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 $\displaystyle x$.

$\displaystyle 2.5=\frac{x-70}{7}$

$\displaystyle 2.5\cdot 7={x-70}$

$\displaystyle 17.5=x-70$

$\displaystyle x=87.5$