Question

consider the standard normal curve given. the mean is the standard deviation is the data point 37 is one standard deviation from the mean.

Explanation:

Step1: Recall the properties of the normal distribution

In a normal distribution, the peak of the curve represents the mean. From the graph, the value at the peak is \(30\).

Step2: Identify the standard deviation

The standard deviation \(\sigma\) is given in the graph as \(\sigma = 5\).

Step3: Calculate the number of standard deviations from the mean

We use the formula \(z=\frac{x-\mu}{\sigma}\), where \(x = 37\), \(\mu=30\) and \(\sigma = 5\).

$$z=\frac{37 - 30}{5}=\frac{7}{5}=1.4$$

But wait, looking at the graph's scale (assuming the intervals are based on standard deviation), if \(\mu = 30\) and \(\sigma=5\), \(30 + 5=35\) (one - standard deviation above the mean), \(30+2\times5 = 40\). But if we consider the problem statement says "one - standard deviation from the mean", and maybe there was a mis - read of the graph. If we assume the intended value:
The mean \(\mu\) is the center value of the normal curve. From the graph, \(\mu = 30\). The standard deviation \(\sigma\) is given as \(5\) (from the label \(\sigma = 5\) in the graph). For the data - point \(x = 37\), if we consider the formula \(x=\mu + z\sigma\), \(37=30+z\times5\), \(z=\frac{37 - 30}{5}=1.4\). But if we assume a mis - print in the problem (maybe the data - point was \(35\), but since the problem says "one - standard deviation from the mean" and we go with the given labels:

Answer:

The mean is \(30\). The standard deviation is \(5\). The data point \(37\) is \(1.4\) standard deviations from the mean (but if we follow the problem's "one - standard deviation" hint and assume a graph - reading error, if we consider the intervals: mean \(30\), standard deviation \(5\)).