Problem 2.
THE PROBABILITY THAT A RANDOMLY CHOSEN SALES PROSPECT WILL MAKE A PURCHASE IS 0.20. IF A SALESMAN CALLS ON SIX PROSPECTS,
A. WHAT IS THE PROBABILITY THAT HE WILL MAKE EXACTLY FOUR SALES?
B. WHAT IS THE PROBABILITY THAT THE SALESMAN WILL MAKE FOUR OR MORE SALES?
C. WHAT IS THE EXPECTED NUMBER OF SALES (AS A LONG-RUN AVERAGE) AND THE VARIANCE ASSOCIATED WITH MAKING CALLS ON 15 PROSPECTS?
Problem 3.
THE NUMBER OF TRUCKS ARRIVING HOURLY AT A WAREHOUSE FACILITY HAS BEEN FOUND TO FOLLOW THE PROBABILITY DISTRIBUTION IN TABLE BELOW. CALCULATE
NUMBER OF TRUCKS ( X ) 0.00 1.00 2.00 3.00 4.00 5.00 6.00
PROBABILITY [P( X )] 0.05 0.10 0.15 0.25 0.30 0.10 0.05
A. THE EXPECTED NUMBER OF ARRIVALS X PER HOUR AND
B. THE VARIANCE OF THIS PROBABILITY DISTRIBUTION FOR THE DISCRETE RANDOM VARIABLE.
Problem 4.
A MAIL-ORDER FIRM HAS A CIRCULAR WHICH ELICITS A 10 % RESPONSE RATE. SUPPOSE 20 OF THE CIRCULARS ARE MAILED AS A MARKET TEST IN A NEW GEOGRAPHIC AREA. ASSUMING THAT THE 10% RESPONSE RATE IS APPLICABLE IN THE NEW AREA, DETERMINE THE PROBABILITIES OF THE FOLLOWING EVENTS:
A. NO ONE RESPONDS
B. A MAJORITY OF THE PEOPLE RESPOND
C. WHAT IS THE EXPECTED NUMBER OF PEOPLE WILL RESPOND?
Problem 5.
THE MAXIMUM NUMBER OF CUSTOMERS ARRIVING DURING RANDOMLY CHOSEN 10-MIN INTERVALS IS 5 AT A DRIVE-IN FACILITY SPECIALIZING IN PHOTO DEVELOPMENT AND FILM SALES. IT HAS BEEN FOUND THE NUMBER OF ACTUAL SALES MADE FOLLOWS THE PROBABILITY DISTRIBUTION IN TABLE BELOW.
NUMBER OF SALES ( X ) 0 1 2 3 4 5
PROBABILITY [P( X )] 0.15 0.25 0.25 ? 0.10 0.05
A. PLEASE FILL IN THE MISSING PROBABILITY FOR X=3 IN THE TABLE ABOVE (CELL WITH “?” MARK)
B. CALCULATE THE EXPECTED NUMBER AND VARIANCE OF ARRIVALS IN 10-MIN INTERVAL.
Did anyone answer this?
To solve these problems, we need to apply the concepts of probability and expected value. Let's go through each problem one by one:
Problem 2:
A. To find the probability that the salesman will make exactly four sales out of six prospects, we can use the binomial distribution formula. The formula is:
P(X=k) = nCk * p^k * (1-p)^(n-k)
where:
- P(X=k) is the probability of getting exactly k successes,
- n is the total number of trials (prospects),
- k is the number of desired successes,
- p is the probability of success in a single trial.
In this case, n = 6 (number of prospects), k = 4 (number of desired successes), and p = 0.20 (probability of success in a single trial).
P(X=4) = 6C4 * 0.20^4 * (1-0.20)^(6-4)
= 15 * 0.20^4 * 0.80^2
= 0.1856
Therefore, the probability that the salesman will make exactly four sales is 0.1856.
B. To find the probability that the salesman will make four or more sales, we need to calculate the cumulative probability from four to six sales:
P(X≥4) = P(X=4) + P(X=5) + P(X=6)
To calculate each term, you can use the same binomial distribution formula as explained above and substitute the values of n, k, and p. Add up the three probabilities to get the cumulative probability.
C. To find the expected number of sales and the variance associated with making calls on 15 prospects, we can use the formulas for expected value (mean) and variance for a binomial distribution:
Expected value (mean):
E(X) = n * p
Variance:
Var(X) = n * p * (1-p)
Here, n = 15 (number of prospects) and p = 0.20 (probability of success in a single trial). Calculate the expected value and variance using these formulas.
Problem 3:
A. To find the expected number of arrivals per hour, you need to multiply the number of trucks (X) by their respective probabilities (P(X)), and then add up these values. For example:
Expected number of arrivals:
E(X) = 0.00 * 0.05 + 1.00 * 0.10 + 2.00 * 0.15 + 3.00 * 0.25 + 4.00 * 0.30 + 5.00 * 0.10 + 6.00 * 0.05
B. To find the variance of the probability distribution for the discrete random variable, you can use the formula:
Variance:
Var(X) = ∑ (X^2 * P(X)) - (E(X))^2
where ∑ denotes the sum over all possible values of X. Calculate the variance using this formula.
Problem 4:
To determine the probabilities in this problem, we can use the binomial distribution formula again. Here, we are given a response rate of 10%, and we want to find the probabilities of different events for 20 circulars mailed.
A. To find the probability that no one responds, use the binomial distribution formula with n = 20, k = 0, and p = 0.10.
B. To find the probability that a majority of people respond, you need to calculate the sum of probabilities for the number of responses greater than or equal to half of 20. Add up the probabilities for k = 10, k = 11, ..., k = 20, using the binomial distribution formula.
C. The expected number of people who will respond can be calculated by multiplying the total number of circulars mailed (20) by the response rate (10%).
Problem 5:
A. To fill in the missing probability for X = 3, you need to subtract the sum of probabilities for X = 0, X = 1, X = 2, X = 4, and X = 5 from 1 (since the sum of all probabilities must equal 1).
B. To calculate the expected number and variance of arrivals in a 10-minute interval, you can follow the same approach as in Problem 3. Multiply the number of sales (X) by their respective probabilities (P(X)), and then add up these values to find the expected number. For variance, use the formula explained in Problem 3.