Question

given \\( \mu = 5 \\) and \\( \sigma = 0.5 \\), find the probability that a random variable, \\( x \\), is between 3.6 and 6.1. (1 point)

\\( \bigcirc 98.4 \\% \\)
\\( \bigcirc - 98.4 \\% \\)
\\( \bigcirc 1.4 \\% \\)
\\( \bigcirc 98.6 \\% \\)

Explanation:

Step1: Calculate z - scores

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\).
For \(x = 3.6\), \(z_1=\frac{3.6 - 5}{0.5}=\frac{-1.4}{0.5}=-2.8\).
For \(x = 6.1\), \(z_2=\frac{6.1 - 5}{0.5}=\frac{1.1}{0.5}=2.2\).

Step2: Use the standard normal distribution table

We know that \(P(3.6<X<6.1)=P(-2.8<Z<2.2)\).
Since \(P(-2.8<Z<2.2)=P(Z < 2.2)-P(Z<-2.8)\).
From the standard normal table, \(P(Z < 2.2)=0.9861\) and \(P(Z<-2.8)=0.0026\).

Step3: Calculate the probability

\(P(-2.8<Z<2.2)=0.9861 - 0.0026=0.9835\approx98.4\%\).

Answer:

98.4%