Flashcards on Statistics: Time Series and Probability

Statistics: Time Series & Probability Explained for Students

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When tossing a fair coin and a fair six-sided die together, how many possible paired outcomes are in the sample space?

12 (2 coin faces × 6 die faces = 12)

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Statistics

58 cards

Card 1

Question: When tossing a fair coin and a fair six-sided die together, how many possible paired outcomes are in the sample space?

Answer: 12 (2 coin faces × 6 die faces = 12)

Card 2

Question: In a normal distribution, where are most data values located and what is the distribution's symmetry?

Answer: Most data values are clustered around the mean, and the distribution is symmetric about the mean.

Card 3

Question: What are the mean and variance of the standard normal distribution?

Answer: Mean = 0 and variance = 1.

Card 4

Question: How does changing the standard deviation affect the shape of a normal distribution?

Answer: A smaller standard deviation yields a narrower and taller curve (data more tightly clustered); a larger standard deviation yields a wider and shorter

Card 5

Question: If X ~ N(μ, σ^2), how do you standardize X to obtain a standard normal variable Z?

Answer: Z = (X − μ) / σ, which follows the standard normal distribution.

Card 6

Question: How can you express P(a ≤ X ≤ b) for X ~ N(μ, σ^2) using the standard normal CDF?

Answer: P(a ≤ X ≤ b) = Φ((b − μ)/σ) − Φ((a − μ)/σ), where Φ is the standard normal CDF.

Card 7

Question: Complete the CLT statement: “Whatever the population distribution, the distribution of the _______ is approximately normal when the sample is _______.

Answer: sample mean; large.

Card 8

Question: Given Z ~ N(0,1), what is the k such that P(Z < k) = 0.0041 (approximate z-value)?

Answer: k ≈ −2.65 (since 0.0041 is in the lower tail around −2.65).

Card 9

Question: If lifetimes of lightbulbs are N(900, 35^2), what is the probability a randomly chosen bulb lasts less than 950 hours?

Answer: Compute Z = (950−900)/35 = 50/35 ≈ 1.4286, so P(X<950) ≈ Φ(1.4286) ≈ 0.9236 (approx).

Card 10

Question: With bulb lifetimes N(900,35^2), what is P(930 < X < 940) for a single bulb?

Answer: Z1=(930−900)/35≈0.8571; Z2=(940−900)/35≈1.1429; P≈Φ(1.1429)−Φ(0.8571) ≈ 0.8729−0.6809 ≈ 0.1920 (approx; closest listed ≈0.1949).