Question

ch 3 what percentage of values of a variable following the standard normal distribution is between z - scores of 0 and 3?
94.5%
90.7%
49.5%
33.2%
question 9
1 pts
ch 3 scores on a university exam are normally distributed with a mean of 78 and a standard deviation of 8. the professor teaching the class declares that a score of 70 or higher is required for a grade of at least \c.\ using the 68 - 95 - 99.7 rule, what percent of students score below 62?
32%
16%
5%
2.5%

Explanation:

Step1: Recall the 68 - 95 - 99.7 rule for normal distribution

The 68 - 95 - 99.7 rule 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 of the mean ($\mu\pm2\sigma$), and about 99.7% lies within 3 standard - deviations of the mean ($\mu\pm3\sigma$).

Step2: Calculate the number of standard - deviations 62 is from the mean

Given $\mu = 78$ and $\sigma=8$. We want to find how many standard - deviations 62 is from the mean. Let $x = 62$. The formula for the z - score is $z=\frac{x - \mu}{\sigma}$. Substituting the values, we get $z=\frac{62 - 78}{8}=\frac{- 16}{8}=-2$.

Step3: Determine the percentage of data below 62

Since 95% of the data lies within $\mu\pm2\sigma$ (i.e., between $z=-2$ and $z = 2$), the remaining percentage of data outside of this range is $100 - 95=5\%$. This 5% is split evenly between the two tails. So the percentage of data in the left - tail (i.e., below $z=-2$) is $\frac{5}{2}=2.5\%$.

Answer:

2.5%