The first step in this problem is calculating the z-score. \(\displaystyle z=( x-\mu)/\sigma\)

\(\displaystyle (95-85)/2.5= 4\)

The next step is to look up 4 in the z-table. The value from the table is \(\displaystyle .0000317\).

\(\displaystyle 1-.0000317\: is \: .9999683.\)

A \(\displaystyle z\)-score indicates whether a particular value is typical for a population or data set.  The closer the \(\displaystyle z\)-score is to 0, the closer the value is to the mean of the population and the more typical it is.  The \(\displaystyle z\)-score is calculated by subtracting the mean of a population from the particular value in question, then dividing the result by the population's standard deviation. 

\(\displaystyle z=\frac{X-\bar{X}}S{}\) 

\(\displaystyle (115-103)/8=1.5\)

A Z-score indicates whether a particular value is typical for a population or data set.  The closer the Z-score is to 0, the closer the value is to the mean of the population and the more typical it is.  The Z-score is calculated by subtracting the mean of a population from the particular value in question, then dividing the result by the population's standard deviation.

To find the z-score, follow the formula

\(\displaystyle z=\frac{x-\mu}{\sigma }, x=score, \mu=mean, \sigma=S.D.\)

\(\displaystyle z=\frac{73-55}{5}\)

or

\(\displaystyle z=3.6\)

A z score is unique to each value within a population.

To find a z score, subtract the mean of a population from the particular value in question, then divide the result by the population's standard deviation.

\(\displaystyle z=\frac{x-\mu}{\sigma}=\frac{19.5-17.2}{3.5}= \frac{2.3}{3.5}=0.6571\)

We need to calculate Natalie's z-scores for both her Spanish and math exams. Calculating z-scores is as follows:

\(\displaystyle \small z=\frac{x-\bar{x}}{s}\)

Her z-score for the Spanish exam is \(\displaystyle \small \frac{82-72}{8}\), which equals 1.25, while her z-score for the math exam is \(\displaystyle \small \frac{86-68}{12}\), which equals 1.50. Since both z-scores are positive, Natalie did above average on both tests, but since her z-score for the math exam is higher than her z-score for the Spanish exam, Natalie did better on her math exam when compared to the rest of her peers taking the exam.

To find the z-score, we follow the formula

\(\displaystyle z=\frac{x-\mu }{\sigma }\) where \(\displaystyle x\) is the given value, \(\displaystyle \mu\) is the mean, and \(\displaystyle \sigma\) is the standard deviation.

For this problem we see that

\(\displaystyle x=111\)

\(\displaystyle \mu = 102\)    and 

\(\displaystyle \sigma =9\)

Substituting for these values we see

\(\displaystyle z=\frac{111-102 }{9 } =1\)

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

\(\displaystyle \mu\) is the average life of the bread maker, in this problem 9 years.

\(\displaystyle \sigma\) is the standard deviation of the break makers life. The standard deviation is the square root of the variance. 

\(\displaystyle \sigma=\sqrt{4}=2\)

X is the test statistic we are looking to find. In this problem it is the probability that the bread maker lasts less than 13 years.

\(\displaystyle z=\frac{13-9 }{2}=\frac{4}{2}=2\)

If you were to find the value associated with a z-score of 2 on the z-table, you would see that

\(\displaystyle Pr(X\leq 13)=.9973\)