Question

determine whether the following graph can represent a variable with a normal distribution. explain your reasoning. if the graph appears to represent a normal distribution, estimate the mean and standard deviation.
could the graph represent a variable with a normal distribution? explain your reasoning. select the correct choice below and, if necessary, fill in the answer boxes within your choice.
a. no, because the graph is skewed left.
b. no, because the graph is skewed right.
c. no, because the graph crosses the x - axis.
d. yes, the graph fulfills the properties of the normal distribution. the mean is approximately 23 and the standard deviation is about 5
(type whole numbers.)

Explanation:

Step1: Recall properties of normal distribution

A normal distribution is symmetric about the mean and does not cross the x - axis (since the probability density function \(y = f(x)=\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(x - \mu)^{2}}{2\sigma^{2}}}>0\) for all \(x\in(-\infty,\infty)\), where \(\mu\) is the mean and \(\sigma\) is the standard deviation). Skewed graphs (left - skewed or right - skewed) are not normal distributions.

Step2: Analyze the options

• Option A: A left - skewed graph is not a normal distribution. A normal distribution is symmetric.
• Option B: A right - skewed graph is not a normal distribution. A normal distribution is symmetric.
• Option C: A graph that crosses the x - axis cannot be a normal distribution because the normal probability density function \(y>0\) for all \(x\).
• Option D: If the graph is symmetric (a key property of a normal distribution), and we assume it does not cross the x - axis (another property). Also, if we estimate the mean (\(\mu\)) as the center of symmetry (in a normal distribution, the mean, median, and mode are equal) and the standard deviation (\(\sigma\)) as a measure of the spread (about 68% of the data lies within \(\mu\pm\sigma\), 95% within \(\mu\pm2\sigma\), 99.7% within \(\mu\pm3\sigma\)).

Answer:

D. Yes, the graph fulfills the properties of the normal distribution. The mean is approximately 23 and the standard deviation is about 5.