Answer:
4
Step-by-step explanation:
7*7-9*5
49-45
4
Part A:
The probability that a normally distributed data with a mean, μ and standard deviation, σ is greater than a given value, a is given by:

Given that the average precipitation in
Toledo, Ohio for the past 7 months is 19.32 inches with a standard deviation of 2.44 inches, the probability that <span>a randomly selected year will have precipitation greater than 18 inches for the first 7 months is given by:

Part B:
</span>The probability that an n randomly selected samples of a normally distributed data with a mean, μ and
standard deviation, σ is greater than a given value, a is given by:

Given that the average precipitation in
Toledo, Ohio for the past 7 months is 19.32 inches with a standard deviation of 2.44 inches, the probability that <span>5 randomly selected years will have precipitation greater than 18 inches for the first 7 months is given by:
</span>
Answer:
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