Normal Distribution, Binomial Distribution, Poisson Distribution
Binomial Probability Distribution
Is the binomial distribution is a continuous
Notation: X ~ B(n,p)
There are 4 conditions need to be satisfied for a
1. There is a fixed number of n trials carried out.
2. The outcome of a given trial is either a
3. The probability of success (p) remains constant
from trial to trial.
4. The trials are independent, the outcome of a
trial is not affected by the outcome of any
Comparison between binomial and normal
If X ~ B(n, p), then
Ten percent of computer parts produced by a
certain supplier are defective. What is the
probability that a sample of 10 parts contains
more than 3 defective ones?
If X is binomially distributed with 6 trials and a
probability of success equal to ¼ at each attempt.
What is the probability of
a)exactly 4 succes.
b)at least one success.
Jeremy sells a magazine which is produced in order
to raise money for homeless people. The probability
of making a sale is, independently, 0.50 for each
person he approaches. Given that he approaches 12
people, find the probability that he will make:
(a)2 or fewer sales;
(b)exactly 4 sales;
(c)more than 5 sales.
In general, when we gather data, we expect to see
a particular pattern to
the data, called a normal distribution. A normal
distribution is one
where the data is evenly distributed around the
mean, which when plotted as a
histogram will result in a bell curve also known as
a Gaussian distribution.
thus, things tend towards the mean – the closer a
value is to the mean, the more you’ll see it; and
the number of values on
either side of the mean at any particular distance
are equal or in symmetry.
with mean and standard deviation of a set of
scores which are normally distributed, we can
standardize each "raw" score, x, by converting it
into a z score by using the following formula on
each individual score:
a) Find the z-score corresponding to a raw score of 132 from a normal distribution with
mean 100 and standard deviation 15.
b) A z-score of 1.7 was found from an observation coming from a normal distribution with
mean 14 and standard deviation 3. Find the raw score.
z = __________ = 2.133
b) We have
1.7 = ________
To solve this we just multiply both sides by the denominator 3,
(1.7)(3) = x - 14
5.1 = x - 14
x = 19.1
a) P(z < 2.37)
b) P(z > 1.82)
a)We use the table. Notice the picture on the table has shaded region
corresponding to the area to the left (below) a z-score. This is exactly what
P(z < 2.37) = .9911
b) In this case, we want the area to the right of 1.82. This is not what is given
in the table. We can use the identity
P(z > 1.82) = 1 - P(z < 1.82)
reading the table gives
P(z < 1.82) = .9656
Our answer is
P(z > 1.82) = 1 - .9656 = .0344
P(-1.18 < z < 2.1)
Once again, the table does not exactly handle this type of area. However, the area
between -1.18 and 2.1 is equal to the area to the left of 2.1 minus the area to the left
of -1.18. That is
P(-1.18 < z < 2.1) = P(z < 2.1) - P(z < -1.18)
To find P(z < 2.1) we rewrite it as P(z < 2.10) and use the table to get
P(z < 2.10) = .9821.
The table also tells us that
P(z < -1.18) = .1190
Now subtract to get
P(-1.18 < z < 2.1) = .9821 - .1190 = .8631
a discrete probability distribution for the count of
events that occur randomly in a given time.
a discrete frequency distribution which gives the
probability of a number of independent events
occurring in a fixed time.
Poisson distribution only apply one formula:
X = the number of events
λ = mean of the event per interval
Where e is the constant, Euler's number (e = 2.71828...)
Births rate in a hospital occur randomly at an average rate of 1.8 births
What is the probability of observing 4 births in a given hour at the
X = No. of births in a given hour
i) Events occur randomly
ii) Mean rate λ = 1.8
Using the poisson formula, we cam simply calculate the distribution.
P(X = 4) =( e^-1.8)(1.8^4)/(4!)
If the probability of an item failing is 0.001, what is the probability
of 3 failing out of a population of 2000?
Λ = n * p = 2000 * 0.001 = 2
Hence, use the Poisson formula
X = 3,
P(X = 3) =
A small life insurance company has determined that on the
average it receives 6 death claims per day. Find the probability
that the company receives at least seven death claims on a
randomly selected day.
1st: analyse the given data.
2nd: label the value of x, λ
At least 7 days, means the probability must be ≥ 7. but
the value will be to the infinity. Hence, must apply the
probability rule which is
P(X ≥ 7) = 1 – P(X ≤ 6)
P(X ≤ 6) means that the value of x must be from 0, 1, 2,
3, 4, 5, 6.
Total them up using Poisson, then 1 subtract the
Ans = 0.3938
The number of traffic accidents that occurs on a particular
stretch of road during a month follows a Poisson distribution
with a mean of 9.4. Find the probability that less than two
accidents will occur on this stretch of road during a randomly
P(x < 2) = P(x = 0) + P(x = 1)