Question

x is a normally distributed random variable with mean 68 and standard deviation 24. what is the probability that x is greater than 20? 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: Find how many standard deviations 20 is from the mean

The mean $\mu = 68$ and standard deviation $\sigma = 24$. We calculate the number of standard deviations $z$ as $z=\frac{X - \mu}{\sigma}=\frac{20 - 68}{24}=\frac{- 48}{24}=- 2$. So 20 is 2 standard deviations below the mean.

Step2: Apply the 0.68 - 0.95 - 0.997 rule

The 0.95 part of the rule says that about 95% of the data lies within 2 standard deviations of the mean, i.e., between $\mu - 2\sigma$ and $\mu+2\sigma$. So the proportion of data outside of this interval (less than $\mu - 2\sigma$ or greater than $\mu + 2\sigma$) is $1 - 0.95=0.05$. Since the normal distribution is symmetric, the proportion of data less than $\mu - 2\sigma$ (which is 20 in this case) is $\frac{0.05}{2}=0.025$.

Step3: Find the probability that $X>20$

The probability that $X>20$ is $1 -$ probability that $X\leqslant20$. We know that $P(X\leqslant20) = 0.025$, so $P(X > 20)=1 - 0.025 = 0.975$.

Answer:

0.975