Normal Distribution.pptx PowerPointPresentation

Normal Distribution

1. GRADE STATISTICS AND

THE NORMAL
DISTRIBUTION AND |/; 5}
» ITS PROPERTIESY I)

o

2. Normal Probability Distribution is a probability
distribution of continuous random variables.

+ Many random variables are either normally distributed

or, at least approximately normally distributed.
Examples: Height, Weights, and examination scores.

« It is easy for mathematical statisticians to work with the
normal curve. A number of hypothesis test and the
regression model are based on the assumption that the
underlying data have normal distributions.

3. Properties of a Normal Curve

Mean
Median
Mode

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* The distribution curve is be//-shaped (the curve is asymptotic to the base line)

* The curve is symmetrical about its center + The area under the curve is 1. thus, it
* The mean, median, and mode coincide at represents the probability or proportion or
ociated with specific sets

the center the percentage a
+ The tails of the curve flatten out indefinitely of measurement values
along the horizontal axis but never touch it 6

4. «The change of value of the mean shifts the
graph of the normal curve to the right or to
the left.

* The standard deviation determines the shape
of the graphs (particularly the height and
width of the curve). When the standard
deviation is large, the normal curve is short
and wide, while a small value for the standard
deviation yields skinnier and taller graph.

5. The standard normal curve is a normal probability distribution that
has a u: = 0 and standard deviation o = 1.
1/X—py?
EAST)

oy2n

where:
Y = height of the curve particular values of X.
X = any score in the distribution
a = standard deviation of the population.
u = mean of the population
m = 3.1416
e = 2.7189

6. EMPIRICAL RULE

* The Empirical Rule is also referred to
as the 68-95-99.7% Rule. What it tells
us is that for a normally distributed
variable, the following are true:

+ Approximately 68% of the data lie
within 1 standard deviation of the
mean. Pru-o<X<u+to)

+ Approximately 95% of the data lie

within 2 standard deviations of the

mean. Pr(u

+ Approximately 99.7% of the data lie
within 3 standard deviations of the

mean.

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7. Example 1: What is the frequency and relative frequency of babies

weights that are within:
[224 | 4.21

5.16 | 5.63 | 6.18 | 6.4 | 6.8 | 7.34 | 8.47
2.93 | 4.38 | 5.26 | 5.84 | 6.19 | 6.56 | 6.83 | 7.35 | 8.6
3.48 | 4.69 | 5.32 | 5.87 | 6.24 | 6.61 | 7.19 | 7.35 | 9.01
3.99 | 4.94 | 5.37 | 6.11 | 6.38
a) One standard b) Two standard
deviation from deviations from the
the mean mean

26 out of 36, or 72%

34 out of 36, or 95%

u = 6.11
o = 1.63

c) Three standard
deviations from the

mean
36 out of 36,or 100%

8. > 2: The scores of the Senior High School students in their Statistics and
Probab ty quarterly examination are normally distributed with a mean of 35
and a standard deviation of 5. (Statistics and Probability, PIVOT) .

Answer the following questions:

a. What percent of the scores are between 30 to 40?

b. What scores fall within 95% of the distribution?

a) The scores that fall between 30 and 40 is approximately 68%, of the
distribution.

9. Example 2: The scores of the Senior High School students in their Statistics and
Probability quarterly examination are normally distributed with a mean of 35
and a standard deviation of 5. (Statistics and Probability, PIVOT) .

Answer the following questions:

a. What percent of the scores are between 30 to 40?

b, What scores fall within 95% of the distribution?

b) The scores corresponding to 95% of the distribution are scores from 25 up to 45.

6

10. Example 3: Use Empirical rule to complete the following table. Write on the
respective column the range or interval of the scores based on the given
parameters.

<X<u+o u-2a0<X<u+20 n—3a<X<u+3a

a) 20 2 20-2=18 20-2(2)=16 20-2(3)=14
) 20+2=22 20+2(2)=24 20+ 2(3)=26

18 to 22 16 to 24 14 to 26

87-55=815 87-55(2)=76 87-55(3)=705
D) 87 53 yss=e2s 87+55(2)=98 874550) = 1035

81.5 to 92.5 16 to 98 ae 103.5

11. ple 4: Illustrate the following distribution through a diagram.
a) u = 20 and o =

b)u =87ando=55

12. All AND
GRADE ° PROBABILITY

IDENTIFYING REGIONS
UNDER NORMAL CURVE
CORRESPONDS TO

, DIFFERENT STANDARD 4,
NORMAL-VALUES le

13. NE

1026} 1064) 1103) 4]
T Y
1406} 1443) 148

14. : Find the area that corresponds to z = 2

15. : Find the area that corresponds to z = 2.47

A G 2017 9

The area that corresponds to z = 2.47 is 0.4932.

16. 3: Find the area that corresponds to z = -2.47

m u m sm nu im =] 40
6 au aa a |

4490840480405 a af

Fa a ut a aa a 498 41

ñ
4
4
4

The area that corresponds to z = -2.47 is 0.4932.

17. Example 4: Determine the area under the standard normal
curve to the right of z= 1.63.

greater than
to the right
at least
more than
above

greater than
to the right
at least negative
more than
above

corresponds to the
given from 0.5
o

0.5 to the
corresponds area

leas than

at most 0.5 to the
no more than corresponds area
not greater than
to the left
leas than
at most
no more than negative
not greater than
to the left

the area
P(z<-a) corresponds to the
given from 0.5

18. Areas Under the One-Taited Standard Normal Curve

Tho value that corresponds
toz = -0.52 is 0.1985

= 05 + 0.1985
0.6985

19. 4 Under the Ore-Tailed Standard Normal Curve

Example 6; Determine the area under the standard normal
curve to the left of z= 1.25.

The valuo that corresponds
toz= 1.29 is 0.3944

= 0.5 + 0.3944

0.8944
h

20. s Under the One-Tailed Standard Normal Curve

‘ Example 7: Determine the area under the standard normal
curve to the left of z = -0.95,

Tho valuo that corresponds
to 2 = -0.95 is 0.3289

= 05-0328)

21. Areas Under the One-Tailed Standard Normal Curve

Example 8: Find the area under the standard normal curve between
03 and z = -0.37,

‚a value that coreponds DEE 0.3485 + 0.1443
2= 1,03 is 0.3485 and x

= 0.4928

22. Example 9: Find the arca under the standard normal curvo between
22 0.32 and z = 2.42,

The value that corresponds to = 0.4922 - 0.1255
2= 0.32 is 0.1255 and
2 "2,42 is 0.4922. = 0.3667

23. Find the area under the normal curve in each of the following cases.
I. between z = Oandz=1.54 2. between z=O andz= 212

4 SI. la

ka 0-4982 ar 43. el.

24. Find the area under the normal curve in each of the following cases.
3. nd 168

ds N

0.4906 0-4535
A = 0-4104 4 0.4535 = 0.9441 or 94.4