If the mean of a normally distributed set of scores is \(\displaystyle \mu = 78.2\) and the standard deviation is \(\displaystyle \sigma = 4.3\), then the \(\displaystyle z\)-score corresponding to a test score of \(\displaystyle X= 85\) is 

\(\displaystyle \frac{X-\mu}{\sigma} =\frac{85-78.2}{4.3} = \frac{6.8}{4.3} \approx 1.58\)

From a \(\displaystyle z\)-score table, in a normal distribution, 

\(\displaystyle P (z< 1.58) = 0.9429\)

We want the percent of students whose test score is 85 or better, so we want \(\displaystyle P (z\geq 1.58)\). This is

\(\displaystyle P (z\geq 1.58) = 1- P (z< 1.58) = 1- 0.9429 = 0.0571\)

or about 5.7 % The correct choice is 6%.

Let X represent the salaries of employees at XYZ Corporation.

We want to determine the probability that X is between 69,000 and 78,000:

\(\displaystyle Pr(69000< X< 78000)\)

To approximate this probability, we convert 69,000 and 78,000 to standardized values (z-scores).

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

\(\displaystyle \frac{69000-60000}{7500}=1.2\)

\(\displaystyle \frac{78000-60000}{7500}=2.4\)

We then want to determine the probability that z is between 1.2 and 2.4

\(\displaystyle P(1.2< z< 2.4)=P(z< 2.4)-P(z< 1.2)\)

\(\displaystyle P(1.2< z< 2.4)=0.9918-0.8849=0.107\)

The proportion of employees who earn between 69,000 and 78,000 is 0.107.

The z-score is a measure of an actual score's distance from the mean in terms of the standard deviation. The formula is:

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

Where \(\displaystyle \large \mu, \sigma\) are the mean and standard deviation, respectively. \(\displaystyle x\) is the actual score.

If we plug in the values we have from the original problem we have 

\(\displaystyle {z = \frac{92-80}{7}}\)

which is approximately \(\displaystyle 1.714\).

The z-score can be expressed as

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

where 

\(\displaystyle \mu= \text{average }=20\)

\(\displaystyle \sigma= \text{standard\: deviation}=2\)

\(\displaystyle x=18\)

Therefore the z-score is:

\(\displaystyle z=\frac{x-\mu }{\sigma }=\frac{18-20}{2}=\frac{-2}{2}=-1\)

Use the formula for z-score:

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

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

\(\displaystyle z=\frac{88-79}{6}=1.5\)

The formula for a z-score is

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

where  \(\displaystyle \mu\) = mean and \(\displaystyle \sigma\) = standard deviation and \(\displaystyle x\)=your test grade.

Plugging in your z-score, mean, and standard deviation that was originally given in the question we get the following.

\(\displaystyle 2.5=\frac{x-70}{7}\)

Now to find the grade you got on the test we will solve for \(\displaystyle x\).

\(\displaystyle 2.5=\frac{x-70}{7}\)

\(\displaystyle 2.5\cdot 7={x-70}\)

\(\displaystyle 17.5=x-70\)

\(\displaystyle x=87.5\) 

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:  \(\displaystyle -1\)