Question
A normal distribution has a mean of 98 and a standard deviation of 6. What is the probability that a randomly selected x-value from the distribution is at least 80?
Answers
Normalize 80 as
(80-98)/6=-3
Look up the probability from a normal distribution table for -3 to ∞.
(It is the same value from -∞ to +3, depending on the availability of the table).
I get 0.9987.
(80-98)/6=-3
Look up the probability from a normal distribution table for -3 to ∞.
(It is the same value from -∞ to +3, depending on the availability of the table).
I get 0.9987.
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