Binomial Probability Distribution

Only available on StudyMode
  • Topic: Sales, Binomial distribution, Door-to-door
  • Pages : 6 (807 words )
  • Download(s) : 287
  • Published : October 9, 2010
Open Document
Text Preview
#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 |...
tracking img