Statistics Examples

Prove that the given table satisfies the two properties needed for a probability distribution.
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A discrete random variable takes a set of separate values (such as , , ...). Its probability distribution assigns a probability to each possible value . For each , the probability falls between and inclusive and the sum of the probabilities for all the possible values equals to .
1. For each , .
2. .
is between and inclusive, which meets the first property of the probability distribution.
is between and inclusive
is between and inclusive, which meets the first property of the probability distribution.
is between and inclusive
is between and inclusive, which meets the first property of the probability distribution.
is between and inclusive
is between and inclusive, which meets the first property of the probability distribution.
is between and inclusive
For each , the probability falls between and inclusive, which meets the first property of the probability distribution.
for all x values
Find the sum of the probabilities for all the possible values.
The sum of the probabilities for all the possible values is .
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Add and .
Add and .
Add and .
For each , the probability of falls between and inclusive. In addition, the sum of the probabilities for all the possible equals , which means that the table satisfies the two properties of a probability distribution.
The table satisfies the two properties of a probability distribution:
Property 1: for all values
Property 2:
The table satisfies the two properties of a probability distribution:
Property 1: for all values
Property 2:
The expectation mean of a distribution is the value expected if trials of the distribution could continue indefinitely. This is equal to each value multiplied by its discrete probability.
Simplify the expression.
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Simplify each term.
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Multiply by .
Multiply by .
Multiply by .
Multiply by .
Simplify by adding numbers.
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Add and .
Add and .
Add and .
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