Question

x is a normally distributed random variable with mean 50 and standard deviation 16. what is the probability that x is between 18 and 66? 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: Calculate z - scores

For \(x = 18\), \(z_1=\frac{18 - 50}{16}=\frac{- 32}{16}=-2\). For \(x = 66\), \(z_2=\frac{66 - 50}{16}=\frac{16}{16}=1\).

Step2: Apply the 0.68 - 0.95 - 0.997 rule

The 0.68 - 0.95 - 0.997 rule states that about 68% of the data is within 1 standard - deviation of the mean (\(z=\pm1\)), about 95% is within 2 standard - deviations of the mean (\(z = \pm2\)), and about 99.7% is within 3 standard - deviations of the mean (\(z=\pm3\)).
The area between \(z=-2\) and \(z = 0\) is half of the area between \(z=-2\) and \(z = 2\), which is \(\frac{0.95}{2}=0.475\). The area between \(z = 0\) and \(z = 1\) is half of the area between \(z=-1\) and \(z = 1\), which is \(\frac{0.68}{2}=0.34\).

Step3: Calculate the probability

The probability that \(X\) is between 18 and 66 is the sum of the area between \(z=-2\) and \(z = 0\) and the area between \(z = 0\) and \(z = 1\), so \(P(18<X<66)=0.475 + 0.34=0.815\).

Answer:

0.815