Learning Objective

to learn just how to find, for a common random change X and also an area a, the worth x* the X so that P(Xx*)=a or that P(X>x*)=a, whichever is required.

Definition

The left tailThe an ar under a thickness curve whose area is one of two people P(Xx*) or P(X>x*) for some number x*. of a thickness curve y=f(x) of a constant random variable X cut off by a value x* of X is the an ar under the curve that is to the left of x*, as displayed by the shading in number 5.19 "Right and also Left Tails the a Distribution"(a). The right tail cut turn off by x* is identified similarly, as indicated by the shading in figure 5.19 "Right and Left Tails the a Distribution"(b).

You are watching: Put the following in order for the most area in the tails of the distribution.


Figure 5.19 Right and also Left Tails of a Distribution

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The probabilities tabulated in figure 12.2 "Cumulative common Probability" are areas of left tails in the standard normal distribution.


Tails the the conventional Normal Distribution

At times it is crucial to have the ability to solve the type of problem depicted by number 5.20. We have actually a certain details area in mind, in this situation the area 0.0125 of the shaded region in the figure, and also we desire to uncover the worth z* of Z that produces it. This is exactly the turning back of the type of difficulties encountered for this reason far. Instead of learning a worth z* that Z and finding a equivalent area, we know the area and want to uncover z*. In the case at hand, in the terminology of the an interpretation just above, us wish to discover the worth z* that cut off a left tail of area 0.0125 in the typical normal distribution.

The idea for solving such a problem is fairly simple, although occasionally its implementation have the right to be a little bit complicated. In a nutshell, one reads the accumulation probability table because that Z in reverse, looking increase the pertinent area in the inner of the table and reading off the value of Z native the margins.


Figure 5.20 Z worth that produces a well-known Area

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Example 12

Find the value z* that Z as established by number 5.20: the worth z* that cuts off a left tail the area 0.0125 in the conventional normal distribution. In symbols, discover the number z* such that P(Zz*)=0.0125.

Solution:

The number the is known, 0.0125, is the area that a left tail, and as already mentioned the probabilities tabulated in number 12.2 "Cumulative typical Probability" are locations of left tails. For this reason to resolve this difficulty we require only search in the internal of figure 12.2 "Cumulative regular Probability" for the number 0.0125. The lies in the row with the heading −2.2 and in the column with the heading 0.04. This method that P(Z z*=−2.24.


Example 13

Find the worth z* that Z as established by figure 5.21: the worth z* that cuts off a best tail that area 0.0250 in the typical normal distribution. In symbols, find the number z* such that P(Z>z*)=0.0250.


Figure 5.21 Z worth that to produce a recognized Area

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Solution:

The important difference between this example and also the vault one is that below it is the area that a right tail the is known. In order to be able to use figure 12.2 "Cumulative typical Probability" us must an initial find the area the the left tail reduced off by the unknown number z*. Since the full area under the density curve is 1, that area is 1−0.0250=0.9750. This is the number we look for in the interior of number 12.2 "Cumulative common Probability". It lies in the row through the heading 1.9 and in the tower with the heading 0.06. As such z*=1.96.


Definition

The value of the standard normal arbitrarily variable Z that cut off a ideal tail that area c is denoted zc. By symmetry, worth of Z that cuts off a left tail the area c is −zc. See number 5.22 "The number ".


Figure 5.22 The number zc and also −zc

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The previous two examples were atypical since the areas we were trying to find in the internal of number 12.2 "Cumulative typical Probability" were actually there. The following example illustrates the instance that is an ext common.


Example 14

Find z.01 and also −z.01, the worths of Z that cut off right and left tails the area 0.01 in the traditional normal distribution.

Solution:

Since −z.01 cut off a left tail the area 0.01 and Figure 12.2 "Cumulative common Probability" is a table the left tails, we look for the number 0.0100 in the internal of the table. It is no there, yet falls in between the 2 numbers 0.0102 and 0.0099 in the row through heading −2.3. The number 0.0099 is closer come 0.0100 than 0.0102 is, so because that the hundredths place in −z.01 we usage the heading of the shaft that consists of 0.0099, namely, 0.03, and also write −z.01≈−2.33.

The answer to the second half of the difficulty is automatic: since −z.01=−2.33, us conclude immediately that z.01=2.33.

We might just as well have resolved this trouble by trying to find z.01 first, and also it is instructive come rework the trouble this way. To begin with, us must an initial subtract 0.01 indigenous 1 to discover the area 1−0.0100=0.9900 the the left tail cut off by the unknown number z.01. See figure 5.23 "Computation that the Number ". Then we find for the area 0.9900 in figure 12.2 "Cumulative regular Probability". That is no there, but falls between the numbers 0.9898 and 0.9901 in the row with heading 2.3. Due to the fact that 0.9901 is closer to 0.9900 보다 0.9898 is, we usage the tower heading above it, 0.03, to obtain the approximation z.01≈2.33. Then ultimately −z.01≈−2.33.


Figure 5.23 Computation that the Number z.01

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Example 15

Find x* such that P(Xx*)=0.9332, where X is a regular random variable with typical μ = 10 and standard deviation σ = 2.5.

Solution:

All the ideas for the solution are depicted in figure 5.24 "Tail the a Normally distributed Random Variable". Because 0.9332 is the area of a left tail, we can uncover z* merely by looking for 0.9332 in the internal of figure 12.2 "Cumulative normal Probability". It is in the row and column with headings 1.5 and 0.00, for this reason z*=1.50. Thus x* is 1.50 traditional deviations over the mean, so

x*=μ+z*σ=10+1.50·2.5=13.75.

Figure 5.24 Tail the a Normally distributed Random Variable

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Example 16

Find x* such the P(X>x*)=0.65, whereby X is a regular random change with average μ = 175 and also standard deviation σ = 12.

Solution:

The situation is portrayed in number 5.25 "Tail that a Normally spread Random Variable". Because 0.65 is the area of a right tail, we an initial subtract the from 1 to obtain 1−0.65=0.35, the area the the safety left tail. We discover z* by looking for 0.3500 in the interior of number 12.2 "Cumulative common Probability". It is not present, yet lies between table entries 0.3520 and also 0.3483. The entry 0.3483 with row and also column headings −0.3 and 0.09 is closer come 0.3500 than the various other entry is, therefore z*≈−0.39. Hence x* is 0.39 traditional deviations below the mean, so

x*=μ+z*σ=175+(−0.39)·12=170.32

Figure 5.25 Tail the a Normally dispersed Random Variable

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Example 17

Scores ~ above a standardized university entrance examination (CEE) are normally spread with mean 510 and also standard deviation 60. A selective university decides to offer serious factor to consider for admission to applicants whose CEE scores are in the optimal 5% of every CEE scores. Discover the minimum score the meets this criterion for serious factor to consider for admission.

Solution:

Let X signify the score make on the CEE through a randomly selected individual. Climate X is normally dispersed with typical 510 and standard deviation 60. The probability that X lie in a certain interval is the exact same as the ratio of all exam scores that lie in that interval. For this reason the minimum score the is in the height 5% of all CEE is the score x* that cuts off a right tail in the circulation of X that area 0.05 (5% expressed as a proportion). See number 5.26 "Tail the a Normally spread Random Variable".


Figure 5.26 Tail that a Normally spread Random Variable

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Since 0.0500 is the area the a best tail, we first subtract that from 1 to attain 1−0.0500=0.9500, the area that the safety left tail. We uncover z*=z.05 by trying to find 0.9500 in the interior of number 12.2 "Cumulative normal Probability". That is not present, and also lies exactly half-way between the 2 nearest entries the are, 0.9495 and also 0.9505. In the situation of a tie favor this, we will constantly average the values of Z corresponding to the 2 table entries, obtaining below the value z*=1.645. Utilizing this value, us conclude the x* is 1.645 conventional deviations above the mean, so

x*=μ+z*σ=510+1.645·60=608.7

Example 18

All guys at a army school have to run a fixed course as rapid as they have the right to as part of a physics examination. Finishing times room normally dispersed with median 29 minutes and standard deviation 2 minutes. The center 75% of all finishing times space classified as “average.” discover the range of times the are mean finishing time by this definition.

Solution:

Let X signify the complete time that a randomly selected boy. Then X is normally distributed with typical 29 and also standard deviation 2. The probability that X lie in a specific interval is the exact same as the proportion of all finish times the lie in that interval. Thus the situation is as shown in number 5.27 "Distribution of time to operation a Course". Because the area in the middle equivalent to “average” time is 0.75, the areas of the 2 tails add up to 1 − 0.75 = 0.25 in all. Through the symmetry of the thickness curve every tail should have fifty percent of this total, or area 0.125 each. Hence the faster time the is “average” has actually z-score −z.125, i m sorry by number 12.2 "Cumulative typical Probability" is −1.15, and the slowest time the is “average” has actually z-score z.125=1.15. The fastest and also slowest time that space still thought about average are

x fast=μ+(−z.125)σ=29+(−1.15)·2=26.7

and

x slow=μ+z.125σ=29+(1.15)·2=31.3

Figure 5.27 distribution of time to operation a Course

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A boy has an median finishing time if he runs the course with a time between 26.7 and also 31.3 minutes, or equivalently between 26 minutes 42 seconds and also 31 minute 18 seconds.


Key Takeaways

for a normal random variable X with average μ and also standard deviation σ, the trouble of detect the number x* so the P(Xx*) is a mentioned value c (or so that P(X>x*) is a mentioned value c) is resolved in two steps: (1) resolve the equivalent problem because that Z v the exact same value of c, thereby obtaining the z-score, z*, of x*; (2) find x* using x*=μ+z*·σ. The worth of Z that cut off a ideal tail of area c in the traditional normal circulation is denoted zc.

Exercises

Basic


Find the worth of z* that yields the probability shown.

P(Zz*)=0.0075 P(Zz*)=0.9850 P(Z>z*)=0.8997 P(Z>z*)=0.0110

Find the value of z* that yields the probability shown.

P(Zz*)=0.3300 P(Zz*)=0.9901 P(Z>z*)=0.0055 P(Z>z*)=0.7995

Find the worth of z* that yields the probability shown.

P(Zz*)=0.1500 P(Zz*)=0.7500 P(Z>z*)=0.3333 P(Z>z*)=0.8000

Find the worth of z* that returns the probability shown.

P(Zz*)=0.2200 P(Zz*)=0.6000 P(Z>z*)=0.0750 P(Z>z*)=0.8200

Find the shown value of Z. (It is easier to uncover −zc and also negate it.)

z0.025 z0.20

Find the indicated value that Z. (It is much easier to uncover −zc and also negate it.)

z0.002 z0.02

The typical finishing time among all high college boys in a specific track event in a details state is 5 minute 17 seconds. Times are normally dispersed with traditional deviation 12 seconds.

The qualifying time in this event for joining in the state satisfy is to be set so that only the more quickly 5% of all runners qualify. Discover the qualifying time. (Hint: convert seconds come minutes.) In the western region of the state the time of all boys running in this occasion are normally spread with standard deviation 12 seconds, yet with average 5 minutes 22 seconds. Discover the relationship of boys from this an ar who qualify to operation in this occasion in the state meet.

Tests that a brand-new tire occurred by a tires manufacturer brought about an estimated mean tread life the 67,350 miles and standard deviation the 1,120 miles. The manufacturer will advertise the lifetime of the tire (for example, a “50,000 mile tire”) making use of the biggest value for which the is supposed that 98% the the tires will certainly last at least that long. Assuming tires life is typically distributed, find that advertised value.


Tests the a brand-new light brought about an estimated mean life the 1,321 hours and also standard deviation of 106 hours. The manufacturer will certainly advertise the lifetime of the bulb using the largest value for which it is intended that 90% the the bulbs will last at the very least that long. Assuming bulb life is typically distributed, find that advertised value.


The weights X of eggs developed at a details farm room normally dispersed with average 1.72 ounces and standard deviation 0.12 ounce. Egg whose weights lie in the middle 75% of the circulation of weights of every eggs space classified together “medium.” uncover the maximum and also minimum weights of together eggs. (These weights room endpoints of an interval the is symmetric about the mean and also in which the weights the 75% the the eggs produced at this farm yard lie.)


The lengths X that hardwood flooring strips are normally distributed with average 28.9 inches and also standard deviation 6.12 inches. Strips whose lengths lie in the center 80% that the circulation of lengths of every strips space classified together “average-length strips.” find the maximum and minimum lengths of such strips. (These lengths are endpoints of one interval that is symmetric about the mean and in i m sorry the lengths of 80% that the hardwood strips lie.)


All student in a large enrollment multiple section course take typical in-class exams and also a typical final, and also submit usual homework assignments. Course grades are assigned based on students" final overall scores, i m sorry are approximately normally distributed. The department assigns a C to students who scores constitute the center 2/3 of every scores. If scores this semester had actually mean 72.5 and also standard deviation 6.14, uncover the interval of scores that will certainly be assigned a C.


Researchers great to inspection the overall health of individuals with abnormally high or low levels the glucose in the blood stream. Mean glucose levels space normally distributed with typical 96 and also standard deviation 8.5 mg/d ℓ, and that “normal” is characterized as the middle 90% that the population. Discover the term of normal glucose levels, that is, the interval focused at 96 that contains 90% of every glucose level in the population.

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Additional Exercises


A device for pour it until it is full 2-liter party of soft drink it is provided an amount to each party that varies from bottle to party according come a normal circulation with standard deviation 0.002 liter and mean everything amount the machine is collection to deliver.

If the maker is set to supply 2 liters (so the typical amount delivered is 2 liters) what relationship of the bottles will contain at the very least 2 liters the soft drink? discover the minimum setup of the median amount yielded by the device so that at least 99% of every bottles will contain at least 2 liters.

A nursery has observed that the mean variety of days it should darken the environment of a varieties poinsettia plant day-to-day in bespeak to have it all set for industry is 71 days. Mean the lengths that such periods of darkening are normally distributed with conventional deviation 2 days. Discover the number of days in advancement of the projected delivery dates of the tree to industry that the nursery must begin the daily darkening procedure in order that at the very least 95% that the plants will certainly be all set on time. (Poinsettias space so long-lived that when ready for sector the plant continues to be salable indefinitely.)