There are three brands of milk for sale in a supermarket. Let's call them brand A, B, and C. A customer who purchases milk will choose brands A,B, and C with probability 0.6, 0.3, 0.1, respectively. Each day, the toal number of customers who purchase milk at this market follow a Binomial(200,0.5) distribution. Let X be the daily number of sales of brand A milk. Find E(X) and V(X)
In a batch of 1000 cartons estimate the number of cartons that contain less than the number of cartons that contain less the advertised volume of milk
To find expected value (E(X)) and variance (V(X)) of the number of sales of brand A milk (X), we need to use the properties of the binomial distribution.
First, let's find E(X).
E(X) = n * p
Where:
n = number of trials (total number of customers who purchase milk)
p = probability of success (probability of choosing brand A milk)
In this case, n = 200 (total number of customers) and p = 0.6 (probability of choosing brand A milk).
E(X) = 200 * 0.6
E(X) = 120
So, the expected number of sales of brand A milk (X) is 120.
Next, let's find V(X).
V(X) = n * p * (1 - p)
In this case, p = 0.6 (probability of choosing brand A milk).
V(X) = 200 * 0.6 * (1 - 0.6)
V(X) = 200 * 0.6 * 0.4
V(X) = 48
So, the variance of the number of sales of brand A milk (X) is 48.
In summary,
E(X) = 120
V(X) = 48
To find the expected value (E) and variance (V) of a random variable, we need to determine the probability distribution of the random variable.
In this case, X represents the daily number of sales of brand A milk. Since the total number of customers who purchase milk follows a Binomial(200, 0.5) distribution, it means each customer has a 0.5 probability of purchasing milk.
To find the probability of selling brand A milk, we multiply the probability of a customer choosing brand A (0.6) by the probability of a customer buying milk (0.5):
P(X = k) = (0.6)(0.5)^(k)(0.5)^(200-k) * C(200,k)
where C(200,k) is the binomial coefficient, given by C(200,k) = 200! / (k!(200-k)!).
Now, let's calculate the expected value (E(X)):
E(X) = ∑[k * P(X = k)]
For each value of k from 0 to 200, we calculate k multiplied by the corresponding probability P(X = k) and sum them all up.
Similarly, let's calculate the variance (V(X)):
V(X) = ∑[(k - E(X))^2 * P(X = k)]
For each value of k from 0 to 200, we calculate (k - E(X))^2 multiplied by the corresponding probability P(X = k) and sum them all up.
By calculating these sums, we can find the expected value and variance of X.
Since n = 200, .6 = 120, .3 = 60, and .1 = 20
Therefore, the expected vale of X is:
E(X) = sum x*p = 120(.6) + 60(.3) + 20(.1) = 92
To get the variance, subtract the mean (expected value) from each X. Next, square each of those values. Finally, add the squares times their respective probabilities. If you need the standard deviation, just take the square root of the variance. That should be it.