Z-Scores, Standardization, and the Standard Normal Distribution (5.3)
Summary
TLDRThis video script offers an insightful exploration of z-scores and the standardization process in statistics. It explains the concept of the standard normal distribution, a bell curve with a mean of 0 and a standard deviation of 1. The script delves into how to calculate exact proportions using this distribution and z-score tables, illustrating the process with examples. It also highlights the benefits of standardization, allowing for the conversion of any normal distribution into the standard form, and demonstrates how to apply this to calculate specific areas under the curve, providing a practical understanding of statistical analysis.
Takeaways
- 📊 The standard normal distribution is a type of normal distribution with a mean of 0 and a standard deviation of 1.
- 🧮 A z-score represents how many standard deviations an observation is from the mean.
- 📉 A negative z-score indicates that the observation is below the mean, while a positive z-score indicates it's above the mean.
- 📑 The z-score table (standard normal table) helps to calculate the area (proportion) to the left of any z-score.
- 🔄 To find the area to the right of a z-score, subtract the table value from 1.
- 🔍 The z-score table can also be used for reverse look-ups to find the z-score associated with a specific area.
- 🔄 Any normal distribution can be standardized, converting it to a standard normal distribution with a mean of 0 and a standard deviation of 1.
- 🧪 The formula for standardization is z = (X - μ) / σ, where X is the observation, μ is the mean, and σ is the standard deviation.
- 📏 Standardization allows you to use the z-score table to calculate areas for any normally distributed population.
- 📐 The proportion of values within a specific range can be calculated by finding the z-scores of the range limits and then subtracting the corresponding areas.
Q & A
What is a standard normal distribution?
-A standard normal distribution is a special type of normal distribution with a mean of 0 and a standard deviation of 1. It is always centered at 0, and each number on the horizontal axis corresponds to a z-score, which indicates how many standard deviations an observation is from the mean.
What does a z-score represent?
-A z-score represents the number of standard deviations an observation is from the mean (mu). For example, a z-score of -2 indicates that the observation is 2 standard deviations below the mean, while a z-score of 1.5 indicates it is 1.5 standard deviations above the mean.
How can we find the exact area associated with a specific z-score?
-We can find the exact area associated with a specific z-score using a z-score table, also known as the standard normal table. This table provides the total amount of area contained to the left side of any z-value.
What is the proportion of Z less than -1.95 according to the standard normal table?
-According to the standard normal table, the proportion of Z less than -1.95 is 0.0256. This means there is 2.56% of the area to the left of a z-score of -1.95.
How do we determine the area to the right of a z-score?
-To determine the area to the right of a z-score, we find the area that corresponds to that z-value in the standard normal table and then subtract it from 1. For example, to find the area to the right of a z-score of 0.57, we subtract the area of 0.7157 (to the left of 0.57) from 1, resulting in an area of 0.2843.
What is the purpose of using a z-score table for reverse look-up?
-The purpose of using a z-score table for reverse look-up is to determine what z-score corresponds to a specific area. For instance, if you want to know what z-score is associated with an area of 0.8446 to the left of it, you find 0.8446 on the table and see the corresponding z-score, which is 1.02.
What is the process of standardization in the context of normal distributions?
-Standardization is the process of converting any normal distribution with any mean (mu) and standard deviation (sigma) into the standard normal distribution, where the mean is zero and the standard deviation is one. This allows the use of the z-score table to calculate exact areas for any given normally distributed population.
What is the standardization formula used to convert a normal distribution to a standard normal distribution?
-The standardization formula is Z = (X - mu) / sigma, where Z is the z-score, X is the observation, mu is the population mean, and sigma is the population standard deviation.
How can we find the proportion of students scoring less than 49 on an exam with a mean of 60 and a standard deviation of 10?
-First, we standardize the distribution using the formula Z = (X - mu) / sigma. For X = 49, the z-score is -1.1. Then, we look up the z-score of -1.1 in the standard normal table to find the proportion of Z less than -1.1, which is 0.1357. This is the proportion of students scoring less than 49.
What proportion of students are between 5.81 feet and 6.3 feet tall if the height distribution has a mean of 5.5 feet and a standard deviation of 0.5 feet?
-We standardize the distribution to find the z-scores for 5.81 feet (z = 0.62) and 6.3 feet (z = 1.6). Then, we subtract the area corresponding to z = 0.62 (0.7304) from the area corresponding to z = 1.6 (0.9452), resulting in a proportion of 0.2148, or 21.48% of students.
Outlines
📚 Introduction to Z-Scores and Standard Normal Distribution
This paragraph introduces the concept of Z-scores and the standard normal distribution. It explains that the standard normal distribution is a special kind of normal distribution with a mean of 0 and a standard deviation of 1, making it symmetrical and centered at zero. The paragraph details how Z-scores indicate the number of standard deviations an observation is from the mean, with examples provided to illustrate this. It also discusses the use of the Z-score table, or the standard normal table, to find the exact area associated with a specific Z-score, and how to use this table to determine areas to the left or right of any Z value. The process of standardization is introduced as a method to convert any normal distribution into the standard normal distribution, using a specific formula that involves subtracting the population mean and dividing by the population standard deviation. An example using final chemistry exam scores is given to demonstrate how to use the standardization formula and Z-score table to find the proportion of students scoring below a certain threshold.
📏 Applying Standardization to Measure Heights and Calculate Proportions
The second paragraph applies the concept of standardization to a real-world example of measuring the heights of university students, which are normally distributed with a mean of 5.5 feet and a standard deviation of 0.5 feet. The paragraph outlines the process of using the standardization formula to convert individual height measurements into Z-scores, which allows for the use of the standard normal table to find the corresponding areas. The example calculates the proportion of students with heights between 5.81 feet and 6.3 feet by finding the Z-scores for these heights, looking up the areas in the standard normal table, and subtracting the smaller area from the larger one to find the difference. The result is a proportion of students within the specified height range, demonstrating the practical application of Z-scores and standardization in statistical analysis.
Mindmap
Keywords
💡Z-scores
💡Standard Normal Distribution
💡Mean (μ)
💡Standard Deviation (σ)
💡Z-score Table
💡Proportion
💡Standardization
💡Area Under the Curve
💡Observation (X)
💡Density Curve
💡Reverse Look-up
Highlights
The standard normal distribution is a unique normal distribution with a mean of 0 and a standard deviation of 1, allowing for easy calculation of exact proportions using the z-score.
Z-scores indicate the number of standard deviations an observation is from the mean, with negative values indicating below the mean and positive values above.
A z-score table, also known as the standard normal table, provides the total area to the left of any z-score, facilitating the calculation of exact areas.
The proportion of Z less than a specific value can be formally stated, for example, Z less than -1.95 corresponds to an area of 0.0256.
The area to the right of any z-score can be determined by subtracting the area to the left from 1, as demonstrated with a z-score of 0.57 leading to an area of 0.2843.
The z-score table can perform a reverse lookup to find the z-score associated with a specific area, such as finding the z-score for an area of 0.8416 to the left.
Any normal distribution can be transformed into the standard normal distribution through a process called standardization, which uses a specific formula.
Standardization benefits from the ability to use the z-score table to calculate exact areas for any given normally distributed population, regardless of the mean (mu) or standard deviation (sigma).
The standardization formula involves subtracting the population mean from an observation and dividing by the population standard deviation.
An example demonstrates converting a normal distribution of final chemistry exam scores with a mean of 60 and a standard deviation of 10 into the standard normal distribution.
The conversion process results in the mean being zero and the standard deviation being one, consistent with the standard normal distribution.
Using the standardization formula, one can determine the proportion of students scoring below a certain value, such as less than 49 on an exam.
The proportion of students scoring below 49 is found to be 0.1357, using the z-score table for a z-score of -1.1.
Another example involves determining the proportion of students with heights between 5.81 and 6.3 feet, using the standardization formula with a mean height of 5.5 feet and a standard deviation of 0.5 feet.
The proportion of students within the specified height range is calculated to be 0.2128 by subtracting the areas corresponding to z-scores of 0.62 and 1.6.
Support for the video creators can be provided through Patreon, and additional study materials can be accessed on their website, Simple Learning Pro.
Transcripts
in this video we'll be learning about
zet scores and standardization by
learning about both of these topics you
will learn how to calculate exact
proportions using the standard normal
distribution what is the standard normal
distribution the standard normal
distribution is a special type of normal
distribution that has a mean of 0 and a
standard deviation of 1 because of this
the standard normal distribution is
always centered at 0 and has intervals
that increase by 1 each number on the
horizontal axis corresponds to a z-score
as zetz core tells us how many standard
deviations an observation is from the
mean mu for example a z-score of
negative 2 tells me that I am 2 standard
deviations to the left of the mean and
as that score of 1.5 tells me that I am
one and a half standard deviations to
the right of the mean most importantly a
set score allows us to calculate how
much area that specific z-score is
associated with and we can find out that
exact area using something called a
z-score table also known as the standard
normal table this table tells us the
total amount of area contained to the
left side of any value of Z for this
table the top row and the first column
correspond to Z values and all the
numbers in the middle correspond to
areas for example according to the table
a z-score of negative 1 point 9 5 has an
area of 0.025 6 to the left of it to say
this in a more formal manner we can say
that the proportion of Z less than
negative one point nine five is equal to
0.025 6 we can also use the standard
normal table to determine the area to
the right of any Z value all we have to
do is take 1 minus the area that
corresponds to that value for example to
determine the area to the right of a
z-score of 0.57 all we have to do is
find the area that corresponds to this
set value and then subtracted from 1
according to the table the z-score of
0.57 has an area of 0.7 157 to the left
of it so 1 minus 0.7 157 gives us an
area of 0.28 43 and that is our answer
the reason why we can do this is because
we have to remember that the normal
distribution is a density curve and it
always has a total area equal to 1 or
percent you can also use the set score
table to do a reverse look-up which
means you can use the table to see what
that score is associated with a specific
area so if I wanted to know what value
of Zed corresponds to an area of 0.8
four six one to the left of it
all we have to do is find zero point
eight four six one on the table and see
what value is that it corresponds to we
see that it corresponds to a set value
of one point zero two the special thing
about the standard normal distribution
is that any type of normal distribution
can be transformed into it in other
words any normal distribution with any
value of mu and Sigma can be transformed
into the standard normal distribution
where you have a mean of zero and a
standard deviation of one this
conversion process is called
standardization the benefit of
standardization is that it allows us to
use the set score table to calculate
exact areas for any given normally
distributed population with any value of
mu or Sigma standardization involves
using this formula this formula says
that the z-score is equal to an
observation X minus the population mean
mu divided by the population standard
deviation Sigma so suppose that we
gather data from last year's final
chemistry exam and found that it
followed a normal distribution with a
mean of 60 and a standard deviation of
10 if we were to draw this normal
distribution we would have 60 located at
the center of the distribution because
it is the value of the mean and each
interval would increase by 10 since that
is the value of the standard deviation
to convert this distribution to the
standard normal distribution we will use
the formula the value of MU is equal to
60 and the value of Sigma is equal to 10
we can then take each value of x and
plug it into the equation if I plug in
60 I will get a value of 0 if I plug in
50 I will get a value of negative 1 if I
plug in 40 I will get a value of
negative 2 if we do this for each value
you can see that we end up with the same
values as a standard normal distribution
when doing this conversion process the
mean of the normal distribution will
always be converted to zero and the
standard deviation will always
correspond to a value of 1 it's
important to remember that this will
happen with any normal distribution no
matter what value of MU and sick
are now if I asked you what proportion
of students score less than 49 on the
exam it is this area that we are
interested in however the proportion of
X less than 49 is unknown until we use
the standardization formula after
plugging in 49 into this formula we end
up with a value of negative 1.1 as a
result we will be looking for the
proportion of Zed less than negative 1.1
and finally we can use the z-score table
to determine how much area is associated
with his Zed score according to the
table there's an area of 0.1 3 5 7 to
the left of this set value this means
that the proportion of Zed less than
negative 1.1 is zero point 1 3 5 7 this
value is in fact the same proportion of
individuals that scored less than 49 on
the exam as a result this is the answer
let's do one more example when measuring
the heights of all students at a local
university it was found that it was
normally distributed with a mean height
of 5.5 feet and a standard deviation of
0.5 feet what proportion of students are
between five point eight one feet and
6.3 feet tall before we solve this
question it's always a good habit to
first write down important information
so we have a meal of 5.5 feet and a
sigma of 0.5 feet we are also looking
for the proportion of individuals
between 5.8 one feet and 6.3 feet tall
this corresponds to this highlighted
area to determine this area we need to
standardize the distribution so we will
use the standardization formula plugging
in 5.8 went to this formula gives us a
Zed square of 0.62 and plugging in 6.3
into the formula gives us a z-score of
1.6
according to the standard normal table
the z-score of 0.62 corresponds to an
area of 0.7 3 to 4 and the z-score of
1.6 corresponds to an area of 0.945 2 to
find the proportion of values between
0.62 and 1.6 we must subtract the
smaller area from the bigger area so
0.945 2 minus 0.7 3 to 4 gives us zero
point 2 1 to 8 as a result the
proportion of students between 5.8 one
feet and 6.3 feet tall is zero point two
one two eight if you found this video
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