Finite Math Examples

Find the Standard Deviation table[[x,P(x)],[149.5-169.5,4],[169.5-189.5,11],[189.5-209.5,15],[209.5-229.5,25]]
xP(x)149.5169.54169.5189.511189.5209.515209.5229.525
Step 1
Prove that the given table satisfies the two properties needed for a probability distribution.
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Step 1.1
A discrete random variable x takes a set of separate values (such as 0, 1, 2...). Its probability distribution assigns a probability P(x) to each possible value x. For each x, the probability P(x) falls between 0 and 1 inclusive and the sum of the probabilities for all the possible x values equals to 1.
1. For each x, 0P(x)1.
2. P(x0)+P(x1)+P(x2)++P(xn)=1.
Step 1.2
4 is not less than or equal to 1, which doesn't meet the first property of the probability distribution.
4 is not less than or equal to 1
Step 1.3
11 is not less than or equal to 1, which doesn't meet the first property of the probability distribution.
11 is not less than or equal to 1
Step 1.4
15 is not less than or equal to 1, which doesn't meet the first property of the probability distribution.
15 is not less than or equal to 1
Step 1.5
25 is not less than or equal to 1, which doesn't meet the first property of the probability distribution.
25 is not less than or equal to 1
Step 1.6
The probability P(x) does not fall between 0 and 1 inclusive for all x values, which does not meet the first property of the probability distribution.
The table does not satisfy the two properties of a probability distribution
The table does not satisfy the two properties of a probability distribution
Step 2
The table does not satisfy the two properties of a probability distribution, which means that the standard deviation can't be found using the given table.
Can't find the standard deviation
 x2  12  π  xdx