6-4D Lecture

Section 6 – 4D:

Finding a Value of x with a Given tail arae

Label the shaded area for both graphs. Find the value for z and label the z axis. Find the value for x for the given area under the normal curve with the specified mean and standard deviation. Round the value for x to 2 decimal places. Example 1

Given: µ x = 3.2 and σ x = .45

Find the x value that has a left tail area of .1131 If the left tail area for the x value is .1131

P ( x < ??? ) = . 1131 then the left tail area for the z curve is also .1131

left tail area 

left tail area 

= .1131

= .1131

x < ???

µ x = 3.2 σ x = .45

X

z < ???

Z

0 1

Find a Z score that has a left tail area as close to .1131 as possible.

Negative Z Scores Standard Normal (Z) Distribution: Cumulative Area to the LEFT of Z Z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 –1.2 0.1151 0.1131 0.1112 0.1093 0.1075 0.1056 0.1038 0.1020

0.08

0.09

0.1003

0.0985

If Z = –1.21 then convert z to x by the following formula x = µ + z(σ ) setup for finding x:

x = µ + z(σ )

x = 3.2 –1.21(.45) = 2.66

x = 2.66

P ( x < 2.66 ) = .1131 English conclusion: About 11% of the x values are less than 2.66

Stat 300 6 – 4D Lecture

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© 2012 Eitel

Example 2 Given:

µ x = 21.9 and σ x = 3.4

Find the x value that has a right tail area of .1134 If the right tail area for the x distribution is .1314 then the left tail area is .8686

P ( x > ??? ) = . 1314 If the left tail area for the x distribution is .8686 then the left tail area for the z curve is also .8686

Ieft tail area 

right tail area 

Ieft tail area 

= .8686

= .1314

= .8686

µ x = 21.9 σ x = 3.4

x = ???

X

0 1

Z

z = ???

Find a Z score that has a left tail area as close to .8686 as possible.

Positive Z Scores Standard Normal (Z) Distribution: Cumulative Area to the LEFT of Z Z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 1.1 0.8643 0.8665 0.8686 0.8708 0.8729 0.8749 0.8770 0.8790

0.08

0.09

0.8810

0.8830

If Z = 1.12 then convert z to x by the following formula x = µ + z(σ ) setup for finding x:

x = µ + z(σ )

x = 21.9 + 1.12(3.4) = 25.71

x = 25.71

P ( x > 25.71 ) = . 1314 English conclusion: About 13% of the x values are more than than 25.71

Stat 300 6 – 4D Lecture

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© 2012 Eitel

Example 3 Given:

µ x = 92.6 and σ x = 7.1

Find the x value that has a right tail area of .9382 If the right tail area for the x distribution is .9382 then the left tail area is .0618

P ( x > ??? ) = . 9382 If the left tail area for the x distribution is .0618 then the left tail area for the z curve is also .0618

Ieft tail area 

right tail area 

= .0618

= .9382

Ieft tail area  = .0618

x = ???

µ x = 92.6 σ x = 7.1

x

z = ???

Z

0 1

Find a Z score that has a left tail area as close to .8686 as possible.

Negative Z Scores Standard Normal (Z) Distribution: Cumulative Area to the LEFT of Z Z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 –1.5 0.0668 0.0655 0.0643 0.0630 0.0618 0.0606 0.0594 0.0582

0.08

0.09

0.0571

0.0559

If Z = –1.54 then convert z to x by the following formula x = µ + z(σ ) setup for finding x:

x = µ + z(σ )

x = 92.6 −1.54(7.1) = 81.67

x = 81.67

P ( x > 81.67 ) = . 9382

English conclusion: About 94% of the x values are more than than 81.67

Stat 300 6 – 4D Lecture

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© 2012 Eitel

Example 4 Kradle to Kindergarten Daycare Center knows that the ages of the children it cares for are normally distributed with a mean age of 3.2 years and a standard deviation of .45 years. The daycare center wants to mail out an advertisement to the parents of the youngest 35% of its members. What is the age that separates the youngest 35% of the children from the older children? Given:

µ x = 3.2 and σ x = .45

Find the x value that has a left tail area of .3500

P ( x < .??? ) = . 3500

If the left tail area for the x value is .3500

then the left tail area for the z curve is also .3500

left tail area  = .3500

left tail area  = .3500 x = ???

µ x = 3.2 σ x = .45

X

z = ???

Z

0 1

Find a Z score that has a left tail area as close to .3500 as possible.

Negative Z Scores Standard Normal (Z) Distribution: Cumulative Area to the LEFT of Z Z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 –0.3 0.3821 0.3783 0.3745 0.3707 0.3669 0.3632 0.3594 0.3557

0.08

0.09

0.3520

0.3483

If Z = –.39 then convert z to x by the following formula x = µ + z(σ ) setup for finding x:

x = µ + z(σ )

x = 3.2 – .39(.45) = 3.024

x = 3.02

English conclusion: If I send an advertisement to the parents of children less than 3.02 years old, the advertisement will be sent to the youngest 35% of the daycare children. Note: The age that separates the youngest 35% of the children from the older children is called P35 . We use a Normal Distribution and the Standard Normal Distribution to find P35 instead of the sorted list and percentile formula technique used in the past. We do not have the values listed for the distribution of x values so we cannot sort the list and find the location of P35 .

Stat 300 6 – 4D Lecture

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© 2012 Eitel

Example 5 Kradle to Kindergarten Daycare Center knows that the ages of the children it cares for are normally distributed with a mean age of 3.2 years and a standard deviation of .45 years. The daycare center wants to send out an add to the parents of the oldest 15% of its members. What is the age that separates the oldest 15% of the children from the younger children? Given:

µ x = 3.2 and σ x = .45

Find the x value that has a right tail area of .1500 If the right tail area for the x values is .1500 then the left tail area is 1 – .1500 = .8500

P ( x < .??? ) = .1500 If the left tail area for the x values is .8500 then the left tail area or the z values is also .8500 left tail area  = .8500

right tail

left tail

area = .15

area = .85

µ x = 3.2 x = ??? σ x = .45

X

Z

z = ???

0 1

If the left tail area for the x values is .8500 then the left tail area or the z values is also .8500 Find a Z score that has a left tail area as close to .8500 as possible.

Positive Z Scores Standard Normal (Z) Distribution: Cumulative Area to the LEFT of Z Z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 1.0 0.8413 0.8438 0.8461 0.8485 0.8508 0.8531 0.8554 0.8577

0.08

0.09

0.8599

0.8621

If Z = 1.04 then convert z to x by the following formula x = µ + z(σ ) setup for finding x:

x = µ + z(σ )

x = 3.2 + 1.04(.45) = 3.668

x = 3.67

English conclusion: If I send an add to the parents of children more than 3.67 years old, the add will be sent to the oldest 15% of the daycare children. Note: The age that separates the oldest 15% from the youngest 15% of the children from the older children is called P15 . Stat 300 6 – 4D Lecture

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© 2012 Eitel