Question

unit 2: the normal distribution
practice converting worksheet
name:
directions: show work by drawing the normal curve and using formulas when necessary. as part of the
learning process, it is important that i see your thinking.

aw data\ to z-scores

1. the music assessment scores for eighth - grade students are approximately normal with a mean of 150 and a

standard deviation of 35.
(a) find and interpret the standardized score
for a music assessment of 180.
(b) find and interpret the standardized score
for a music assessment of 100.

Explanation:

Step1: Recall the z - score formula

The z - score formula is \(z=\frac{x-\mu}{\sigma}\), where \(x\) is the raw score, \(\mu\) is the mean, and \(\sigma\) is the standard deviation. Given \(\mu = 150\) and \(\sigma=35\).

Step2: Calculate the z - score for \(x = 180\)

Substitute \(x = 180\), \(\mu = 150\), and \(\sigma = 35\) into the formula: \(z=\frac{180 - 150}{35}=\frac{30}{35}\approx0.86\).
The interpretation: A z - score of approximately \(0.86\) means that the score of \(180\) is \(0.86\) standard deviations above the mean.

Step3: Calculate the z - score for \(x = 100\)

Substitute \(x = 100\), \(\mu = 150\), and \(\sigma = 35\) into the formula: \(z=\frac{100 - 150}{35}=\frac{- 50}{35}\approx - 1.43\).
The interpretation: A z - score of approximately \(-1.43\) means that the score of \(100\) is \(1.43\) standard deviations below the mean.

Answer:

(a) The z - score is \(z=\frac{180 - 150}{35}\approx0.86\). Interpretation: The score of \(180\) is \(0.86\) standard deviations above the mean.
(b) The z - score is \(z=\frac{100 - 150}{35}\approx - 1.43\). Interpretation: The score of \(100\) is \(1.43\) standard deviations below the mean.