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NBT Statistics and Probability Flashcards

6 cards from real NBT practice questions. Tap to flip, then mark Knew It or Still Learning — missed cards come back until you master them.

Read the first 6 NBT Statistics and Probability flashcards as text

  1. A dataset has a mean of 50 and a standard deviation of 10. A second dataset has a mean of 50 and a standard deviation of 25. Which statement is correct about the two distributions, assuming both are approximately normal?

    Answer: The second dataset has more variability and a flatter curve

    Standard deviation measures spread. A larger standard deviation (25 vs. 10) means greater variability. In a normal distribution, greater spread corresponds to a flatter, wider bell curve — not a taller, more peaked one. The mean being equal tells us nothing about variability.

  2. A bag contains 4 red, 3 blue, and 5 green marbles. Two marbles are drawn without replacement. What is the probability that the second marble is blue, given that the first marble drawn was NOT blue?

    Answer: 3/11

    If the first marble is not blue, it is one of 4+5=9 non-blue marbles. After removing one non-blue marble, 11 marbles remain in the bag, of which 3 are still blue. The conditional probability is 3/11.

  3. For the data set {2, 4, 4, 6, 8, 10, 10, 12}, which measure of central tendency would be MOST affected if the value 12 were replaced with 120?

    Answer: Mean

    The mean sums all values and divides by n, so replacing 12 with 120 increases the sum by 108, dramatically shifting the mean. The median is the average of the 4th and 5th values (6 and 8), which remains 7 regardless of how large the maximum is. The mode (4 and 10) is also unaffected.

  4. Events A and B are such that P(A) = 0.5, P(B) = 0.4, and P(A ∪ B) = 0.7. What is P(A | B)?

    Answer: 0.2

    Using the addition rule: P(A ∩ B) = P(A) + P(B) − P(A ∪ B) = 0.5 + 0.4 − 0.7 = 0.2. Then P(A | B) = P(A ∩ B) / P(B) = 0.2 / 0.4 = 0.5. Wait — that gives 0.5. Let me restate: P(A|B) = 0.2/0.4 = 0.5. The correct answer is 0.5 (index 0).

  5. A symmetric distribution has Q1 = 20 and Q3 = 40. A data point has a value of 65. Using the IQR method, is this point an outlier, and why?

    Answer: Yes — 65 exceeds Q3 + 1.5 × IQR, making it a mild outlier

    IQR = Q3 − Q1 = 40 − 20 = 20. The upper outlier fence is Q3 + 1.5 × IQR = 40 + 30 = 70. Since 65 < 70, this point is NOT beyond the outer fence. However, the upper inner fence is Q3 + 1.5×IQR = 70. Since 65 < 70, it is actually NOT an outlier by the standard 1.5×IQR rule. The correct answer is A: 65 ≤ 70, so it is not an outlier.

  6. The probability that a student passes Mathematics is 0.6 and the probability that the same student passes Science is 0.5. If the two events are independent, what is the probability that the student passes Mathematics but FAILS Science?

    Answer: 0.30

    P(fail Science) = 1 − 0.5 = 0.5. Since the events are independent, P(pass Maths AND fail Science) = P(pass Maths) × P(fail Science) = 0.6 × 0.5 = 0.30.