This question relates to the \(\displaystyle 68-95-99.7\) rule of normal distribution. We know that \(\displaystyle 95\%\) of the data are within \(\displaystyle 2\) standard deviations from the mean.

In this case this means that \(\displaystyle 95\%\) of the students are between

\(\displaystyle 15-2\cdot 0.5\) and \(\displaystyle 15+2\cdot 0.5\) 

\(\displaystyle 15-1\) and \(\displaystyle 15+1\)

\(\displaystyle 14\) and \(\displaystyle 16\)

Therefore we have \(\displaystyle 5\%\) of the students that are outside of this range. Since the normal distribution is symmetric, the proportion of students below \(\displaystyle 14\) is the same as the proportion of students above \(\displaystyle 16\).

Thus the right answer is \(\displaystyle \frac{5}{2}\) or \(\displaystyle 2.5\%\).

Use the 68-95-99.7 rule which states that 68% of the data is within 1 standard deviation (in either direction) of the mean, 95% is within 2 standard deviations, and 99.7% is within 3 standard deviations of the mean.

In this case, 95% of the students' scores were between:

75-(2 x 4) and 75+(2 x 4)

or between a 67 and a 83, with equal amounts of the leftover 5% of scores above and below those scores. This would mean that 2.5% of the students scored below a 67% on the test.

Use the 68-95-99.7 rule which states that 68% of the data is within 1 standard deviation (in either direction) of the mean, 95% is within 2 standard deviations, and 99.7% is within 3 standard deviations (in either direction) of the mean.

In this case, 99.7% of the students' scores were between 3 standard deviation above the mean and 3 standard deviations below the mean:

78-(2 x 3) and 78+(2 x 3)

or between a 72 and an 84. 

The graph of a normally-distributed data set is symmetrical.

The graph of a normally-distributed data set has a single, central peak at the mean of the data set that it describes.

The graph of a normally-distributed data set will vary based only upon the mean and the standard deviation of the set that it describes.

The graph of a normally-distributed data set with a higher standard deviation will be wider than the graph of a normally-distributed data set with a lower standard deviation.

The question asks us to find the statement that is not true; however, all statements are true so the correct response is "All of these are true."

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 \(\displaystyle x\).

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

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

\(\displaystyle 17.5=x-70\)

\(\displaystyle x=87.5\)