Of 200 adults, 176 own one TV set, 22 own two TV sets, and 2 own three TV sets. A person
is chosen at random.
a. What is the probability function of X?
b. What is the expected value for the probability distribution?
c. What are the variance and standard deviations for the probability distribution?
To determine the probability function of X, we need to calculate the probabilities for each possible outcome.
a. Calculation of the probability function:
Let's define X as the number of TV sets owned by a randomly selected person.
There are three possible outcomes: 1 TV set, 2 TV sets, or 3 TV sets.
First, let's calculate the probability of owning 1 TV set:
P(X = 1) = Number of adults owning 1 TV set / Total number of adults
P(X = 1) = 176 / 200
P(X = 1) = 0.88
Next, let's calculate the probability of owning 2 TV sets:
P(X = 2) = Number of adults owning 2 TV sets / Total number of adults
P(X = 2) = 22 / 200
P(X = 2) = 0.11
Finally, let's calculate the probability of owning 3 TV sets:
P(X = 3) = Number of adults owning 3 TV sets / Total number of adults
P(X = 3) = 2 / 200
P(X = 3) = 0.01
Therefore, the probability function of X is:
P(X = 1) = 0.88
P(X = 2) = 0.11
P(X = 3) = 0.01
b. Calculation of the expected value:
To find the expected value (E), multiply each possible outcome by its respective probability and sum them up.
E = (X1 * P(X = X1)) + (X2 * P(X = X2)) + (X3 * P(X = X3))
E = (1 * 0.88) + (2 * 0.11) + (3 * 0.01)
E = 0.88 + 0.22 + 0.03
E = 1.13
Therefore, the expected value for the probability distribution is 1.13.
c. Calculation of the variance and standard deviation:
The variance (Var) can be calculated as follows:
Var = [ (X1 - E)^2 * P(X = X1) ] + [ (X2 - E)^2 * P(X = X2) ] + [ (X3 - E)^2 * P(X = X3) ]
Var = [ (1 - 1.13)^2 * 0.88 ] + [ (2 - 1.13)^2 * 0.11 ] + [ (3 - 1.13)^2 * 0.01 ]
Var = (0.0144 * 0.88) + (0.7056 * 0.11) + (1.8049 * 0.01)
Var = 0.0127 + 0.0776 + 0.0180
Var = 0.1083
The standard deviation (SD) is the square root of the variance:
SD = √Var
SD = √0.1083
SD ≈ 0.3294 (rounded to four decimal places)
Therefore, the variance is approximately 0.1083 and the standard deviation is approximately 0.3294.
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