University:
Boston University
Course:
CGS MA 121 | Calculus
Academic year:

2024
Views:

337

Pages:

2
Author:

fiziokrat16l

Question 1 Assume the random variable X is normally distributed with mean μ = 50 and standard deviation σ = 7. What is the probability P(X>42)? Answer Step 1 We must standardise the Random Variable X with the Standardised Normal Distribution Z Variable using the relationship: Z = (X - μ) / σ And we will use Normal Distribution Tables of the function: Φ(z) = P(Z < z) And so we get: P(X>42) = P(Z > (42 - 50) / 7) = P(Z > -8/7) = P(Z > -1.1429) If we look at this graphically it is the shaded part of this Standardised Normal Distribution: (Graph) By symmetry of the Standardised Normal Distribution it is the same as this shaded part: (Graph) So: P(Z>42) = P(Z > -1.1429) = 1 − P(Z < 1.1429) = 1 – Φ(1.1429) = 1 - 0.8729 (from tables) = 0.1271 Question 2 What is the mathematical definition of conditional probability? Answer A probability that can only be possible after another condition exists. See also "Bayesian Statistics". Explanation: An example would be that "B" has a probability of 40%, but ONLY if "A" occurs. If the probability of "A" occurring is 60% the conditional probability of B occurring is 0.60 * 0.40 = 0.24, or 24%. Question 3 What is the probability of drawing 2 cards in succession (without replacement) from a standard deck of playing cards and having them both be hearts? Answer Step 1 Because the results of the first draw have an impact on the second, the events are interconnected. Thus, we have: P(heart and heart) = P(heart) * P(heart then heart) The probability that a heart card is draw on the first draw is: P(heart) = 13/52 = 1/4 The probability that a heart card is draw on the second draw is: P(heart) = 12/51 Step 2 Therefore, the probability of drawing 2 hearts in succession without replacement is: P(heart and heart) = (1/4) * (12/51) = (1/4) * (4/17) = 1/17 P(heart and heart) = 1/17 Question 4 How likely is it that a seven-digit phone number will include the number 7? Answer We start by locating the complement of the number 7, which is not a phone number with the number 7, since we assume it must appear at least once. The probability of not having a 7 is 9/10 (there are 10 possible digits). Hence, P(no 7) = (9/10) * (9/10) * (9/10) * (9/10) * (9/10) * (9/10) * (9/10) = (9/10)^7 Hence, the probability that there is at least 1 number 7 is: P(at least one 7) = 1 - P(no 7) P(at least one 7) = 1 - (9/10)^7 P(at least one 7) ≈ 0.522 → 52.2%