Find probability with random variable

Question

Assume the random variable x is normally distributed with mean μ=82 and standard deviation σ=4.

Find the indicated probability.

​P(71<x<74​)

Answer 1

Question Summary: We want to find the probability that X is between 71 and 74 if it's part of a normal distribution with a mean of 82 and a standard deviation of 4.

At it's heart, this is an interval problem which can be solved by finding p(x <74) - p(x <71). Assuming you don't have a computer that calculates this for you, or a nice TI calculator with probability functions, we're going to walk through the math on how to solve this problem.

Step 1: Convert our range into Z-Scores

Z = ( X - µ)/σ <in this case our X is the end points of our potential X range (71 and 74)>

Z₁ = (71 - 82)/4 Z₂ = (74 - 82)/4

Z₁ = (-11)/4 <solve the values in the parenthesis> Z₂ = (-8)/4

Z₁ = -2.75 <divide by the SD> Z₂ = -2

Step 2: Lookup the p-values for our z-scores

If you pull up a table showing our Z-scores you'll find the following values

Z = 2, Mean to Z = 0.4772, Larger Portion = 0.9772, Smaller Portion = 0.0228

Z = 2.75, Mean to Z = 0.4970, Larger Portion = 0.9970, Smaller Portion = 0.0030

It doesn't matter that our Z-scores were negative, because the probabilities are mirrored on both sides of your mean. Similarly this means that it doesn't matter if you subtract the larger portions from each other or the smaller portions from each other because the probability that x will be within the intervals is equal in both groups.

Step 3: Calculate the Probability

p(Z₁<x<Z₂) = p(Z₂) - p(Z₁)

p = 0.0228 - 0.0030 <plug in our values> p = 0.9970 - 0.9772

p = 0.0198 <subtract the overlapping areas> p = 0.0198

A 100% probability is p = 1.00. If we multiply our probabilities we can see the percentage chance that X falls within the specified range.

0.0198 * 100 = 1.98%

There is a 1.98% chance that X falls within the specified range