# Normal Distribution Question. Need help!

[MENTOR note] Post moved from General Math forum hence no template.

Assume that a random variable follows a normal distribution with a mean of 80 and a standard deviation of 24. What percentage of this distribution is not between 32 and 116?
My approach is to calculate the Probability for (mean - 2*σ < X < mean + 1.5*σ), not sure how to solve this further.

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fresh_42
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Ray Vickson
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Dearly Missed
[MENTOR note] Post moved from General Math forum hence no template.

Assume that a random variable follows a normal distribution with a mean of 80 and a standard deviation of 24. What percentage of this distribution is not between 32 and 116?
My approach is to calculate the Probability for (mean - 2*σ < X < mean + 1.5*σ), not sure how to solve this further.
Use $P(a < X < b) = P(X < b) - P(X < a)$, and find both $P(X < a)$ and $P(X < b)$ from standard normal tables.

Use $P(a < X < b) = P(X < b) - P(X < a)$, and find both $P(X < a)$ and $P(X < b)$ from standard normal tables.
The solution I have with me is:
Prob(mean - 2*σ < X < mean + 1.5*σ) = (0.5 - 0.0228) + (0.5 - 0.0668)
My question is how we got (0.5 - 0.0228) + (0.5 - 0.0668) ?

Try to break the problem into smaller pieces. What probability do you believe 0.5 - 0.0228 is specifying? (i.e. Probability that X is in what range?).

RUber
Homework Helper
The solution I have with me is:
Prob(mean - 2*σ < X < mean + 1.5*σ) = (0.5 - 0.0228) + (0.5 - 0.0668)
My question is how we got (0.5 - 0.0228) + (0.5 - 0.0668) ?
Try both your method and the one suggested by Ray. They should give the same result. Drawing a picture might help as well.
It might also be interesting to note that (0.5 - 0.0228) + (0.5 - 0.0668) = 1 - 0.0228 - 0.0668. Do you know why those two numbers (0.0228 and 0.0668) are significant in this problem?