Binomial Probability Distribution

Topics: Sales, Binomial distribution, Door-to-door Pages: 6 (807 words) Published: October 9, 2010
#1 True or false: Even if the sample size is more than 1000, we cannot always use the normal approximation to binomial.

Solution:

If a sample is n>30, we can say that sample size is sufficiently large to assume normal approximation to binomial curve.

Hence the statement is false.

#2

A salesperson goes door-to-door in a residential area to demonstrate the use of a new Household appliance to potential customers. She has found from her years of experience that after demonstration, the probability of purchase (long run average) is 0.30. To perform satisfactory on the job, the salesperson needs at least four orders this week. If she performs 15 demonstrations this week, what is the probability of her being satisfactory? What is the probability of between 4 and 8 (inclusive) orders?

Solution

p=0.30

q=0.70

n=15

k=4

[pic]

Using Megastat we get

| | |15 |
| |0.3 | P |
| | | |
| | |Cumulative |
|k |p(k) |Probability |
|0 |0.00056 |0.00056 |
|1 |0.00503 |0.00559 |
|2 |0.02154 |0.02713 |
|3 |0.05848 |0.08561 |
|4 |0.11278 |0.19838 |
|5 |0.16433 |0.36271 |
|6 |0.18781 |0.55052 |
|7 |0.17248 |0.72299 |
|8 |0.12936 |0.85235 |
|9 |0.08008 |0.93243 |
|10 |0.04118 |0.97361 |
|11 |0.01765 |0.99126 |
|12 |0.00630 |0.99756 |
|13 |0.00187 |0.99943 |
|14 |0.00046 |0.99989 |
|15 |0.00009 |0.99998 |
|16 |0.00001 |1.00000 |
|17 |0.00000 |1.00000 |
|18 |0.00000 |1.00000 |
|19 |0.00000 |1.00000 |
|20 |0.00000 |1.00000 |
|21 |0.00000 |1.00000 |
| |1.00000 | |

1- {p (k=0)+p(k=1)+p(k=2)+p(k=3}) ≥0.9

At n=21

1-p (k=0) +p (k=1)+p(k=2)+p(k=3) = 0.91439

As 0.91439> 0.9

So a sales person has to make 21 demonstrations in order to be at least 90% confident that she will

get a satisfactory evaluation this week.

iii)

As the performance is increased due to training and the probability of success has increased to 0.35

So we expect a lower number of demonstrations to attain at least 4 sales and thus a satisfactory evaluation

Using megastat we get to know that

| |18 | N |
| |0.35 | P |
| | | |
|  |  |Cumulative |
|K |p(k) |Probability |
|0 |0.00043 |0.00043 |
|1 |0.00416 |0.00459 |
|2 |0.01903 |0.02362 |
|3 |0.05465 |0.07827 |
|4 |0.11035 |0.18862 |
|5 |0.16638 |0.35500 |
|6 |0.19411 |0.54910 |
|7 |0.17918 |0.72828 |
|8 |0.13266 |0.86094 |
|9 |0.07937 |0.94031 |
|10 |0.03846 |0.97877 |
|11 |0.01506 |0.99383 |
|12 |0.00473 |0.99856 |
|13 |0.00118 |0.99974 |
|14 |0.00023 |0.99996 |
|15 |0.00003 |1.00000 |
|16 |0.00000 |1.00000 |
|17 |0.00000 |1.00000 |
|18 |0.00000 |1.00000 |
| |1.00000 |...
Continue Reading

Please join StudyMode to read the full document

You May Also Find These Documents Helpful

  • Datastor Case Study
  • Probability distribution Essay
  • Essay on Distribution Assignment
  • Probability Theory and Ans Essay
  • Probability Theory Research Paper
  • Essay on Normal Distribution
  • Lab 03 Normal Distribution Essay
  • Essay about Normal Distribution

Become a StudyMode Member

Sign Up - It's Free