Question

Probability and Statistics
Could you please solve the following questions with specifying the answers

A. What is the probability of getting a sum of exactly 11 when rolling a pair of dice given that the sum is at least 9

1. 4/36        b. 2/10        c. 9/36        d. 4/9

B. Scores on a test are normally distributed with a mean of 520 and a standard deviation of 50. George scores 622 on the test. He was in which percentile approximately?

1. 84^th percentile      b. 98^th percentile  

c. 95^th percentile      d. 2^nd percentile

C. On a multiple choice test with three possible answers for each of the 5 questions, what is the probability that a student will get exactly 3 right just by guessing?

a. 40/81     b. 40/243       c. 2/5       d. 80/243

Answer 1

(A) A Sum of exactly 11 is obtained by (5,6) and (6,5) = 2 outcomes

A Sum of at least 9 is obtained by getting a 9, 10, 11 or 12 = (4,5) (5,4) (3,6) (6,3) (6,4) (4, 6) (5,5) (5,6) (6,5) and (6,6) = 10 outcomes

By Bayes Theorem) P(A given B) = P(A and B) / P(B)

Probability = Favorable Outcomes / Total Outcomes

P(Getting a 11 given he got at least a 9) = P(Exactly 11 and At least 9) / P(At least 9) = (2/36) / (10/36) = 2 / 10 Option b

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(B) The z score is (X - ) / = (622 - 520) / 50 = 2.04

The probability from the normal distribution tables is 0.9793

The percentage of students 0.9793 * 100 = 97.93 or Option b: The 98th Percentile

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(C) This is answered using the Binomial Probability

Please note

Binomial Probability = ⁿC_x * (p)^x * (q)^n-x, where n = number of trials and x is the number of successes.

Here n = 5, p = out correct anser out of 3 choices = 1/3 , q = 1 – p = 2/3.

P(X = 3) = ⁵C₃ * (1/3)³ * (2/3)^5-3 = 10 * (1 / 27) * (4 / 9) = 40 / 243 Option b

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