Probability Normal Distribution

What is Normal Distribution? Any event can have at least one possible outcome. A trial is a single event. An experiment consists of the same trial being performed repeatedly under the same conditions. If an experiment is performed with enough trials, the populations of each possible outcome can be distributed according to different patterns. this is a typical Poisson Distribution. Note the lack of symmetry. This is the symmetrical Gaussian, or Normal Distribution. Learn this! notice the numbers. We’ll deal with them later

How it works... The Normal Distribution is characterised by grouped continuous data. The typical graph is a histogram of the populations of each grouped range of possible outcome values For example, if we returned to the era of norm-standardised testing for NCEA, then the distribution of test scores, as percentages, would look something like this: To pass, you would need a score of 50% or greater. Notice about 50% of all candidates achieved this score.

that 50% pass-rate is quite important. For any Normally-Distributed data, the central peak is the mean, μ. AND 50% of all data is <μ; which means 50% of all data is >μ

a quick bit of revision... the standard deviation now comes into its own. Recall: For a set of continuous data, the mean, μ, is a measure of central tendancy - it is one value that represents the peak data value population. The majority of the data does not equal μ. The standard deviation, σ, is analogous to the mean difference of every data value from μ. and now, back to the graph and it’s numbers...

features of Normal Distribution... the peak is the mean, μ. 50% of the data lie either side of μ. 68% of all data lie within 1σ of μ 95% of all data lie within 2σ of μ 99% of all data lie within 3σ of μ the distribution is symmetrical about μ the x-axis is asymptotic

summary: the x-axis is asymptotic the peak is the mean, μ. the distribution is symmetrical about μ ; 50% of the data lie either side of μ. 68% of all data lie within 1σ of μ 95% of all data lie within 2σ of μ 99% of all data lie within 3σ of μ } these percentages are rounded this distribution lets us calculate the probability that any outcome will be within a specified range of values

Ah, the wonder that is u The Normal Distribution of outcome frequencies is defined in terms of how many standard deviations either side of the mean contain a specified range of outcome values. In order to calculate the probability that the outcome of a random event X will lie within a specified multiple of σ either side of μ, we use an intermediate Random Variable, Z. Z X - μ σ Z = For this relationship to hold true, for Z, σ = 1 and μ = 0; hence, for a Normally distributed population, the range is from -3σ to +3σ

Z, PQR, and You Probability calculations using Z give the likelihood that an outcome will be within a specified multiple of σ from the mean. There are three models used: P( t ) = the probability that an outcome t is a ny value of X up to a defined multiple of σ beyond μ P(Z< t ) ≡ P( μ <Z< t ) Q( t ) = the probability that an outcome t is a ny value of X between μ and a defined multiple of σ P( μ <Z< t ) R( t ) = the probability that an outcome t is a ny value of X greater than a defined multiple of σ below μ P(Z> t )

Solving PQR Problems Read the problem carefully. Draw a diagram - sketch the Bell Curve, and use this to identify the problem as P, Q or R You could now use the Z probability tables to calculate P, Q or R, or use a Graphic Calculator such as the Casio fx-9750G Plus 1.Enter RUN mode. 2.OPTN 3.F6 4.F3 5.F6 This gives the F-menu for PQR. 1. Choose the function (P, Q or R) appropriate to your problem; 2. enter the value of t, and EXE.

Calculating Z from Real Data The PQR function assumes a perfectly-symmetrical distribution about μ. Real survey distributions are rarely perfect. For any set of real data, we can calculate μ and σ, and therefore Z. For example, if μ=33 and σ=8, then to find P(X<20): P(X<20) = P(Z < ) [ ] 20-μ σ = P(Z < ) [ ] = P(Z < ) Now, use the R function, and subtract the result from 1. SO... use to calculate Z, and then use PQR. X - μ σ Z =

continuity Normally-distributed data is often continuous. If asked to calculate probability for continuous data above a value q, apply the principle of measurement error, and take 0.5 the basic unit above the stated value. This is because any value in the range q-0.5 to q+0.5 will be recorded as q.

Inverse Normal This is the reverse process to finding the probability. Given the probability that an event’s outcome will lie within a defined range, we can rearrange the Z equation to give X = Z σ + μ But... we cannot define X, as it represents the entire range of values of all possible outcomes. What the equation will give us is the value k. k is the upper or lower limit of the range of X that is included in the P calculations

Use the PQR model, and sketch a bell curve to identify the regions being included in the p range. Use the ND table to find the value of Z: Z range is from -1 to +1, so find = Gives Z = So, k = 4 x = or, the short way... using a graphic calculator, for example the trusty Casio fx9750G; MODE: STATS F5 ➜ F1 ➜ F3 Area = probability, as a decimal σ = μ = EXECUTE an example... if X is a normally-distributed variable with σ=4, μ=25, and p(X<k) = 0.982, what is k? The long way...

Combinations of Variance The real world is rarely a simple place. However, apparently complex relationships can be rationalised to form straightforward equations. In addition to functional relationships that involve a single Random Variable, there can be interactions between two or more independent random variables, X and Y.

Sum of Random Variables For each Random Variable there is a calculable variance - this is true irrespective of the number of possible outcomes. Sums of Random Variables occur when we want the likelihood of a specific pair of outcomes (T) from two independent events; X + Y = T Simply, VAR(T) = VAR(X + Y) = VAR(X) + VAR(Y) and VAR(X - Y) = VAR(X) + VAR(Y) Got it? Whether adding or subtracting, you always add the independent Variances!

Linear Combinations of Random Variables We know that for a single Random Variable X, the linear function is E(aX + b) = aE(X) + b, and VAR(aX + b) = a 2 VAR(X) If we introduce a second Random Variable Y, E(aX + bY) = aE(X) + bE(Y) so VAR(aX + bY) = a 2 VAR(X) + b 2 VAR(Y) NOTE: this only holds true if X and Y are independent

an example... A hydroponic lettuce grower has her weekly costs expressed by two random variables - the number of plants X, and the liquid fertiliser concentrate costs Y. Both variables are independent. The standard deviation of X is 150, and the standard deviation of Y is 4 litres. Each lettuce costs $0.50 to irrigate and each litre of concentrate costs $10. Find the standard deviation of her costs. X: σ = 150, so VAR(X) = = Y: σ = 4, so VAR(Y) = 4 2 = 16 VAR(0.5X + 10Y) = VAR(X) VAR(Y) = (0.25 x 22500) + (100 x 16) = = 7225 so, σ = 7225 ½ = $85