Concept explainers
Check whether each row of given distribution is valid or invalid.
Answer to Problem 39CE
The
Explanation of Solution
From the given grade distribution, row (a) values are 0.3, 0.3, 0.3, 0.2, and 0.1.
The valid
Here, the sum of probability is calculated as follows:
Thus, row (a) is an invalid probability distribution since the total probability is greater than 1.
From the given grade distribution, row (b) values are 0, 0, 1, 0, and 0.
The valid probability distribution should have total probability equal to 1 and probability values should be greater than or equal to zero.
Here, the sum of probability is calculated as follows:
Thus, row (b) is a valid probability distribution since the total probability is equal than 1.
From the given grade distribution, row (c) values are 0.3, 0.3, 0.3, 0, and 0.
The valid probability distribution should have a total probability equal to 1.
Here, the sum of probability is calculated as follows:
Thus, row (c) is an invalid probability distribution since the total probability is less than 1.
From the given grade distribution, row (d) values are 0.3, 0.5, 0.2, 0.1, and –0.1.
The valid probability distribution should have a total probability equal to 1 and probability values should be greater than or equal to zero.
Here, the negative probability value is given.
Thus, row (d) is an invalid probability distribution since the negative probability value is found.
From the given grade distribution, row (e) values are 0.2, 0.4, 0.2, 0.1, and 0.1.
The valid probability distribution should have a total probability equal to 1 and probability values should be greater than or equal to zero.
Here, the sum of probability is calculated as follows:
Thus, row (e) is a valid probability distribution since the total probability is equal than 1.
From the given grade distribution, row (f) values are 0, –0.1, 1.1, 0, and 0.
The valid probability distribution should have a total probability equal to 1 and probability values should be greater than or equal to zero.
Here, the negative probability value is given.
Thus, row (f) is an invalid probability distribution since negative probability value is found.
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