Question

x is a normally distributed random variable with mean 27 and standard deviation 1.
what is the probability that x is between 26 and 28?
use the 0.68 - 0.95 - 0.997 rule and write your answer as a decimal. round to the nearest thousandth if necessary.

Explanation:

Step1: Recall the 68 - 95 - 99.7 rule

The 68 - 95 - 99.7 rule states that for a normal distribution \(X\sim N(\mu,\sigma)\), \(P(\mu-\sigma<X<\mu + \sigma)\approx0.68\), \(P(\mu - 2\sigma<X<\mu+2\sigma)\approx0.95\), \(P(\mu-3\sigma<X<\mu + 3\sigma)\approx0.997\).
Given \(\mu = 27\) and \(\sigma=1\).

Step2: Calculate the interval in terms of \(\mu\) and \(\sigma\)

We want to find \(P(26<X<28)\). Since \(26=\mu-\sigma\) (because \(27 - 1=26\)) and \(28=\mu+\sigma\) (because \(27+1 = 28\)).
By the 68 - 95 - 99.7 rule, when \(X\sim N(\mu,\sigma)\) and the interval is \((\mu-\sigma,\mu+\sigma)\), the probability \(P(\mu-\sigma<X<\mu+\sigma)\) is approximately \(0.68\).

Answer:

\(0.68\)