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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.)

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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%.

This problem has been solved

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