Write the formula to find the z-score.  Z-scores are defined as the number of standard deviations from the given mean.

\(\displaystyle z=\frac{x-\mu}{\sigma }\)

Substitute the values into the formula and solve for the z-score.

\(\displaystyle z=\frac{x-\mu}{\sigma } = \frac{9.3-8}{2.5} =0.52\)

Write the formula for z-score where \(\displaystyle x\) is the data, \(\displaystyle \mu\) is the population mean, and \(\displaystyle \sigma\) is the population standard deviation.

\(\displaystyle z=\frac{x-\mu}{\sigma}\)

Substitute the variables.

\(\displaystyle z=\frac{65-72}{8} =- \frac{7}{8} = -0.875\)

The z-score is:  \(\displaystyle -0.875\) 

Write the formula for z-scores.  This tells how many standard deviations above the below the mean.

\(\displaystyle z=\frac{x-\mu}{\sigma}\)

Substitute the known values into the equation.

\(\displaystyle z=\frac{50-75}{3} =-\frac{25}{3}\)

The z-score is:  \(\displaystyle -\frac{25}{3}\)

Write the formula to determine the z-scores.

\(\displaystyle z=\frac{x-\mu}{\sigma}\)

Substitute all the known values into the formula to determine the z-score.

\(\displaystyle z=\frac{6-8}{2}\)

Simplify this equation.

\(\displaystyle z=\frac{-2}{2} = -1\)

The answer is:  \(\displaystyle -1\)

Write the formula for z-scores.  Z-scores determine the number of standard deviations below or above the mean.

\(\displaystyle z=\frac{x-\mu}{\sigma}\)

Substitute the values into the formula to determine the z-score.

\(\displaystyle z=\frac{45-60}{8}= -\frac{15}{8}\)

The z-score is \(\displaystyle -\frac{15}{8}\).

Write the formula for z-scores.  Z scores determine how many standard deviations a score is above or below the mean.

\(\displaystyle z=\frac{X-\mu}{\sigma}\)

Substitute the known values in order to determine the z-score.

\(\displaystyle z=\frac{X-\mu}{\sigma} = \frac{54-66}{3} = -4\)

The answer is:  \(\displaystyle -4\)

Write the formula for z-scores.  Z-scores are indicators of how many standard deviations above or below the mean.

\(\displaystyle z=\frac{x-\mu}{\sigma}\)

\(\displaystyle x=\textup{score}\)

\(\displaystyle \mu = \textup{mean}\)

\(\displaystyle \sigma =\textup{ standard deviation}\)

Substitute the known values.

\(\displaystyle z=\frac{8.5-9.2}{1.3} = \frac{-0.7}{1.3} = -0.538\)

The answer is:  \(\displaystyle -0.538\)

Use the formula for z-score:

\(\displaystyle z=\frac{x-\mu }{\sigma }\)

Where \(\displaystyle x\) is your score, \(\displaystyle \mu\) is the mean, and \(\displaystyle \sigma\) is the standard deviation.

\(\displaystyle z=\frac{26-22}{3}=1.33\)

If \(\displaystyle y\) varies directly with the square root of \(\displaystyle x\), then for some constant of variation \(\displaystyle K\), 

\(\displaystyle \small y = K \sqrt{x}\)

If \(\displaystyle x = 25\), then \(\displaystyle y = 48\); therefore, the equation becomes 

\(\displaystyle \small \small 48 = K \sqrt{25}\), 

or

\(\displaystyle 5K = 48\).

Divide by 5 to get \(\displaystyle K = 9.6\), making the equation 

\(\displaystyle \small \small y = 9.6 \sqrt{x}\).

If \(\displaystyle x = 81\), then \(\displaystyle \small \small \small y = 9.6 \sqrt{81} = 9.6 \cdot 9 = 86.4\).

Since \(\displaystyle y\) varies directly with \(\displaystyle x^{3}\), \(\displaystyle y=Kx^{3}\) where \(\displaystyle K\) is a constant.

1. Solve for \(\displaystyle K\) when \(\displaystyle x=5\) and \(\displaystyle y=61.125\).

\(\displaystyle y=Kx^{3}\)

\(\displaystyle 61.125=K(5^{3})\)

\(\displaystyle K=\frac{61.125}{125}=0.489\)

2. Use your equation to solve for \(\displaystyle y\) when \(\displaystyle x=11\).

\(\displaystyle y=0.489x^{3}\)

\(\displaystyle y=0.489(11^{3})=650.859\)