shade the correct answers in PART I: Multipl orthe follopwing maltiple- choice q
ID: 384208 • Letter: S
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shade the correct answers in PART I: Multipl orthe follopwing maltiple- choice questions (from no. 8 to mo. 15) The University Registration and Admission Department provided information on the College of Business Administration's 2016 graduates. The following table gives the probability distribution of the number of semesters graduates were on academic probation during their years of study and the corresponding probabilities Based on this table, answer questions 8 and 9. boxes given in page 2 above: 060 0.157 0.08 0.05 8. Find the expected value of this distribution. b. 0.83 .0.166 d. 1.461 a. 2.0 9 Calculate the variance of this distribution. a. 1.461 b. 083 c. 1.209 d. 1.05 10. On average, 4 car accidents occur per week at Awali roundabout. If these accidents follow a Poisson distribution what is the probability that at most 2 car accidents will occur at that roundab per week? a. 0.1465 b. 0. 2198 e. 0.0916 d. 0.2381 Stress on the job is a major concern of a large number of people who go into managerial positio It is estimated that 80% of managers of all companies suffer from job-related stress. A sample 6 managers of companies were taken at random. Use the binomial distribution to answer questi no. 11 and no.12 11. What is the probability that at least 4 out of this sample of managers suffer from 12. Find the probability that exactly 5 of this sample will not suffer from job-related stress. 3. Suppose p (F)-0.35 and P (KIF)-0.40, what is the joint probability of K and F? Assume event A is·Abdulla arrives between 9 am, and 4 p.m., while event B is . Abdu job-related d. 0.754 a. 0.099 b. 0.246 c. 0.901 a. 0.393 b. 0.0015 c. 0.655 d. 0.9985 a. 0.75 b. 0.14 C. d. 0.875 in his new Toyota car'. In this case, events A and B are a. Mutually exclusive b. Complementary d. Not mutually exclusive he listing of all the outcomes of an experiment and the probability associated with tcome is known as a a. Probability distribution c. Conditional probability b. Sample space d. Joint probability 6Explanation / Answer
8.
Option B
So following the above formula:
Option A
10.
P(x; ) = (e-) (x) / x!
So using the above:
P(2; 4) = 0.14652
Option A
11.
P(X)= n!*P^x* Q^n-x/(n-X)!*X!
P(X) = 0.246
Option B
12.
similarly just putting x =5
P(x) = 0.393
Option A
No of Sems 0 1 2 3 4 Sum of all prob Prob 0.6 0.15 0.08 0.05 0.88 So = 1-0.88 0.12 Ans 8 So expected value No of Sem *Prob 0.83Related Questions
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