PART 1

Answer to the question 1

The characteristics of Binomial experiment are as below (Shirazi et al., 2016)

• The number of trails is fixed.
• Each of the trail independent from the others.
• The number of outcomes is two.
• Each of the outcome has a constant probability from trail to trail.

Answer to the question 2

In Poisson distribution the random variable X represent the number of success in a specified region and which is actual (Daskalakis, Diakonikolas & Servedio, 2015).

Answer to the question 3

In Hyper geometric distribution N represents the number of items in the population, n represents the number of item in the sample and r represents the number of success (Hu, Cui & Yin, 2013).

Answer to the question 4

When every outcome in a sample space is equally likely then the uniform distribution has been applied to a variable (Stephens, 2017).

Answer to the question 5

In a standard normal curve mean is 0 and standard deviation is 1. If the data set is normal then about 68% observations lie within the standard deviation and mean and it is denote (-1, 1). Again about 95% observations lie within 2 standard deviations of the mean and it is denote (-2, 2). Similarly about 99.7% observations lie within 3 standard deviations of the mean and it is denote (-3, 3) (Bell, 2015).

Answer to the question 6

The poison distribution deals with the occurrences number in a fixed period of time. On the other hand the exponential distribution deals with the time of successive events when the time continuously flow. Both are related to queuing theory .Moreover both are under Poisson process. The waiting time Poisson distribution is an exponential distribution (Rossi et al., 2015).

PART 2

Answer to the question 1

Part (a)

Let

A denote the event that the individual is African American

B denote the event that the individual is a women.

C denote the event that the individual is earned MBA.

Yes, the event A, B and C are independent.  Because an MBA degree individual has to be man women. Moreover, women has or not the MBA degree and also they are or are not from African American. All the independent with each other.

Part (b)

Let  denote the white male individual.

Given that A,  and C are independent.

P (A) = 0.08

P (B) = 0.13

P (C) = 0.17

P () =1- P (B)

= 0.87

Therefore the required probability that the randomly selected manager with white male and earned MBA degree from top 10 graduate business school is as below

 P (A∩  ∩ C) = 0.08*0.87*0.17

= 0.012

Part (c)

If A, B and C are not independent then the required probability that the randomly selected manager with white male and earned MBA degree from top 10 graduate business school is not determined.  To calculate this relation without independence then the following data required. These are P (B/A), P(C/A∩B).

Answer to the question 2

Part (a)

Given that

P= 0.53

N= 100,000,000,000

When X= 1500, then it follows binomial probability distribution.

P (X=1500) = * (0.53)¹⁵⁰⁰ * (1-0.53)^{100000000000-1500}

=* (0.53)¹⁵⁰⁰ * (1-0.53)^99999998500

Part (b)

When the number of sample voters (X) = 6000

P (X=6000) = * (0.53)⁶⁰⁰⁰ * (1-0.53)^{100000000000-6000}

= = * (0.53)⁶⁰⁰⁰ * (0.47)^99999994000

Answer to the question 3

Part (a)

Given that the mean length = 1

Standard deviation = 0.01

Hence the required probability P (X > 2.05) = P (Z> (2.05-1)/0.01)

= P (Z> 105)

= 1- P (Z 105)

= 1- 1

= 0

Part (b)

Now P (1.96 < X < 2.02) = P (((1.96-1)/0.01) < Z < ((2.02-1)/0.01))

     = (96 < Z < 102)

     = 1-1

     = 0

Answer to the question 4

Given that

The average employment (n) = 2000

x = 20

n1 = 100000

p = 0.0002

Hence the required probability P(X  10) =1-(P (X=9)+ P (X=8) + (X=7)+ ……….+ P (X=0)

     = 1- (* (0.0002)⁹ * (1-0.0002)^2000-9+ ………+ ( * (0.0002)⁰ * (1-0.0002)^2000-0)

=1-1

=0

The success probability is very close to zero. Thus this means that the probability of at least 10 brain tumors during 1982 to 1991 is zero.

Answer to the question 5

Part (a)

Given that

n = 200

p= 0.93

Hence the required probability that at least 15 invoices which over charges the customer is as below

P(X  15) = 1- (P (X=14) + P(X=13) + …….+ P (X=0)

     =1- (* (0.93)⁹ * (1-.93)^200-14+ ………+ ( * (0.93)⁰ * (1-0.93)^200-0)

=1

It is a certain probability. The probability that would not find any that under change is (1-1) =0

Part (b)

P (X k) = 0.15

Hence the invoices is 0.15.

PART 3

The study is based on GO Bananas Breakfast cereal case problem. The company GGG produce breakfast with the help of rice flakes and banana flavored marshmallows. In general the company produces 25 boxes per week which contains 1.6 to 2.4 ounces.

When the production process is properly working then 0.7 of probability sample will result in shutdown. The GGG management policy is decide to production on shutdown Moreover they design the production process only 8% among the 16 boxes. Also if at least 5 boxes is fail in a weekly then the sample will not meet the standard weight.

The management of GGG shut down the production process less than 1% time production process is working properly. Then the GGG produces only 2 boxes per week which fail to meet the standard weight of banana-flavored marshmallows.

It is given that at least 5 of the sampled boxes fail to meet the standard weight. Ms. Finkel expect that to meet their target up to level (5/16). Moreover the number of boxes is 16, probability is 0.01 and at least five of the sample boxes fail.

P(X5) = 1- (P(X=4) + P(X=3) +P(X=2) + P(X=1) + P(X=0))

=1- (* (0.01)¹⁴* (1-0.01)^16-4+………+ * (0.01)⁰ * (1-0.01)^16-0  

=1

References and Bibliography

Bell, J. (2015). A simple and pragmatic approximation to the normal cumulative probability distribution. Available at SSRN 2579686.

Collins, D. J., Neild, A., DeMello, A., Liu, A. Q., & Ai, Y. (2015). The Poisson distribution and beyond: methods for microfluidic droplet production and single cell encapsulation. Lab on a Chip, 15(17), 3439-3459.

Daskalakis, C., Diakonikolas, I., & Servedio, R. A. (2015). Learning poisson binomial distributions. Algorithmica, 72(1), 316-357.

Hu, D. P., Cui, Y. Q., & Yin, A. H. (2013). An improved negative binomial approximation for negative hypergeometric distribution. In Applied Mechanics and Materials (Vol. 427, pp. 2549-2553). Trans Tech Publications Ltd.

Kumagai, W., & Hayashi, M. (2016). Second-order asymptotics of conversions of distributions and entangled states based on Rayleigh-normal probability distributions. IEEE Transactions on Information Theory, 63(3), 1829-1857.

Rossi, R., Prestwich, S., Tarim, S. A., & Hnich, B. (2014). Confidence-based optimisation for the newsvendor problem under binomial, Poisson and exponential demand. European Journal of Operational Research, 239(3), 674-684.

Shirazi, M., Lord, D., Dhavala, S. S., & Geedipally, S. R. (2016). A semiparametric negative binomial generalized linear model for modeling over-dispersed count data with a heavy tail: characteristics and applications to crash data. Accident Analysis & Prevention, 91, 10-18.

Stephens, M. A. (2017). Tests for the uniform distribution. In Goodness-of-Fit-Techniques (pp. 331-366). Routledge.

p	0.93
n	200
x	15
X>=15	16,17,…200
Answer	1
Answer	0.148803826