Examples of zvalue in the following topics:

 A common mistake is to look up a $z$value in the table and simply report the corresponding entry, regardless of whether the problem asks for the area to the left or to the right of the $z$value.
 The table only gives the probabilities to the left of the $z$value.
 There is another note of caution to take into consideration when using the table: The table provided only gives values for positive $z$values, which correspond to values above the mean.
 What if we wished instead to find out the probability that a value falls below a $z$value of $0.51$, or 0.51 standard deviations below the mean?
 This table can be used to find the cumulative probability up to the standardized normal value $z$.

 To find the kth percentile when the zscore is known: k = µ + ( z ) σ

 The standard normal distribution is a normal distribution of standardized values called zscores.
 A zscore is measured in units of the standard deviation.
 x = µ + ( z ) σ = 5 + ( 3 )( 2 ) = 11 (6.1)
 The transformation z = (x − µ)/σ produces the distribution Z ∼ N ( 0,1 ) .
 The value x comes from a normal distribution with mean µ and standard deviation σ.

 A normal probability table, which lists Z scores and corresponding percentiles, can be used to identify a percentile based on the Z score (and vice versa).
 Generally, we round Z to two decimals, identify the proper row in the normal probability table up through the ﬁrst decimal, and then determine the column representing the second decimal value.
 We can also ﬁnd the Z score associated with a percentile.
 For example, to identify Z for the 80th percentile, we look for the value closest to 0.8000 in the middle portion of the table: 0.7995.
 We determine the Z score for the 80th percentile by combining the row and column Z values: 0.84.

 Thus, a positive $z$score represents an observation above the mean, while a negative $z$score represents an observation below the mean.
 $z$scores are also called standard scores, $z$values, normal scores or standardized variables.
 The use of "$z$" is because the normal distribution is also known as the "$z$ distribution."
 The absolute value of $z$ represents the distance between the raw score and the population mean in units of the standard deviation.
 Define $z$scores and demonstrate how they are converted from raw scores

 The zscore tells you how many standard deviations that the value x is above (to the right of) or below (to the left of) the mean, µ.
 Values of x that are larger than the mean have positive zscores and values of x that are smaller than the mean have negative zscores.
 The zscore for y = 4 is z = 2.
 The values 6 and 6 are within 1 standard deviation of the mean 50.
 The values 12 and 12 are within 2 standard deviations of the mean 50.

 That is, Z 1, Z 2, Z 3, and Z 4 must be combined somehow to help determine if they – as a group – tend to be unusually far from zero.
 Z 1  + Z 2  + Z 3  + Z 4  = 4.58
 However, it is more common to add the squared values:
 The test statistic X 2 , which is the sum of the Z 2 values, is generally used for these reasons.
 Using this distribution, we will be able to obtain a pvalue to evaluate the hypotheses.

 For each significance level, the $Z$test has a single critical value (for example, $1.96$ for 5% two tailed) which makes it more convenient than the Student's ttest which has separate critical values for each sample size.
 We then calculate the standard score $Z = \frac{(T\theta)}{s}$, from which onetailed and twotailed $p$values can be calculated as $\varphi(Z)$ (for uppertailed tests), $\varphi(Z)$ (for lowertailed tests) and $2\varphi(\leftZ\right)$ (for twotailed tests) where $\varphi$ is the standard normal cumulative distribution function.
 To calculate the standardized statistic $Z = \frac{(X − μ_0)} {s}$ , we need to either know or have an approximate value for $\sigma^2$, from which we can calculate $s^2 = \frac{\sigma^2}{n}$.
 For larger sample sizes, the $t$test procedure gives almost identical $p$values as the $Z$test procedure.
 $Z$tests focus on a single parameter, and treat all other unknown parameters as being fixed at their true values.

 Observations above the mean always have positive Z scores while those below the mean have negative Z scores.
 SAT score of 1500), then the Z score is 0.
 One observation x1 is said to be more unusual than another observation x2 if the absolute value of its Z score is larger than the absolute value of the other observation's Z score: Z1 > Z2.
 3.4: (a) Its Z score is given by Z = (x−)/σ = 5.19−3 2 = 2.19/2 = 1.095. ( b) The observation x is 1.095 standard deviations above the mean.We know it must be above the mean since Z is positive.
 3.6: Because the absolute value of Z score for the second observation is larger than that of the ﬁrst, the second observation has a more unusual head length.

 In order to do this, we use a $z$score table.
 However, this is the probability that the value is less than 1.17 sigmas above the mean.
 The difficulty arrises from the fact that our table of values does not allow us to directly calculate $P(Z\leq 1.16)$.
 This table gives the cumulative probability up to the standardized normal value $z$.
 Interpret a $z$score table to calculate the probability that a variable is within range in a normal distribution