For Any Distribution, What Is the Z-score Corresponding to the Mean?

Given any information value, we tin can identify how far that data value is away from the mean, only by doing a subtraction x – μ . This value volition be positive if your data value lies above (to the right) of the mean, and negative if information technology lies beneath (to the left) of the mean. But what we'd really like to know is, relative to the spread of our information set, how far is x from μ ? Call up that the standard deviation σ gives the states a measure of how spread out our entire set of individual data values is.

The z-score for any single data value tin exist constitute by the formula (in English):

or with symbols (as seen earlier!):

Obviously a z-score will be positive if the information value lies above (to the correct) of the mean, and negative if the data value lies below (to the left) of the mean.

Example half-dozen.1: Computing and Graphing z-Values

Given a normal distribution with μ = 48 and due south = 5, convert an x-value of 45 to a z-value and indicate where this z-value would be on the standard normal distribution.

Solution

Brainstorm by finding the z-score for x = 45 every bit follows.

At present draw each of the distributions, marking a standard score of z = −0.60 on the standard normal distribution.

The distribution on the left is a normal distribution with a hateful of 48 and a standard deviation of five. The distribution on the correct is a standard normal distribution with a standard score of z = −0.threescore indicated.

Z -scores measure out the distance of any information bespeak from the mean in units of standard deviations and are useful because they permit us to compare the relative positions of data values in unlike samples. In other words, the z-score allows u.s. to standardize ii or more normal distributions, or more than accordingly, to put them on the same calibration. Therefore, we'll be able to compare relative positions of information values within their own distribution to decide which data values are closer to or farther from the mean. A prime example for this is to compare the exam scores for two students, one who scored a 28 on the ACT (scores range from one – 36) and some other who scored a 1280 on the Sat (scores range from 400 – 1600). Who, relative to their associated exam, scored better?

Example

Suppose you are enrolled in 3 classes, statistics, biology, and kayaking, and you just took the first exam in each. You receive a grade of 82 on your statistics exam, where the mean grade was 74 and the standard deviation was 12. You receive a course of 72 on your biological science exam, where the mean grade was 65 and the standard deviation was x. Finally, you receive a grade of 91 on your kayaking exam, where the mean form was 88 and the standard deviation was six. Although your highest test score was 91 (kayaking), in which class did y'all score the all-time, relative to the residuum of the course ? We can respond this using a z-score!

Your statistics exam score was 0.67 standard deviations ameliorate than the course boilerplate; your biology score was 0.7 standard deviations better than the class average; your kayaking score was only 0.5 standard deviations better than the class average.  Therefore, fifty-fifty though your bodily score on the biology exam was the lowest of the three test scores, relative to the distribution of all class exam scores , your biological science exam score was the highest relative course.

Finding an Area (Proportion) Given a Specific Z-Value

To determine the area under the N (0, 1) bend for any data value that does not fall exactly i, 2, or three standard deviations above or beneath the hateful really requires some calculus. Lucky for united states of america, areas nether the Due north(0, one) curve tin be obtained in numerous other ways, including technology (TI-83/84, Excel) and a tabular array of values. Search the Net for "standard normal table" and you'll observe hundreds of tables illustrating z-scores and their associated areas. The bulk of these methods report the area to the left of the specified z-score z, no matter where it lies. This comes from a calculus operation of integration, which finds an surface area from the offset of a distribution (i.e., the far left-tail) up to the z-score. Two images are provided.

In that location are three types of area calculations that you volition exist performing, each requiring slightly different work:

• For areas to the left of z: just use the area provided by a table or technology.
• For areas to the right of z: considering the total area under a density curve is 1 (100%), simply calculate: one − area to the left of z ₀ .
• For areas between two z-values, say z₀ andz_i  (where z ₀< z _i): find the area to the left ofz ₁ and subtract from it the expanse to the left of z ₀ .

Finding a Z-Value Given an Expanse

This is a slightly more challenging chore than calculating an surface area, because you basically work "backwards" from an algebraic standpoint. Information technology's important to realize that a Standard Normal Table has two parts: (1) the acme and side margins, which form the tenths and hundredths of a z-score, and (2) the torso of the tabular array, which are all the area (probability) values. Also, call up that the Standard Normal Tabular array merely provides united states information on the surface area (probability) to the left of a z-score. A small excerpt of Table B from Appendix A is shown below.

Notice that the z-values given in the table are rounded to two decimal places. The commencement decimal place of each z-value is listed in the left column, with the second decimal place in the top row. Where the appropriate row and column intersect, we find the amount of area under the standard normal curve to the left of that detail z-value.

Example : Finding Surface area to the Left of a Positive z-Value Using a Cumulative Normal Table

Find the surface area under the standard normal curve to the left of z = 1.37.

Solution

To read the table, we must pause the given z-value (one.37) into two parts: ane containing the first decimal place (one.3) and the other containing the second decimal identify (0.07). And so, in Table B from Appendix A, look across the row labeled i.3 and down the column labeled 0.07. The row and column intersect at 0.9147. Thus, the area under the standard normal curve to the left of z = 1.37 is 0.9147.

Using a TI-83/84 Plus estimator, we tin can find a value of the surface area to the left of a z-score. To obtain the solution using a TI-83/84 Plus estimator, perform the following steps.

• Printing 2nd and then Vars to access the DISTR menu.
• Cull option two:normalcdf( .
• Enter lower bound, upper bound, µ , σ. Note If you want to find surface area under the standard normal curve, equally in this case, so you lot do non need to enter µ or σ.
• Since we are asked to find the surface area to the left of z, the lower bound is -∞. From the empirical dominion nosotros know that afterwards about 3 standard deviations away from the hateful nosotros have accounted for near all of the information, so for our lower bound we will merely apply a very negative number.We cannot enter -∞ into the calculator, so we will enter a very small value for the lower endpoint, such as -10⁹⁹. This number appears as -1E99 when entered correctly into the estimator. To enter -1E99, printing(-) 1 [ii^nd][ , ]99. This appears on the screen as normcdf(-1E99,one.37,0,1).

If we are given an area (or probability) value, we need to get-go locate it in the trunk of a table, then track our way up and to the left in order to piece together the z-score that relates to the specified area. Keep in mind that y'all may non detect the verbal surface area value in the body of the tabular array…and so just use the closest value y'all tin can observe, and and so identify the proper z-score.

One calculation that volition be used frequently in the coming chapters is to identify the two z-scores that dissever a specific expanse in the middle of the standard normal distribution.

Instance

Suppose we want to know which ii z-scores separate out the middle 95% of the data. From the empirical rule, we already know the z-scores that do this are ±ii (ii standard deviations on either side of the mean). In reality, it's not exactly ±two, but close enough for rough calculations.

To find the exact ii z-scores, we utilize the following logic: If the middle portion is 95% = 0.95, so how much area lies outside of the middle (to the left and right)? A simple subtraction solves this! 1 – 0.95 = 0.05. The "exterior" surface area, 0.05, must be split as betwixt the ii tails (because of symmetry!). Therefore, dividing 0.05 by ii gives us an expanse of 0.025 in each tail .

Using a standard normal table "backwards," nosotros outset look through the body of the table to find an area closest to 0.025. The z-score corresponding to a left-tail area of 0.025 isz = −i.96. Now, therefore, the upper z-score will be z = ane.96, by the symmetry property of the standard normal distribution. Y'all could also discover the upper z-score by looking upwards the surface area/probability value 0.025 + 0.95 = 0.975 in the body of the table and finding the associated z-value. By the end of the class, you lot will be extremely familiar with z-scores that define a central 90% (z = ± 1.645), 95% (z = ± 1.96), and 99% (z = ± 2.576).

Case: Find and interpret the probability of a random Normal variable

Suppose you simply purchased a 2005 Honda Insight with automatic manual. Using world wide web.fueleconomy.gov you determine for the 2005 Honda Insights have mean highway gas milage is 56 miles per gallon with a standard deviation of 3.two. The distribution of this data has a bell-shape and is normal. Y'all want to know the following:

a) How probable is information technology that your Honda Insight with automatic transition will get better than 60 miles per gallon on the highway.

b) How likely is it that your Honda Insight with automatic transition will get less than 50 miles per gallon on the highway.

c) How likely is it that your Honda Insight with aoutomatic transition volition go between 52 and 62 miles per gallon on the highway.

Solution

This problem deals with data that is unremarkably distributed with mean 56 and standard divergence iii.two, i.e., .

(a)

In symbols, nosotros are asked to calculate P(X > 60). Sketching a normal bend and shading the area respective to greater than lx, gives us the graph shown. In order to calculate the appropriate surface area in the upper (correct) tail, we must first convert our information to the standard normal distribution. The z-score for x = 60 is:

This ways that 60 is 1.25 standard deviations to a higher place the mean. Notice how lining the two normal curves up as shown illustrates how the 2 areas are the same: P(X > 60) = P(Z > 1.25).

Using z = one.25, nosotros become to Table IV (or use normcdf(1.25,1E99,0,1)) to find the area to the left of z = 1.25 is 0.8943. Since we need the area to the correct, we simply accept 1 – 0.8943 = 0.1057.

Therefore, P(Ten > 60) = 0.1057 = 10.57%. There are a couple ways to interpret this answer:

• Of all the model year 2005 Honda Insight cars produced with an automatic transmission, x.57% will become over threescore miles per gallon on the highway.
• If you went to a car lot and purchased a new model twelvemonth 2005 Honda Insight cars produced with an automatic manual, there is a ten.57% chance that your car volition get over 60 miles per gallon on the highway.

(b)

In symbols, we are asked to summate P(Ten < 50). Sketching a normal curve N(56, 3.ii)and shading the area respective to less than 50, gives us the graph shown to the right.

In gild to calculate the advisable expanse in the lower (left) tail, we must beginning catechumen our information to the standard normal distribution. The z-score for x = 50 is:

Thus, the value 50 MPG is 1.88 standard deviations below the mean. In symbols nosotros meet:P(X < l) = P(Z < −1.88).

Using z = -one.88, we go to Table IV (or utilize normcdf(-1E99,-1.88,0,ane)) to find the area to the left of z = -1.88 is 0.0301. Therefore, P(10 < 50) = 0.0301 = 3.01%. There are a couple ways to translate this reply:

• Of all the model year 2005 Honda Insight cars produced with an automatic transmission, three.01% will get less than fifty miles per gallon on the highway.
• If yous went to a car lot and purchased a new model year 2005 Honda Insight cars produced with an automated transmission, there is a 3.01% adventure that your auto volition go less than 50 miles per gallon on the highway.

(c)

In symbols, we are asked to summate P(58 < X < 62). Sketching a normal bend North(56, 3.ii)  and shading the expanse corresponding to greater than 58 simply less than 62, gives usa the graph shown. In order to calculate the appropriate surface area, we must first convert both data to the standard normal distribution.

The z-score for X= 58 is:

and thez-score forx = 62 is:

In terms of probability, nosotros tin can at present say: P(58 < X < 62) = P(0.63 < Z < 1.88).

Using z = ane.88, we go to Table Four (or use applied science) to find the area to the left of z = i.88 is 0.9699. Now, we demand to remove (decrease) the area left of z = 0.63, which is 0.7357. Therefore, P(58 < Ten < 62) = 0.9699 – 0.7357 = 0.2342, or 23.42%. There are a couple ways to interpret this answer:

• Of all the model year 2005 Honda Insight cars produced with an automatic transmission, 23.42% will get between 58 and 62 miles per gallon on the highway.
• If you went to a car lot and purchased a new model yr 2005 Honda Insight cars produced with an automatic transmission, there is a 23.42% chance that your auto will get betwixt 58 and 62 miles per gallon on the highway.

This calculation tin be done with both normcdf(0.63,1.88,0,1) and normcdf(58,62,56,iii.2), which will exist the same.

Detect the Value of a Random Variable Knowing a Probability Value

In these types of problems, we need to work "backwards." Starting with a specified probability, find the specified z-score, so work our way back to the random variable. The tables of standard normal values are not a "i-manner" tool! What practise we hateful past that? So far you've started with a value for a random variable (like a gas mileage value in the previous problem), turned it into a z-score, and and then looked up the associated probability value for that z-score. We can employ this tabular array to piece of work backwards! We can start with a known probability value in the body of a table, place the z-score corresponding to that area by moving your fingers to the associated row and cavalcade, the reverse the algebra transformation from a z-score to a random variable.

If this sounds confusing, retrieve back to the steps we took in the preceding example:

If, withal, we are given an surface area/probability, then to work our way back to the original data value, we must first identify the appropriate z-score, and then "united nations-standardize" the z-score to make it (finally!) back at the data value. How practice we algebraically "undo" the z-score? Easy…just solve for the data value 10:

Multiply both sides past σ to remove information technology from the denominator on the left side:

X – μ = Z⋅σ

Finally, add the value of μ to both sides to isolate the value of the random variable X:

X = Z⋅σ + μ

Instance: Finding the value of a normal random variable

Instead you want to know a gas mileage for a particular probability. Detect what gas mileage for your 2005 Honda Insight will get improve gas mileage than 97% of all other 2005 Honda Insights with automatics transmission.

Solution

This problem once more deals with data that is ordinarily distributed with mean 56 and standard deviation iii.2, i.e., N(56, three.2).

To observe the 97% percentile gas mileage, we need to find the specific miles per gallon X that separates the bottom 97% of all gas mileages from the top 3%. And then for this problem nosotros are given a percentage/surface area. Sketching the normal curve gives the graph shown.

Using Tabular array Four, we find 0.97 in the torso of the tabular array, and then identify the z-score of 1.88. Notice that the exact area 0.97 is not in the table, just the closest area of 0.9699 has the z-score of ane.88. At present we un-standardize the z-score of 1.88. In English this means we need to identify the specific gas mileage that is 1.88 standard deviations higher up the mean of 56. Solving for X in the Z transform gives:

Therefore, if your 2005 Honda Insight cars with an automatic transmission gets 62 mpg, information technology gets better miles per gallon than 97% of all 2005 Honda Insight cars with an automatic transmission.

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Source: https://mat117.wisconsin.edu/4-the-z-score/

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