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 standard - normal table property

The standard - normal distribution has a mean of 0 and a standard deviation of 1. The total area under the curve is 1 or 100%. The area to the left of $z = 0$ is 0.5 or 50%, and the area to the left of $z=3$ is approximately 0.9987 from the standard - normal table.

Step2: Calculate the area between $z = 0$ and $z = 3$

We use the formula $P(0

Step3: For the second question, apply the 68 - 95 - 99.7 rule

The 68 - 95 - 99.7 rule states that for a normal distribution: 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 ($\mu\pm3\sigma$). Given $\mu = 78$ and $\sigma = 8$. The value 62 is $\mu-2\sigma$ (since $78-2\times8=78 - 16 = 62$). The area outside of $\mu\pm2\sigma$ is $1 - 0.95=0.05$. Since the normal distribution is symmetric, the area below $\mu - 2\sigma$ is $\frac{0.05}{2}=0.025$ or 2.5%.

Answer:

Question 8: C. 49.5%
Question 9: D. 2.5%