Question

x is a normally distributed random variable with mean 27 and standard deviation 25. what is the probability that x is between 52 and 77? 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

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $\mu$ is the mean, $\sigma$ is the standard deviation, and $x$ is the value from the data set.
For $x = 52$, $z_1=\frac{52 - 27}{25}=\frac{25}{25}=1$.
For $x = 77$, $z_2=\frac{77 - 27}{25}=\frac{50}{25}=2$.

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 lies within 1 standard deviation of the mean ($z=\pm1$), about 95% lies within 2 standard deviations of the mean ($z = \pm2$), and about 99.7% lies within 3 standard deviations of the mean ($z=\pm3$).
The proportion of data between $z = 0$ and $z = 1$ is $\frac{0.68}{2}=0.34$, and the proportion of data between $z = 0$ and $z = 2$ is $\frac{0.95}{2}=0.475$.
The probability that $X$ is between 52 and 77 is the probability corresponding to the $z$ - scores. The probability between $z = 1$ and $z = 2$ is $P(1$P(1

Answer:

$0.135$