Binomial Probability Distribution
#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
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