(a)
To find out what is the
(a)
Answer to Problem 16.26AT
The probability that any one seed weighs more than
Explanation of Solution
In the question, it is given that the distribution of seed weights for a variety of winged bean. This distribution is approximately Normal with,
And we are considering selecting at random four seeds of this variety. Thus, the probability that any one seed weighs more than
And the probability that any one seed weighs between
(b)
To find out what is the probability that all four seeds in the sample weighs more than
(b)
Answer to Problem 16.26AT
The probability that all four seeds in the sample weighs more than
Explanation of Solution
In the question, it is given that the distribution of seed weights for a variety of winged bean. This distribution is approximately Normal with,
And we are considering selecting at random four seeds of this variety. Thus, theprobability that all four seeds in the sample weighs more than
And the probability that the average weight of the four seeds is greater than
By looking at the above two probabilities we can see that the probability that all four seeds in the sample weighs more than
(c)
To find out what is the probability that all four seeds in the sample weighs between
(c)
Answer to Problem 16.26AT
The probability that all four seeds in the sample weighs between
Explanation of Solution
In the question, it is given that the distribution of seed weights for a variety of winged bean. This distribution is approximately Normal with,
And we are considering selecting at random four seeds of this variety. Thus, the probability that all four seeds in the sample weighs between
And the probability that the average weight of the four seeds is between
Both of these probabilities are smaller relative to the corresponding answer from part (b). Since the probability an individual seed weighs between
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Chapter 16 Solutions
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