The document discusses the binomial distribution, which models the number of successes in a fixed number of independent yes/no experiments, called Bernoulli trials, where the probability of success p is the same for each trial. It provides examples of binomial experiments and calculations. Key points covered include the definition of a binomial random variable X with parameters n (number of trials) and p (probability of success), and that the mean and variance of X are np and np(1-p) respectively. Examples calculate probabilities of outcomes and compare values to means and standard deviations.
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Determination_potassium_by_ammonium_acetate_extraction_method_zahid_sau_sylhet.
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Detail Description about Probability Distribution for Dummies. The contents are about random variables, its types(Discrete and Continuous) , it's distribution (Discrete probability distribution and probability density function), Expected value, Binomial, Poisson and Normal Distribution usage and solved example for each topic.
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Chapter 5: Discrete Probability Distribution
5.2 - Binomial Probability Distributions
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The Roman Empire, a vast and enduring power, stands as one of history's most remarkable civilizations, leaving an indelible imprint on the world. It emerged from the Roman Republic, transitioning into an imperial powerhouse under the leadership of Augustus Caesar in 27 BCE. This transformation marked the beginning of an era defined by unprecedented territorial expansion, architectural marvels, and profound cultural influence.
The empire's roots lie in the city of Rome, founded, according to legend, by Romulus in 753 BCE. Over centuries, Rome evolved from a small settlement to a formidable republic, characterized by a complex political system with elected officials and checks on power. However, internal strife, class conflicts, and military ambitions paved the way for the end of the Republic. Julius Caesar’s dictatorship and subsequent assassination in 44 BCE created a power vacuum, leading to a civil war. Octavian, later Augustus, emerged victorious, heralding the Roman Empire’s birth.
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Culturally, the Romans were eclectic, absorbing and adapting elements from the civilizations they encountered, particularly the Greeks. Roman art, literature, and philosophy reflected this synthesis, creating a rich cultural tapestry. Latin, the Roman language, became the lingua franca of the Western world, influencing numerous modern languages.
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We all have good and bad thoughts from time to time and situation to situation. We are bombarded daily with spiraling thoughts(both negative and positive) creating all-consuming feel , making us difficult to manage with associated suffering. Good thoughts are like our Mob Signal (Positive thought) amidst noise(negative thought) in the atmosphere. Negative thoughts like noise outweigh positive thoughts. These thoughts often create unwanted confusion, trouble, stress and frustration in our mind as well as chaos in our physical world. Negative thoughts are also known as “distorted thinking”.
2. Bernoulli Trials
Definition
A Bernoulli trial is a random experiment in which there are only
two possible outcomes - success and failure.
1 Tossing a coin and considering heads as success and tails as
failure.
The Binomial Distribution
3. Bernoulli Trials
Definition
A Bernoulli trial is a random experiment in which there are only
two possible outcomes - success and failure.
1 Tossing a coin and considering heads as success and tails as
failure.
2 Checking items from a production line: success = not
defective, failure = defective.
The Binomial Distribution
4. Bernoulli Trials
Definition
A Bernoulli trial is a random experiment in which there are only
two possible outcomes - success and failure.
1 Tossing a coin and considering heads as success and tails as
failure.
2 Checking items from a production line: success = not
defective, failure = defective.
3 Phoning a call centre: success = operator free; failure = no
operator free.
The Binomial Distribution
5. Bernoulli Random Variables
A Bernoulli random variable X takes the values 0 and 1 and
P(X = 1) = p
P(X = 0) = 1 − p.
It can be easily checked that the mean and variance of a Bernoulli
random variable are
E(X) = p
V (X) = p(1 − p).
The Binomial Distribution
6. Binomial Experiments
Consider the following type of random experiment:
1 The experiment consists of n repeated Bernoulli trials - each
trial has only two possible outcomes labelled as success and
failure;
The Binomial Distribution
7. Binomial Experiments
Consider the following type of random experiment:
1 The experiment consists of n repeated Bernoulli trials - each
trial has only two possible outcomes labelled as success and
failure;
2 The trials are independent - the outcome of any trial has no
effect on the probability of the others;
The Binomial Distribution
8. Binomial Experiments
Consider the following type of random experiment:
1 The experiment consists of n repeated Bernoulli trials - each
trial has only two possible outcomes labelled as success and
failure;
2 The trials are independent - the outcome of any trial has no
effect on the probability of the others;
3 The probability of success in each trial is constant which we
denote by p.
The Binomial Distribution
9. The Binomial Distribution
Definition
The random variable X that counts the number of successes, k, in
the n trials is said to have a binomial distribution with parameters
n and p, written bin(k; n, p).
The probability mass function of a binomial random variable X
with parameters n and p is
The Binomial Distribution
10. The Binomial Distribution
Definition
The random variable X that counts the number of successes, k, in
the n trials is said to have a binomial distribution with parameters
n and p, written bin(k; n, p).
The probability mass function of a binomial random variable X
with parameters n and p is
f (k) = P(X = k) =
n
k
pk
(1 − p)n−k
for k = 0, 1, 2, 3, . . . , n.
The Binomial Distribution
11. The Binomial Distribution
Definition
The random variable X that counts the number of successes, k, in
the n trials is said to have a binomial distribution with parameters
n and p, written bin(k; n, p).
The probability mass function of a binomial random variable X
with parameters n and p is
f (k) = P(X = k) =
n
k
pk
(1 − p)n−k
for k = 0, 1, 2, 3, . . . , n.
n
k counts the number of outcomes that include exactly k
successes and n − k failures.
The Binomial Distribution
12. Binomial Distribution - Examples
Example
A biased coin is tossed 6 times. The probability of heads on any
toss is 0.3. Let X denote the number of heads that come up.
Calculate:
(i) P(X = 2)
(ii) P(X = 3)
(iii) P(1 < X ≤ 5).
The Binomial Distribution
13. Binomial Distribution - Examples
Example
(i) If we call heads a success then this X has a binomial
distribution with parameters n = 6 and p = 0.3.
P(X = 2) =
6
2
(0.3)2
(0.7)4
= 0.324135
The Binomial Distribution
14. Binomial Distribution - Examples
Example
(i) If we call heads a success then this X has a binomial
distribution with parameters n = 6 and p = 0.3.
P(X = 2) =
6
2
(0.3)2
(0.7)4
= 0.324135
(ii)
P(X = 3) =
6
3
(0.3)3
(0.7)3
= 0.18522.
(iii) We need P(1 < X ≤ 5)
The Binomial Distribution
15. Binomial Distribution - Examples
Example
(i) If we call heads a success then this X has a binomial
distribution with parameters n = 6 and p = 0.3.
P(X = 2) =
6
2
(0.3)2
(0.7)4
= 0.324135
(ii)
P(X = 3) =
6
3
(0.3)3
(0.7)3
= 0.18522.
(iii) We need P(1 < X ≤ 5)
P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)
= 0.324 + 0.185 + 0.059 + 0.01
= 0.578
The Binomial Distribution
16. Binomial Distribution - Example
Example
A quality control engineer is in charge of testing whether or not
90% of the DVD players produced by his company conform to
specifications. To do this, the engineer randomly selects a batch of
12 DVD players from each day’s production. The day’s production
is acceptable provided no more than 1 DVD player fails to meet
specifications. Otherwise, the entire day’s production has to be
tested.
(i) What is the probability that the engineer incorrectly passes a
day’s production as acceptable if only 80% of the day’s DVD
players actually conform to specification?
(ii) What is the probability that the engineer unnecessarily
requires the entire day’s production to be tested if in fact 90%
of the DVD players conform to specifications?
The Binomial Distribution
17. Example
Example
(i) Let X denote the number of DVD players in the sample that
fail to meet specifications. In part (i) we want P(X ≤ 1) with
binomial parameters n = 12, p = 0.2.
P(X ≤ 1) = P(X = 0) + P(X = 1)
=
12
0
(0.2)0
(0.8)12
+
12
1
(0.2)1
(0.8)11
= 0.069 + 0.206 = 0.275
The Binomial Distribution
18. Example
Example
(ii) We now want P(X > 1) with parameters n = 12, p = 0.1.
P(X ≤ 1) = P(X = 0) + P(X = 1)
=
12
0
(0.1)0
(0.9)12
+
12
1
(0.1)1
(0.9)11
= 0.659
So P(X > 1) = 0.341.
The Binomial Distribution
19. Binomial Distribution - Mean and Variance
1 Any random variable with a binomial distribution X with
parameters n and p is a sum of n independent Bernoulli
random variables in which the probability of success is p.
X = X1 + X2 + · · · + Xn.
The Binomial Distribution
20. Binomial Distribution - Mean and Variance
1 Any random variable with a binomial distribution X with
parameters n and p is a sum of n independent Bernoulli
random variables in which the probability of success is p.
X = X1 + X2 + · · · + Xn.
2 The mean and variance of each Xi can easily be calculated as:
E(Xi ) = p, V (Xi ) = p(1 − p).
The Binomial Distribution
21. Binomial Distribution - Mean and Variance
1 Any random variable with a binomial distribution X with
parameters n and p is a sum of n independent Bernoulli
random variables in which the probability of success is p.
X = X1 + X2 + · · · + Xn.
2 The mean and variance of each Xi can easily be calculated as:
E(Xi ) = p, V (Xi ) = p(1 − p).
3 Hence the mean and variance of X are given by (remember
the Xi are independent)
E(X) = np, V (X) = np(1 − p).
The Binomial Distribution
22. Binomial Distribution - Examples
Example
Bits are sent over a communications channel in packets of 12. If
the probability of a bit being corrupted over this channel is 0.1 and
such errors are independent, what is the probability that no more
than 2 bits in a packet are corrupted?
If 6 packets are sent over the channel, what is the probability that
at least one packet will contain 3 or more corrupted bits?
Let X denote the number of packets containing 3 or more
corrupted bits. What is the probability that X will exceed its mean
by more than 2 standard deviations?
The Binomial Distribution
23. Binomial Distribution - Examples
Example
Let C denote the number of corrupted bits in a packet. Then in
the first question, we want
P(C ≤ 2) = P(C = 0) + P(C = 1) + P(C = 2).
The Binomial Distribution
24. Binomial Distribution - Examples
Example
Let C denote the number of corrupted bits in a packet. Then in
the first question, we want
P(C ≤ 2) = P(C = 0) + P(C = 1) + P(C = 2).
P(C = 0) =
12
0
(0.1)0
(0.9)12
= 0.282
The Binomial Distribution
25. Binomial Distribution - Examples
Example
Let C denote the number of corrupted bits in a packet. Then in
the first question, we want
P(C ≤ 2) = P(C = 0) + P(C = 1) + P(C = 2).
P(C = 0) =
12
0
(0.1)0
(0.9)12
= 0.282
P(C = 1) =
12
1
(0.1)1
(0.9)11
= 0.377
P(C = 2) =
12
2
(0.1)2
(0.9)10
= 0.23
The Binomial Distribution
26. Binomial Distribution - Examples
Example
Let C denote the number of corrupted bits in a packet. Then in
the first question, we want
P(C ≤ 2) = P(C = 0) + P(C = 1) + P(C = 2).
P(C = 0) =
12
0
(0.1)0
(0.9)12
= 0.282
P(C = 1) =
12
1
(0.1)1
(0.9)11
= 0.377
P(C = 2) =
12
2
(0.1)2
(0.9)10
= 0.23
So P(C ≤ 2) = 0.282 + 0.377 + 0.23 = 0.889.
The Binomial Distribution
27. Binomial Distribution - Examples
Example
The probability of a packet containing 3 or more corrupted bits is
1 − 0.889 = 0.111.
Let X be the number of packets containing 3 or more corrupted
bits. X can be modelled with a binomial distribution with
parameters n = 6, p = 0.111. The probability that at least one
packet will contain 3 or more corrupted bits is:
The Binomial Distribution
28. Binomial Distribution - Examples
Example
The probability of a packet containing 3 or more corrupted bits is
1 − 0.889 = 0.111.
Let X be the number of packets containing 3 or more corrupted
bits. X can be modelled with a binomial distribution with
parameters n = 6, p = 0.111. The probability that at least one
packet will contain 3 or more corrupted bits is:
1 − P(X = 0) = 1 −
6
0
(0.111)0
(0.889)6
= 0.494.
The Binomial Distribution
29. Binomial Distribution - Examples
Example
The probability of a packet containing 3 or more corrupted bits is
1 − 0.889 = 0.111.
Let X be the number of packets containing 3 or more corrupted
bits. X can be modelled with a binomial distribution with
parameters n = 6, p = 0.111. The probability that at least one
packet will contain 3 or more corrupted bits is:
1 − P(X = 0) = 1 −
6
0
(0.111)0
(0.889)6
= 0.494.
The mean of X is µ = 6(0.111) = 0.666 and its standard deviation
is σ = 6(0.111)(0.889) = 0.77.
The Binomial Distribution
30. Binomial Distribution - Examples
So the probability that X exceeds its mean by more than 2
standard deviations is P(X − µ > 2σ) = P(X > 2.2).
As X is discrete, this is equal to
The Binomial Distribution
31. Binomial Distribution - Examples
So the probability that X exceeds its mean by more than 2
standard deviations is P(X − µ > 2σ) = P(X > 2.2).
As X is discrete, this is equal to
P(X ≥ 3)
The Binomial Distribution
32. Binomial Distribution - Examples
So the probability that X exceeds its mean by more than 2
standard deviations is P(X − µ > 2σ) = P(X > 2.2).
As X is discrete, this is equal to
P(X ≥ 3)
P(X = 1) =
6
1
(0.111)1
(0.889)5
= 0.37
P(X = 2) =
6
2
(0.111)2
(0.889)4
= 0.115.
So P(X ≥ 3) = 1 − (.506 + .37 + .115) = 0.009.
The Binomial Distribution