chapter9 the normal curve distribution

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Chapter 9 THE NORMAL CURVE DISTRIBUTION Prepared by : Nenevie D. Villando

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Chapter 9THE NORMAL CURVE

DISTRIBUTION

Prepared by :Nenevie D. Villando

Learning Objectives

By the end of this chapter, you will be able to:1.Define and explain the concept of

the normal curve.2.Convert empirical scores to Z scores

and use Z scores and the normal curve table to find areas above, below and between points on the curve.

3. Express areas under the curve in terms of probabilities.

Uses of the Normal Curve 1. The normal curve can be used for describing distribution of scores. 2. The normal curve can be used in interpreting the standard deviation. 3. The normal curve can be used in making statements of probability. 4.The normal curve is an essential ingredient of statistical decision making, whereby the researcher generalizes her or his results from sample to populations.

Summary of the Properties of the Theoretical Normal Distribution

1.The normal distribution curve is bell- shaped.2. The mean, median, and mode are equal and located at the center of the distribution.3. The normal curve is unimodal (i.e., it has only one mode).4. The curve is symmetrical about the mean, which is equivalent to saying that its shape is the same on both sides of a vertical line passing through the center.

5. The curve is continuous – i.e., there are no gaps or holes. For each value of X, there is a corresponding value of Y.6. The curve never touches the x axis. Theoretically, no matter how far in either direction the curve extends, it never meets the X axis- but it gets increasingly closer. This is so, because N is infinite.7. The total area under the normal distribution curve is equal to 1.00, or 100%.

8. The area under the normal distribution curve that lies within one standard deviation of the mean is approximately 0.68 or 68%; within two standard deviations, about 0.95 or 95%; and within three standard deviations about 0.997, or 99.7%.

9. The basic unit of measurement is expressed in sigma unit () or standard deviations along the baseline. The sigma units are called the z-scores.

10. Two parameters are used to describe the curve. One is the parameter mean, which is equal to zero, and the other is the standard deviation, which is equal to 1.

11. Standard deviation or z-scores departing away from the mean towards the right of curve or above the mean are expressed in positive values, while the scores departing from the mean to the left of the curve or below the mean are in negative values.

50% or 0.05

- -2 -3 X +1

Figure 9.1 The Normal Probability Curve

50% or0.05

Areas under the Normal CurveIn making use of the properties of the

normal curve to solve certain types of statistical problems, one must learn how to find areas under the normal curve.

Steps to follow in Finding Areas under the Standard Normal Distribution Curve :

Step 1 Draw a picture.

Step 2 Shade the area desired.

Step 3 Find the correct figure in the following procedure table.

Step 4 Follow the directions given in the appropriate block of the procedure table to get the desired area.

Procedure TableFinding the Area under the Normal Distribution Curve1. Between 0 and any z

value: Look up the z value in the table to get the area.

2. In any tail :a. Look up the z value to get the area.b. Subtract the area from 0.5000.

0 +z -z 0 0 +z -z 0

3. Between two z values on the same side of the mean :a. Look up both z values to

get the areas.b. Subtract the smaller area

from the large area.

4. Between two z values on opposite sides of the mean:a. Look up both z values to

get the areas.b. Add the areas.

0 z1 z2 -z1–z2 0 -z 0 +z

5. To the left of any z value, where z is greater than the mean:a. Look up the z value to get

the area.b. Add 0.5000 to the area.

6. To the right of any z value, where z is less than the mean:a. Look up the z value in the

table to get the area.b. Add 0.5000 to the area.

0 +z -z 0

7. In any two tails:a. Look up the z value in the

table to get the areas.b. Subtract both areas from

0.5000.c. Add the answers.

Procedure 1. Draw the picture. 2. Shade the area desired. 3. Find the correct figure. 4. Follow the directions.

-z 0 +z

Situation 1 Find the area under the normal curve between 0 and any z value.

Example 1Find the area under the normal

distribution curve between z= 0 and z= 1.5

Solution Draw the figure and represent the area as shown below. 0.4357 or 43.57%

0 1.52

Since Table A gives the area between 0 and any z value to the right of 0.,one need only to look up the z value in the table. Find 1.5 in the left column and 0.02 in the top row. The value where the column and row meet in the table is the answer.

.4357

Z .00 .01 .02 .03 .04 .05 .07 .08 .09

0.00.10.2

1.31.41.51.6

Example 2 Find the area between z= 0 and z=-1.52Solution

Table A does not give the areas for negative values of z. But since the normal distribution is symmetric about the mean, the area to the left of the mean is the same as the area to the right of the mean. Hence one need only to look up the area for z= +1.52, which is 0.4357 or 43.57%Note : Remember that area is always a

positive number, even if the z value is negative.

Situation 2 Find the area under the normal curve in either tail.

Example 3 Find the area to the right of z= 1.52Solution Draw the figure and present the areas shown in the figure below.

0.4357

0.5000-0.4357 0.0643

1.52

Example 4Find the area to the left of z= -1.52

SolutionThe desired area is shown below.

-1.52

Again Table A gives the area for positive z values. But from the symmetric property of the normal distribution, the area to the left of -1.52 is the same as the area to the right of +z= 1.52. You will notice that the area will be derived as in example 3, 0.0643.

Situation 3 Find the area between z=1.52 and z=2.00Example 5

Find the area between z=1.52 and z=2.00

SolutionThe desired area is shown below.

1.52

◦For this situation, look up the area from z=0 to z=2.00 and the area from z=+1.52 . Then subtract the two areas. From Table A, the areas from z=0 to z=2.00 is 0.4772 and the area from z=0 to z=1.52 is 0.4357. Subtract the two areas and the resulting area is 0.0415.

Example 6 Find the area between z=-1.52 and -2.00

Solution The desired area is shown below.

-2.00

In this case the same result 0.0415 will be obtained since the normal curve is symmetrical meaning the half of the curve is 0.5000 and the other half is also 0.5000 irregardless of whether they are located below or above the mean.

Situation 4 Find the area under the curve between any two z values on opposite sides of the mean.Example 7

Find the area between z=+1.52 and z=-1.52

Solution The desired area is shown below.

-1.52

Now, since the two areas are on opposite sides of z=0, one must find both areas and add them. The area between z=0 and z=1.52 is 0.4357. The area between z=0 and z= -1.52 is also 0.4357. Hence, the total area between z= +1.52 and z=-1.52 is 0.4357+0.4357=0.8714 or 87.14%.

Situation 5 Find the area under the curve to the left of any z value, where z is greater than the mean.Example 8

Find the area to the left of z=+1.52

Solution The desired area is shown below.

+1.52

Since Table A gives only the area between z=0 and z=+1.52, one must add 0.5000 to the table area, since 0.5000 (half) of the area lies to the left to z=0. The area between z=0 and z=+1.52 is 0.4357, and the total area is 0.4357+0.5000=0.9357.

Situation 6 Find the area under the normal curve to the right of any z value, where z is less than the mean.Example 9

Find the area to the right of z=-1.52

Solution The desired area is shown below.

-1.52

In this case, the same result will be obtained as in example 8 since the normal curve is symmetrical and z=0 to z=-1.52 is 0.4357+0.5000, the half of the area, then the total area to the right of z=-1.52 is 0.9357.

Situation 7 Find the total area under the curve of any two tails.Example 10

Find the area to the right of z=+1.52 and to the left of z= -1.52

SolutionThe desired area is shown below.

-1.52

The area to the right of z=+1.52 is 0.5000- 0.4357= 0.0643. The area to the left of z=-1.52 is also 0.0643 since the same area was used. The total area, then, is 0.4357+0.4357=0.8714 or 87.14%.

Application of the Normal Probability Curve

In a mathematics test, with a sample of 50 students, the mean score is 30 and standard is 4.0. Assuming that the distribution is normal.

1. What percentage of the students falls between the mean and the score of 38?

2. What is the probability that a score picked at random will lie above the score of 38?

3. What is the probability that a score will lie below score of 38?

4. How many cases fall between scores of 32 and 36?

5. What limits will include the middle 60% of cases?

6. What is the minimum score a pupil should get to belong to the upper 10% of the group?

7. What are the two extreme scores, outside of which only 5% of the group can be expected to fall?

Solutions1. What percentage of the pupils falls

between the mean and the score of 38?a. Convert raw score to z-score.

z = - = 38 – 30 = 8 = 2.0

b. Refer to the table for areas under the normal curve in Appendix A. z= 2 corresponds to 0.4918

S 4 4

c. Convert 0.4918 to percent: 0.4918= 49.18%

So the percentage of the pupils that falls between the mean and the score of 38 is 49.18%.

0.4918 or 49.18%

2. What is the probability that a score picked at random will lie above the score of 38?

a. Convert raw score to z-score.

z= x – x = 38 -30 = 8 = 2.0

b. z=2 corresponds to 0.4918 in the table in Table A.

S 4 4

c. 0.4918 is the proportion of the area from the mean to z=2. What we are interested in this problem is the area above z=2, so get the proportion of the area above z=2.

d. We know that the proportion of the area above the mean is 0.5000, so to determine the area above z=2 , we should subtract 0.4918 from 0.5000

0.5000 – 0.4918 = 0.0082 (the probability above the score of 38)

0.5000 or 50.00%-0.4918 or 49.18%

0.0082 or 0.82%

0 2

3. What is the probability that a score will lie below score of 38?a. Convert raw score to z-score

z= x – x = 38 -30 = 8 = 2.0

b. z=2 corresponds to 0.4918 in Table A.

c. Proportion below the mean is 0.5000 plus the proportion from the mean to z=2 which is 0.4918= 0.9918 (probability below the score of 38).

0.5000 or 50.00%0.4918 or 49.18%

0.9918 or 99.18%

0 2

4. How many cases fall between scores of 32 and 36?

a. Convert each score to z-score

z= x – x = 32 -30 = 2 = 0.50

z= x – x = 36 -30 = 6 = 1.50

b. From Table A z= 0.50 = 0.1915 z= 1.50= 0.4432

c. Since we are looking for the area between z= 0.50 which is 0.1915 and z= 1.50 which is 0.4432 what we are going to do is to subtract 0.1915 from 0.4432.

0.4432- 0.1915= 0.2517 since we are looking for the cases that falls within this area and the resulting value is 0.2517, we need to convert this into percent and round it off to a whole number because we are dealing with pupils. Therefore, the number of cases is 25. This means that there are 25 students with scores between 32 and 36 including 32 and 36,respectively.

0.4432 or 44.32%0.1915 or 19.15%0.2517 or 25.17%

0.5

5. What limits will include the middle 60% of cases?a. Divide 60% into 2. That gives 30% below

the mean and 30% above the mean.b. From Table A, 30% or 0.3000 in the area

column corresponds to a z-score of 0.85c. So 60% of the cases falls between -0.85

and +0.85. Convert ±0.85 to a raw score by transforming the formula for z-score into a raw score.

Therefore the middle 60% of cases falls between limits of 26.66 and 33.34

30%30%

-.85 0 +.85

6. What is the minimum score a pupil should get to belong to the upper 10% of the group?a. 10% is at the end of the right side of the

curve, So find 40% of the scores from the mean.

b. From Table A, column for area. There is no 40% or 0.4000 but the value nearest to it is 0.3997

c. 0.3997 corresponds to a z-score of 1.28d. Convert z= 1.28 to a raw score.

x = zs + = (1.28) (4) +30 = 5.12 + 30 = 35.12So, the minimum score is 35.

𝑥

40%

10%

0 1.28

7. What are the two extreme scores, outside of which only 5% of the group can be expected to fall?a. Two extreme score suggests high and low scores. Hence 5% refers to top 2.5% and bottom 2.5%b. 2.5% or 0.025 leaves 47.5% or 0.0475 from the mean.c. 47.5% or 0.475 in Table A corresponds to +1.96 above the mean and -1.96 below the mean.

d. Convert 1.96 to raw score

X = zs + x = (1.96)(4)+ 30 = 7.84 +30 = 22.16 and 37.84

e. Therefore, 2.5% of the cases is expected to fall above 37.84 or score of 38 and 2.5% can be expected to fall below 22. (The two extreme scores are 38 and 22).

47.5% 47.5%

2.5%2.5%

-1.96

Exercise 9.1Areas Under the Normal Curve Distribution

A. Find the proportion of area under the normal curve.1. Between the mean and z= 1352. To the right of z= 2.543. To the left of z= 1.324. Between z= -1.35 and z=1.855. To the right z=1.61

6. Between z= -2.25 and z=2.0

7. Between the mean and z= -.75

8. To the left of z= 1.76

9. To the left of 2.32

10. Between z= -.53 and z= -.75

Exercise 9.2Areas Under the Normal Curve Distribution

In a class, 25% of the students are expected to fail. Examination marks are roughly normally distributed, with a mean of 75 and a standard deviation of 6.1. What mark must a student make to

pass?2. What percent of the class is included

between marks of 78 and 80?3. What is the probability of getting a mark

higher than 80?

4. Between the mean and what mark below it is include 30% of the class?5. Within what 2 marks can we expect the middle 95% of the cases to be included?6. What 2 marks are so extreme that only 1% of the class is expected to fall beyond them?