NORMAL PROBABILIY DISTRIBUTION Given that 𝜇 = 6.11 and 𝜎 = 1.

61, what is the

 A probability distribution of continuous frequency and relative frequency of babies’ µ 68% 95% 99.7
random variable. weight that are within: 25 3 22-28
 GAUSSIAN DISTRIBUTION 2.24 4.69 5.63 6.24 6.8 7.35 88 6.5 81.5- 75- 68.5-
 Most of the data cluster towards the 2.93 4.95 5.84 6.38 6.83 7.68 94.5 101 107.5
middle (average) while the rest either 3.45 5.16 5.87 6.4 7.19 8.47
extremely low or extremely high 3.99 5.26 6.11 6.56 7.29 8.6 Add mean and standard deviation to get value
 It is most important probability 4.21 5.32 6.18 6.61 7.34 9.01 of 1, 2 and 3 standard deviations from the
distribution in statistics 4.38 5.37 6.19 6.76 7.35 9.47 mean to the right and subtract to find the
Properties of a Normal Curve values on the left.
- The distribution is bell shaped A. One standard deviation; AREAS UNDER THE NORMAL CURVE
- The curve is symmetrical at the center - Find the area that corresponds to z= 2
- The mean, median and mode coincide = 0.4772
at the center. - Find the area that corresponds to z=
2.47
=mean is the average value of the - =0.4932
given set - Find the area that corresponds to z= -
=median is the middle number of a 2.47
given set 26 out of 36 or 73% =0.4932
=mode is the frequent number Determine the area under the standard normal
curve to the right of 𝑧 = 1.63
- The tails of the curve flatten out B. Two standard deviations: The area to the right of 𝑧 = 1.63 is 0.0516
indefinitely, always approaching the Let 𝐴1= area of 𝑧 = 1.63
axis but never touching it. Let A = area to the right of 𝑧 = 1.63
- The area under the curve is 1. 𝐴1 = 0.4484
Standard Normal Curve 𝐴 = 0.5 − 𝐴1
- A normal Probability distribution that 𝐴 = 0.5 − 0.4484
has µ=0 and = 1 𝑨 = 𝟎. 𝟎𝟓16
2 FACTORS THAT AFFECTS THE NORMAL
CURVE: 34 out of 36, or 94% : if the direction is to the right and its positive
µ= mean = standard deviation you subtract 0.5 to the are given
- The change of the value of the mean C. Three standard deviations: : if the direction is to the right and its negative
shifts the graph to the right or to the then you add the given area to 0.5.
left
- The standard deviation determines the Determine the area under the standard normal
shape of the graphs (height and curve to the right of 𝑧 = −1.8
width)
EMPIRICAL RULE The area to the right of 𝑧 = −1.8 is 0.9641.
- This is also referred to as the 68-95-
99.7% rule. 36 out of 36, 100% Let 𝐴1= area of 𝑧 = −1.8
= 68% percent of the data lies within Let A = area to the right of 𝑧 = −1.8
1 standard deviation from the mean Complete the table. Write on the 𝐴1 = 0.4641
= 95% of the data lie within 2 respective column the range or interval of 𝐴 = 0.5 + 𝐴1
standard deviations from the mean the scores based on the given 𝐴 = 0.5 + 0.4641
= 99.7% of the data lie within 3 parameters. Then illustrate through a 𝑨 = 𝟎. 𝟗𝟔𝟒1
standard deviations from the mean diagram.
VALIDATE THE EMPIRICAL RULE:
Determine the area under the standard normal Let 𝐴2= area of 𝑧 = −0.37 P95 means locating an area before (or below)
curve to the left of 𝑧 = 1.25 Let A = area between 𝑧 = 1.03 and 𝑧 = −0.37 the point. We want to know what the z-value is
The area to the left of 𝑧 = 1.25 is 0.8944. 𝐴1 = 0.3485 𝐴2 = 0.1443 at this point.
𝐴 = 𝐴1 + 𝐴2 95% or 0.95
Let 𝐴1= area of 𝑧 = 1.25 𝐴 = 0.3485 + 0.1443 ? = 95 – 0.50 = 0.45
Let A = area to the left of 𝑧 = 1.25 𝑨 = 𝟎. 𝟒𝟗𝟐𝟖 0.4495 to 0.4505
𝐴1 = 0.3944 The area between 𝑧 = 1.03 and 𝑧 = −0.37 is z = 1.64 (0.4495)
𝐴 = 0.5 + 𝐴1 0.4928. z = 1.65 (0.4505)
𝐴 = 0.5 + 0.3944 𝑧 = 1.64 + 1.65/ 2
𝑨 = 𝟎. 𝟖𝟗𝟒4 STANDARD SCORE = 1.645
➢ Measures how many standard deviation a
: if the direction is to the left and the sign is given value (x) is above or below the mean. Example 2: Find the value of the 40th
positive then you add the area to 0.5 ➢ A positive z-score indicates that the percentile of a standard normal curve.
: if the direction is to the left and the sign is observed value is above the mean, whereas a Solution: P40 means locating an area before
negative then subtract 0.5 from the given area. negative z-score indicates the observed value (or below) the point. We want to know what
is below the mean. the z-value is at this point.
Determine the area under the standard normal ➢ Formula: 𝑧 = 𝑥−𝜇 0.4 40% or 0.40
curve to the left of 𝑧 = −1.52 given minus mean all over standard deviation 0.0987 to 0.1026
z = -0.25 z = -0.26
The area to the left of 𝑧 = −1.52 is 0.0643. a. greater than 48? 𝑧 = −0.25 + (−0.26)/ 2
Let 𝐴1= area of 𝑧 = −1.52 𝐴1 = 0.3849 = −𝟎. 𝟐𝟓𝟓
Let A = area to the left of 𝑧 = −1.52 𝐴 = 0.5 − 𝐴1
𝐴1 = 0.4357 𝐴 = 0.5 − 0.3849
𝐴 = 0.5 − 𝐴1 𝑨 = 𝟎. 𝟏𝟏𝟓𝟏 RANDOM SAMPLING
𝐴 = 0.5 − 0.4357 Solution: POPULATION
𝑨 = 𝟎. 𝟎𝟔𝟒 𝑧 = 𝑥−𝜇 /𝜎 ➢ population refers to the whole group under
𝑧 = 48−42 /5 study or investigation
Determine the area under the standard normal 𝑧 = 1.2 42 SAMPLE
curve between 𝑧 = 0.32 and 𝑧 = 2.42. ➢ a subset taken from a population; a
Let 𝐴1= area of 𝑧 = 0.32 Hence, 11.51% of the scores is greater than 48 representation of the population
Let 𝐴2= area of 𝑧 = 2.42 RANDOM SAMPLING
Let A = area between 𝑧 = 0.32 and 𝑧 = 2.42 PERCENTILE ➢ selection of elements derived from the
𝐴1 = 0.1255 𝐴2 = 0.4922 ➢ A measure used in statistics indicating the population, where each equal chance of being
𝐴 = 𝐴2 − 𝐴1 value below which a given percentage of selected using the appropriate sampling
𝐴 = 0.4922 − 0.1255 observations in a group of observations fall. technique.
𝑨 = 𝟎. 𝟑𝟔𝟔𝟕 ➢ A measure of relative standing. It is a
The area between 𝑧 = 0.32 and 𝑧 = 2.42 is descriptive measure of the relationship of a
0.3667 measurement to the rest of the data. Types of Random Sampling Techniques
➢ Your score in Statistics and Probability exam 1. Simple Random Sampling
is at the 95th percentile. That means you fared ➢ most basic random sampling wherein
: if the sign of the area is the same SUBTRACT every member of the population has an
compared to the 95% of your class, while 5%
the bigger area to the smaller area equal chance of being chosen to be part of
of the class fared above you
:if the sign isn’t the same add them together the sample
Example 1: Find the value of the 95th
Determine the area under the standard normal Simple Random Sampling
percentile of a standard normal curve.
curve between 𝑧 = 1.03 and 𝑧 = −0.37. 𝒏 = 𝑵 𝟏 + 𝑵𝒆/ 1+ Ne
Solution:
Let 𝐴1= area of 𝑧 = 1.03
Where: Step3:
n = number of samples needed Create a sampling distribution of the
N = population size sample means
e = margin of error

2. Systematic Sampling
➢ Members of the population are listed
and samples are selected at intervals. In
this technique, every kth item in the list
will be selected from a randomly selected
starting point.

Where:
k = sampling interval
N = population size
n = sample size
𝒌=𝑵/n

3. Stratified Random Sampling

➢ The population is partitioned into
several subgroups called strata, based on
some characteristics like year level, age,
gender, ethnicity, etc.

Where:
s: strata
N: population
n: sample needed
𝒙 = 𝒔 /𝑵 × n

4. Cluster Sampling
➢ The entire population is broken into
smaller groups, or clusters, and then some
of the clusters are randomly selected. All
the elements from the selected clusters
will make up the sample.

SAMPLING DISTRIBUTION OF THE MEAN

Formula: nCr

Step1:
– Find out the number of possible
outcomes
Step2:
List all the possible outcome and get the
mean of every sample