SENIOR HIGH SCHOOL
STATISTICS AND
PROBABILITY
QUARTER 3
Module 3 – Means and Variances
Module 4 – The Normal Distribution and Its Properties
Name of the Learner:
Grade and Section:
Teacher: Richard B. Sucgang
Contact number: 09106743561_________________________________________
LEARNING COMPETENCIES
At the end of the lesson, learners should be able to:
•
•
•
•
interprets the mean and the variance of a discrete random variable.;
solves problems involving mean and variance of probability distributions.;
illustrates a normal random variable and its characteristics.;
identifies regions under the normal curve corresponding to different
standard normal values.;
• converts a normal random variable to a standard normal variable and vice
versa.;
• computes probabilities and percentiles using the standard normal table.
LESSON 1: MEANS AND VARIANCES
INTRODUCTION/MOTIVATION: Importance of Mean and Variance (and
Standard Deviation) / The Chebyshev’s Inequality
You may wonder why the mean and standard deviation are by far the two
most important summary measures of a distribution (whether a list of data, or
for probability distribution, including a probability density function).
There is a mathematical result derived by a Russian mathematician named
Pafnuty Chebyshev, called Chebyshev’s Inequality, that says that for a
distribution,
(i) at least three fourths of the distribution is within two standard
deviations from the mean;
(ii) at least eight ninths of the data are within three standard deviations
from the mean. These bounds may be conservative though.
MAIN LESSON: PROPERTIES OF MEANS AND VARIANCES
In previous lessons, you were shown that adding or subtracting a constant
from data shifts the mean but it does not change the variance and standard
deviation. This is also the case for random variables.
E( X ± c ) = E(X) ± c
Var( X ± c ) = Var(X)
If a teacher decides to give extra points to everyone in an exam, the average
in the exam increases by the number of extra points given by the teacher, but
the variability of the new increased scores stays the same.
If a company (or the government) decides to double the income of its
employees, this would double the average income, and increase it by four times
the variability in income. (The latter is the reason why government should be
careful of doubling incomes, as this would increase income inequality). Learners
may have observed that multiplying or dividing data by a constant changes both
the mean and the standard deviation by the same factor. Variance, being the
square of standard deviation, would be affected even more, by the square of this
constant. This is also the case for random variables.
E( aX ) = a E(X)
Var( aX ) = a2 Var(X)
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To make it simple, consider a case of two independent random variables,
X and Y. The expected value of the sum of independent random variables X and
Y is the sum of the expected values:
E(X + Y) = E(X) + E(Y)
while the expected value of the difference of X and Y is the difference of the
expected values:
E(X - Y) = E(X) - E(Y)
How about the variance? Explain to learners that if the random variables
are independent, then there is a simple Addition Rule for variances (for a sum
of random variables):
Var( X + Y ) = Var(X) + Var(Y)
What about the variances of a difference? Surprisingly, variance also adds
up for a difference of random variables:
Var( X - Y ) = Var(X) + Var(Y)
Variances are added for both the sum and difference of two independent
random variables because the variation in each variable contributes to the
variation in each case. The variability of the differences increases as much as
the variability of sums.
To illustrate this notion about sums (or differences of random variables),
consider a team of four swimmers that are supposed to perform 4 medley relay
events and swimming 100 meters. The swimmers’ performances are
independent, having the following means and standard deviations of the times
(in seconds) to finish 100 meters
What would be the mean and standard deviation would be for the relay
team’s total time in the relay. If the team practice’s best time was 201.62
seconds, would it be likely for the team to swim faster than this during actual
competition?
You should be able to obtain the mean of the team’s total time in the relay
as the sum of the means 45.02+50.02 +51.45 +56.38 = 203.1 seconds, with a
variance equal to the sum of the variances, i.e. 0.202 + 0.262 + 0.242 + 0.222
= 0.2136, so that the standard deviation is the square root of 0.2136=0.46
seconds. The best time of 201.62 seconds is 3.2 standard deviations below the
mean, thus, it would be very likely for the team to swim faster than this best
time.
The crucial assumption is independence of the random variables.
Suppose the amount of money spent by learners for lunch is represented by
the random variable X, and the amount of money the same group of learners
spends on afternoon snacks is represented by variable Y. The variance of the
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sum X + Y is not the sum of the variances, since X and Y are not independent
random variables.
Consider tossing a fair coin 10 times and then ask learners: What would
be the number of heads expected? Likely, you will answer 5 derived from10
times one half. If you answered 5, your intuition is correct.
Here’s why:
Define Xi as 1 if the ith toss comes up heads, and 0 if the ith toss comes
up tails, and assuming in general that the coin has a chance p of yielding heads
(with p=1/2 when the coin is fair), then tell them that the probability mass
function for Xi is
For all values of i: i=1, up to 10 (or whatever number of tosses we make).
Here the mean of Xi is
while the variance of Xi is
For tossing a fair coin ten times, the expected value of the number of heads
is
while the variance here is
and thus a standard deviation of approximately 1.58
Using Chebyshev’s Inequality, we know that when tossing a fair coin ten
times (and repeating this coin tossing process many, many times), at least three
fourths of the time, we would have the number of heads range between 5 heads
(the expected value) and, give or take, 3 heads ( 3 = 2 times the standard
deviation 1.58 ).
In general, when we have a sequence of independent random variables X1,
X2, X3, …, Xn, with a common mean m, and a common standard deviation s,
then the sum
will have an expected value of (n m) and a variance of (n s 2).
If we were to toss a fair coin 100 times, then expected value of the number
of heads obtained is 100 (1/2)=50 , while the variance is =100 (1/2) (1/2) =25.
According to Chebyshev’s Inequality, at least three fourths of the distribution of
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the number of heads in 100 tosses of a fair coin is within 50 – 2(5) = 40 heads
to 50 + 2 (5) = 60 heads.
For tossing a coin n times where the probability of getting a head is p, if S
is the number of heads, then E(S) = n (p) while Var (S) = n (p) (1-p).
REMINDER!
Variances of independent random variables are the ones that add up (not
the standard deviations: variances have squared units, so the intuition here is
the underlying use of the Pythagorean theorem: the square of the hypotenuse
is the sum of squares of the legs). In addition, remind them that variances of
independent random variables add even when we are considering differences
between them.
ASSESSMENT
1. A grade 12 student uses the Internet to get information on temperatures in
the city where he intends to go for college. He finds information in degrees
Fahrenheit.
Determine the summary statistics equivalents in Celsius scale given °C =(°F-32)
(5/9)
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LESSON 2: THE NORMAL DISTRIBUTION AND ITS PROPERTIES
Many continuous random variables, such as IQ scores, heights of people,
or weights of M&Ms, have histograms that have bell-shaped distributions.
the most important distribution in statistical science is a normal
distribution, which has a "bell-shaped" curve. Explain that there are many
reasons why the normal distribution is considered the most important curve in
statistics.
(a) Many random variables are either normally distributed or, at least,
approximately normally distributed. Heights, weights, examination scores,
the log of the length of life of some equipment are among a few random
variables that are approximately normally distributed. Although the
distributions are only approximately normal, the approximation is usually
quite close.
(b) It is easy for mathematical statisticians to work with the normal curve.
A number of hypothesis tests and the regression model are based on the
assumption that the underlying data have normal distributions. (Extra
note: There are, however, other kinds of continuous distributions that are
used in practice. For instance, the distribution that has been found
convenient for modeling the length of life of an equipment is the Weibull
distribution.)
Stress that the normal distribution is a continuous distribution just like
the uniform and triangular distribution. However, the left and right tails of the
normal distribution extend indefinitely but come infinitely close to the x-axis.
This is a picture of the normal (bell-shaped) curve
The graph of the normal distribution depends on two factors: the mean m
and the standard deviation σ. In fact, the mean and standard deviation
characterize the whole distribution. That is, we can get areas under the normal
curve given information about the mean and standard deviation.
The mean determines the location of the center of the bell-shaped curve.
Thus, a change in the value of the mean shifts the graph of the normal curve to
the right or to the left.
For symmetric distributions with a single peak, such as the normal curve,
assist learners to remember that in this case: Mean = Median = Mode.
The standard deviation determines the shape of the graphs (particularly,
the height and width of the curve). When the standard deviation is large, the
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normal curve is short and wide, while a small value for the standard deviation
yields a skinnier and taller graph.
the curve above on the left is shorter and wider than the curve on the right,
because the curve on the left has a bigger standard deviation.
a normal curve is symmetric about its mean and is more concentrated
in the middle rather than in the tails. Aside from that, observe that normal
curves differ in how spread out they are (and that the spread or variability is
measured by the standard deviations).
when a random variable has a normal distribution with mean m and
variance σ2, we denote this as X~N(μ,σ2).
Technical Note: The height of a normal curve at some value x is a formidable
looking expression that depends on the mean m and standard deviations:
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KEY POINTS
• The normal distribution, a special continuous distribution, is extremely
important in statistics because many random variables that occur in real
applications have normal distributions (or approximately normal distributions).
• The normal distribution, characterized by its mean m and its standard
deviations., has a graph that is bell-shaped. It is also symmetric about the mean
so that in consequence, the mean is the median and is also the mode (since the
curve is highest at the mean).
• The normal curve satisfies the Empirical Rule: (a) Approximately 68% of the
area under the normal curve is within one standard deviation from the mean;
(b) Approximately 95% of the area under the normal curve is within two
standard deviations from the mean; and (c) nearly everything, approximately
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99.7% of the area under the normal curve, is within three standard deviations
from the mean.
ASSESSMENT
1. The data below and the accompanying histogram give the weights, to the
nearest hundredth of a gram, of a sample of 100 coins (each with a value of
P10). The mean weight is 8.69 grams and the standard deviation s is
approximately 0.055 gram.
a. Compare the mean and median.
b. What percentage of the data is within one standard deviation of the mean?
Within two standard deviations? Within three standard deviations?
c. Suppose you were to randomly select a coin from this collection. What is the
chance that its weight would be within one standard from the mean? Two
standard deviations? Three standard deviations?
d. What percentage of the data is below the mean?
e. Suppose you were to randomly select a coin from this collection. What is the
chance that its weight would be below the mean?
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LESSON 3: AREAS UNDER A STANDARD NORMAL DISTRIBUTION
The Standard Normal Curve
Define the standard normal distribution to be the normal distribution with
a mean of 0 and a standard deviation of 1, and draw a standard normal curve:
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the table’s rows show the whole number and tenths place of the z-score,
while the table’s columns show the hundredths place, and finally, the
cumulative probability Φ(z) appears in the cell of the table.
For example, a section of the standard normal table is reproduced below.
To find the cumulative probability of a z-score equal to -1.31, explain to
students that they should cross-reference the row of the table containing -1.3
with the column containing 0.01. The table shows that the probability that a
standard normal random variable will be less than -1.31 is 0.0951; that is,
Φ(1.31) = P(Z ≤ -1.31) = 0.0951.
Practice this table of cumulative probabilities under a standard normal
curve. Assume that we have a random variable Z that has a standard normal
distribution. Ask them what would be: (a) P( Z ≤ 0 ): Answer should be 0.5 since
the first entry of the first line (of the second page) for the Table of values of Φ(z)
reads so.
(b) P( Z ≤ -1.54 ) ; As per Table of values of Φ(z), answer is 0.0618
(c) P(-1.54 ≤ Z ≤ 1.54 ) = 0.8764. Get a graph of the pertinent area of interest,
and show that the area between -1.54 and 1.54 can be obtained from the
difference of the area to the left of 1.54 and the area to the left of -1.54:
= P( Z ≤ 1.54 ) - P( Z ≤ -1.54 ) = 0.9382 - 0.0618 (as per the table entries) =
0.8764
(d) P(Z ≥ 1.54) = 0.0618 P(Z ≥ 1.54) is an upper tail area, but the total area under
the curve is 1, so P( Z ≥ 1.54 ) is the difference of 1 and the area to the left of
1.54, i.e.
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Alternatively, P( Z ≥ 1.54 ) = P( Z ≤ - 1.54 ) = 0.0618
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KEY POINTS
• The standard normal distribution is a normal distribution with a mean of 0
and a standard deviation of 1.
• Tables of the Cumulative Distribution Function of a Standard Normal
Distribution can be used to generate various areas of a standard normal curve
as well as percentiles of the distribution.
ASSESSMENT
1. The standard normal distribution
a. has a mean of zero (0) and a standard deviation of 1.
b. has a mean of 1 and a variance of zero (0).
c. has an area equal to 0.5.
d. cannot be used to approximate discrete probability distributions.
2. If Z has a standard normal distribution, and P (0 < Z < z ) is 0.3770, then the
value of z is
a. 0.18
b. 0.81
c. 1.16
d. 1.47
3. True or False: The probability that a standard normal random variable, Z,
falls between – 1.50 and 0.81 is 0.7242. __________
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4. Suppose Z has a standard normal distribution with a mean of 0 and standard
deviation of 1. The probability that Z is less than 1.15 is __________
5. Suppose Z has a standard normal distribution with a mean of zero (0) and
standard deviation of 1. The probability that Z values are larger than __________
is 0.3483.
6. Suppose Z has a standard normal distribution with a mean of zero (0) and
standard deviation of 1. 85% of the possible Z values are smaller than
__________.
HANDOUT
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LESSON 10: AREAS UNDER A NORMAL DISTRIBUTION
Given a normally distributed random variable: X~N(μ,σ2), we often wish to
find various probabilities pertaining to where an arbitrary measurement may
lie. For instance, we may want to find P(a ≤ X ≤ b), which is the probability that
a random measurement X lies between a and b.
Standard Scores (or Z-scores)
Whatever the value of the mean and standard deviation of a normal curve,
we can transform the whole normal curve into a standard normal curve (as
illustrated in the following figure).
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This entails transforming the all data in a normal curve into standard
units:
An observation is in standard unit (or z-score) if we see how many standard
deviations it is above or below the average. That is, if x, m, and s respectively
represent the observation, its mean, its standard deviation, then the
standardized form (or z-score) of x is
Remember that a Z-score indicates how many standard deviations a
certain data element is from the mean. For instance, if examination scores
in Statistics and Probability have an average of 75 and a standard deviation of
5, then an exam score of 90 has a z-score of (90-75)/5 = 3, while a score of 70
has a z-score of (70-75)/5 =-1. To interpret these z-scores, we note that 90 is 3
standard deviations above the mean (75), while 70 is one standard deviation
“below’ the mean. Z scores have a very good way of making variables
comparable.
The Z-scores may also be used for normal random variables to transform
them into standard normal random variables, and this, in turn, can help us
relate probabilities for any normal distribution to areas under a standard
normal curve, as the following example on the time to walk a dog illustrates.
Illustration for Finding Areas Under a Normal Curve
Assume that the distribution of heights of all female Grade 11 students
can be modeled well by a normal curve with a mean of 1620 mm and a standard
deviation of 50 mm. Further, we wish to determine (a) the proportion of female
Grade 11 students shorter than 1550 mm; (b) the proportion of female Grade
11 students taller than 1650 mm; (c) the proportion of female Grade 11 students
between 1600 and 1675 mm; (d) the height of a female Grade 11 student for
which 10 percent of female Grade 11 students are shorter than it; (e) the height
of a female Grade 11 student for which 75% of female Grade 11 students are
taller than it.
For computing the answer to (a) first, transform 1550 to its z-score,
yielding (1550-1620)/50 =-1.4 so that we can associate the area to the left of
1550 (under a normal curve with mean 1620 and standard deviation 50) with
that of the area to the left of z = -1.4 under a standard normal curve. Reading
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from the table of Cumulative Distribution Function of a Standard Normal Curve,
we find Φ(-1.4) = 0.0808,
For (b), transform the height value 1650 to its standard units, (16501620)/50 = 0.6, and then note that the area to the right of z = 0.6 under the
standard normal curve is the difference between the total area under a standard
normal curve (100%) and the area to the right of z=0.6, Φ (0.6) = 0.7257. In
consequence, the desired probability (and area) is 1- 0.7257=0.2743.
For (c), learners should mention they need to firstly transform 1600 and
1675 into their respective standardized forms, namely (1600-1620)/50 = -0.4
and (1675- 1620)/50 = 1.1, and then generate the area between these two zscores as the difference between Φ (1.1) and Φ (-0.4), i.e. 0.86430.3446=0.5197.
For (d), draw the figures on the board to illustrate what needs to be done:
The 10th percentile of the height distribution may be obtained by firstly
getting the 10th percentile of the standard normal curve, which can be read off
from the table as –1.282. This means that the 10th percentile of the height
distribution is 1.282 standard deviations below the mean. This required value
for the height is – 1.282(50)+1620 =1555.9.
Finally, for (e), suggest to learners that we want the 25th percentile as this
is the value for which 75 percent of the height distribution would be above it.
Similar to (d), tell students they can find the 25th percentile first of a standard
normal curve (– 0675), then yield the required height as:
–0.675(50)+1620 =1586.25.
KEY POINTS
• To obtain probabilities or percentiles under a normal curve, perform two steps:
Transform the normal curve into a standard normal curve by way of “z-scores”
(which involves subtracting the mean and dividing the result by the standard
deviation)
z = (X - μ) / σ.
Then, use the tables of the Cumulative Distribution Function of a
Standard Normal Distribution to obtain the required areas of a standard normal
curve to find the probabilities associated with the z-scores.
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ASSESSMENT
1. If a particular batch of data is approximately normally distributed, we would
find that approximately
a) 2 of every 3 observations would fall between ±1 standard deviation
around the mean.
b) 4 of every 5 observations would fall between ±1.28 standard deviations
around the mean.
c) 19 of every 20 observations would fall between ±2 standard deviations
around the mean.
d) All the above.
For problems 2 to 4 consider the following case.
The length of time it takes a Grade 11 student to play the Candy Crush
computer app follows a normal distribution with a mean of 3.5 minutes and a
standard deviation of 1 minute
2. The probability that a randomly selected Grade 11 student will play one game
of Candy Crush in less than 3 minutes is
a) 0.3551
b) 0.3085
c) 0.2674
d) 0.1915
3. The probability that a randomly-selected grade 11 student will take between
2 and 4.5 minutes to play Candy Crush is:
a) 0.0919
b) 0.2255
c) 0.4938
d) 0.7745
4. The point in the distribution of times to play Candy Crush in which 75.8% of
the Grade 11 students exceed when playing Candy Crush.
a) 2.8 minutes
b) 3.2 minutes
c) 3.4 minutes
d) 4.2 minutes
5. Rodrigo earned a score of 940 on a national achievement test. The mean test
score was 850 with a standard deviation of 100. What proportion of students
had a higher score than Rodrigo? (Assume that test scores are normally
distributed.) If there were 100,000 students who took the test, how many would
be expected to have a higher score than Rodrigo?
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Workbooks in Statistics 1: 11th Edition, Institute of Statistics, UP Los Baños,
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4031
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