Suppose X is normally distributed with a mean of 8 and a variance of 1. What is the probability that X is greater than 0 but less than 8? (Your answer should be correct to one decimal place, and you should be able to answer without the use of Normal tables.)
Question
Suppose X is normally distributed with a mean of 8 and a variance of 1. What is the probability that X is greater than 0 but less than 8? (Your answer should be correct to one decimal place, and you should be able to answer without the use of Normal tables.)
Solution
To solve this problem, we need to understand the properties of a normal distribution.
A normal distribution is symmetric around its mean. This means that 50% of the values are less than the mean and 50% of the values are greater than the mean.
In this case, the mean of the distribution is 8. So, the probability that X is less than 8 is 0.5 or 50%.
The probability that X is greater than 0 but less than 8 is the same as the probability that X is less than 8, because the probability that X is greater than 0 is 1 (since we are dealing with a normal distribution, which extends from negative infinity to positive infinity).
So, the probability that X is greater than 0 but less than 8 is 0.5 or 50%.
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