Podcast
Questions and Answers
Match the following scenarios with the type of random variable they represent:
Match the following scenarios with the type of random variable they represent:
The number of cars in a parking garage at noon = Discrete The heights of students in a high school = Continuous The time it takes for a customer to complete an online purchase = Continuous Number of defective light bulbs in a sample = Discrete
Match the reasons why a variable can be considered binomial:
Match the reasons why a variable can be considered binomial:
Fixed number of trials = Number of lightbulbs tested are fixed Two possible outcomes = Success (functional) or failure (non-functional) Independent trials = One light bulb's functionality doesn't impact another Constant probability of success = Each bulb has the same 90% success rate
Match the calculations for a binomial distribution with their interpretations:
Match the calculations for a binomial distribution with their interpretations:
μ = np = Expected number of successful bulbs σ = √(np(1-p)) = Standard deviation of the bulb sample P(X = k) = nCk * p^k * (1-p)^(n-k) = Probability of exactly k successful bulbs P(X ≥ x) = Probability of at least x successful bulbs
Match the elements of a probability distribution with their properties:
Match the elements of a probability distribution with their properties:
Match the characteristics of a histogram with their description:
Match the characteristics of a histogram with their description:
Match the formula and description of the mean of a discrete random variable:
Match the formula and description of the mean of a discrete random variable:
Match each definition/description to it's appropriate name.
Match each definition/description to it's appropriate name.
Match the following probabilities that at least one pet is owned to its description.
Match the following probabilities that at least one pet is owned to its description.
Match the distribution name that best fits the description.
Match the distribution name that best fits the description.
Match the follow concepts to ther proper names.
Match the follow concepts to ther proper names.
Flashcards
Discrete Random Variable
Discrete Random Variable
A variable whose value is obtained by counting. It can only take on distinct, separate values.
Continuous Random Variable
Continuous Random Variable
A variable whose value is obtained by measuring. It can take on any value within a given range.
Binary Random Variable
Binary Random Variable
A random variable with two possible outcomes: success or failure.
Same Probability
Same Probability
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Independent Events
Independent Events
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Binomial Probability Formula
Binomial Probability Formula
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Mean of a Random Variable
Mean of a Random Variable
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Standard Deviation
Standard Deviation
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Histogram of Probability Distribution
Histogram of Probability Distribution
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Skewed Right Distribution
Skewed Right Distribution
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Study Notes
- A discrete random variable takes on integer values.
- A continuous random variable can take on many different values.
Light Bulbs
- A factory produces light bulbs with a 90% success rate, and a worker tests 30 bulbs.
Binomial Random Variable
- Success is defined as a functional light bulb.
- Failure is defined as a non-functional light bulb.
- Each light bulb's functionality is independent of others.
- Number of trials: n = 30
- Probability of success: p = 0.9
Mean of X
- µx = n * p = 30 * 0.9 = 27
Standard Deviation of X
- σχ = √(n * p * (1-p)) = √(27 * 0.1) = √2.7 ≈ 1.643
Binomial Probability Formula
- P(X=k) = nCk * p^k * (1-p)^(n-k)
- P(X=27) = 30C27 * (0.9)^27 * (0.1)^3 = 0.236
- There is a 23.6% chance that exactly 27 of 30 light bulbs work
Probability of Machine Approval
- The machine is approved if at least 28 of the 30 bulbs are functional: P(X≥28)
- P(X≥28) = P(X=28) + P(X=29) + P(X=30)
- P(X≥28) = 30C28 * (0.9)^28 * (0.1)^2 + 30C29 * (0.9)^29 * (0.1)^1 + 30C30 * (0.9)^30 (0.1)^0
- P(X≥28) = 0.2277 + 0.1413 + 0.0424 = 0.4114
Number of Pets
- Y = the number of pets owned by a randomly selected household
- Y is a discrete variable because the number of pets will be integers.
Probability Distribution
- Y: 0, 1, 2, 3, 4, 5+
- Prob: 0.05, 0.3, 0.29, 0.2, 0.11, 0.05
- This is a valid probability distribution because all probabilities are between 0 and 1 and add up to 1.
- Histogram shape: Skewed right, with a peak at Y=1
Probability of Owning at Least One Pet
- P(Y≥1) = P(Y=1) + P(Y=2) + P(Y=3) + P(Y=4) + P(Y=5+)
- P(Y≥1) = 1 - P(Y=0) = 1 - 0.05 = 0.95
- P(Y>3) is the probability that a household owns more than 3 pets.
- P(Y>3) = P(Y=4) + P(Y=5+) = 0.11 + 0.05 = 0.16
Mean of Y
- µx = Σxi * pi = 0(0.05) + 1(0.3) + 2(0.29) + 3(0.2) + 4(0.11) + 5(0.05) = 2.17
- If many households are selected, about 2.17 pets are expected.
Standard Deviation
- The standard deviation is 2.163.
- The number of pets would vary by 2.163 pets from the mean of 2.17.
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