Understanding Normal Distribution in Statistics
Explore the fascinating world of normal distribution, its origins from the binomial distribution, mean and standard deviation calculation, and visual representations through density functions. Learn how to manipulate the mean and standard deviation to observe shifts in the distribution curve.
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6.3 NORMAL RANDOM VARIABLES MAT 1372 Stat w/ Prob NYCCT (CUNY) Ezra Halleck
The normal distribution Like Poisson, it was originally discovered as an approximation to the binomial, this time with large n and p not too large or small (say .1<p<.9). What is n and p in the binomial distribution depicted? What is the mean and standard deviation for this distribution?
Description as a stand-alone distribution A normal random variable is a continuous random variable with a density function that is bell-shaped symmetric about the mean Its variability is measured by the standard deviation , Which can be found by finding distance from peak to where nature of curve changes (from being an upside down parabola to exponential decay).
standard standard normal distribution = 0 and = 1 the density function is 2 x 1 2 = y e 2 Exercise: Use Excel to plot this density function. Create a table with two columns, x and y. In the x column, begin at -2, increment by .1 and finish at 2 (41 rows). In the y column, use above formula, referencing appropriate x-value. For the exponential function, use EXP. For pi, use PI().
Changing the mean Changing the mean ( )2 x 1 2 amounts to a horizontal shift and we get = y e 2 Exercise: on the same plot as =0, plot density function with =1 Use the same x-values. Make new column for y-values
Changing the standard deviation Changing the standard deviation A more complicated change: horizontal compression and vertical stretch if 0 < < 1 or vertical compression and horizontal stretch if > 1. The density function is 2 x 1 = 2 y e 2 2 Exercise: on the same plot as = 1 plot the density function with = 2.
Changing both mean Changing both mean & standard deviation & standard deviation Simultaneous horizontal shift & vertical/horizontal compression/stretch. The density function is 1 y ( )2 2 x = e 2 2 (note the error in the book, p. 267)
keep fixed at 2, but vary (a) when is diminished, the height of the peak is increased to compensate for the smaller width (area under density function must always be same, namely 1). (c) When is increased, the height of the peak is diminished to compensate for the larger width.
Analyzing the area under the curve Analyzing the area under the curve Many applications amount to finding the area under different parts of the curve. Picture on right is important. (introduced earlier as the empirical rule ) 68% of the area is within one SD of the mean: ( ) = .68 P X 95% of the area is within two SD of the mean: ( ) = 2 .95 P X 99.7% of the area is within three SD of the mean: ( ) = 3 .997 P X Z is used to indicate the standard normal distribution ( = 0 and = 1):
4. P{Z > 1} is approximately (a) 0.50 (b) 0.95 (c) 0.84 (d) 0.16 6. P{Z > 3} is approximately (a) 0.30 (b) 0.05 (c) 0.002 (d) 0.99
17. If X is normal with = 100 and = 2, and Y is normal with = 100 and = 4, is X or Y more likely to (a) Exceed 104 (b) Exceed 96 (c) Exceed 100 From the diagrams, we see (a) Y is more likely to exceed 104 (What is P(Y>104)?) (b) X is more likely to exceed 96 (What is P(X>96)?) (c) X and Y have equal probability of exceeding 100. (Which is?) X: Y: 9698 100 102 104 96 100 104
18. If X is normal with = 100 and = 2, and Y is normal with = 105 and = 10, is X or Y more likely to Exceed 105 (b) Be less than 95 From the diagrams, we see (a) Y is more likely to exceed 105 (What is P(Y>105)?) (b) Y is more likely to be less than 95 (What is P(Y<95)?) X: Y: 9698 100 102 104 95 105 115
Example: Example: Overbooking on Airlines An aircraft has 168 seats. On average only 90% of all ticket holders on flights actually show up. The airline sells 178 tickets. What is chance that not everyone who arrives at the gate can be accommodated? p=.9, =np=178(0.9)=160.2 and = (npq)= (178*0.9*0.1)=4.00 Let X represent #of ticket holders who show up. P(X > 168) 168 is ~8 units from the mean which is about 2 standard deviations from the mean, so we get about 2.5% We get 1.3% when using binomial distribution directly: =1-BINOM.DIST(168,178,0.9,TRUE). The discrepancy can be ascribed to the somewhat low n and high p. 160 164 168
Exercise: Use Poisson for Exercise: Use Poisson for Overbooking on Airlines An aircraft has 168 seats. On average only 90% of all ticket holders on flights actually show up. The airline sells 178 tickets. What is chance that not everyone who arrives at the gate can be accommodated? Can you figure out how to use Poisson to answer the question? Hint: instead of considering the arrival of a passenger as a success which has a high p, work with the nonarrival of a particular passenger which has a low p (Using Poisson to model binomial requires a high n and a low p.) How does Poisson do? Does it give a better estimate than the normal distribution?
