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Statistics Assignment

?Business Analytics ?Chapter 8 ?Continuous probability

•Business Analytics
•Chapter 8
•Continuous probability distributions
•Chapter outline
8.1   Probability density functions
8.2   Uniform distribution
8.3   Normal distribution
•Learning objectives
LO1   Discuss the basic differences between discrete and continuous random variables
LO2 Identify and calculate probabilities using a uniform distribution
LO3   Convert a normal random variable into a standard normal random variable, and know how to use the table of standard normal probabilities
•Introduction
•Point Probabilities are Zero
Thus, we can determine the probability of a range of values only.
Note: With a discrete random variable like tossing a die, it is meaningful to talk about P(X=5), say.
In a continuous setting (e.g. with time as a random variable), the probability the random variable of interest, say task length, takes exactly 5 minutes is infinitesimally small, hence P(X=5) = 0.
It is meaningful to talk about P(X ≤ 5).
•8.1  Probability Density Function
A function f(x) is called a probability density function (over the range a ≤ x ≤ b if it meets the following requirements:
1) f(x) ≥ 0 for all x between a and b, and
2) The total area under the curve between a and b is 1.0.
•Probability Density Function…
The probability that x falls between a and b is found by calculating the area under the graph of f(x) between a and b.
Shaded area is P(a ≤ X ≤ b)


•8.2  Uniform Distribution
•Uniform Distribution…
•Example 1
The time elapsing between the placement of an order and the delivery time is uniformly distributed between 100 and 180 minutes.
i.Define the graph and the density function.
ii.What proportion of orders takes between 2 and 2.5 hours to be delivered?
•Example 1 - Solution
  Let X = The time elapsing between the placement of an order and the delivery time.
  i.





  f(x) = 1/80        100 £ x £ 180
  ii.  P(120 £  x £ 150) = (150-120)(1/80) = 0.375

•8.3 Normal Distribution 
This is the most important continuous distribution considered here.
•Many random variables can be properly modelled as normally distributed.
•Many distributions can be approximated by a normal distribution.
•Normal distribution is the cornerstone distribution of statistical inference.
•Normal Distribution … 
The normal distribution is fully defined by two parameters:
•its mean m
•its variance s2

If a random variable X is normally distributed:
Mean = E(X) = m
Variance = V(X) = s2
•Normal Distribution … 
A random variable X with mean m and variance s2 is normally distributed if its probability density function is given by
Unlike the range of the uniform distribution   (a ≤ x ≤ b), normal distributions range from minus infinity to plus infinity.
•The shape of the normal distribution
•Calculating Normal Probabilities
Two facts help calculate normal probabilities:
•Normal distribution is symmetrical.
•Any normal distribution can be transformed into a specific normal distribution called the
`STANDARD NORMAL DISTRIBUTION’.
•Calculating Normal Probabilities
We can use the following function to convert any normal random variable to a standard normal random variable…
•Calculating Normal Probabilities
We can use the following function to convert any normal random variable to a standard normal random variable…
•Calculating Normal Probabilities…
We can use the following function to convert any normal random variable to a standard normal random variable…
•Calculating Normal Probabilities…
If we know the mean and standard deviation of a normally distributed random variable, we can always transform the probability statement about X into a probability statement about Z.

Consequently, we need only one table, Table 3 in Appendix B, the standard normal probability table.
This standard normal table is similar to the ones we used for the binomial distribution.
That is, this table lists cumulative probabilities
  P(Z < z)
for values of z ranging from −3.09 to +3.09


Suppose we want to determine the following probability: P(Z < −1.52)
 
We first find −1.5 in the left margin. We then move along this row until we find the probability under the .02 column heading. Thus,

  P(Z < −1.52) = 0.0643


P(Z < −1.52) = 0.0643


As was the case with Tables 1 we can also determine the probability that the standard normal random variable is greater than some value of z.
For example, we can find the probability that Z is greater than 1.52, P(Z > 1.52). This probability can be obtained by determining the probability that Z is less than 1.52 and subtracting that value from 1.
   P(Z > 1.52)  =  1- (P(Z < 1.52)
This is depicted in graphical form below.
Applying the complement rule we get
  P(Z > 1.52) = 1 – P(Z < 1.52)
                      = 1 – 0.9375 = 0.0643
Alternatively, using symmetry,
  P(Z > 1.52) = P(Z < -1.52) = 0.0643

We can also easily determine the probability that a standard normal random variable lies between two values of z. For example, we want to find the probability
  P(−1.30 < Z < 2.10).
This can be obtained by finding the two cumulative probabilities and calculating their difference,
  P(−1.30 < Z < 2.10) = P(Z < 2.10) − P(Z < −1.30)
From the z table,
  P(Z < −1.30) = 0.0968   
and   P(Z < 2.10)   = 0.9821

Hence,
  P(−1.30 < Z < 2.10)   =   P(Z < 2.10) − P(Z < −1.30)
  =  0.9821 −0.0968
  =   0.8853

Notice that the largest value of z in the table is 3.09, and that P( Z < 3.09) = 0.9990. This means that
  P(Z > 3.09) = 1 − 0.9990 = 0.0010
However, because the table lists no values beyond 3.09, we approximate any area beyond 3.10 as 0. That is,
  P(Z > 3.10) = P(Z < −3.10) ≈ 0


Recall that in Tables 1 we were able to use the table to find the probability that X is equal to some value of x, but that we won’t do the same with the normal table.
Remember that the normal random variable is continuous and the probability that a continuous random variable X equal to any single value x is 0.


•Example 2
Suppose that at a country town petrol station, the daily demand for regular petrol is normally distributed with a mean of 1,000 litres and a standard deviation of 100 litres.
The station manager has just opened the station for business and notes that there is exactly 1,100 litres of regular petrol in storage. The next delivery is scheduled later today at the close of business. The manager would like to know the probability that he will have enough regular petrol to satisfy today’s demands.
•Example 2 - Solution
The demand for petrol is normally distributed with mean µ = 1,000 and standard deviation σ = 100. We want to find the probability
  P(X < 1,100)
Graphically we want to calculate the shaded area:


•Example 2 - Solution
The first step is to standardize X. However, if we perform any operations on X we must perform the same operations on 1,100. Thus,

  P(X < 1,100)   =                                           

                     =   P(Z < 1.00)



•Example 2 - Solution
The figure below graphically depicts the probability we seek.

•Example 2 - Solution
The values of Z specify the location of the corresponding value of X. A value of Z = 1 corresponds to a value of X that is 1 standard deviation above the mean. Notice as well that the mean of Z, which is 0 corresponds to the mean of X.
The probability that we seek is 
  P(X < 1,100) = P( Z < 1.00) = 0.8413




•Example 2 - Solution
•Example 3: Finance application:
Try it yourself
Consider an investment whose return is normally distributed with a mean of 10% and a standard deviation of 5%.
a.   Determine the probability of losing money.
b. Find the probability of losing money when the standard deviation is equal to 10%.
•Example 3 - Solution
a. The investment loses money when the return is negative. Thus we wish to determine
 
  P(X < 0)

  The first step is to standardize both X and 0 in the probability statement.

  P(X < 0) =                             = P(Z < –2.00)























•Example 3 - Solution
From Table 3 we find
 
  P(Z < − 2.00) = 0.0228

Therefore, the probability of losing money is 0.0228.





















•Example 3 - Solution
b. If we increase the standard deviation to 10% the probability of suffering a loss becomes

  P(X < 0)    =                                    
  = P(Z <  –1.00)  
  = 0.1587

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