Math Problem Statement
solve number 6
Solution
Let's break down and solve problem number 6 from the image:
Problem:
A salesman sells goods by telephone. The probability that any particular call achieves a sale is 0.08, independently of all other calls. In a particular one-hour period, the salesman makes a total of 12 calls. We are asked to compute:
- (i) The expected number of sales.
- (ii) The probability that the number of sales is 2 or fewer.
- (iii) The probability that the number of sales is 2 or more.
Key Points:
- The number of calls is 12.
- The probability of a successful sale on any given call is .
- This follows a Binomial Distribution , where is the number of successful sales.
Solution:
(i) Expected number of sales:
For a binomial distribution , the expected value is given by:
Substituting and :
So, the expected number of sales is 0.96.
(ii) Probability that the number of sales is 2 or fewer:
We need to calculate , where . This can be calculated using the binomial probability mass function (PMF):
Summing for :
Let me calculate this numerically.The probability that the number of sales is 2 or fewer is approximately 0.935.
(iii) Probability that the number of sales is 2 or more:
We can use the complement rule to find this probability:
Let's calculate .The probability that the number of sales is 2 or more is approximately 0.249.
Summary of results:
- Expected number of sales: 0.96
- Probability that the number of sales is 2 or fewer: 0.935
- Probability that the number of sales is 2 or more: 0.249
Would you like further clarification or details on any of the steps?
Here are some related questions to explore:
- How does the probability change with a different number of calls?
- What happens to the expected value if the success probability changes?
- How would the results change if the sales calls were dependent on each other?
- What is the variance of the number of sales in this scenario?
- Can you derive the cumulative distribution for other values of sales?
Tip: For small probabilities, the binomial distribution often approximates a Poisson distribution for easier calculation.
Ask a new question for Free
By Image
Drop file here or Click Here to upload
Math Problem Analysis
Mathematical Concepts
Binomial Distribution
Expected Value
Probability
Formulas
E(X) = n * p
P(X = k) = C(n, k) * p^k * (1 - p)^(n - k)
P(X <= 2) = P(X = 0) + P(X = 1) + P(X = 2)
Complement Rule: P(X >= 2) = 1 - P(X <= 1)
Theorems
Binomial Distribution Theorem
Complement Rule in Probability
Suitable Grade Level
Grades 10-12
Related Recommendation
Expected Number of Sales Contracts Using Binomial Distribution
Sample Space for Sales and No Sales with Stopping Conditions
Poisson Distribution: Probability of Receiving At Least 6 or 4 Calls per Hour
Salesman's Commission and Expected Income Problem
Binomial Distribution: Expected Value, Standard Deviation, and Probability for Fax Messages