Question

1. The file name Exam-2.xlsx, sheet named Part 1is kept blank. Use Part 1sheet to answer this question. Put a box around your answers. For (a) and (b), round your answers to 4 decimal places.

The Wall Street Journalreported some interesting statistics on the job market. One statistic is that 40% of all workers say they would change jobs for “slightly higher pay.” Suppose 20 workers are randomly selected and asked if they would change jobs for “slightly higher pay.”

1. What is the probability that nine or more say yes?
2. What is the probability that exactly six say yes?
3. What is the expected [mean] number of workers that would say they would change jobs for slightly higher pay?

[Hint: Use binomial distribution to answer question 1]

Answer 1

The probability distribution is given by

(a+b)²⁰ where a is the probablity of worker saying yes ( 0.4)

1 Probability of 9 or more saying yes

1 - [ b²⁰ + 20C1* b¹⁹*a+20C2*b¹⁸*a²+20C3*b¹⁷*a³+........+20C8*b¹²*a⁸]

1- [ 0.00003656+0.000487+0.00308+0.0123+0.03499+0.0746+0.1244+0.1612+0.1796]

1-(0.5906) =0.4093

2.

Probability of exactly six saying yes

20C6*a⁶*b¹⁴ = 0.1244

3 Mean of a probability distribution ( probability of saying yes)

mu = n*p = 20x0.4 =8