Question

the scores of a national standardized test have a normal distribution. if the mean score is 1030 with a standard deviation of 112, what is the approximate probability that a test - taker chosen at random would have a score that is less than 1142? use the empirical rule and give your answer as a percent.

Explanation:

Step1: Calculate the z - score

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $x = 1142$, $\mu=1030$, and $\sigma = 112$. So $z=\frac{1142 - 1030}{112}=\frac{112}{112}=1$.

Step2: Apply the Empirical Rule

The Empirical Rule for a normal distribution states that about 68% of the data lies within 1 standard - deviation of the mean ($\mu\pm\sigma$), about 95% lies within 2 standard - deviations ($\mu\pm2\sigma$), and about 99.7% lies within 3 standard - deviations ($\mu\pm3\sigma$). The area to the left of $z = 1$ includes the 50% of the data to the left of the mean and half of the 68% that lies within 1 standard - deviation of the mean. So the probability is $50\%+\frac{68\%}{2}=50\% + 34\%=84\%$.

Answer:

84%