Question

A researcher records the number of buckets of popcorn purchased by patrons during one night at the movies. She finds that the probability that a patron purchased 0 buckets of popcorn is p = .27; 1 bucket is p = .51; 2 buckets is p = .17; and 3 buckets is p = .05. How many buckets of popcorn can we expect a patron to purchase per night at the movies in the long run?

Answer 1

Answer:

X         0          1            2         3  

P(X)    0.27    0.51     0.17     0.05

$E(X) = \sum_{i=1}^n X_i P(X_i)$

And replacing we got:

$E(X) = 0*0.27+ 1*0.51 +2*0.17 +3*0.05= 1$

Step-by-step explanation:

Let X the random variable that represent "the number of buckets of popcorn purchased by patrons during one night at the movies"

And we have the following distribution for X

X         0          1            2         3  

P(X)    0.27    0.51     0.17     0.05

In statistics and probability analysis, the expected value "is calculated by multiplying each of the possible outcomes by the likelihood each outcome will occur and then summing all of those values".

The variance of a random variable Var(X) is the expected value of the squared deviation from the mean of X, E(X).

And the standard deviation of a random variable X is just the square root of the variance.  

For this case we can calculate the expected value with the following formula:

$E(X) = \sum_{i=1}^n X_i P(X_i)$

And replacing we got:

$E(X) = 0*0.27+ 1*0.51 +2*0.17 +3*0.05= 1$