Next find the mean: μ = E(X) = ∑_(i=1)^n [x_i * f(x_i)] = 0*f(0) + 1*f(1) + 2*f(2) + 3*f(3) + 4*f(4) = 0(0.341) + 1(0.233) + 2(0.021) + 3(0.176) + 4(0.229) = 1.719 Then, find the variance: σ^2 = V(x) = ∑_(i=1)^n [(x_i)^2 * f(x_i)] − μ^2 = {∑_(i=1)^n [(x_i)^2 * f(x_i)]} − μ^2 = [0^2 * f(0) + 1^2 * f(1) + 2^2 * f(2) + 3^2 * f(3) + 4^2 * f(4)] - μ^2 = [0^2 * (0.341) + 1^2 * (0.233) + 2^2 * (0.021) + 3^2 * (0.176) + 4^2 *(0.229)] - (1.719)^2 = 2.610039 Finally, find the standard deviation: σ = sqrt(σ^2) = sqrt(2.610039) = 1.62

Let X denote the lifetime of a component (in hours) with the following probabilities: Be sure you are using the values given in this question. P(X ≤ 6000) = 0.057 P(X > 9000) = 0.338 Find P(6000 < X ≤ 9000)

Find the total probability: P(X ≤ 6000 OR X > 8000) =0.057 + 0.338 = 0.395 Find the probability that neither of these conditions are met, i.e. X is less than 6000 AND less than or equal to 9000, via the complement rule. P(6000 < X ≤ 9000) = 1 - P(X ≤ 6000 OR X > 8000) = 1 - 0.395 = 0.605

Consider the following PMF. You will need to find the missing probability before answering the question (or you will need to approach this question using the complement rule). Probability Mass Function x | 1 | 2 | 3 | 4 | 5 | 6 | -----------------|--------|---------|--------|--------|-------|-------|--- f(x) = P(X = x) | 0.042 | 0.113 | | 0.175 | 0.331 | 0.18 | Question: Find P(2 ≤ X < 5). Do not round. Be sure you are using the values given in this question. Hint: Which values will be included in this probability? Consider both endpoints carefully based upon the inequalities stated in the question. Hint: The sum of all of the probabilities must equal 1.

First we find the missing probability, f(2) = P(X = 2) f(2) = P(X = 2) = 1 - (0.042 + 0.113 + 0.175 + 0.331 + 0.18) = 0.159 Then we find the probability that X is between 2 (inclusive) and 5. Meaning that X can either be 2, 3, OR 4 (not 5). P(2 ≤ X < 5) = f(2) + f(3) + f(4) = 0.113+ 0.159 + 0.175 = 0.447

Consider the following PMF. You will need to find the missing probability before answering the question (or you will need to approach tμ = E(X) = ∑^(i=1)^n [x_i * f(x_i)]his question using the complement rule). Probability Mass Function x | 1 | 2 | 3 | 4 | 5 -----------------|--------|----------|--------|--------|------- f(x) = P(X = x) | 0.033 | 0.054 | | 0.376 | 0.174 Question: Find the mean of X. Round your answer to 2 decimal place. Be sure you are using the values given in this question. You should be using the formula: μ = E(X) = ∑^_i=1)^n [x_i * f(x_i)]

First we find the missing probability: f(3) = P(X = 3) = 1 - (0.033 + 0.054 + 0.376 + 0.174) = 0.363 Then we use the formula for mean: μ = E(X) = ∑^(i=1)^n [x_i * f(x_i)] = 1*f(1) + 2*f(2) + 3*f(3) + 4*f(4) + 5*f(5) = 1(0.033) + 2(0.054) + 3(0.363) + 4(0.376) + 5(0.174) = 3.6

From the Fundamental Theorem of Calculus, we have: P(a < X < b) = F(b) - F(a). So we need to find this in R. round(pnorm(12.06, mean = 12.05,sd = 0.02) - pnorm(12.04, mean = 12.05, sd = 0.02), 4) Note: We are looking at the left area, so we leave lower.tails on.

Let X denote the life of a semiconductor laser (in hours) with the following probabilities: P(X ≤ 5000) = 0.121 P(5000 < X ≤ 7000) = 0.605 What is the probability that the life is greater than 5000 hours? Round your answer to 3 decimal places. Note: Be sure to use the values given in THIS question.

If a semiconductor laser's life is greater than to 5000 hours, then the life can be between 5000 and 7000 hours, OR greater than 7000 hours. Meaning that it CANNOT be less than or equal to 5000 hours. This is true because both events are mutually exclusive (a laser cannot have a life less than 5000 hours and simultaneously equal to 6000 hours). We'll use the complement rule with the first probability here. P(X > 5000) = P(5000 < X ≤ 7000) + P(X > 7000) = P(X ≤ 5000)' = 1 - P(X ≤ 5000) = 1 - 0.121 = 0.879

Let X denote the life of a semiconductor laser (in hours) with the following probabilities: P(X ≤ 5000) = 0.102 P(5000 < X ≤ 7000) = 0.646 What is the probability that the life is less than or equal to 7000 hours? Round your answer to 3 decimal places.

If a semiconductor laser's life is less than or equal to 7000 hours, then the life is either in between 5000 and 7000, or less than or equal to 5000. This is true because both events are mutually exclusive (a laser cannot have a life less than 5000 hours and simultaneously equal to 6000 hours). P(X ≤ 7000) = P(5000 < X ≤ 7000) + P(X ≤ 5000) = 0.102 + 0.646 = 0.748

Why is the name of the operator on duty during production of a circuit board not a Random Variable?

It's a name. It's got characters (letters), not numbers. You can have gaps in-between numbers, but not names.

For discrete random variables, the distribution contains a list of the possible values along with the probability of each of these values. This distribution is a function called the probability ____ function.

Mass

Consider a discrete random variable, X. For each of the notations given below, select the correct term. The notation μ denotes the ____.

Mean

For a given continuous random variable, X: The notation μ denotes the ______.

Mean

Suppose the random variable X has pdf f(x) = (3x^2)/16 for −2 ≤ X ≤ 2. Choose the correct response for each of the following integrals. ∫_(−2)^(2) [x * (3x^2)/16 dx] represents the __________.

Mean or Expected Value

Suppose the random variable X has pdf f(x) = (3x^2)/16 for −2 ≤ X ≤ 2. Choose the correct response for each of the following integrals. ∫_(−2)^(2) [x^2 * (3x^2)/16 dx] represents the __________.

Mean or Expected Value of x-squared

For each description below, select the correct probability term which applies. When the intersection of all pairs are empty in a set of events.

Mutual Exclusivity

Suppose the fill amounts from a bottling process follow a normal distribution with a mean of 12.05 ounces and a standard deviation of 0.02 ounces. Question: Between what two values would we find the middle 50% of possible fill amounts? Note that this is equivalent to finding Q1 and Q3. Write lines of code in R to find these TWO values, which identify the middle 50%, and round the results to 3 decimal places.

Need to find Q1 and Q3 here to find the range of values Q2 (Median/50th percentile) lies. round(qnorm(0.25, mean = 12.05, sd = 0.02), 3) # Q1 round(qnorm(0.75, mean = 12.05, sd = 0.02), 3) # Q3 Note: For Q1, we are looking at the 25th percentile, which means we analyze the left area of the graph (leave lower.tails on). Similarly, for Q3 we are looking at the 75th percentile (left area), so we leave lower.tails on.

Consider a random variable, X, which follows a normal distribution with a mean of 79 and a standard deviation of 5.3. Write a line of code in R to find the probability that X is between 79 and 82. Round your answer to 4 decimal places. Review the examples involving finding probabilities in Lesson 7. Pay particular attention to examples which involve calculating P(a<X<b), i.e. between a and b.

Need to use F(b) - F(a), where b = 82, and a = 79 round(pnorm(82, mean = 79, sd = 5.3) - pnorm(79, mean = 79, sd = 5.3),4) Note: Looking at the left area of the graph, so we leave lower.tails on.

The Cumulative Density Function for CRVs has the following properties: As x increases, F(x) is (7) [ Select ] ["non-decreasing", "non-increasing", "decreasing", "increasing"]

Non-Decreasing

The Cumulative Distribution Function has the following properties: As x increases, F(x) is ______________.

Non-Decreasing

The Cumulative Density Function for CRVs has the following properties: F(x) is ____________ for all x.

Non-Negative

The Cumulative Distribution Function has the following properties: F(x) is ____________ for all x.

Non-Negative

The Cumulative Distribution Function has the following properties: As x tends to infinity, F(x) tends to _____.

One

Suppose the random variable X has pdf f(x) = (3x^2)/16 for −2 ≤ X ≤ 2. Choose the correct response for each of the following integrals. ∫_(−1)^(1) [(3x^2)/16 dx] represents the __________________________________.

P(-1 < X < 1)

Say X denotes the number of cars in a ChickFil-A parking lot, with the following probabilities: P(X ≤ 10) = 0.283 P(10 < X ≤ 30) = 0.198 P(10 < X ≤ 30) mean, in layman's terms?

P(10 < X ≤ 30) is the probability that the number of cars in a ChickFil-A parking lot is greater than 10 AND less than or equal to 30. This probability is equal to 0.198, or 19.8%. This means that there is a 19.8% chance that the greater than 10 AND less than or equal to 30. The number of cars could be 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, OR 30. NOT 10, because of the "10 < X" in the probability statement. However, 30 IS included, due to the "X ≤ 30" in the statement.

Suppose P(A) = 0.223. Find the probability of the complement of A = P(A′). Be sure you are using the values given in this question.

P(A') = 1 - P(A) P(A') = 1 - 0.223 = 0.777

Say X denotes the number of cars in a ChickFil-A parking lot, with the following probabilities: P(X ≤ 10) = 0.283 P(10 < X ≤ 30) = 0.198 What does P(X ≤ 10) mean, in layman's terms?

P(X ≤ 10) is the probability that the number of cars in a ChickFil-A parking lot is less than or equal to 10. This probability is equal to 0.283, or 28.3%. This means that there is a 28.3% chance that the number of cars in a ChickFil-A parking lot is less than or equal to 10. The number of cars could be 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, OR 10. To is included because of the "≤ 10" in the probability statement.

For discrete random variables, the graph of the CDF is a piecewise continuous function where the size of the jump at a given point equals the ___________ at that point.

Probability

For each description below, select the correct probability term which applies. Quantifies the likelihood or chance.

Probability

And only continuous random variables can be specified using __________________________________.

Probability Density Function (PDF)

For a given continuous random variable, X: The notation f(x) denotes the _________________________________.

Probability Density Function (PDF)

When using histograms to display data for a continuous random variable, the shape of the histogram approximates the shape of the ________________________________.

Probability Density Function (PDF)

Consider a discrete random variable, X. For each of the notations given below, select the correct term. The notation f(x) denotes the _________________________.

Probability Mass Function

However, only discrete random variables can be specified using _______________________________.

Probability Mass Function (PMF)

What's another way to think about probability?

Probability is the percent chance that an outcome occurs. Say there is a probability of 80 that there will be traffic on I4 as you head to work. We say that there is an 80% chance there is traffic on I4.

Probability that X exceeds 16.5

Probability that X is less than 16.5

The Cumulative Distribution Function has the following properties: F(x) is defined for all ____________, x∈R.

Real Numbers

Give an example of the complement rule.

Say there is a 30% probability that your mother is late for work. The probability that she won't is 1 - 30%, which is 70%. This is because all probabilities for a given event must add up to 1!

Choose the correct description of how to interpret the Normal Q-Q plot below.

Since the points fall reasonably on the line, the distribution of the data is approximately normally distributed.

Consider a discrete random variable, X. For each of the notations given below, select the correct term. The notation σ denotes the __________________.

Standard Deviation

For a given continuous random variable, X: The notation σ denotes the __________.

Standard Deviation

Which of the following represents a discrete random variable? Hint: Remember that discrete random variables must be numeric with gaps between the possible values.

The number of rafters used in each unit in a large condominium complex.

Consider the PMF below. Probability Mass Function x | 1 | 2 | 3 | 4 | 5 -----------------|--------|---------|--------|--------|-------- f(x) = P(X = x) | | 0.268 | 0.174 | 0.127 | 0.038 Question: Find f(0) = P(X = 0). Do not round. Be sure you are using the values given in this question. Hint: The sum of all of the probabilities must equal 1.

The total probability for an event must equal 1. So, the sum of the probabilities in the table must equal 1 ! Use the complement rule to find the missing probability. f(0) = P(X = 0) = 1 - (0.268 + 0.174 + 0.127 + 0.038) = 0.393

What happens when Discrete Random Variables have an infinite number of posible values?

The variable is countably infinite. The possible values match up 1 to 1 with the reals, meaning while there are an infinite number of values, we can count them (given enough time, of course).

Why is the weight of a circuit board in grams a CRV?

The weight of a board picked at random from a population can take on any range of values within a specified interval. If we increase the precision of our measuring device, we can find more values (there's always a number between 2 other ones: 1.5, 1.75, 2)

A random sample of US adults is selected and their blood types are recorded. A frequency table summarizing the results is provided below. Be sure you are using the values given in this question. Blood Type AB A B O Frequency 16 113 27 124 Based upon this data, estimate the probability that a randomly selected US adult will have blood type AB or O, in notation: Find P(AB or O).

Total Frequency = 16 + 113 + 27 + 124 = 280 Events are mutually exclusive, i.e. a patient cannot have both type AB AND O blood. P(AB OR O) = P(AB) + P(O) = (16/280) + (124/280) = ((16 + 124)/280) = 0.5

Consider the following table summarizing recent orders for a particular anti-fatigue mat. It comes in 3 colors and 2 sizes. Blue Green Black Small 303 359 379 Large 205 269 374 Question: Based upon the data, estimate the probability that a randomly selected order will be for a green mat. Round your answer to 3 decimal places. Note: Be sure to use the values given in THIS question.

Total Number of Mats = 303 + 359 + 279 + 205 + 369 + 374 = 1889 Events are mutually exclusive, i.e. a mat can either be green OR blue OR black P(Mat = Small) = (359 + 269)/1889 = 0.332

Consider the following table summarizing recent orders for a particular anti-fatigue mat. It comes in 3 colors and 2 sizes. Blue Green Black Small 322 297 200 Large 313 252 377 Question: Based upon the data, estimate the probability that a randomly selected order will be for a small mat. Round your answer to 3 decimal places. Note: Be sure to use the values given in THIS question.

Total Number of Mats = 322 + 297 + 2000 + 313 + 252 + 377 = 1761 Events are mutually exclusive, i.e. a mat can either be small OR large, not both. P(Mat = Small) = (322 + 297 + 200)/1761 =

Consider the following table summarizing recent orders for a particular anti-fatigue mat. It comes in 3 colors and 2 sizes. Blue Green Black Small 347 275 394 Large 380 343 288 Question: Based upon the data, estimate the probability that a randomly selected order will be small or green. Round your answer to 3 decimal places. Hint: Are these events mutually exclusive? Remember not to count any cell twice. Note: Be sure to use the values given in THIS question.

Total Number of Mats = 347 + 380 + 275 + 343 + 394 + 288 = 2027 Events are NOT mutually exclusive, i.e. a mat can be small AND green, or small OR green P(Mat = Small OR green) = ((347 + 275 + 394) + (275 + 343) - 275)/2027 = 0.670 Note: the value of 275 is counted twice, because it is the number of mats that are green AND small. So we subtract it, as we should only count it ONCE.

True or False: For continuous random variables which are not defined on the entire real line, we must write the CDF as a piece-wise function.

True

True or False: Random variables must be numeric.

True

Consider a discrete random variable, X. For each of the notations given below, select the correct term. The notation σ^2 denotes the ________.

Variance

For a given continuous random variable, X: The notation σ^2 denotes the __________.

Variance

Suppose the random variable X has pdf f(x) = (3x^2)/16 for −2 ≤ X ≤ 2. Choose the correct response for each of the following integrals. ∫_(−2)^(2) [(x - μ)^2 * (3x^2)/16 dx] represents the __________.

Variance

Let E_1 denote the event that a structural component fails during a test and E_2 denote the event that the component shows some strain but does not fail. Given P(E_1)= 0.146 and P(E2)= 0.248, what is the probability that a component does not fail during a test? Round your answer to 3 decimal places. Hint: Be sure you are correctly considering these events. Try using the complement rule. Note: Be sure to use the values given in THIS question.

We are looking for the probability that Event 1 does not happen. Use complement rule. P(E_1)' = 1 - P(E_1) = 1 - 0.246 = 0.854

Let E_1 denote the event that a structural component fails during a test and E_2 denote the event that the component shows some strain but does not fail. Given P(E_1)= 0.177 and P(E_2)= 0.159, what is the probability that a component neither fails nor shows strain but does not fail? Round your answer to 3 decimal places. Hint: Be sure you are correctly considering these events. What is the only remaining possibility besides the two events defined? Note: Be sure to use the values given in THIS question.

We are looking for the probability that NEITHER Event 1 NOR Event 2 happens. Use the complement rule. P(E_1 AND E_2)' = 1 - (P(E_1) + P(E_2)) = 1 - (0.177 + 0.159) = 0.664

Let E_1 denote the event that a structural component fails during a test and E_2 denote the event that the component shows some strain but does not fail. Given P(E_1)= 0.136 and P(E_2)= 0.303, what is the probability that a component either fails or does not fail but shows strain during a test? Round your answer to 3 decimal places. Hint: Be sure you are correctly considering these events. Are these two events mutually exclusive? Note: Be sure to use the values given in THIS question.

We are looking for the probability that either of the 2 events happen. P(E_1 OR E_2) = P(E_1) + P(E_2) = 0.136 + 0.303 = 0.439

Say X denotes the number of cars in a ChickFil-A parking lot, with the following probabilities: P(X ≤ 10) = 0.283 P(10 < X ≤ 30) = 0.198 What type of variable is X?

X is a Discrete Random Variable. You can't have 13.5 or 29.67 cars, but you can have 12 or 27. Say the parking lot can have a MAXIMUM of 45 cars, then there are 46 DISCRETE values X can have (0 to 45 cars).

Why is the number of solder joints on a circuit board a DRV?

You cannot have 2.5 solder joints (what defines a "fractional solder joint"???). Its either 2, or 3. Sure, you could technically have an infinite number of joints (given a sufficiently large board), but then this number would be countably infinite as the data points would be 1 to 1 with the reals (we can count all the numbers, up to infinity, given enough time).

Why is the number of rafters used in each unit in a large condominium complex a DRV?

You cannot have 5.455 rafters, only a discrete number like 3 or 12.

To find probabilities for a continuous random variable with probability density function f(x), we must find the ________________________________________________________.

area under the density curve over the region of interest

The Cumulative Density Function for CRVs has the following properties: F(x) is defined for all ___________, x ∈ R.

real numbers

round(pnorm(12.01, mean = 12.05, sd = 0.02), 4) Note: We are looking at the left area of the normal distribution graph, so we leave the lower.tails on.

Consider a random variable, X, which follows a normal distribution with a mean of 78 and a standard deviation of 3.6. Write a line of code in R to find the probability that X is less than 75.2. Round your answer to 4 decimal places. Review the examples involving finding probabilities in Lesson 7. Pay particular attention to examples which involve calculating P(X<a), i.e. less than a.

round(pnorm(75.2, mean = 78, sd = 3.6),4) Note: Looking at the left area of the normal distribution function, so we leave lower.tails on. This should be obvious to you because the problem is asking us to find the "probability that X is less than..."

MATH 3020: Exam 2 Study Set 1

Kaugnay na mga set ng pag-aaral

QUIZ 2 chapter 4

Practice Test 10

PEDS EXAM 3

Uworld All

AC chemistry Unit 4/5 Chemical Quantities and Reactions and Stoichometry Test

MNO 3370 Exam 3

Chem/Phys and Bio/Biochem MCAT Full Length 3-6

Course 2 - W3: All Quiz

Mortar Types

FIN Test 3

chapter 7 (wade)

Gov test

Strategic Management Chapter 13

EAQ 2 Nurs 101

Nutrition 241 Quiz 1

Economics Ch. 8

Ch 4 Whole Life Insurance

What Was the Declaration of Independence questions

Conceptual physics chapter15,16,17 questions

Appendix 7,8,9