use the information to answer the question. scores on a video games are normally distributed with a mean…

use the information to answer the question. scores on a video games are normally distributed with a mean score of 190 points and a standard deviation of 15 points. using the 68 - 95 - 99.7% rule, what percentage of players would be expected to score higher than 235 points? enter the answer in the box.

use the information to answer the question. scores on a video games are normally distributed with a mean score of 190 points and a standard deviation of 15 points. using the 68 - 95 - 99.7% rule, what percentage of players would be expected to score higher than 235 points? enter the answer in the box.

Answer

Explanation:

Step1: Calculate z - score

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $x = 235$, $\mu=190$, and $\sigma = 15$. So $z=\frac{235 - 190}{15}=\frac{45}{15}=3$.

Step2: Apply the 68 - 95 - 99.7 rule

The 68 - 95 - 99.7 rule states that for a normal distribution, about 99.7% of the data lies within 3 standard deviations of the mean, i.e., $P(- 3<Z<3)=0.997$. The total area under the normal - distribution curve is 1. The area outside of $(-3,3)$ is $1 - 0.997 = 0.003$. Since the normal distribution is symmetric, the area to the right of $z = 3$ is $\frac{1 - 0.997}{2}$.

Step3: Calculate the percentage

$\frac{1 - 0.997}{2}=0.0015$, and as a percentage, $0.0015\times100 = 0.15%$.

Answer:

0.15%