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x is a normally distributed random variable with mean 58 and standard d…

Question

x is a normally distributed random variable with mean 58 and standard deviation 20. what is the probability that x is greater than 18? 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-score

$$z = \frac{X - \mu}{\sigma} = \frac{18 - 58}{20} = -2$$

Step2: Apply 0.68-0.95-0.997 rule

The rule states 95% of data lies within $\mu \pm 2\sigma$. So the remaining 5% is split equally in the two tails: $\frac{0.05}{2} = 0.025$ in the lower tail.

Step3: Find $P(X > 18)$

Subtract the lower tail probability from 1: $1 - 0.025 = 0.975$

Answer:

0.975