Question

guided practice activity #13 - distributions and z - scores
the grades in a class are {40, 60, 65, 72, 72, 75, 78, 78, 78, 83, 84, 87, 92, 95}. the mean $overline{x}=75.6$ and the standard deviation $s = 14.1$. in this worksheet we are going to continue working with z - scores and explore how they relate to data distributions and histograms. recall that:
$z=\frac{x - overline{x}}{s}$

1. what x - value would have a z - score of + 1? - 1? - 2? 0?

1. using the dot plot below, put tick marks below the x - axis at $z = 0,pm1, - 2$ and write these numbers below their respective tick marks.

dot plot grades

now we are going to draw a histogram of the data. first you must choose intervals or \bins\ into which the data values will fall. with grades it is typical for the bins to be intervals of length 10.

Explanation:

Step1: Solve for \(x\) when \(z=-2\)

Given \(z=\frac{x - \overline{X}}{s}\), substitute \(z = - 2\), \(\overline{X}=75.6\), and \(s = 14.1\) into the formula.

$$
LATEXBLOCK0
$$

Step2: Solve for \(x\) when \(z=-1\)

Substitute \(z=-1\), \(\overline{X}=75.6\), and \(s = 14.1\) into \(z=\frac{x - \overline{X}}{s}\)

$$
LATEXBLOCK1
$$

Step3: Solve for \(x\) when \(z = 0\)

Substitute \(z = 0\), \(\overline{X}=75.6\), and \(s = 14.1\) into \(z=\frac{x - \overline{X}}{s}\)

$$
LATEXBLOCK2
$$

Step4: Solve for \(x\) when \(z = +1\)

Substitute \(z = 1\), \(\overline{X}=75.6\), and \(s = 14.1\) into \(z=\frac{x - \overline{X}}{s}\)

$$
LATEXBLOCK3
$$

Answer:

\(x\)-value	\(47.4\)	\(61.5\)	\(75.6\)	\(89.7\)